A sheet that turns with the twist

Every panel holds a frame explained why a folded pattern can give different directional shrink factors depending on which panel is held down, which the folded strip lies at its own slant had found on half the patterns it measured. Holding a panel does not change the folded paper; it turns the frame the paper is measured in, by twice the alternating sum of the crease directions crossed to reach that panel. So a pattern’s panels impose a set of frames, read off its creases before anything is folded, and the placements agree exactly when the folded state is the same width in every frame. A width cannot tell a shape from its half-turn, so the symmetry that matters is the folded outline’s with its half-turn averaged in.

The table that result explained had the twists at their most awkward. On the unit square this series draws every pattern on, the three-sided twist gave two pairs of factors, the six-sided twist two and the five-sided twist five — one pair for every frame its panels hold. The essay’s closing suggestion was the direct test: cut the twists from sheets that match them. A six-sided twist on a hexagon should close the gap between its frames; a five-sided twist on a pentagon should collapse five answers to one.

Both do. And the reason they do is a count that says, for any twist on any regular sheet, how many answers there will be before anything is folded.

Each twist on its own sheet #

The twists are the same twists as before: the same central polygon, of the same size, with the same two pleats leaving each corner at the same angles. Only the paper round them changes. Each pleat runs until it leaves the sheet, and on a regular sheet of the twist’s own number of sides it leaves through the side facing it, so the pattern has the twist’s full rotational symmetry — the square sheet gave the triangle, pentagon and hexagon twists only as much of it as a square shares with them.

Every one of the four gives one pair of factors. The triangle twist holds its folded state in three frames and gives one answer; the square twist two frames, one answer, as it did on the square; the pentagon five frames, one answer; the hexagon three frames, one answer. The frames are unchanged — they are written in the crease directions, and the crease directions are the same on any paper — and the widths in them now agree.

The pentagon twist shows it most plainly. On the square its five frames need boxes from 0.64 by 0.62 to 0.81 by 0.80 of the sheet, and every panel held gives one of five pairs of factors. On its own sheet all five frames need a box of 0.441 by 0.432, the same box turned by the frame’s angle, and every panel held gives the same pair.

A count of rotations #

Why the widths agree is a matter of which rotations the folded state has, and it can be stated without folding anything.

A twist’s frames are fixed by its creases. A $K$-sided twist’s crease pattern is unchanged by a turn of $360/K$ degrees, and the frames its panels impose — twice the alternating sum of the crease directions crossed, modulo a half-turn — come out as the multiples of $f = \gcd(360/K, 180)$ degrees: 60 for the triangle and hexagon twists, 90 for the square, 36 for the pentagon. Those are the frames the table showed: three for the triangle and hexagon, two for the square, five for the pentagon.

A sheet’s shared rotations fix which frames must agree. A regular $M$-sided sheet placed with a corner on the direction of one of the twist’s corners shares the twist’s rotations by $360/g$ degrees, $g = \gcd(K, M)$. The crease pattern and outline are then unchanged by those rotations, so the folded state is too, and its width in any frame equals its width in that frame turned by $360/g$. Add the half-turn a width cannot see, and the widths are the same in every frame that a multiple of $s = \gcd(360/g, 180)$ degrees reaches.

So every panel gives the same pair when $s$ divides $f$, and otherwise the frames fall into $s/f$ classes that may each give a different pair. On its own sheet $g = K$ and $s = f$, and a twist always agrees with itself.

On twenty-two pairs of twist and sheet the count is exactly right — not just an upper bound that happens to be met, but equal to the number of answers on every pair measured. A hexagon twist on a triangle gives one answer: the triangle shares its turns by 120 degrees, a half-turn added makes 60, and 60 is the twist’s frame step. A triangle twist on a hexagon gives one answer for the same reason read the other way. A pentagon twist gives five answers on a square, a triangle, a hexagon and a twelve-sided sheet, and one on a pentagon or a decagon, since only five and ten share a rotation with it. A square twist gives two answers on a triangle, pentagon, hexagon and decagon and one on a square or a twelve-sided sheet.

