The floor at degree four

Where two bands part turned the ground a vertex creases twice into a closed form. Every crease sits in a band of paper, and near a vertex the bands overlap: they share a disc, and beyond it each neighbouring pair of creases shares a kite whose area is $k(\theta) = h^2\left(\cot\tfrac{\theta}{2} - \tfrac{\pi - \theta}{2}\right)$ for the sector $\theta$ between them and a band half-width $h$. A vertex is its disc, plus its kites, less the overlaps of neighbouring kites — which are themselves kites, for the two sectors taken together. In band widths squared, with $h$ a half,

$$D = \tfrac14\Bigl(\pi + \sum_i k(\theta_i) - \sum_i k(\theta_i + \theta_{i+1})\Bigr),$$

where $k$ is nought at a half-turn or more.

It then found a floor by searching. At every degree it tried, the vertex with equal sectors doubled the least ground — exactly one band width squared at degree four, $\sqrt3$ at six, $2\sqrt2$ at eight — and twenty thousand random vertices at each degree never went below. It said plainly that this was a finding, not a proof, and closed on the case that might be elementary: a flat-foldable vertex of degree four has only two free sectors, and it is the vertex every Miura-type pattern is built from.

It is elementary. The proof is three identities and two derivatives, and each step says something about what a designer pays for.

Two free sectors #

A degree-four vertex folds flat only if its opposite sectors add to a half-turn — Kawasaki’s condition — so going round it the sectors are $\alpha$, $\beta$, $\pi - \alpha$ and $\pi - \beta$. Two numbers describe every flat-foldable degree-four vertex there is.

The first identity is about a sector and its supplement. Their kites add to

$$k(\theta) + k(\pi - \theta) = \cot\tfrac{\theta}{2} + \tan\tfrac{\theta}{2} - \tfrac{\pi}{2} = \frac{2}{\sin\theta} - \tfrac{\pi}{2},$$

so the four kites of a flat-foldable degree-four vertex are $2/\sin\alpha + 2/\sin\beta - \pi$, and the $\pi$ cancels the disc’s. The disc is not in the answer at all: what a degree-four vertex doubles beyond the square grid’s is entirely a matter of how the four sectors fall.

The second is about the overlaps. The four pair sums are $\alpha + \beta$, $\pi - \alpha + \beta$, $2\pi - \alpha - \beta$ and $\pi + \alpha - \beta$, and they come in two opposite pairs that each add to a full turn. So in each pair at most one sum is under a half-turn, and only that one has a kite. Of the four overlaps, at most two are there, and they are the kites of $\sigma$, the smaller of $\alpha + \beta$ and its complement, and of $\tau = \pi - |\alpha - \beta|$.

The ground as a function of two numbers #

The vertex does not change if its two free sectors are swapped, or if both are replaced by their supplements — that is the same vertex read the other way round — so it can be arranged that $\alpha \le \beta$ and $\alpha + \beta \le \pi$. Write $x$ for the mean of the two and $y$ for half their difference, so $\alpha = x - y$ and $\beta = x + y$, with $x$ at most a right angle and $y$ between nought and $x$. The third identity, $\sin(x + y) + \sin(x - y) = 2\sin x\cos y$ with $\sin(x + y)\sin(x - y) = \sin^2 x - \sin^2 y$, puts the two sines over one denominator, and the doubled ground is $f(x, y)/4$ with

$$f(x, y) = \frac{4\sin x\cos y}{\sin^2 x - \sin^2 y} - \cot x - x + \frac{\pi}{2} - \tan y + y.$$

Four right angles are $x = \pi/2$, $y = 0$, and there $f = 4$ and the ground is one band width squared, as the search found.

The table reads the function on vertices a designer might draw. The Miura’s printed vertex, sectors near 70 and 110 degrees, doubles 1.06 band widths squared. Keep the mean at 70 and part the free sectors — 60 and 80, 50 and 90, 35 and 105 — and the ground climbs to 1.08, 1.15 and 1.36. A vertex of 20 and 60 degrees doubles nearly two.

One family, two edges #

The plane has two edges on which vertices of a familiar kind sit, and they turn out to be one family. Along the bottom edge the free sectors are equal: $x$, $x$, $\pi - x$, $\pi - x$ in turn. Along the right edge the mean is a right angle and the sectors are $\tfrac{\pi}{2} - y$, $\tfrac{\pi}{2} + y$, $\tfrac{\pi}{2} + y$, $\tfrac{\pi}{2} - y$. Read round the vertex from a different crease, the second is the first with $x = \tfrac{\pi}{2} - y$: both are the Miura’s vertex, two equal acute sectors side by side. The reduction has drawn the Miura’s family twice, once as a mean sector and once as a lean, and the function agrees on both: $f(\tfrac{\pi}{2}, y) = 4\sec y - \tan y + y$, which is $f(x, 0)$ at $x = \tfrac{\pi}{2} - y$.

