Every reach set ever drawn is too small

Seven rungs of this anchor build reach sets — a range ring, a catchment, a corridor, a set under a cost that depends on direction — and score them for area, for boundary length, for compactness and for asymmetry. Every one of them draws the set the same way, and none of them says so.

The way is this. Pick a number of bearings. From the centre, walk out along each bearing until the cost reaches the stated budget. Join the points. That polygon is what appears on the page, it is what every area in this anchor was measured from, and it is not the reach set.

The direction of that error is the fact worth having, because it is a sign rather than a size. It does not average out, no amount of care with the walk removes it, and it is the same sign for every convex reach set anybody has ever drawn.

Why it is always short and never long #

The argument is one sentence and it needs no calculus.

A reach set under a cost that grows along every path is convex in the relevant sense: the boundary is a single closed curve, and the straight line between any two points of it lies inside. So a polygon whose vertices are on the boundary has every edge inside, encloses less than the boundary does, and is short.

That is not a claim about the sphere. It is a claim about chords, and it holds in a plane, on a sphere, on an ellipsoid and on any surface where the boundary is convex. What the surface changes is the size of the shortfall and not its sign.

The other half of the argument is what stops it being trivial. If the reach set is not convex — a flow strong enough to fold the boundary back on itself, an obstacle carving a bite out of it — then a chord can pass outside, and the polygon can be too large somewhere. A route that must go round prices what an exclusion does to a route; what it does to a reach set is put a re-entrant corner in the boundary, and a fan through it can cut across the bite. So the sign is guaranteed for a convex set and only for a convex set, and every set in this anchor except the ones with obstacles in them is convex.

The rate is the square of the count #

The sign says the drawn set is short. The rate says by how much, and it is the rate that decides whether anybody should care.

The exponent comes from the shape of what is lost. Each chord cuts off a circular segment whose height is quadratic in the chord’s length and whose area is therefore cubic in it, and there are n of them, so the total falls as n × n⁻³ = n⁻². This is the same arithmetic flying a curve in straight legs finds for a great circle flown as constant-heading legs, where the gap between plan and curve falls as the square of the number of legs — a chord approximation is a chord approximation, and the exponent does not know what it is approximating.

The two curves in that figure are parallel and are not the same curve, which is the second finding.

What “the true set” means when nothing computes it #

There is a step in the argument above that is doing quiet work, and it is worth exposing before the numbers are used.

For a geodesic disc the true area is known: 2πR²(1 − cos t), a closed form, and the polygon is measured against it. For anything else there is no such thing, and a deficit has to be measured against something.

The something used here is the set the finest fan draws — 1,024 bearings — and it is stated as that rather than called the truth. That is not a dodge and it is not free either. It biases every deficit in the flow figure downwards by the finest fan’s own deficit, which the square law puts at about 4.6 × 10⁻⁶: negligible against the 1.8 × 10⁻¹ of the coarsest row, and not negligible against the finest rows in the same table.

So the flow figure’s fitted exponent is measured over rows whose deficits are between a hundred and forty thousand times the reference’s own, and it stops being trustworthy below that. The disc figure has no such limit and that is why the exponent is quoted from it.

This is the same problem the score is not stable at any scale meets from the other end, where a perimeter has no limit at all and so no reference exists at any resolution. Here a limit exists and cannot be computed, which is a better position and not a comfortable one.

A cap loses less than a circle, and by exactly how much #

A circle in a plane has a closed form for this. A polygon of n sides inscribed in a circle has area (n/2π)·sin(2π/n) of it, so the deficit is 1 − (n/2π)sin(2π/n), which is 2π²/3n² to leading order — the constant is 6.5797.

The sphere does not follow it. At 4,000 kilometres the cap’s deficit is 0.7322 times the planar circle’s at every count above about twenty-four, and 0.7322 is not a constant of nature — it is a function of the cap’s size.