The count, worked twice #

The count is short enough to do in the head, and doing it twice shows both of its outcomes. A pentagon twist on a decagon: five and ten share $g = 5$, so the sheet shares the twist’s turns by 72 degrees; a half-turn added makes steps of $\gcd(72, 180) = 36$ degrees; the twist’s frames are the multiples of $\gcd(72, 180) = 36$ too. Every frame is reached, and every panel agrees. The same twist on a square: five and four share only $g = 1$, the identity; a half-turn added makes steps of 180 degrees, which reach nothing a width does not already ignore; the five frames at multiples of 36 degrees fall into five classes, and the measurement finds five pairs of factors.

The half-turn is what makes the count more generous than symmetry alone. A hexagon twist on a triangle shares only turns by 120 degrees, and its frames are sixty apart, so by rotations alone the frames at 60 and 180 degrees would be left out. But a frame is a direction modulo a half-turn, since a box has no front, and 240 degrees is 60 modulo 180. The triangle reaches all three of the hexagon twist’s frames because a width treats a shape and its half-turn alike — which is exactly the property every panel holds a frame found the even Miura exploiting, now doing the same work for a sheet.

What the unit square was doing #

Read with the count, the earlier table’s twist rows stop being a list of cases. The unit square shares a quarter-turn with the square twist, and the square twist’s frame step is a quarter-turn, so it agrees with itself — one answer. The square shares only its half-turn with the five-sided twist, and a half-turn adds nothing a width does not already ignore, so all five frames are free to differ, and they do. For the triangle and hexagon twists the count allows three answers on a square and the measurement found two: the square’s half-turn predicts nothing, but something the count does not use — presumably a mirror of the folded outline — makes two of the three frames agree. That is the one place in the whole comparison where the count is a bound rather than the number — and it is on the axis-aligned square, the one sheet that is not placed with a corner on a twist’s corner.

So the sufficiency runs one way. Enough shared rotation makes every panel agree; too little makes disagreement possible, not certain, and another symmetry of the folded state, a reflection or an accident of the outline, can still merge frames the rotations leave apart.

One pair of factors, per twist #

On their own sheets the twists shrink by one pair of factors each: the triangle twist 1.64 along and 1.52 across, the square twist 2.94 both ways, the pentagon 2.16 and 2.09, the hexagon 1.52 and 2.00. The pairs differ between along and across only because a triangle, a pentagon and a hexagon are not as wide as they are tall; the folded state’s widths are the same in every frame. The square twist on its own sheet — a square turned so its sides face the twist’s, a diamond on the page — shrinks by nearly three in both directions, against 1.52 on the axis-aligned square — partly because a factor divides by the sheet’s span on the page, and a diamond spans its full diagonal while holding half the square’s paper.

That is also the answer to the earlier essay’s first open question: which frame is the natural one for a shrink factor? For a twist on its own sheet the question does not arise, since every frame gives the same pair. For a twist on a sheet that shares too little, the essay proposed reporting the factors as a set with the number of panels holding each. The count says how large that set can be before anything is folded, and the measurement says whether it is.

A hexagon twist on a triangle #

The most useful case in practice is the mismatched one. A hexagon twist does not need a hexagon of paper to measure consistently: a triangle shares enough of its rotations, since three turns of 120 degrees with a half-turn added reach every multiple of 60, which is all the hexagon twist’s frames. On the triangle its three frames need a box of 0.433 by 0.495 each. On the square two frames need 0.777 by 0.856 and the third 0.706 by 0.577.

That connects this result to a sheet rewards the multiples of its sides, where the regular polygon a regular sheet holds best is decided by whether the polygon repeats all the sheet’s directions. Here it is the other way round: the sheet has to repeat enough of the twist’s directions, and the half-turn a width ignores lets a sheet with half the twist’s symmetry suffice when that half is the right half. In both, a measurement made on a sheet sees only the symmetry the sheet and the shape share, and the question to ask first is how much that is.

