Where two bands part

A disc as deep as the degree found that every flat-foldable vertex creases a disc half a band across as many times as it has creases, and that round the disc lies a fringe creased twice whose size the angles decide. It read the fringe off a sampling grid for the eight kinds of vertex on the printed patterns — 0.99 band widths squared at a right-angled vertex, 2.86 at the preliminary base’s centre — and left the obvious question standing: what is the fringe as a function of the sectors?

It has an exact answer, and the answer is a shape before it is a number. Two creases leaving a vertex share a kite of paper. A vertex is its kites less the places where neighbouring kites overlap. The area of the kite is one line of trigonometry, and a whole vertex follows by addition.

The same kite answers a second question that nobody had asked, because an older essay had already answered it: how far along a crease its band goes on sharing ground with a neighbour’s. A vertex creases the paper twice charged that distance as $w/\sin\theta$ and built a bound on the finest pattern a paper can carry out of it. The kite shows where the two bands actually part, and it is nearer the vertex than that — so the bound moves, by half as much again on the Miura, and the account of what a fine Miura does to its paper changes with it.

The two edges that touch the disc #

Put the vertex at the origin, give each crease a band of half-width $h$, and take a point at distance $r > h$ from the vertex. It lies in a crease’s band exactly when its direction is within $\arcsin(h/r)$ of that crease: nearer in angle than that and its perpendicular distance to the crease is under $h$, further and it is over. Two creases at a sector $\theta$ therefore both cover the point when its direction is within $\arcsin(h/r)$ of each, and that window has width $2\arcsin(h/r) - \theta$ — open while $r < h/\sin(\theta/2)$ and shut beyond.

The region those windows sweep out has a boundary anyone can draw. The edge of the first band on the side facing the second crease is a straight line parallel to the first crease at distance $h$, and it touches the disc. So does the second band’s facing edge. The two edges meet on the bisector, at $h/\sin(\theta/2)$ from the vertex, and between the two points where they touch the disc the ground both bands cover is a kite less a slice of the disc. The kite is two right triangles with legs $h$ and $h\cot(\theta/2)$; the slice has an angle of $\pi - \theta$. So the ground two creases share outside the disc is

$$k(\theta) = h^2\left(\cot\frac{\theta}{2} - \frac{\pi - \theta}{2}\right).$$

At seventy degrees that is 0.117 band widths squared. At a right angle it is 0.054, the four small corner pieces the square grid’s doubled ground was known to have. At 110 degrees it is still 0.022, and it is gone only at a half-turn, where the two edges are one line and touch the disc at the same point. The slider on the figure opens the sector through that whole range; the kite shrinks toward the disc’s rim and does not reach it until the creases are collinear.

The distance at which the bands part #

The kite has a second reading that matters more than its area. Its tip projects onto each crease at $h\cot(\theta/2)$ from the vertex, and that is where a crease’s band stops sharing ground with its neighbour’s. Beyond it, no cross-section of one band meets the other.

Inside the disc every band contains the vertex, so the reach is never less than half a band. At and above a right angle the kite’s tip projects inside the disc, and the reach is exactly $h = w/2$. Below a right angle it is the kite’s:

$$x^{*} = \frac{w}{2}\cot\frac{\theta}{2} \quad (\theta \le 90^\circ), \qquad x^{*} = \frac{w}{2} \quad (\theta \ge 90^\circ).$$

The earlier essay’s reach was $w/\sin\theta$ and $w$. Its derivation walked along one crease and asked when the other crease’s centre line came within a band width — which is the test for two parallel bands, where the nearest points of the two strips lie on a common perpendicular. For bands at an angle it is not: the part of the first band nearest the second crease is its edge, and that edge is offset from the centre line by $h\cos\theta$ in the second crease’s direction as well as by $h$ across. The centre-line test overstates the reach by a factor of $2/(1 + \cos\theta)$, which is half as much again at the Miura’s 69.9°, a sixth at 45° and double at a right angle. At very narrow sectors the two agree, which is why the error never showed on the steep part of the curve. The dashed curve is the old one and the solid curve is the kite’s.

