A crease held the other way

A poster that has spent a year in a tube is flattened by rolling it the other way, and left there too long it curls the wrong way instead. A folder who wants a crease to go limp folds it back on itself. Neither of them holds the paper flat, because flat is slow. Holding a fold moves the force gave the law behind this and stopped one step short of it: a hinge’s remembered turn creeps toward the turn it is held at, at a rate set by a relaxation time $\tau$, and every case it drew held the sheet further into its own fold. It ended on the other case. A structure held on the far side of the flat sheet should have a memory that crosses zero, and the essay predicted that at that moment the barrier between the two states would become something else, a moment with no preferred branch that a sheet could be programmed through.

The time is easy, and the prediction turned out to depend on an assumption the earlier essays had made twice in opposite directions.

A time with no stiffness in it #

Write $\varphi_0$ for the turn a corrugation was creased to, and hold it at $-\varphi_h$, the same pattern with every crease bent the other way by $\varphi_h$. The same law as before, with the held turn now negative, gives a remembered turn

$$\varphi^*(T) = -\varphi_h + (\varphi_0 + \varphi_h)\,e^{-T/\tau},$$

which reaches zero at

$$T^* = \tau \ln\left(1 + \frac{\varphi_0}{\varphi_h}\right).$$

No stiffness and no fold count appear, since the law is a statement about one angle creeping toward another. Creased to 0.8 radians and held at 1.2 the other way, the memory is gone after 0.511 relaxation times. Held at 0.8 it takes the logarithm of two, 0.693. Held at 2.4 it takes 0.288.

Compare holding the sheet flat. The memory then decays toward zero and never arrives: it takes $\tau \ln 20$, about three relaxation times, to reach five per cent of where it started. Holding it the other way reaches zero exactly and in a finite time, because the curve is aimed past zero rather than at it. That is what the poster and the back-creased fold are both doing, and it is a fact about aiming an exponential, independent of what the paper is made of.

Two hinges under one name #

So far so good, and now the earlier prediction has to be read carefully. It said that when the memory crosses zero the flat sheet stops being a barrier. For that to mean anything the flat sheet has to be a barrier before then, and on that two earlier essays disagree.

Springs that disagree do not offer a choice drew a corrugation with two wells: the creased turn on one side, its mirror on the other, and the flat sheet the hump between. A corrugation has one resting state put the same four springs on a corrugation and counted one well, because every crease folds by one angle and the sum of parabolas in one variable is a parabola. Both are computed correctly. They differ in what a hinge is taken to remember.

The two-well drawing needs a hinge to rest at $+\varphi^*$ when bent one way and $-\varphi^*$ when bent the other, so the memory is a size with no direction. The one-well count takes the memory to be a signed angle. The crossing can only happen to the second kind. A memory with no sign held on the far side of flat just learns to rest at a larger size on both sides, the two wells stay level for ever, and no moment arrives at which a branch has been erased.

Taking the signed hinge, which is what the question needs, the result is plainer than the prediction and less interesting. A corrugation has one well at every moment, and all the holding does is slide it. Released at time $T$, the sheet goes to $\varphi^*(T)$ and stays. Before $T^*$ it opens back on its old side by a little less than it was creased; after $T^*$ it comes to rest on the other side by a little; at $T^*$ it lies flat. Nothing snaps, because there was never anything to snap across. The earlier essay’s barrier of 8.41 times the creased value belongs to the two-well account only. Its accounting of the torque is untouched: that sum contains no well at all.

Which hinge is the real one is a measurement this argument does not have. A reversed paper crease can plainly be pressed to rest folded the other way, which a purely signed spring would not do, and nothing says it rests at the original’s size, so paper presumably lies somewhere between the two curves. What the choice settles is what can be claimed. A corrugation cannot be programmed through a snap under a law that lets its memory cross zero.

Where a second well really comes from #

A second resting state with a signed memory has to come from the geometry, and the wall is the flat sheet has one. A degree-four vertex has two branches meeting only at the flat state, so four roads leave flat and a sprung vertex usually rests somewhere on two of them. There the barrier is a real barrier, and it can be held and asked the question.

Put a vertex of sectors 60°, 90°, 120° and 90° on four signed springs that remember a state on its first branch, at a drive of 1.6 radians. Hold it instead on its second branch at the same drive, and let every rest angle creep. The four rest angles together are a point in a space of four dimensions, and under this law the point moves in a straight line from the creased state toward the held one.

