Plain weaves (each strip alternates going over and under cross strips) in origami have been done for a while. Some are simple, such as the well-known square weave. Some are more complicated, like Courteous Anarchy by Robert Lang (still a plain weave!)

A closed square weave with a pleat width of half the strip width

However, there’s other types of weaves. Like the (2,2)-twill (each strip goes over 2 cross strips and under 2 cross strips, and adjacent strips have their over-unders offset by 1). So I challenged myself to make one. I’ll be making a closed (2,2)-twill, because it looks cool1

A(n open) (2,2)-twill, for reference
A closed (2,2)-twill. This is what I'm targeting.

Setting Up the Pleat Shift

First, we lay out the rectangles, and add pleats so the rectangles are set to go to their target positions. Aiming for a pleat width of half the width (short side) of a rectangle, if we call the strip width 1 (and thus the pleat width 1/2), we want to shift a pleat by 1/2 and its perpendicular pleat by 3/2, while twisting them (a blog post about pleat shifting is coming soon…)

The pleat setup for the (2,2)-twill. Note how much the pleats are shifted by, and that they're twisted.
Some labelling done on the pleat shift. Quadrilateral $$ABCD$$ is critical; no creases can cross it.

Looking at the above figure, note that no creases can cross the rectangles, as they will show. The closest point \(A\) can get to point \(B\) is a vertical distance of 3/2, making \(\overline{AB}\) critical: the distance between them in the crease pattern is the distance between them in the final folding, so no creases can cross \(\overline{AB}\). Similarly, \(\overline{CD}\) is critical. Note that \(\overline{BC}\) and \(\overline{DA}\) are critical by virtue of being adjacent, so the entire quadrilateral becomes a no-crease zone. With that in mind, let’s try constructing this twist and see what happens!

The twist. Unfortunately...
it doesn't quite work due to a pesky little triangle.

Ugh, that stings. We can try moving \(A\) farther away from \(B\) and seeing if we can fix the overlapping problem. But at the time, I didn’t realize this and gave up on this problem for a good while.2

Rotation to the Rescue

I decided that every twist needed to be a pure quadrilateral twist, with no creases crossing it. It is possible to do this with just one unit, by bringing some rectangles closer together. However, this won’t tile.

A pure quadrilateral twist that gives the intended folded state
But it doesn't tile. Pleats of different widths want to match up.

So how do we solve this? We can try rotating some of the tiles, turning the pleats into angled crimps. This would look something like this:

Like this. And it tiles. (Yes, I over-rotated the tiles, but we'll get into that)
A single "repeating unit" labelled with important values.

The “repeating unit” has 180° rotational symmetry. For it to tile, the lengths of the green line segments must equal. We’ll call that value \(x\). We fix the crimp “width” \(h\) so it’s constant, since this is a parameter we have. For the unit to fold flat, the four vertices in the twist must satisfy Kawasaki’s theroem, so \(\alpha + \beta = \gamma + \delta = 180^\circ\). When \(\theta = 45^\circ\), as shown in the diagram, \(\alpha + \beta > 180^\circ\) and \(\gamma + \delta < 180^\circ\). When \(\theta = 0^\circ\) (no rotation), then the inequalities flip. By the intermediate value theorem, there’s a value for \(\theta\) where \(\alpha + \beta = 180^\circ\). Due to the symmetries of the unit, we know that \(\beta + \delta = 180^\circ\) and \(\alpha + \gamma = 180^\circ\) regardless of \(\theta\). Thus, when \(\alpha + \beta = 180^\circ\), we also have \(\gamma + \delta = 180^\circ\), satisfying the condition. Now we just have to find \(\theta\).

So I ran FreeCAD (a constraint solver, among other things) to find this value, which ended up being about 79.187683036437065°3, and then used Oriedita to construct the tessellation, which worked (resulting in the crease pattern at the top).

