A gearing reflects stiffness squared

Two drivers and one freedom put two actuators on a folded sheet that has a single degree of freedom and asked where the sheet goes when they disagree. It cannot satisfy both, so it settles where the energy stored in the disagreement is least, and with equal stiffnesses the answer was a function of the gearing alone: a disagreement $\delta$ between creases geared $g$ moves the first crease by $g\delta/(1+g^2)$ and leaves $\delta/(1+g^2)$ standing at the second. The loosely geared pairs kept nearly all of it.

It named the assumption most obviously false. Two actuators are almost never equally stiff, and removing that assumption gives a designer a second lever — the stiffness of each actuator — beside the choice of which creases to drive. Whether the stiffness should be matched to the gearing, and in what way, was a computation left for later. It is short.

Where the sheet settles when the springs differ #

Write the first actuator’s stiffness $k_1$ and the second’s $k_2$. The sheet has one freedom, so moving the first crease by $x$ from its command moves the second crease by $gx$, and the second crease’s error from its own command is $gx - \delta$. The stored energy is

$$E = k_1 x^2 + k_2 (g x - \delta)^2,$$

and the sheet settles where it is least:

$$x = \frac{k_2 g\,\delta}{k_1 + k_2 g^2}, \qquad gx - \delta = -\frac{k_1 \delta}{k_1 + k_2 g^2}, \qquad E = \frac{k_1 k_2\,\delta^2}{k_1 + k_2 g^2}.$$

Every one of those can be rewritten with $k_1$ and $k_2$ appearing only as $k_2 g^2 / k_1$. The second actuator acts on the first crease as a spring of stiffness $k_2 g^2$. A crease that turns $g$ times as far as the first for the same motion of the sheet multiplies both the displacement its spring sees and the leverage that spring has back on the sheet, and the two factors of $g$ make a square.

That answers the question as it was left. The earlier account suggested the product $kg$ might be what should be matched. It is not; $kg^2$ is. A second actuator balances the first — the sheet settles exactly halfway between their commands, measured from the first crease — when $k_2 g^2 = k_1$, which for a pair geared 0.3 means the second actuator must be eleven times stiffer than the first just to hold its own.

The square, checked #

The closed form is easy to derive and also easy to get wrong by a factor of $g$, so it is checked two ways. The settled position is found by minimising $E$ directly at four combinations of gearing and stiffness ratio — a weak pair soft and stiff, a strong pair soft and stiff — and agrees with the formula to the precision of the arithmetic. And two pairs with the same $k_2 g^2$ are compared: a pair geared one half with a second actuator four times stiffer, and a pair geared one with equal actuators. They leave exactly the same disagreement standing at the second crease, which a matching rule in $kg$ would not predict.

The second figure shows what the square does on a loosely geared pair. At equal stiffness the sheet barely moves from the first command — 0.014 of a 0.05 disagreement — and leaves almost all of it at the second crease. To pull the sheet halfway the second actuator has to be eleven times stiffer, and to pull it all the way it would have to be stiffer without limit, at which point the sheet has moved $\delta/g$ from the first command — 3.3 times the disagreement itself, because a crease geared 0.3 has to swing the sheet a long way to close a small error of its own.

On a strongly geared pair the square works the other way. At a gearing of 2 the balance is at $k_2 = k_1/4$: a second actuator a quarter as stiff already holds its own, and one of equal stiffness pulls the sheet most of the way to its command. A strongly geared actuator dominates a sheet it barely has the stiffness to hold, and a loosely geared one hardly affects a sheet it could, in principle, crush.

Which actuator fights, and how hard #

The energy is the part a designer pays for — in heat, in wear, in the force each actuator exerts against the other for as long as they disagree. Written in units of $k_1 \delta^2$ it is

$$\frac{E}{k_1\delta^2} = \frac{k_2/k_1}{1 + (k_2/k_1)\,g^2},$$

which rises with the second actuator’s stiffness at every gearing and levels off at $1/g^2$. So a softer second actuator always fights less; the question is how much less, and what the softness costs.

On the mesh two drivers and one freedom measured — sixteen quadrilaterals with no two vertices alike, one crease driven at 0.6 radians — the driven crease reaches the others at six gearings, 0.316 to 1.761.

On the loosest pair, geared 0.316, the energy at equal stiffness is 0.91; with the second actuator ten times stiffer it is 5.00, and ten times softer 0.099 — a factor of fifty between the two. On the tightest pair, geared 1.761, the same three are 0.24, 0.31 and 0.076, a factor of four. A stiff actuator on a loosely geared crease is the arrangement that fights hardest, which is the rule the earlier essay guessed and the table now prices.

Why softness is cheap on the loose pair #

Softening an actuator is not free. An actuator exists to hold its crease where it is told against whatever pushes on it — the sheet’s own weight, a gust, a snag — and a soft one holds less firmly. The question is how much less, and the gearing answers it too.

A crease is held by both actuators. The second crease is held directly by the second actuator, with stiffness $k_2$, and indirectly by the first, through the sheet: to move the second crease by $y$ the sheet must move the first crease by $y/g$, which the first actuator resists with stiffness $k_1$, so it contributes $k_1/g^2$ at the second crease. The total holding stiffness there is $k_2 + k_1/g^2$ — the same square, seen from the other end.

On the loosest pair $k_1/g^2$ is ten times $k_1$. The first actuator already holds the second crease ten times as firmly as an equal second actuator could, and cutting the second actuator to a tenth of its stiffness keeps 92 per cent of the crease’s hold while cutting the fight fifty-fold. On the tightest pair $k_1/g^2$ is a third of $k_1$, the second actuator is doing most of the holding, and cutting it to a tenth keeps only 32 per cent.