So the whole result has a closed form:

deficit = [1 − (n/2π)·sin(2π/n)] · cos t · (1 + cos t)/2

with t the cap’s angular radius. It agrees with the measured polygon to 8.8 × 10⁻⁶ at a thousand kilometres and 128 bearings, to 3.6 × 10⁻⁴ at eighty-nine degrees, and to 5.6 per cent at six bearings — which is what a leading term in 1/n² does and is stated as such rather than as an identity.

Two limits of that expression are worth reading off, because both are checks rather than results.

At small t the factor is one. cos t → 1 and (1 + cos t)/2 → 1, so a small cap loses exactly what a circle loses. It has to: a small enough patch of a sphere is a plane to any stated precision, which is how small is flat enough arriving as a limit of a formula rather than as a tolerance.

At ninety degrees the factor is zero. A cap of angular radius ninety degrees is a hemisphere, its rim is a great circle, and the “chords” between points of a great circle lie along it. So a polygon through any number of points of the rim encloses the hemisphere exactly, and the deficit is not small but nil. Measured at 89.03 degrees it is 8.6 × 10⁻³ of a planar circle’s, falling to zero at ninety.

That second limit is the one that makes the factor interesting rather than a correction. A polygon inscribed in a circle always loses area; a polygon inscribed in a spherical cap loses less and less as the cap grows, and at half the sphere loses nothing at all.

The exponent generalises and the constant does not #

A geodesic disc has a closed form. A reach set in a flow has none — its boundary is the level set of a travel time computed by walking a grid, and there is no expression for its area at all.

That is the useful separation and it is the same one how wrong a flat picture has to be draws for a different quantity. The exponent belongs to the fact that a chord cuts a segment, which is true of any convex boundary anywhere. The constant knows what shape is being cut, and has to be measured for each one.

Which means the practical form of the result needs only the exponent, because the constant enters under a square root.

What a tolerance costs #

Inverting the law gives the number a person drawing an isochrone actually wants: how many bearings for a stated accuracy.

Every reach set drawn anywhere in this collection uses between seventy-two and a thousand and twenty-four bearings, which puts them all between a thousandth and a millionth of the area. So the answer to “does this matter” is no, for every figure on this site — and the reason it is worth an essay anyway is that nobody had checked, and that the sign of the error is the same for all of them.

The place where a coarse fan is used deliberately is the page rather than the calculation. A polyline of a thousand points costs a thousand pairs of coordinates in the file, and this collection’s own figures are eighty per cent polyline coordinates by weight; drawing a boundary at seventy-two points rather than at a thousand is a fiftyfold saving in bytes for a nine-hundredfold increase in an error nobody can see. That is the right trade and it is worth making knowingly rather than by default.

A signed error that never averages out is a different object from an unsigned one of the same size. Comparing two reach sets drawn from different numbers of bearings compares two different underestimates, and the coarser one is smaller by a predictable amount rather than by a random one. The score is not stable at any scale makes the same distinction for a compactness score, where the numerator settles and the denominator does not; here both settle, and what does not settle is which of two settled numbers is being compared.

The one place a denser fan is not the answer #

The square law says the cure for a coarse fan is a finer one, and there is a class of reach set where that is false.

A fan finds, for each bearing, the first place along that bearing where the budget runs out. That is the right answer only if the reach set is star-shaped about its centre — if every ray from the centre crosses the boundary exactly once. Where it crosses three times, the fan takes the first crossing and misses the ground beyond, and the missed ground is not a chord’s worth of segment. It is a whole region, and no number of bearings recovers it, because the failure is in what the walk reports rather than in how many walks there are.

The distinction is worth stating in the same units as the rest of the rung. The chord deficit is O(n⁻²) and goes to zero. The star-shape deficit is O(1) and does not. A figure that is short by a millionth because of the first and by ten per cent because of the second reports one number, and the square law says nothing about which it is.

