Two mechanisms at one point

Flat-foldability and rigid-foldability are two different properties with two different tests, and most of the interesting cases in this subject satisfy one and not the other. A crossing — two straight creases drawn across each other — is the case where they come apart as far as they can go. It has no flat folded state at any angle and under any lettering, and as a mechanism it works: four flat panels, hinged, moving.

It moves in two ways rather than one, and the two have nothing to do with each other.

The solver says so by refusing #

The vertex solver here treats a vertex as a closed polygon on a small sphere around it: each crease pierces the sphere at a point, each sector becomes an arc of fixed length, and folding moves the points without changing the arcs. Given three creases it solves for the fourth by intersecting two cones.

That construction has a documented failure, and it is exactly this vertex. When the first and third creases are collinear the two cones impose the same condition, the fourth crease is free to sit anywhere on a whole circle, and the solve is singular. The solver refuses rather than returning a configuration that quietly is not one — and a crossing is the case where both pairs are collinear, so it is refused twice over.

The refusal is correct and it is not an answer. What the vertex actually does needs a different route.

Asking the flat state which way it can move #

The other route is the composition of rotations. Take the four crease directions in the flat sheet, rotate about each in turn by its fold angle, and multiply; the vertex is in a legal configuration exactly when the product is the identity. That statement is about the fold angles rather than about the crease positions, and it is what the mesh solvers here use, because it has no singular construction in it.

The flat state satisfies it trivially: every fold angle nought, every rotation the identity. So the flat state says nothing about which way a vertex can move, and the useful question is asked a little way out — in which directions can the fold angles leave nought while the product stays at the identity?

Those directions are found rather than derived. Each crease in turn is driven a thousandth of a radian either way, the other three are solved by Newton from eight spread starting points, and the solutions whose closure residual is at rounding are kept and normalised. What comes back is a small set of rays, and a ray is a mode: the ratios in it are the ratios the four fold angles keep as the vertex begins to fold.

What an ordinary vertex does #

Take a developable vertex with sectors 60°, 90°, 120° and 90°. Two rays come back, and in both of them all four fold angles are non-zero.

The first is (0.683, −0.183, 0.683, 0.183): the first and third creases moving together at nearly four times the rate of the second and fourth, which move against each other. The second is (0.183, −0.683, −0.183, −0.683), which is the same numbers with the roles exchanged. Those two rays are the vertex’s two modes, and the ratio between the fast and slow creases — 3.73 here — is the gear ratio that makes a degree-four vertex a mechanism worth building with: push one crease and every other crease responds, in a proportion the angles fixed before anything moved.

Try to move only two of its creases and the closure fails outright: setting the first and third to 0.4 radians and the others to nought leaves a residual of 0.195, which is a fifth of a radian of disagreement rather than a rounding error. An ordinary vertex has no partial motions. Everything at it moves or nothing does.

What a crossing does instead #

Sectors 40°, 140°, 40°, 140° — two straight lines crossing at forty degrees. Two rays come back, as before. They are (1, 0, 1, 0) and (0, 1, 0, 1).

Read those as instructions. The first says: bend the sheet along the first line, by whatever amount, and leave the second line flat. The second says the opposite. Each is a simple fold of a flat sheet along a straight line — the oldest move in the subject and the only one a machine with one head can make — and the crossing supports both of them because a straight crease through a vertex is a straight crease across the whole sheet.

The residuals confirm it exactly rather than approximately. Folding one line to 0.4 radians with the other flat leaves a closure residual of 5 × 10⁻¹⁷, which is zero. Folding both lines to 0.4 radians leaves 0.095, which is not.

So the two motions exist, and they cannot run at once. The crossing is two mechanisms sharing a point, and the point is where they refuse each other.

The gear ratio has a formula, and it diverges here #

The 3.73 measured on the ordinary vertex is not an empirical number, and identifying it turns the crossing’s decoupling into the same statement pushed to a limit.

For a degree-four vertex satisfying the alternating-angle condition, with two adjacent sectors $\alpha$ and $\beta$, the ratio between the fold-angle rates of the two crease families at the flat state is

$$\mu = \frac{\sin\big((\beta+\alpha)/2\big)}{\sin\big((\beta-\alpha)/2\big)}$$

At sectors of 60° and 90° that is $\sin 75° / \sin 15° = 0.9659/0.2588 = 3.732$, against the measured 3.73. The mode’s numbers are a trigonometric identity, not a fit.