The wall does not fall to nothing. It falls from 18.98 to 4.38, less than a quarter, and recovers a little. There is no moment at which the flat sheet stops being a barrier; the rest vector goes past the origin and misses it. What there is, at 1.017 relaxation times, is a moment when the two resting states are equally deep. That is the “no preferred branch” the earlier essay was looking for, and it is a tie between two wells rather than the absence of a wall.

The branches are at right angles #

Why that time, and why does the wall stop where it does? The answer is a small fact about flat-foldable vertices that none of the earlier essays needed.

On the first branch the creases fold as a pair equal and a pair opposite: $(\rho_1, \rho_2, \rho_3, \rho_4) = (a, b, a, -b)$. On the second the roles swap: $(c, d, -c, d)$. The dot product of any point on one with any point on the other is $ac + bd - ac - bd$, which is zero. In the space of its four fold angles, a flat-foldable vertex’s two branches are at right angles to one another, at every fold, for any sectors, with the stiffness equal in opposite pairs.

So a rest vector moving from a creased state $\mathbf{c}$ toward a held state $\mathbf{h}$ on the other branch is the hypotenuse of a right triangle. Writing $\lambda = e^{-T/\tau}$, its squared length is

$$|\mathbf{r}|^2 = \lambda^2 |\mathbf{c}|^2 + (1-\lambda)^2 |\mathbf{h}|^2,$$

which is the wall, and it has a floor at the harmonic combination $|\mathbf{c}|^2|\mathbf{h}|^2 / (|\mathbf{c}|^2 + |\mathbf{h}|^2)$. For the vertex above, 18.98 and 5.70 give 4.38, the number the figure stopped at. If the roads were straight lines, the climb out of each state would be the squared length of its own part of the rest vector, $\lambda^2|\mathbf{c}|^2$ and $(1-\lambda)^2|\mathbf{h}|^2$, and the two would be equal at

$$T_{=} = \tau \ln\left(1 + \frac{|\mathbf{c}|}{|\mathbf{h}|}\right).$$

This is the corrugation’s formula with lengths where the angles were, and for the same reason. The roads are not straight, since a vertex’s fold angles do not keep their proportions as it folds, so this is a law to test and not a theorem.

On these nine holds they agree to within 8.6 per cent, and on three of them to under two. A shallow crease held deep becomes indifferent soonest: creased at a drive of 1 and held at 2.2, the first vertex ties at 0.701 relaxation times, its wall never falling below 49 per cent. A deep crease held shallowly takes twice as long and its wall falls to 8 per cent, because its rest vector passes much closer to the origin. The moment of no preference and the height of the wall at that moment are both set by the ratio of two lengths.

Over a hundred and sixty random vertices and random drives the median miss is under one per cent. Two sit well off the line, at 46 and 37 per cent, and both are nearly kites: sectors of 76°, 104°, 104°, 76° and 112°, 68°, 68°, 112°. There one pair of creases hardly folds along most of one road, the least straight road a flat-foldable vertex has. So the law’s failures can be found in advance from the sectors, and nothing else tested fails it by as much as one part in seven.

The one hold that goes through zero #

One point on the second branch is special. Driven the other way, the far end of that branch at the same drive is exactly the creased state with every crease reversed: if $\mathbf{c} = (a, b, a, -b)$ then $\mathbf{h} = -\mathbf{c}$. The straight line from one to the other runs through the origin.

Here the earlier essay’s prediction comes true, on the one configuration where it can. At 0.693 relaxation times every rest angle is zero together, the wall has fallen to nothing, and the sheet lies flat with no energy in it and no preference for anything. This is the logarithm of two, the case of the first formula where the two angles are equal.

It is also why the mirror works and nothing near it does, because the kinematics on the road is not linear. A sheet held on the same road at a different drive does not have rest angles in the same proportions as the crease’s, since the two pairs of creases keep different ratios at different folds.

