Five bearings of twelve, and only eight ways to draw them

A flat picture has one direction between two places, and the Earth has two put a floor under every flat picture’s bearings. The bearing from New York to Tokyo and the bearing back miss being reverses by 128 degrees, one straight line can only split the difference, and so no picture containing both cities can have a worst bearing error below 64 degrees. On a region the best picture sits exactly on that floor; on the world it lifts well clear of it.

A floor is a statement about the worst bearing. It says nothing about the best ones, and a picture is often wanted for a few directions rather than for all of them — a home port’s bearings to its destinations, a network of radio beacons, the lines a surveyor actually sighted. The question that essay left is the one five distances of six, and never more answered for distances: how many of a set’s bearings can a flat picture hold exactly, and which ones.

The count turns out to be the same as for distances. What differs is everything else: how the picture is found, what it is free to do, and the way it fails.

A bearing is one straight-line equation #

A one-way bearing is the direction the shortest route from one place to another sets out in, measured clockwise from north. On a flat picture with east along $x$ and north along $y$, saying that the line from dot $i$ to dot $j$ is drawn at bearing $\beta_{ij}$ is saying

$$(x_j - x_i)\cos\beta_{ij} - (y_j - y_i)\sin\beta_{ij} = 0 .$$

That is a linear equation in the coordinates of the dots. A distance is not: saying that two dots are a stated distance apart is a quadratic equation — the kind four cities that cannot be drawn to scale tests with a determinant of squared distances — which is why every picture of distances in the best flat picture is not a map had to be searched for. A set of bearings is solved, like any system of linear equations, and the picture that holds them comes out of the arithmetic in one step.

Counting is then a matter of counting what the equations cannot see. A picture of $n$ places has $2n$ coordinates. The equations are unchanged if every dot is moved by the same amount in either direction, and unchanged if the whole picture is enlarged or shrunk about any point, because neither changes the direction of any line. That is three free motions, so a picture can satisfy at most $2n - 3$ independent bearings: five for four places, thirteen for eight, twenty-nine for sixteen.

The distances count is $2n - 3$ as well, and the coincidence is exact rather than approximate. The difference is which three motions are free. Distances are blind to moving and turning a picture and fix its size. Bearings are blind to moving and enlarging a picture and fix its orientation. The same number of equations pins the same number of coordinates, and what is left over is a different shape of freedom.

One more bearing, and the picture is a point #

Every count was checked by elimination rather than taken on trust. On all four sets used here — five towns in Britain, four cities, eight and sixteen — a choice of $2n - 3$ bearings grown one place at a time, each new place joined by two bearings to places already present, has exactly $2n - 3$ independent equations. Adding any one further bearing raises that to $2n - 2$, and the only picture satisfying $2n - 2$ independent bearings is the one with every dot on the same spot.

The word independent is doing work. Two hundred choices of $2n - 2$ bearings made purely at random included 58 that were not independent: they crowded several bearings onto a few places, left others joined by fewer than the count needs, and so had a free part left over. A picture’s capacity for bearings is $2n - 3$, and a choice that spends it unevenly holds fewer.

Turning the north arrow buys nothing #

A map is free to put its north anywhere — north cannot be up everywhere is the essay about why a whole-sphere map cannot keep it pointing one way — and it is natural to expect that freedom to be worth one more bearing. It is worth none. A picture whose bearings are measured from a north arrow turned by some angle is exactly the same picture, turned by that angle, with its bearings measured from the page’s own north. Turning every bearing in a choice by seven tenths of a radian leaves the number of independent equations and the drawability of the choice exactly where they were, on every set.

So $2n - 3$ is the whole of it. For four places that is five of the six pairs, or five of the twelve one-way bearings, since a pair’s bearing out and bearing back can never both be held: they are not reverses, and one line cannot point two ways.

The equations cannot tell a line from its reverse #

The linear equation above says the line from $i$ to $j$ is parallel to the direction $\beta_{ij}$. A line from $j$ to $i$ is parallel to it too.

So the picture that solves a choice of bearings has every held line at the right slope, and nothing in the solution decides which way along its line each arrow points. That is settled afterwards, by looking. A picture may be enlarged by a negative amount — turned through half a revolution about a point — and that flips every arrow at once, so the useful question is whether some sign makes every arrow point forwards together.

When none does, the picture has drawn at least one place on the wrong side of another. It holds that bearing wrong by 180 degrees, which is as wrong as a bearing can be, and it does so while the equations report every held line as exactly satisfied. In the figure above the picture of Tokyo, Sydney, London and New York has every slope right and puts Sydney north of Tokyo.

