Concept

Symmetry — where it appears

A transformation a configuration is unchanged by, which makes an objective stationary in the parameter that breaks it and hides improvements from any local search. It is why an aspect search's third parameter is sometimes worth nothing: the objective is stationary in the rotation a symmetric region is unchanged by.

Named by 17 essays across 7 fields — each of them below, with the objects they name alongside it.

Where two equal geodesics meet, from 15° north. A geodesic leaving at azimuth α and its mirror image at −α have the same length wherever they meet, and by symmetry they meet on the antipodal meridian. On a sphere they all meet at one point, the antipode, to 1.4e-7° — the flat line. On the ellipsoid the meeting latitude moves with the azimuth, from -15.563° to -15.009°, so the set of points with two shortest routes is an arc 61 km long, and the distance to it varies by 31 km along its own length.

Where the shortest route stops being the only one

On a sphere there is exactly one point with no shortest route from a given place: the antipode. On the ellipsoid the Earth actually is, that point is an arc — sixty-six kilometres of the antipodal meridian for a point on the equator, half a kilometre for one at 85°, and every point of it reachable by two different geodesics of exactly equal length.

paths · Paths
Six projections, at 40° north. Flexion (solid) and skewness (light) against direction of travel, drawn about a zero circle at one point of each projection. The three conformal ones have curves of exactly equal size, offset by exactly 90°, because on a conformal map both quantities are components of one vector — the gradient of the log of the scale. The others have neither property: ratios run from 0.72 to 2.72.

Bending and stretching are one failure

The ladder was written expecting flexion and skewness to be two independent ways for a map to be wrong. On a conformal projection they are not independent at all: the two extremes are the same size to four figures and sit exactly ninety degrees apart, because both are components of a single vector.

distortion · Flexion
The set that can be reached is not the set that can reach. Two sets under a steady westerly: everywhere reachable from the marked place in a stated time, and everywhere from which the marked place can be reached in the same time. They have the same area to a fraction of a per cent — reversing a uniform flow is a reflection — and they overlap on only 32.3 per cent of their union. 67.7 per cent of the ground in one of them is not in the other.

The set that can be reached is not the set that can reach

The moment a cost stops being symmetric, two questions that read alike stop having the same answer. Under a flow at 45 per cent of a vehicle's own speed the set reachable from a place and the set from which the place is reachable have the same area to five significant figures and share 32 per cent of their union — so 68 per cent of the ground in one of them is not in the other.

paths · Reach
Where the projection's pole should go, and the answer the third rotation moves it to. Every dot is a pole position the search tried, sized by the best score it can reach there when the third rotation is also free — the score is the distortion of Robinson over Europe by Kavrayskiy's criterion, so smaller is better and the large green dots are the good regions. The circled mark is the two-parameter optimum and the square is the three-parameter one: they are 74 pixels apart on this map, which is a different aspect rather than a refinement of the same one. The search costs 8 times the evaluations of the two-parameter one. Drawn in Mollweide.

The third parameter, run

An aspect has three numbers and this site has been searching two of them, with a note admitting it. Searching all three is worth up to 2.1 times — and the obvious way to do it, starting from the two-parameter answer and letting the third move, finds a fraction of that or nothing at all.

choosing · Choosing
One slice of the aspect objective, at the best γ. The Kavrayskiy score for Robinson over Japan, as the pole is moved over the whole sphere with the third rotation held at the value the search settled on. Dark is good. The marks are local minima of the full three-dimensional grid that happen to lie in this slice: there are 6 of them here and 58 in the cube, and a search that walks downhill from a random start reaches the best of them 7 per cent of the time.

The landscape the search walks on

The three-parameter aspect search was run and its answer recorded with a note admitting nothing proved it global. Mapping the objective finds 26 to 34 local minima for every projection and region tried, a downhill walk from a random start reaching the best of them 6 to 35 per cent of the time — and one seed from the coarse grid the search already uses reaching it in all four cases. The score is reproducible to two per cent across a sevenfold refinement; the pole it names moves 60 degrees.

choosing · Choosing
What is left after the best rigid motion of the page. Every projection in the library, rotated about its own symmetry axis by five degrees, with the best rigid motion of the page fitted and the leftover measured against the picture's own size — on a logarithmic scale, because the answers span four orders of magnitude. Fourteen sit at  2e-6 or below, which is the axis search's own floor. Seven sit between 9e-3 and 3e-2. There is nothing in between, so the split is a fact rather than a threshold — and the fourteen are exactly the cylinders, the cones and the planes.