Which paper to cut #

The count is also a recipe, and it is worth stating as one because the smallest sheet that works is not always the obvious one. For a $K$-sided twist, the sheets whose every panel agrees are the regular $M$-gons with $\gcd(360/\gcd(K, M), 180)$ dividing $\gcd(360/K, 180)$. For the triangle twist those are the triangle, the hexagon and every multiple of three; for the square twist the square and every multiple of four; for the pentagon twist the pentagon and every multiple of five; for the hexagon twist the triangle as well as the hexagon, and every multiple of three. The hexagon twist is the one twist in this list that can be measured consistently on paper with fewer sides than itself.

A folder wanting a twist’s shrink factors without the ambiguity of frames therefore has a short list of sheets, and among the twists here the square is on it only for the square twist. The count also reaches twists nobody measured: an eight-sided twist’s frames are 45 degrees apart, and a square, sharing only quarter-turns with it, would leave them in two classes, so the count predicts that an octagonal twist is sure of one answer only on an octagon or a sheet of sixteen or twenty-four sides, and may give two on a square. That is a prediction made without folding anything, and it is the kind the count is for. A square that turns introduced the square twist on a square sheet, and it was the one twist whose table row had never needed explaining; the count says it was also the one twist for which the square was the right paper.

What the count takes as given #

The widths are the flat folded state’s. Every panel is placed by reflections across the creases between it and a fixed panel, and a width is a width of that geometry; the lettering, which decides which panel lies on top, does not move any panel, and nothing here depends on it. The count therefore says nothing about whether the twist folds without its layers crossing on each sheet; which polygons twist is where that question lives, and it is answered for the twists on the square.

The twist is the same size on every sheet, a central polygon of radius 0.17 in a sheet inscribed in the unit square. A larger twist on the same sheet would change every width, and the pleats of a much larger one would leave through different sides of a small sheet; the count assumes only that the pattern and outline share the rotations, which holds for any size that keeps each pleat leaving through the side it faces.

And the sheet is placed with a corner on the direction of a twist’s corner. Turn the sheet relative to the twist and the shared rotations are lost unless the turn is itself one of them. A regular sheet with the right number of sides, cut carelessly, is a sheet with less symmetry than the count needs.

How the twists were measured #

Each twist is built on its sheet by running every pleat to the first side of the sheet it meets, lettered by the same search as on the square, and folded flat; for every panel the frame is read from the panel’s own turn and the factors from the folded state’s widths in that frame, and holding each panel is required to give the factors its frame predicts. On twenty-two pairs of twist and sheet the number of different pairs of factors is required never to exceed the frame classes the shared rotations and half-turn leave, and to be one wherever the count says one.

Still open: twists that tile #

A tessellation of twists repeats in a lattice, and a lattice has its own rotations. A single twist on its own sheet gives one answer; a patch of square twists on a square grid, or of hexagon twists on a triangular lattice, has the symmetry of the lattice as well as of each twist, and the count would say whether a patch cut to the lattice’s own outline also gives one pair of factors. Any tiling makes a twist builds such patches from any tiling, and the ones on tilings with no rotations at all would be the hard case.

The other direction is the corrugations. The turn a column costs found the Yoshimura carrying one column onto the next by a turn of 240 degrees, which is a frame of 60 modulo a half-turn, and what a corrugation costs measured covered area, the one shrink number no frame changes. A Yoshimura cut from a sheet sharing its column turn should agree with itself the way the hexagon twist on a triangle does, and it is a pattern people fold for its shrinking.

Sideways from here, the symmetry the letters cannot keep found a folded object keeping only the symmetries its lettering keeps. The widths here keep every rotation the pattern and paper share whatever the lettering, which is the difference between where a panel lies and which panel is on top.

The habit worth carrying is about measurements that depend on a choice. Before choosing the convention, count the symmetries that make the choice not matter. The frames were always going to disagree on a sheet that shares too little with the twist, and the count of shared rotations says, in advance, how many answers to expect.