Which says what a crossing is #

Now evaluate it at the one crossing the alternating condition admits — four right angles.

$$\mu = \frac{\sin 90°}{\sin 0°} = \frac{1}{0}$$

The gear ratio is infinite, and an infinite ratio is exactly the mode $(1, 0, 1, 0)$: one family of creases moves and the other does not move at all, whatever the first is driven to.

So the crossing’s two independent motions are not a different phenomenon from an ordinary vertex’s gearing. They are the gearing at its singularity, reached when the two adjacent sectors become equal and the denominator vanishes.

That also says how a vertex approaches the crossing. Take sectors of 89° and 91° — a vertex a hair from being a crossing — and the ratio is $\sin 90°/\sin 1° = 57.3$. At 89.9° and 90.1° it is 573. The gearing does not fail suddenly; it grows without bound as the vertex flattens toward two straight lines, and the crossing is where it has finished growing.

And it explains why nothing is built on it #

A vertex with a gear ratio of 573 is not a useful mechanism, and the reason is not that the number is large. It is that a rate of 573 to one means the slow crease has to be placed to a five-hundredth of the fast one’s precision for the assembly to reach the state it was designed for — and a vertex at exactly the crossing has no relation between them at all.

So the crossing’s uselessness is continuous with its neighbours’ rather than categorical. A vertex near a crossing is a mechanism with an unusable gear ratio; a crossing is one with none. The essay’s sentence about half-planes is the geometric reading of the same divergence, and the two agree because $\alpha = \beta$ is exactly the condition for the two creases to be one straight line.

Why the two cannot be combined #

The reason is visible without any of the arithmetic once the right question is asked, which is what the paper on either side of a line is.

A straight crease through the vertex separates the sheet into two half-planes, and the paper on each side of it is one piece — the second line lies inside those pieces rather than between them. Folding the first line therefore folds two rigid half-sheets about a hinge, and nothing at all is asked of the second line, because it is not on the boundary between the moving pieces.

Fold both and that stops being true. The four panels are now hinged in a ring, and going round the ring the rotations have to compose back to the identity, which is the condition that gives an ordinary vertex its gear ratio. At a crossing the ring’s constraint has only the two trivial solutions — one line or the other — because the sectors come in two equal opposite pairs and any non-trivial mixture leaves the residual above.

That is the same fact Maekawa’s count expresses in the flat case, arriving in three dimensions instead of by winding: a crossing’s four spokes are two creases, and two creases through a point cannot be made to cooperate.

The same closure runs the whole subject #

The rotation composition is not a device invented for this vertex. It is what every rigid-folding computation here is built on, which is why the crossing’s answer can be trusted as far as it can.

Solving a whole quadrilateral mesh is this same condition written once per interior face and solved together: the residual is a vector with one entry per face, the crease lengths are the unknowns, and a damped Gauss–Newton step closes them all at once. Following a mesh through its motion is the same condition continued from one fold angle to the next. The vertex here is the smallest instance of the machinery that does all of that — one face, one loop, four creases.

So the two rays are produced by the same code that produces the mesh results, at a size where the answer can be read off directly. And the check the other way round holds too: on vertices the spherical solver can handle, the two agree — that comparison is run on every build, and it is what makes the composition trustworthy on the vertices the spherical solver refuses.

Why neither motion reaches flat #

The crossing’s motions are real and they are also useless for flat folding, and the reason is a sentence rather than a theorem.

A flat folded state has every fold angle at plus or minus a half turn. On the mode that bends only the first line, the second line’s fold angle is nought for the whole motion and stays nought however far the first is driven. So the motion approaches a state in which one line is fully folded and the other is not folded at all, which is a sheet folded once — not the flat state of the crossing pattern, which would need both lines fully folded together.

There is no path from one mode to the other except through the flat state itself, and at the flat state nothing is folded. So the pattern as drawn has two motions, both of which lead somewhere that is not its own flat folded state, and that is the strongest possible version of the difference between the two properties.