The sketch made in FreeCAD. The boundary is a little different than described, but it still works. Here is the FreeCAD file

But what’s the exact value? Well, it turns out that when looking for the exact value, I realized that given \(x\), we can derive the whole “repeating unit” without needing \(\theta\), so we’ll be discussing the exact value of \(x\) instead. I used Maxima because the equations get complicated, turning into a degree-10 polynomial on \(x\). Which can thankfully be factored into quadratics. The result is that \(x\) satisfies:

\[(x+2)^2\cdot(x^2-hx+h^2+1)\cdot(x^2-hx-h)\cdot(x^2+(-h+1)x-2h)\cdot((2h+1)x^2+(-2h^2+h+1)x-2h^2)=0\]

We can set \(h = \frac{1}{2}\), a nice value for it to be, we have:

\[(x+2)^2\cdot\left(x^2-x+\frac{5}{4}\right)\cdot\left(x^2-\frac{x}{2}-\frac{1}{2}\right)\cdot\left(x^2+\frac{x}{2}-1\right)\cdot\left(2x^2+x-\frac{1}{2}\right) = 0\]

Solving this, we get:

\[x = \left\{-2, \frac{1}{2}\pm i, -\frac{1}{2}, 1, \frac{-1\pm\sqrt{17}}{4}, \frac{-1\pm\sqrt{5}}{4}\right\}\]

Well, that’s a lot of solutions. Let’s see. Like half of these are negative and thus invalid. The non-real-number solutions are also invalid. Then, note that \(h\) is the crimp “width”, so \(x > h\) doesn’t make sense. This leaves us with just \(x = \frac{-1+\sqrt{5}}{4}\). Wait, hold up, \(\cos\left(\frac{1}{5}\tau\right)\)? Also known as \(\frac{1}{2\varphi}\)? Interesting. But probably doesn’t really mean anything, especially when, since all coordinates happen to be \(a + b\sqrt{5}\) where \(a\) and \(b\) are rational numbers, \(\sin\left(\frac{1}{5}\tau\right)\) can’t show up.

The (\(k\), \(k\))-Twill

We can generalize this to the (\(k\), \(k\))-twill, where each strip goes over \(k\) strips and under \(k\) strips. In this case, our rectangles are \(k\) long (instead of 2 long), and the equation for \(x\) becomes (this time removing bad quadratics):

\[(2h+1)x^2 + (hk + k - 2h^2 - h - 1)x - h^2k = 0\]

And when \(h = \frac{1}{2}\), we get:

\[2x^2 + \frac{3k - 4}{2}x - \frac{k}{4}\]

resulting in (omitting the negative solution)

\[x = \frac{-3n + 4 + \sqrt{9n^2 - 16n + 16}}{8}\]

We can make a table of the first few values (remember, \(h = \frac{1}{2}\)).

\(k\) \(x\) Approximation
1 \(\frac{1}{2}\) 0.5
2 \(\frac{-1 + \sqrt{5}}{4}\) 0.3090169943749474
3 \(\frac{1}{4}\) 0.25
4 \(\frac{-2 + \sqrt{6}}{2}\) 0.2247448713915889
5 \(\frac{-11 + \sqrt{161}}{8}\) 0.21107219255619
6 \(\frac{-7 + \sqrt{61}}{4}\) 0.2025624189766635
7 \(\frac{-17 + \sqrt{345}}{8}\) 0.1967719526258387
8 \(\frac{-5 + \sqrt{29}}{2}\) 0.1925824035672518
9 \(\frac{-23 + \sqrt{601}}{8}\) 0.1894126680328156

Interestingly enough, \(k=3\) gives a rational value for \(x\)4. unfortunately, due to the glide reflections, the denominators of the coordinates tend to be multiples of 17.

The (3, 3)-twill crease pattern

Final Thoughts

This is a highly-efficient way to fold a (\(k\), \(k\))-twill, on par with the square weave. Unfortunately, it’s not on nice grid coordinates, though perhaps using a different value of \(h\) would help with this. This paper by Helena Verrill includes a (3, 3)-twill folded from a grid, but it’s not as efficient. This video by Madonna Yoder shows how to fold a… (2.5, 4.5)-twill? (technically not a proper weave because parallel strips have inconsistent layer order)

  1. and also because the open (2, 2)-twill has extra polygons I don’t want to worry about. This is an important difference from the plain weave. ↩

  2. Spoiler alert, this doesn’t work either. ↩

  3. Yes, you do need lots of precision when plugging this value into Oriedita. In particular, 5 decimal places isn’t enough. ↩

  4. Other values of \(k\) may give rational values for \(x\) for values of \(h\) other than 1/2. ↩