The count of $2n - 3$ is a count of slopes. How many of those slopes can be made into arrows pointing the right way is a separate question, and it has a different answer.

Of 192 choices, eight #

For four places the question can be settled completely, because there are few enough choices to try every one. Leave out one of the six pairs, pick a direction for each of the other five, solve, and look.

Eight of the 192 can be drawn. Forty-eight put one arrow backwards and 136 put two.

The eight are not scattered. Every one of them leaves out the pair London–New York, the two closest of the four cities, and every choice that leaves out any other pair is undrawable, whichever directions are picked. A picture of four world cities that holds five bearings exactly exists, but the five are almost forced: the pair that is not held is decided by the places, and only the directions on the other five are free, and even among those only eight of 32 combinations survive.

Why that pair and not another is not something this measurement explains. London and New York are the closest of the four, and it is tempting to say that the shortest line is the one a picture can most afford to let go; but that is a story told after the fact, not a reason, and the only evidence here is the enumeration itself.

The price the other seven pay #

The hero picture is the drawable choice whose worst error on the seven bearings it does not hold is smallest, and that worst error is 165.0 degrees, on the bearing from London to New York. The eight drawable choices put it between 165.0 and 169.6.

Set that against the numbers from a flat picture has one direction between two places. The floor for these four cities is 63.9 degrees, and the best flat picture, which holds no bearing exactly and minimises the worst of all twelve, reaches 116.1 — the directional counterpart of the least-error picture how wrong a flat picture has to be finds for distances. Holding five exactly adds nearly fifty degrees to that, and puts one direction within fifteen degrees of pointing straight backwards.

On a region, every choice can be drawn #

The backward arrows are a property of places spread over the world, and shrinking the places shows where they set in.

The four cities shrunk towards their centre stay mostly undrawable while they span more than about fifty degrees — 4.2 per cent at their full span of 153 degrees, 7.3 at 55 — then climb to a third at 40 degrees and to every choice at 10. Five towns in Britain, spanning 5.6 degrees, are drawable on every one of two hundred choices grown at random.

Eight and sixteen cities need to be much smaller. Eight cities reach a third of their choices at 31 degrees and four fifths at 7.7. Sixteen reach none at 32 degrees and 13 per cent at 8, where a choice still puts four and a half arrows backwards on average.

More places fail more at the same size for a reason the counting makes plain. A choice for sixteen places holds twenty-nine arrows, and a picture is drawable only if every one of them points forwards. Each extra place brings two more arrows that can flip, so the chance that none does falls as the set grows even where each arrow on its own is nearly certain to be right.

Every bearing from two places #

A choice made at random is not how a picture of bearings would actually be commissioned. The natural choice is the one a navigator with two home ports would ask for: every bearing out of the first, and every bearing out of the second except the one back to the first. That is $(n - 1) + (n - 2) = 2n - 3$, exactly the capacity, and it has a construction a pencil can carry out — each place goes where the ray from one home port meets the ray from the other.

On five towns in Britain every one of the twenty pairs of home places works. On the world sets none does. The best pair for four cities, London and Sydney, leaves one arrow of five backwards; the best for eight leaves three of thirteen; the best for sixteen, Santiago and Buenos Aires, leaves eight of twenty-nine, and no pair of the 240 leaves fewer.

The failure is a ray meeting a ray behind its starting point. The bearing from London to Moscow and the bearing from Tokyo to Moscow are two rays, and a picture that holds both must put Moscow where they cross; when they cross behind one of the two cities, the picture holds both slopes and draws Moscow on the wrong side of it.

There is a projection with a similar-sounding name, the two-point azimuthal, and it does not do this. Its equidistant cousin is built from exactly this kind of pair of conditions in a projection written as a condition, and both keep a property from two places at once. It keeps the angles at each of two places between the directions to every other place, which is a different property: its meridians are straight lines that converge, so north at one of its two places and north at the other point in different directions on the page, and a reader with one north arrow cannot read bearings off it. The picture here has one north, which is what a bearing needs, and on the world it cannot have it for two home ports at once.

What holding them exactly costs #

Exactness is not free even when it is possible, and the price is worth setting against the two numbers the previous essay established for every set.

On a region the price is small if the choice is good. Five towns in Britain have a floor of 2.13 degrees, the best picture reaches it, and the best of two hundred drawable choices holding seven bearings exactly gets every other bearing within 3.53. But the choice matters even there: the median of the two hundred has a worst error of 7.3 degrees and the worst of them 75.7. Which bearings are held decides what the rest can do, on a region as well as on the world, and a region only makes a good choice easy to find.