The family is a symmetry, not a shape

Rung seven dissolves the taxonomy: eleven named maps fitted to one perspective formula, six reproduced exactly and five refused, so the developable surface describes how the projections were discovered rather than what they are. What is left is the question it does not ask. If the shapes are not the classification, what is?

families · Families
The set of aspects within a stated distance of the best, for Robinson over Japan. Each row takes every point of a 36 × 19 × 24 grid in the three aspect parameters that scores within (1 + t) of the best, joins neighbouring points, and identifies the pieces the exact degeneracy relates. At t = 3 it is one connected piece spanning 170° of pole; by t = 1 it has broken into 2 pieces; and by t = 0.3 the largest of them spans 11°. So it is not one valley and it is not one basin — it is a sheet that fractures.

The shape of the valley

An aspect search returns three numbers, two searches return triples that differ by a hemisphere, and the maps they produce agree. One cause is an exact degeneracy and the rest was called a valley and left unmeasured. Sampled densely, it is neither a valley nor a basin: a connected sheet spanning 170° of pole that splits in two at twice the best score and into thirteen pieces by 1.7 times it.

choosing · Choosing
The pass between the basins, measured rather than bracketed. The height of the lowest path from one near-optimal basin to the other, for regions of growing size. It is found by sorting the score surface and joining cells in order, so it is the minimax path's own height rather than the level at which a bisection stops finding two pieces. The fitted exponent is -1.34 at a coefficient of determination of 0.986. The earlier level-set measurement on the same projection gave -1.57 and the argument predicts −1, so measuring the height directly moves the answer towards the prediction, though not onto it.

The height of the pass between two basins

One thing was left owed: the fracture threshold falls faster than the argument predicts, and the difference was supposed to be the level set's blindness to the height of the pass. Measured directly, the pass between the two deepest basins falls as the −1.34 power at R-squared 0.986 — nearer the predicted −1 than the level set's −1.57 on the same projection, and not onto it. It is not the pass that breaks the set first, and at five of six sizes the second basin is the first one's mirror twin.

choosing · Choosing
The corner against where along the seam the feature crosses. A feature crossing the seam of a cube at right angles, drawn on each face's own map and unfolded, with the crossing moved along the edge. Every curve starts at zero, because the face's mirror through the edge's own midpoint reverses the along-edge direction and forces the shear term in the Jacobian to be odd. Away from it the gnomonic — the map measured earlier as having no corner at all — reaches 44.4°, three times the equal-area map's and far past the conformal one's. The edge's half-length is 35.3°, so the right-hand end is still well inside it.

The corner is not at the midpoint

An earlier measurement took the corner a feature gets crossing a polyhedral seam, and found none at all under the gnomonic face map. It crossed at the edge's own midpoint every time — the one point on the edge where a face's own mirror symmetry forces the corner to vanish. Two fifths of the way to the vertex the gnomonic gives 20.15°, the equal-area map 7.28° and the conformal map under a degree, which reverses the ordering entirely.

families · Polyhedral
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

A datum's rotations were expected 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
The test the conjecture asked for, and it refutes the conjecture. How much better the best asymmetric projection is than the best symmetric one, under weightings of four different symmetries, with every symmetric map allowed to re-aim its axis at twelve candidate poles. The conjecture was that the seven earn their place by PLACING distortion where a symmetric map cannot, so their advantage should collapse under a criterion with no place preference. It does the opposite: the advantage is largest at 1.343 under the uniform weighting and smallest at 1.144 under a band, with the fully asymmetric concentration at 1.204 in between. The winner is named on each row and the map it beat is Equirectangular throughout. The seven are simply better maps.

The maps with no family are simply better

Seven projections have no continuous symmetry, and they are almost exactly the set anybody would choose for a world map. That coincidence came with a conjecture and a test attached: their advantage should collapse under a criterion that does not care where anything is. Run, it does the opposite — 1.343 times under a uniform weighting and 1.204 under a concentration. The conjecture is refuted.

families · Families
Three lines of position, and a cocked hat that misses the ship. Three bodies whose azimuths from the ship are 20°, 140°, 250°, each observed with an independent error of about 2 km along its azimuth, give three lines that do not meet. They bound a triangle of 7.5 km², outlined, and the ship — the dot — lies outside it. With independent errors that is the usual outcome: three times in four, whatever the triangle's shape.