Held at drives of 1 and 2.4 on the same road, the rest vector passes within 6 and 8 per cent of the creased length and misses. That is still better than flat manages in three relaxation times, and it gets there in well under one. Erasing a crease by overshooting it is about four times as fast as holding it flat, at the same residue, and it costs a tolerance on the clock. Held at the mirror, a ten per cent error in the holding time leaves 6.7 per cent of the crease; on the corrugation creased to 0.8, that same ten per cent leaves 5 per cent at a hold of 0.4, 7.5 at 1.2 and 8.5 at 2.4, so the harder hold is quicker and less forgiving. Flat has no such trouble, since holding it longer only ever helps. Rolling a poster the other way has exactly this character, and people do overshoot it.

What this says about choosing a branch #

Paper that folds itself found the hard half of self-folding to be the choice between two equally downhill outcomes. A held memory is a way of making that choice in advance, and the numbers say how much of one is available.

It is cheap. Holding the sheet on the wanted road gives that road a resting state almost at once, shallow at first, and at the tie it is as deep as the old one. A sheet creased at one drive and pressed for one relaxation time into a chosen other state has reached indifference, and by two it is firmly committed, with the climb out of the old state down to about two per cent of the original wall by the straight-road account: $\lambda^2|\mathbf{c}|^2$ at $\lambda = e^{-2}$. What is not available is the clean slate the earlier essay hoped for. Erasing the old choice altogether requires the exact mirror, which means already holding the whole pattern in the reverse of its crease, and for one moment. In practice a sheet would not be erased and rewritten. It would be overwritten, with the old state still there, ever shallower, for as long as anybody cares to look.

This also separates the four places the same corrugation is found more than the materials did. A plain pleat in a leaf has one well that slides, so a leaf bent the wrong way in a bud would come out bent the wrong way by however much it had learned, with no snap in it. A wing’s degree-four vertex has two states and a wall, and if its hinges relax at all, the wing’s choice of state would be written gradually into its rest angles, with the wall dipping but not gone. Whether either organism’s hinges do that over a timescale that matters is the measurement the sheet remembers has been asking for.

The assumptions, and what the drawings cannot show #

The law is linear and has one relaxation time, and the earlier essay’s account of mixed hinges applies here unchanged: hinges relaxing at different rates have rest vectors that do not move in straight lines, and an exact crossing through zero is then impossible at any hold, even the mirror. So the erasing figure is a best case. The stiffness is equal in opposite pairs, which is what makes the branches exactly perpendicular; with unequal pairs the inner product is not zero, the triangle leans, and the closed forms need that correction. The tie times were measured with equal stiffnesses only.

A held sheet is taken to be held exactly at a single point of a road, with every rest angle free to creep toward it. A real press lets some creases slip and holds others hard, and contact between layers holds a packed sheet as well as the springs do. Nothing here is a kinetic account. Equally deep is not equally likely, and which state a released vertex actually reaches depends on where it was released and how fast it moves; this is an account of the landscape only. The drawings do not show the sheet: each one is a set of energies over time, and a reader wanting the fold itself needs the vertex essays’ drawings of the two roads.

A right angle nobody needed until now #

The surprise is that the fact doing the work is a small piece of arithmetic about the vertex that has nothing to do with time. Two branches meeting at the flat state are perpendicular in the space of the fold angles, for any sectors, so their fold-angle vectors decompose like the legs of a right triangle, and a rest vector moving from one to the other is a hypotenuse. The wall’s floor, the tie’s time and the fact that nothing but the mirror reaches zero are all that triangle. The biology is a law of creep, one line long; the rest was there in the geometry of the vertex, since the first essay that drew both branches.

Still open: a sheet of them, and a reversed crease measured #

A single vertex has been held here. A sheet of vertices held in a chosen state shares its creases, so a rest angle that creeps at one vertex creeps at its neighbour as well. Whether a whole Miura’s relaxed rest angles stay on one family of states, and whether a loop that takes choices away makes a sheet’s tie sharper or spreads it across the sheet, is the same straight line in a much larger space, with the same perpendicularity available at each vertex and no guarantee that it survives the sharing.

The other is a number: the rest angle of a paper crease bent the other way, as a share of the original. It places real paper between the two curves of the second figure, and it decides which of these accounts, or what mixture, a real corrugation obeys.

The habit worth carrying is about drawings of landscapes. Before asking what happens when a barrier vanishes, check what the barrier was made of. The corrugation’s was one sentence about a hinge’s memory, and the question could not be asked until that sentence was written down.