A vertex is its kites less their overlaps #

A vertex has a kite for every neighbouring pair of creases. Where two sectors together are less than a half-turn, the kites of the pairs on either side of the middle crease overlap — close to the disc, a point can be within reach of three creases — and the overlap is a kite of the same shape for the summed sector. So subtracting the overlaps of neighbouring kites counts every point once. The ground creased at least twice at a vertex with sectors $\theta_1, \dots, \theta_d$ is

$$A_2 = \pi h^2 + \sum_i k(\theta_i) - \sum_i k(\theta_i + \theta_{i+1}),$$

with $k$ taken as zero at a half-turn or beyond. The subtraction is exact rather than a first correction, because a point within reach of two non-neighbouring creases is within reach of every crease between them, so the pair-kites along the circle behave like overlapping intervals in a row, whose union is their total less their pairwise overlaps. The one way that could fail is a point within reach of every crease at once, which would need every crease inside a half-plane. Kawasaki’s condition forbids exactly that, as it forbade the gap the disc argument needed; the same theorem underwrites both.

For an evenly spread vertex of degree $d$ the sum collapses. Every sector is $2\pi/d$, every pair-sum $4\pi/d$, and the $\pi$ that the disc contributes is cancelled exactly by the slices the kites leave out, so what remains is algebraic:

$$A_2 = \frac{d}{4}\csc\frac{2\pi}{d}\ \text{band widths squared}.$$

At degree four that is exactly one; at six, $\sqrt{3}$; at eight, $2\sqrt{2}$; at twelve, exactly six. The sampling grid had read 0.99, 1.74 and 2.86.

The cancellation has a picture. The doubled ground of an evenly spread vertex of degree five or more is a star polygon with straight sides — points on the bisectors at $h/\sin(\pi/d)$, waists on the creases at $h/\sin(2\pi/d)$ — and the disc lies wholly inside it, so no arc of the disc appears on its boundary and no $\pi$ appears in its area. It is $2d$ triangles, each half the product of those two distances times the sine of the half-sector between them. At degree four the star degenerates to the square the four kites make, and the disc touches it at four points. Past degree eight the star keeps its shape and grows: sixteen creases evenly spread double 10.45 band widths squared, within three per cent of $d^2/8\pi$. Doubled ground grows as the square of the degree, while the disc creased as deep as the degree stays $\pi/4$ at every degree.

The table was right to a per cent, and one sentence was not #

The formula can be checked against two things that do not use it: the sampling grid the earlier table came from, and a polar integral of the doubled ground’s boundary, which reads, in every direction, how far out the second-nearest band still reaches and assumes nothing about kites.

The formula and the polar integral agree to the fourth decimal on all eight kinds. The grid was within 1.3 per cent everywhere — low by 0.9 per cent on the right-angled vertex and by 1.3 on the Miura’s, high by 1.2 on the preliminary base’s centre — which is what a grid of square cells does to a boundary made of slanted lines, and none of the earlier essay’s conclusions rested on the third figure. The waterbomb’s degree-six vertex, with sectors of 45, 45 and 90 degrees twice over, comes out at exactly two.

The Miura’s vertex decomposes cleanly. Its sectors run acute, acute, obtuse, obtuse — 69.9, 69.9, 110.1, 110.1 at a slant of 0.35 — and it carries the disc, 0.785, two acute kites of 0.117 each, two obtuse kites of 0.022 each, less the overlap of the two acute kites, 0.004, whose sectors together make 139.8 degrees. Total 1.061.

The obtuse kites are the sentence that was wrong. The disc essay said that between two creases at a sector of a right angle or more the bands part at the disc’s rim. They do not: at 110.1 degrees they share a kite reaching 1.22 half-bands out along the bisector, and the Miura’s two obtuse sectors account for 0.044 of its 1.061 band widths squared, four per cent of its doubled ground. The error was harmless to that essay’s numbers, which came from sampling rather than from the sentence, and it is the same error as the reach: both treated the bands as parting where their centre lines diverge, when a band is parted from its neighbour only once its far edge has cleared the other’s.

What a lean costs #

Every degree-four vertex on the printed patterns is of the Miura’s type — two equal acute sectors side by side and their supplements opposite — so a single curve prices all of them.

It falls all the way to the right angle, where it is one band width squared, and it rises slowly at first: 1.015 at 80 degrees, 1.061 at the Miura’s 69.9, 1.141 at the hexagon twist’s 60. Past 45 degrees it climbs steeply, because a kite’s area grows as $\cot(\theta/2)$ and that is $2/\theta$ at a narrow sector; at 30 degrees it is 1.83 and at 20 degrees 2.54. The dashed curve shows what the overlap subtraction is worth. At the Miura’s sectors it is a third of a per cent; at 30 degrees, where the two acute sectors together make a narrow 60, it is nine per cent.