Every other point of the plane is a vertex with no two equal sectors next to each other — the general flat-foldable degree-four vertex, which the family the Miura belongs to found in quadrilateral meshes that fold with one freedom like the Miura’s and are not Miuras. The two steps of the proof are two ways of walking from such a vertex to the corner: first along the half-difference to the Miura’s family, then along the family to the square grid. The ground falls on both walks, and that is the whole argument.

Parting the sectors always costs #

The first derivative is in $y$, at fixed mean. Differentiating,

$$\frac{\partial f}{\partial y} = \frac{4\sin x\,\sin y\,(\cos^2 x + \cos^2 y)}{(\sin^2 x - \sin^2 y)^2} - \tan^2 y.$$

Write $a = \cos^2 y$ and $b = \cos^2 x$; since $y \le x \le \pi/2$, $a \ge b$, and $\sin^2 x - \sin^2 y = a - b$. The derivative is at least nought exactly when $4\sin x\,(a + b)\,a \ge \sin y\,(a - b)^2$, and because $\sin y \le \sin x$ it is enough that $4a(a + b) \ge (a - b)^2$, which is $3a^2 + 6ab - b^2 \ge 0$ — true, since $a \ge b$. So at any mean, the doubled ground rises as the two free sectors part, and it is least when they are equal.

Equal free sectors are the Miura’s family: sectors $x$, $x$, $\pi - x$ and $\pi - x$, the acute pair side by side. The derivative says a designer who wants a vertex at a given mean sector should draw it that way, and the premium for not doing so is the dashed rise — small for a small parting, since the derivative is nought at $y = 0$, and steep once the smaller sector is narrow.

Opening the mean always saves #

The second derivative is along the equal-sector edge. There $f(x, 0) = 4/\sin x - \cot x - x + \pi/2$, and

$$\frac{d}{dx} f(x, 0) = \frac{-4\cos x + 1 - \sin^2 x}{\sin^2 x} = \frac{\cos x\,(\cos x - 4)}{\sin^2 x},$$

which is negative for every $x$ below a right angle, since $\cos x$ is positive and less than four. So along the Miura’s family the doubled ground falls all the way to four right angles.

The reason is visible in the terms. Along the family the two acute sectors are equal and the two obtuse ones are their supplements, so only one overlap survives, the kite of the two acute sectors taken together, $2x$. As $x$ opens, every kite shrinks — the kite of a sector falls as its cotangent does — and the surviving overlap shrinks with them until at a right angle the pair of acute sectors makes a half-turn and has no kite at all. What is left is the four kites of right angles, each $1 - \tfrac{\pi}{4}$ half-bands squared, and the disc, and they add to exactly one band width squared. The square grid is the vertex whose kites are as small as kites can be while still being four of them, and every other degree-four vertex has at least one sector narrower than a right angle, whose kite grows faster than any overlap can take back.

Put the two steps together and the proof is done: $f(x, y) \ge f(x, 0) \ge f(\pi/2, 0) = 4$. Every flat-foldable degree-four vertex creases at least one band width squared twice, and only four right angles crease exactly that. The bound the search reported is a theorem at degree four.

The Miura pays twice over one square #

The Miura-type curve was drawn before, in where two bands part, as a curve that happened to fall all the way to the square grid. It is now the edge of a proved bound, and it says what the Miura’s lean costs in the only terms that matter at a vertex. The printed Miura’s acute sector of 69.9 degrees doubles 1.061 band widths squared, six per cent above the floor. Leaning the pattern further to fold it smaller costs more: a Miura at 45 degrees doubles 1.36, and the steepness near small sectors is the cotangent in the kite showing through.

That connects the floor to the finest pattern a paper carries. A vertex creases the paper twice charged each vertex its doubled ground and asked how many vertices a sheet can hold before the doubled ground crowds the single-creased paper out; a disc as deep as the degree found the Miura’s finest at 207 cells a side on copier paper once the reach was corrected. The floor says the square grid’s vertices are the cheapest a flat-foldable degree-four pattern can have, so every other degree-four pattern at the same spacing reaches its paper’s limit at a coarser grid than the square grid does — a bound a designer can now quote rather than a table they have to trust.

The lean of every zigzag, priced #

The right edge gives the doubled ground of a corrugation’s vertex directly as a function of its slant: $\tfrac14\left(4\sec a - \tan a + a\right)$ band widths squared for a zigzag leaning $a$ from upright. At the printed slant of about twenty degrees that is 1.06; at thirty degrees 1.14; at forty-five, 1.36. The curve is flat at first — the derivative is nought at an upright zigzag — and steepens as the lean grows, because $\sec a$ does.