Everything measured on this ladder is star-shaped, and the fan is the right instrument for all of it. What tells the two cases apart is a different measurement — walking each ray to the far side of the sphere and counting crossings — which nothing here does.

Where this leaves the anchor’s own numbers #

Seven rungs of measured areas were all computed from polygons, and it is worth saying which of them the correction touches.

The ones reported as ratios are untouched. A circle of a distance is not a circle reports that the same ground distance comes out 4.17 times longer in one direction than another; both distances are read from the same fan, so the deficit divides out. The same is true of every compactness score, every asymmetry ratio and every share of the sphere in the anchor.

The ones reported as absolute areas are short by a known amount. At the seventy-two bearings this collection’s reach figures use, the deficit is 0.0929 per cent for a cap of this size, or nine hundred and twenty-nine parts per million. Every absolute area in the anchor carries that, in the same direction, and none of the conclusions depends on the fourth digit.

And one is not a polygon at all. The corridor and disc areas in the third rung come from a latitude–longitude quadrature rather than from a fan, so they carry a different error with a different sign, and the two should not be compared without saying which is which.

The perimeter is short too, and by a quarter as much #

The area is not the only quantity a reach figure reports. Every boundary length in this anchor is the perimeter of the same polygon, and it has a deficit of its own with a different constant.

A chord is shorter than the arc it subtends, so a polygon’s perimeter is short as well — same sign, same square law, fitted exponent −2.0126. What differs is the constant: on a 4,000-kilometre disc the area deficit is 4.4710 times the perimeter deficit at every count above about twenty-four, converging to that figure from below. Eight bearings lose 7.53 per cent of the area and 1.71 of the boundary; a hundred and twenty-eight lose 0.0294 and 0.0066.

The ratio matters because compactness is an area divided by a squared perimeter, so the two deficits partly cancel and do not fully. A set short by δ in area and by δ/4.471 in perimeter scores 1 − δ + 2δ/4.471 = 1 − 0.553δ of the true score, and the fan therefore reports a compactness lower than the truth rather than higher. Measured directly on the disc, whose true score is exactly one: eight bearings give 0.96476, twenty-four give 0.99624, a hundred and twenty-eight give 0.99987.

At the seventy-two bearings this collection’s compactness scores were computed at, the understatement is five parts in ten thousand — three orders of magnitude below the 4.7 per cent separation the most compact shape depends on the paper is about, so nothing there moves. It is recorded because it is a bias rather than a noise, in a known direction, of a size that can be written down.

Why nobody noticed #

A signed error of a millionth of the area is easy to miss and it is worth asking why, because the answer is about how this kind of measurement is usually checked.

The check anybody would run is convergence: draw the set at 72 bearings, again at 144, and see whether the number stops moving. It does — the deficit falls by four each doubling, so by a few hundred bearings the last printed digit is stable and the reading looks converged. What convergence does not say is which side of the truth it converged from, and a monotone approach from below looks exactly like a monotone approach from above.

The way to see the sign is not a finer run but a different instrument, and there are two here. The disc has a closed form, so its polygon can be measured against something that is not a polygon. And the argument about chords settles the sign for every shape without computing anything at all.

That pairing — one case with an independent answer, plus a proof that generalises the sign — is what this collection means by giving a claim a test it could fail. The most compact shape depends on the paper uses the same disc for the same reason: it is the one shape in the anchor whose true numbers are known, so it is the one that can catch an instrument rather than merely be measured by it.

What the fan cannot see at all #

Everything above is about how well a finite fan resolves a boundary it can reach. It says nothing about a boundary it cannot.

A fan from a centre finds, for each bearing, the first place along that bearing where the budget runs out. If the reach set is not star-shaped about its centre — if there is ground within the budget that lies beyond ground that is not — then the fan finds the near edge and misses everything past it, and no number of bearings recovers it, because the failure is in the walk rather than in the count.

That is not a hypothetical. A flow strong enough to carry a vehicle round an obstacle, or a cost surface with a barrier in it, produces exactly that shape.