So as area, the lean is cheap. A Miura at 69.9 degrees doubles six per cent more ground at each vertex than a square grid would, and that is the price of the single folding freedom a sheet with one freedom describes. As reach it is not cheap at all, which is where this account arrives at the bound.

No vertex spreads its damage thinner than the even one #

The star suggests a question the formula can answer by search even where it cannot yet answer it by proof: among all flat-foldable vertices of a given degree, which doubles the least ground?

On the four hundred degree-six vertices drawn, none doubles less than the evenly spread vertex’s 1.732, and the cloud falls toward that line as the narrowest sector opens toward sixty degrees, which is as wide as the narrowest of six sectors can be. A wider search, twenty thousand vertices at each of degrees four, six and eight, found the same floor at each degree: 1, 1.733 against $\sqrt{3}$, and 2.832 against $2\sqrt{2}$. Equal sectors appear to be the cheapest arrangement of a given number of creases, and the degree then fixes the price. That is measured and not proved. The formula is a difference of convex terms, so the obvious inequality does not apply to it directly, and a search that has not found a counterexample has not shown there is none.

If it holds, it sets a floor any designer can read off a crease pattern. A vertex of degree $d$ doubles at least $(d/4)\csc(2\pi/d)$ band widths squared however its sectors are chosen, and every departure from equal sectors costs more. The preliminary base’s centre is already at its floor. The smallest sector decides whether a vertex folds; here it decides, on this evidence, how far above its floor a vertex’s damage sits.

The bound, charged at the reach the kite allows #

The earlier essay’s third bound on fineness charged each crease for the ground near its two ends that its band shares with a neighbour’s, and asked when a crease is shared from end to end. The argument is sound, and only the reach was wrong. Charged at the kite’s reach, it moves.

The Miura’s tightest crease is a zigzag crease 1.05 cells long, flanked at both ends by one acute and one obtuse sector, so each end reaches 0.715 band widths and the two together 1.43. That is 0.734 of a cell for each band width of shared ground, where the old reach gave 0.477. On a 170 mm sheet of copier paper the Miura is creased as fine as the vertices allow at 207 cells a side, not 135. On kami it is 296, on washi 519, on foil-backed tissue 799. The waterbomb’s tightest crease is a full cell flanked by 45-degree sectors at both ends, reaching 1.21 band widths each, so it carries 0.414 of a cell per band width and reaches its bound at 117 cells rather than 82.

The vertex bound still binds before the spacing bound, and it is no longer beside the density bound. The Miura’s density bound is 139 and its spacing bound 262; the waterbomb’s are 74 and 141. The earlier essay found the density bound “had the wrong argument and nearly the right number”, within three per cent for the Miura and eleven for the waterbomb. That agreement was the overstated reach coinciding with the density arithmetic. Charged correctly, the vertices allow half as much again as the density argument did, and the waterbomb 58 per cent more.

The lean is where the correction bites hardest. With no lean at all a Miura’s vertices are right angles, every reach is half a band, and a crease a cell long runs out at one cell per band width. At a slant of 0.35 it runs out at 0.734. The lean costs the Miura 27 per cent of its fineness, not the under five per cent the old reach put on it, because the old reach was flat at a band width above a right angle and nearly flat just below, while the kite’s reach doubles between a right angle and 53 degrees. At a slant of 0.5 the cost is a third, and at 0.7 over two fifths.

A sheet read before it is folded #

With the reach and the fringe both exact, the share of a fine Miura creased twice can be predicted from one vertex. While neighbouring vertices’ kites do not meet, the doubled share is the vertex’s $A_2$ over the area of a cell, $1.061\,(w/\ell)^2$, and nothing else on the sheet is doubled: the creases that do not meet stay more than a band apart until the spacing bound, well past the vertex bound.

The prediction and the sampling agree to within a point at every size drawn. At 64 cells a side the formula gives 5.5 per cent and the block reads 5.3; at 135 it gives 24.2 and the block 24.3; at 207 it gives 56.8 and the block 57.5; even at 240, past the bound, it gives 76 against 75.6. The vertex sum holds right up to the bound, and the reason is the bound itself. A crease becomes shared ground from end to end exactly when the kites at its two ends meet along it, so until then no two vertices’ doubled ground touches, and the sheet’s doubled share is the vertices’ added up.