A cocked hat holds the ship one time in four

Three lines of position with any error in them bound a small triangle, and the instinct is that the ship is inside. With independent errors it is inside one time in four, whatever the triangle's shape. A common error in every sight makes that almost always when the stars surround the ship and almost never when they do not. And on a sheet that forgets the latitude, the triangle keeps its size and moves off the ship altogether.

paths · Position fix
The pass that fails first stands above the pass between the two deepest basins. For a region grown from 6° to 40° of span at 38° N, the threshold at which the near-optimal aspect set first comes apart — the highest merge in the whole filtration — against the merge between the two deepest basins, which is what the earlier measurement of this quantity reported. The first is higher at every size, by a factor of 1.02 to 2.09. At every size the deciding merge is the last basin joining the optimum's own — a basin whose floor stands 0.019 to 0.191 above the best score — so the set is held together by a near-rival to the best placement, reached over higher ground than the second-best basin is.

The pass that fails first is not the one that was measured

Three attempts to explain why the near-optimal aspect set comes apart when it does have fitted the height of the pass between the two deepest basins. That is not the pass that fails first. At every region size the set is held together by the last near-rival to join the best placement's own basin — a basin between 2 and 19 per cent worse than the best — over ground 1.02 to 2.09 times higher than the deep pair's pass. The two passes fall at nearly one rate, −1.42 and −1.31 powers of the region's span, so what separates them is a factor, not a law.

choosing · Choosing
A sea with two islands in it, and the three circles it already is. A stated sea 60° across with two islands 12° across, their centres 12° from its own, drawn on the globe and drawn again under the stereographic projection centred on the sea. Stereographic projection carries every circle on the sphere to a circle in the plane, so the flat picture is bounded by three exact circles with no fitting and no iteration: the sea's edge at radius 0.5359, and the two islands at radius 0.1060 with their centres 0.2108 from the origin. Normalised, the region's three conformal invariants are 0.2361, 0.6848, 0.1221 — three numbers, which is what a region with two holes carries and no angle-keeping map can change.

Two islands are three numbers, and the proof does not survive them

A sea with one island at its centre has one conformal invariant and a forced best map. Put two islands in it and there are three invariants, readable in closed form because stereographic projection carries circles to circles — and the forcing is gone. The theorem that picked the map needed the region to be unchanged by rotation, not to have one hole, and a single harmonic term two hundredths in size beats the stereographic by 0.19 per cent the moment an island moves off centre.

impossibility · Topology
Two of six projections have a curve. The depth of the pass that decides when a near-optimal aspect set comes apart, against how much of its own scale variation the region sees, for six projections at nine latitudes and four region sizes each, with each projection's own power law. A curve, R² 0.8 or better: Mollweide (0.47, R² 0.81), Hammer (1.35, R² 0.95). Partial: Eckert IV (0.34, R² 0.52). No curve: Robinson (0.02, R² 0.01), Winkel tripel (-0.06, R² 0.05), Sinusoidal (0.28, R² 0.15). Pooled over all six the fit is R² 0.03, because the six sit at different levels.

Six projections have two curves, and the refusal has company

Three projections at four latitudes gave three pass-depth exponents, and the question was what sets them. Six at nine latitudes say that most of them have none to set: the Hammer's pass depth rises with what the region sees at R-squared 0.95 and Mollweide's at 0.81, Eckert IV's half-follows, and Robinson's, the Winkel tripel's and the sinusoidal's do not follow it at all. Neither candidate explanation predicts a left-out projection's exponent better than guessing the average — and the sinusoidal, once the lone refusal, is one of three.

choosing · Choosing
A parent and its children, twice. An aperture-7 hexagonal hierarchy beside a square one. The heavy outline is the parent and the light ones are its children. On the right every child is wholly inside and the four of them tile the parent exactly. On the left the child lattice is turned by 19.107° relative to the parent's, only the central child is wholly inside, and 7.14% of the parent is covered by no child of its own. The two families have exactly the same total area — a hexagon cannot be tiled by smaller hexagons at any ratio at all, which is why the mismatch is a construction rather than an approximation.

A cell's children do not fit inside it

Ten rungs price one cell system at one resolution, and every one of them is used hierarchically. A hexagonal hierarchy does not nest: at the aperture-seven scheme the discrete global grids use, one fourteenth of a parent is covered by no child of its own, exactly, and each of the six ring children is eleven twelfths inside.

applied · Cells

The orientation is a policy

A polyhedral cell system has three free angles nobody scores. They cannot improve it: rotating the solid rotates every cell rigidly, so the distribution of cell areas is identical for every orientation there is. What they decide is who stands on the bad cells — and the eight cities measured here get a spread of cell area of 1.00 under the best turn and 1.50 under the worst.

applied · Cells

Named alongside it

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

VerificationAspectOptimisationPurposeConformalityObjective functionReproducibilityConvergenceDegeneracyExponentIdentifiabilityInvariant

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