The right-angled crossing behaves the same way #

The one crossing Kawasaki’s condition allows is the square one, and it might be expected to behave differently. It does not.

Sectors 90°, 90°, 90°, 90° return the same two rays — (1, 0, 1, 0) and (0, 1, 0, 1) — and the same refusal of any mixture, with a residual of 0.152 when both lines are driven together. Being square buys the crossing an angle condition and nothing kinematic; it still has two decoupled motions and no others.

That is worth having because it separates two arguments that could be confused. Kawasaki’s condition is about reaching a flat state and says nothing about the way there; the modes are about the way there and say nothing about the destination. A vertex can satisfy the first and have the second look like a crossing’s, which is exactly what the square case does.

The two rays are also the reason the square crossing is not a special case worth building on. A vertex whose modes are (1, 0, 1, 0) and (0, 1, 0, 1) transmits nothing: driving one crease leaves the other pair exactly where it was, whatever the sector angles happen to be, so the gearing that makes an ordinary degree-four vertex worth assembling into a mesh is not merely small here but absent. At 60°, 90°, 120° and 90° the ratio between the fast and slow creases was 3.73; at 40°, 140°, 40°, 140°, and at four right angles, there is no ratio to quote, because the creases it would relate are not moving. That is the arithmetic form of the sentence about half-planes above, and it holds at every angle a crossing can be drawn at.

What a crossing would do inside a mesh #

A crossing has never appeared in a mesh here, and it is worth saying what would happen if one did, because that is where the cost of the drawing fault would actually be paid.

A quadrilateral mesh folds because the vertices are geared to one another: each interior vertex turns its neighbours’ fold angles into its own, and driving any crease drives the whole sheet. A crossing among them would be a vertex that does not transmit. Driving one of its lines would move that line and nothing else, and the two halves of the mesh on either side would be free of one another — so the sheet would have two motions where it should have one, and neither of them would reach the other’s territory.

That is not a small defect in a deployable. A structure that opens by being pushed at one point relies on every panel being carried by that push; a vertex that decouples is a hinge the actuator does not reach. Where an error goes in a mesh is already a question with a careful answer here, and a crossing is not an error of that kind — it is not a small departure from a solved mesh but a place where the solve has a different structure.

The practical consequence is the one this collection reached from the flat side too. A crossing must be found in the drawing, before anything is solved, because every instrument downstream of it either refuses it or answers a different question about it.

What a folder feels #

Any of this can be checked with a sheet of paper in about a minute, and the checking is more convincing than the arithmetic.

Draw two lines across a square and crease along one of them only. The paper folds; there is nothing remarkable about it, because a straight crease across a sheet is a fold whether or not other lines are drawn on it. Unfold, crease the other line only, and the same thing happens. Both motions are there and both are ordinary.

Now try to bring both up at once. The paper resists in a particular way — not the way it resists a fold made against the grain, which is a matter of force, but by refusing to choose: the sheet buckles somewhere off the lines, or one line takes the fold and the other flattens out, or a cone forms at the crossing point and the flat state never arrives. What is happening is that the four panels cannot all stay flat, so the material does the only other thing available and bends.

That is the difference between the model and the paper, stated at the one vertex where it is easiest to feel. Panels instead of paper is an idealisation that gives clean answers, and its clean answer here is no. Real paper answers not quite, and the not-quite is a cone of curvature a few millimetres across at the crossing — which is the same radius that turns every crease in this subject from a line into a region.

Where the reading is careful #

The rays are computed, not proved. They come from a Newton solve at a thousandth of a radian with eight starting points, kept when the residual is at rounding. That is enough to find the modes a vertex has and not enough to prove it has no others — a mode reachable only from a starting point outside the eight would be missed. The count agrees with the two the subject expects for a degree-four vertex on every case tried here, which is evidence rather than a theorem.

It is about degree four. A crossing has four creases by construction, and the comparison vertices do too. What three lines through a point do — which fold flat, unlike two — is a six-crease question and is not answered here.

And the panels are rigid, which paper is not. Every panel here is flat and unbending, which is the idealisation the whole rigid half of this subject rests on, and at a crossing it is the difference between a clean refusal and the cone of curvature a real sheet produces instead. The model’s answer is exact and the paper’s answer is a few millimetres wide.