That prices a choice the corrugations in this collection now have. The slant belongs to the line found that each zigzag of a corrugation may lean its own amount, and that a tapered corrugation stacks in nearly one depth if each zigzag leans for the width of the columns beside it — nearly upright at the broad middle, forty-five degrees at the narrow ends. The narrow ends’ vertices then double about 1.36 band widths squared each, against 1.06 at a single printed slant, a third more ground creased twice at every vertex along those zigzags. The even pile is paid for in crease density, and the vertices where it is paid are exactly the ones on the narrowest columns, which have least paper between their creases to spare.

The same curve is the reason the Miura’s lean was cheap to begin with. A Miura leaning twenty degrees gives up six per cent of its vertex floor for a pattern that folds in two directions at once; most of what a Miura costs its paper is in its spacing, which a count is not a length and where the length sits priced by the metre of crease, not in the angles at its vertices. The vertex floor adds only a few per cent to that until the lean passes thirty degrees.

Where the proof stops #

The argument used one property that degree four has and degree six does not. At degree four the pair sums come in opposite pairs that each add to a full turn, so at most two overlaps survive and the formula collapses to two numbers. At degree six the six pair sums have no such pairing: on 2,397 of 3,000 random flat-foldable degree-six vertices all six pairs are under a half-turn, and all six overlaps are present. The vertex has four free sectors after Kawasaki, and the function of four variables does not visibly separate into monotone steps. The search’s result stands — none of the 3,000 falls below the even vertex’s $\sqrt3$, the least at 1.736 — and it is still a search.

What a bound is for #

A floor on the doubled ground at a vertex matters because density bounds in this series are built out of per-vertex charges. How much line is on the paper began with crease length alone; the vertex bound added what creases cost where they meet, and every essay since has charged a pattern its vertices’ doubled ground from a table of the vertex kinds it contains. A table is only as good as its rows. The proof turns the table’s first row into a lower bound for every row at degree four: any degree-four vertex, however its sectors fall, charges at least what the square grid’s does, so a pattern made of them can never be charged less than a square grid with the same number of vertices.

That is a statement a designer can use without drawing anything. A crease pattern with $V$ interior degree-four vertices, at band width $w$ and with its vertices far enough apart that their doubled grounds do not meet, doubles at least $V w^2$ of paper at its vertices alone — before any spacing charge, and whatever the pattern is. The square grid meets that bound and nothing else of degree four does.

What the proof takes for granted #

The closed form itself. The proof is about the function $D$, and $D$ is the area of the doubled ground only as far as the kite account is right: that two neighbouring bands share exactly a kite beyond the disc, that overlapping kites of neighbours are the kite of the summed sector, and that no point lies in three kites except inside the middle one. That account was checked against a polar integral of the doubled region with no formula in it, and is checked again here on the vertices tabulated; the proof inherits whatever that check covers.

Straight creases of one width meeting at a point. A real crease has a width set by the paper and a vertex is a small region, not a point, as the shortest crease is not a crease found; the bound is for the idealised vertex that every density calculation here uses. And a vertex alone: past 207 cells a side on the Miura, neighbouring vertices’ kites meet along the short creases and the ground is no longer a sum of vertices, so the floor bounds each vertex’s own ground and not the sheet’s.

How the steps were checked #

On four thousand flat-foldable degree-four vertices drawn at random, the closed form and $f(x, y)/4$ from the reduced sectors agree to $3 \times 10^{-14}$, and on the first twenty-four the polar integral of the doubled region agrees with both; no vertex drawn doubles less than one. The derivative in $y$ is compared with a difference quotient at seven means from 15 to 90 degrees, at every half degree of the half-difference, and is positive everywhere tested; the derivative along the equal-sector edge is compared at 79 means and is negative everywhere.

Still open: degree six, and the kites that meet #

Degree six is the next proof, and it needs a different idea. Its floor, $\sqrt3$, is the Yoshimura’s and the waterbomb’s corner, and it holds on every vertex searched. But the pairing of overlaps that made degree four separable is gone, and a proof would have to find some other way of splitting four free sectors into steps that each cost ground. The obvious candidate is to fix Kawasaki’s two alternating sums and move one sector at a time, but whether each such move is monotone is exactly what the formula does not reveal on sight.

The other direction is past the vertex bound, where the kites of neighbouring vertices meet along the shortest creases and the ground stops being a sum. The proof here bounds each vertex; what the sheet pays once they touch is a question about two kites facing each other along one crease, and the density a paper allows is where its answer would be compared with what printed patterns actually use.

The habit worth carrying is about searches that keep finding the same floor. When a search never beats a symmetric answer, reduce the problem by its symmetries before searching wider. Twenty thousand vertices at degree four found no exception; two identities and two derivatives explain why none exists, and they also say how much each kind of departure costs, which a search that finds nothing cannot.