On the world sets the price is most of what a bearing can be wrong by. Four cities go from 116.1 degrees at best to 165.0 when five bearings are held. Eight go from 128.1 to 152.2. Sixteen go from 158.4 to 172.6, which is to say that some place on the picture is drawn within seven and a half degrees of exactly the wrong direction.

Twenty-nine arrows, built rather than chosen #

None of two hundred choices of twenty-nine bearings made at random for the sixteen cities can be drawn. That is not the same as none existing, and a drawable one can be built.

The construction is the pencil method run with a choice at every step. Put two cities down along the bearing between them. For each city not yet placed, look for two cities already on the page whose bearings to it — or its bearings to them — cross in front of both, and put it there. That adds two bearings per city and keeps every arrow pointing forwards by construction, and when it succeeds for every city it has held exactly $2n - 3$. It was run from every ordered pair of starting cities, and the one drawn, from Nairobi and Tokyo, is the one whose worst error on the other bearings is smallest.

So a picture of sixteen world cities holding twenty-nine bearings exactly and honestly exists, and it has to be looked for, and it draws Los Angeles almost due east of New York. Holding the most bearings a flat picture can hold is compatible with drawing every one of them forwards only by deciding very carefully which ones, and even then it is paid for in full by the rest.

Distances and bearings on the same pairs #

The two counts combine in one more way. A pair held for both its distance and one of its bearings fixes the vector between two dots completely, and the only freedom left to the whole picture is where it sits on the page — two numbers. So a picture of $n$ places can hold distance and bearing together on at most $n - 1$ pairs.

Those pairs cannot close a loop. Round a loop, exact vectors on the page have to add to nothing, and the corresponding legs on the sphere do not: a direction carried round a loop is the essay about how a direction comes back turned, and a set of legs whose lengths and bearings are each exact reaches somewhere other than its start. So the $n - 1$ pairs must form a tree, and any tree will do. An azimuthal equidistant map holds exactly one of them: the star of pairs from its centre. For four places there are sixteen trees, and for sixteen places there are $16^{14}$ — about $7.2 \times 10^{16}$ — every one of which some flat picture holds exactly, with no backward arrows possible, because a tree has no loop to close.

That is the one version of the question where drawability never fails, and it is the version with the least in it: $n - 1$ pairs against $2n - 3$ for either quantity alone.

Where the equations are not independent at all #

The count has one refusal, and it is the case that should break it. Five places along the equator have great-circle bearings of exactly 90 and 270 degrees between every pair, because the equator is itself a great circle and the heading along it never turns.

For those places all twenty one-way bearings, in both directions at once, give only four independent equations, not seven, and those four say nothing except that the dots lie on one horizontal line. The order of the dots along it is left entirely free. Put them in their true order from west to east and every one of the twenty bearings is held, both ways, exactly. A count that stayed at $2n - 3$ there, or a picture that failed to hold every bearing, would mean the arithmetic was about something other than the sphere; the place where bearings go out and back as exact reverses is the one place where a flat picture can hold them all.

What the enumeration settles and what it does not #

The sphere is a sphere, and a bearing is the initial great-circle bearing. On the ellipsoid the bearings between world cities change by fractions of a degree, and every result above is about backward arrows and errors of tens of degrees.

The four-city result is complete; the others are samples. Every one of the 192 choices for four cities was solved. For eight and sixteen cities the drawable shares come from two hundred choices grown one place at a time, which is one natural family of independent choices and not all of them; a different way of choosing at random would give different percentages, though not, on this evidence, the opposite conclusion.

The built picture is the best of one construction. It holds twenty-nine bearings honestly and its worst error elsewhere is 172.6 degrees. A different construction might find a drawable choice with a smaller worst error, and nothing here bounds how much smaller.

And the London–New York pattern is measured, not explained. Every drawable choice for four cities leaves that pair out; why the places force it is described above as a plausible account and left there.

Still open: which pair has to be let go #

For four world cities there was exactly one pair a drawable picture could leave unheld, and the enumeration found it by trying everything. For sixteen cities there are a hundred and twenty pairs, and a picture holding twenty-nine of their bearings leaves ninety-one of them out; the built picture found one arrangement that works without saying which pairs any drawable arrangement must avoid.

The error belongs to a few of the places found the same shape of question for distances, where Sydney alone held a picture of sixteen cities at its worst and leaving it out halved the error. Whether bearings have a counterpart — a pair, or a place, whose bearings no drawable picture can hold, identifiable from the geometry of the places rather than by enumeration — and whether it is the same place distances single out, is a question the count cannot answer and the four-city enumeration can only hint at.