That turns the disc essay’s most quotable number into a different one. It found that at each paper’s finest Miura a quarter of the sheet was creased twice — 24.3 per cent on all four papers, the same to the digit — because the bound fixes the ratio of band to cell and the share is a function of that ratio alone. The argument stands, and the ratio it fixes is now $w/\ell = 0.734$ rather than 0.477. At its finest, a Miura on any paper is 57.5 per cent creased twice and 42 per cent creased four deep. The finest a paper allows is a sheet more than half of which is creased more than once, and the reading $1.061 \times 0.734^2$ gives it before anything is folded.

What the kite takes for granted #

A band is a strip with sharp edges, six sheet thicknesses wide by the convention these essays use, and the paper outside it is untouched. The crease has a radius measured a real fold as an arc whose disturbance falls off rather than stopping, so a kite here is the ground inside a stated width, not the ground where anything physical changes. The shape is exact for the idealisation and only for it.

Creases run straight at least to the tip. A kite reaches $h/\sin(\theta/2)$ out along its bisector, which is 1.75 half-bands at the Miura’s acute sector and 2.61 at 45 degrees. A crease shorter than that has its kite cut off by its far end, and the formula then overstates what that vertex doubles. On every printed pattern the creases are many times longer; on the finest Miura the bound is precisely where the kites of neighbouring vertices begin to reach one another.

Vertices are counted alone. The formula is one vertex’s doubled ground, and a sheet’s share is the sum only while those regions stay apart — which the argument above shows is true up to the vertex bound and not beyond it.

And the floor under the even vertex is a finding from a search. Nothing here proves that equal sectors minimise the doubled ground of a given degree. The disc, the kite, the formula and the corrected reach are exact; the floor is not.

Where the circle goes #

The surprising thing is where the disc goes. Every band contains the disc and the disc is round, so it seems that any area built from bands must carry a $\pi$. The ground creased as deep as the degree does — it is the disc, $\pi/4$, at every vertex. The ground creased twice, at an evenly spread vertex of degree five or more, does not: the kites leave out slices of the disc that add up to exactly the disc, and the doubled ground is a polygon whose area is $\sqrt{3}$ or $2\sqrt{2}$ or six. The deepest damage at a vertex is round and the widest is straight-sided, and the same bands make both.

It also joins two errors that were one. The reach the earlier bound charged and the sentence about obtuse sectors both measured where two creases’ centre lines diverge instead of where their bands do. That is a natural slip, because the centre line is what a crease pattern draws and the band is what the paper does; the kite is simply the band’s view of the same vertex. How much line is on the paper priced a pattern by its lines, and it was right to, because length is a property of lines. Shared ground is a property of bands, and a band’s edge is set off from its centre line in two directions at once.

Still open: the floor, and the kites that meet #

Whether equal sectors minimise the doubled ground at every degree is the proof this account owes. The formula is explicit, the constraint is Kawasaki’s, and the search found no exception at degrees four, six and eight; a proof would turn the floor $(d/4)\csc(2\pi/d)$ into a bound a crease-pattern designer could quote. At degree four it may be elementary, since a flat-foldable degree-four vertex has only two free sectors, and it is the case every Miura-type pattern lives in.

The other direction is past the bound. Beyond 207 cells a side the kites of neighbouring vertices meet along the shortest creases, and the sampled share stops being the vertices’ sum — at 240 cells it runs 0.7 of a point under the prediction, the first sign of it. Where two kites meet, a crease has no single-creased stretch at all, and the ground between them is the region the shortest crease is not a crease was looking at from the drawing’s side. Its shape is computable from the same geometry — two kites facing each other along a crease — and it is the next thing a fine Miura’s paper is asked to carry.

Sideways from here, the corrected bound should be set against the density a paper allows, which compared the densest printed pattern with the paper’s limit and found a factor of nineteen between them. That factor was measured against the density ceiling, the most pessimistic of the three bounds, and the vertices now allow half as much again as that ceiling on the Miura and more on the waterbomb. The gap between what is printed and what a paper carries is wider than nineteen, and the vertex bound is the one to measure it against.

The habit worth carrying is about which line a width is measured from. When two strips meet at an angle, the distance between their centre lines is not the distance between the strips. For parallel strips the two coincide, which is how the test became habit; at an angle the edge of one strip moves toward the other as well as away from its own centre, and the difference — half as much again at seventy degrees — is the whole of the correction here.