The trade can be put as one exchange rate. On the loose pair, softening removes 0.81 of the equal-stiffness energy, in units of $k_1\delta^2$, and surrenders 8 per cent of the hold: about a tenth of a unit of fight for each per cent given up. On the tight pair it removes 0.16 and surrenders 68 per cent, about a four-hundredth for each. The same softening is some forty times better value at one end of the mesh than at the other, and because the energy saved falls with the gearing while the hold lost rises with it, every pair in between sits on that range in the order of its gearing.

So the rule a designer can carry is sharper than “make one actuator soft”. Make the actuator on the loosely geared crease soft: there, the fight it causes is largest and the hold it provides is least needed, because the other actuator reaches its crease through a gearing that multiplies stiffness by $1/g^2$. On a tightly geared crease, softness saves less and costs more, and the two actuators should be closer to balance.

What this means for a sheet that folds itself #

Paper that folds itself describes what a self-folding sheet’s actuators actually are: not motors commanding positions but materials that change shape — a shape-memory polymer, a swelling hydrogel, a bimorph strip — each of which behaves like a spring whose rest length moves. A shape-change actuator is intrinsically compliant, and its stiffness is a property of the material and the hinge geometry rather than a setting.

That turns the stiffness analysis into a placement rule. A designer with one stiff actuator material and one soft one, and a sheet with two creases to drive, should put the soft material on the crease with the smaller gearing to the other, and the stiff material on the crease that moves most. The gearing is computable from the pattern before anything is built — which crease to push measures it crease by crease — so the placement is a design decision rather than a tuning.

It also explains a habit of self-folding designs that has no obvious reason: making most hinges passive and a few active. A passive hinge is an actuator of zero stiffness and zero rest-length change, and zero stiffness is the limit of the softening argument. A passive crease on a loosely geared position costs the sheet almost none of its hold, for exactly the reason softening does.

A second actuator as insurance, priced again #

The reason anybody puts two actuators on a sheet with one freedom is reliability. The crease count is a reliability budget found that a deployment needing every hinge is a product of their reliabilities, and a second actuator looks like the obvious hedge against the first failing to drive. The stiffness analysis changes what that hedge costs.

A backup actuator that is stiff fights the primary for as long as both are working, which is almost always, and on a loosely geared pair it stores up to ten times the energy of an equal pair. A backup that is soft barely fights — but a soft backup on a loosely geared crease is also a weak driver of the whole sheet, since to move the sheet it must turn its own crease through the gearing, and a crease geared 0.3 must swing more than three times as far as the primary crease for the same motion. So the backup that is cheap to carry is the backup least able to take over.

The resolution is the same square read a third way. A backup’s authority over the sheet, if the primary fails and goes slack, is its stiffness reflected to the primary crease, $k_2 g^2$. To be a useful backup it wants a large $g$; to be a cheap passenger it wants a small $k_2 g^2$ while the primary works. Both cannot be had from one crease, which is why a backup belongs on a tightly geared crease and must be engaged only when needed — a clutch rather than a second spring. The table’s tight pair, geared 1.761, is the one where a backup would be worth carrying.

That is a design statement with a mechanism in it rather than a preference. One crease decides the sheet established that on a rigid mesh any crease determines every other, which is what makes a single actuator sufficient in principle; the gearing says how well each crease does it, and a sheet with one freedom is the reason there is a single number for two actuators to disagree about in the first place.

The Miura, where the rule needs care #

The mesh in the table has no two vertices alike, and its gearings are six distinct values. Only four creases decide a Miura found the pattern everybody builds behaving differently: most of its creases leave more than one folded state when driven, because the pattern repeats one vertex and its symmetry offers the sheet a choice.

On a Miura the gearing between two creases is defined only once the branch is chosen, and a pair of actuators on creases that do not decide the sheet could, in principle, agree on the angle and disagree about the branch. The stiffness analysis assumes the sheet is on one branch and stays there; on a Miura that assumption needs one of the deciding creases driven, and the placement rule applies within the branch it selects.

The gearing moves, and so does the rule #

The six gearings in the table are measured at one fold angle, 0.6 radians. The hardest instant followed the amplification of an error along the whole motion and found it worst near the flat sheet on twenty of twenty-four creases, which means the gearings between creases change as the sheet folds, and the balancing stiffness $1/g^2$ changes with them.

An actuator pair balanced at one angle is not balanced at another. A soft actuator placed on the loosely geared crease at 0.6 radians is on the loosely geared crease only for as long as that crease stays loosely geared; a pattern whose gearings cross during the motion would reverse the rule partway through. The sweep above shows the amplification along the motion for every crease, and where two curves cross is where the placement rule changes sides.

What the model assumes #

Each actuator is a linear spring about its commanded angle. A real actuator’s stiffness depends on how far it is displaced, and a shape-change actuator’s rest angle drifts with temperature and time; the analysis is the small-disagreement limit of any of them.

The sheet has exactly one freedom and is otherwise rigid. Panels that bend absorb some of the disagreement themselves, which is a third spring in series with neither actuator, and it would lower every energy here.

The disagreement is static. Two actuators that disagree by an amount that changes during folding store an energy that changes too, and a sheet folding quickly enough could overshoot the settled position this computes.

And the mesh is one mesh. The six gearings come from a single sixteen-quadrilateral pattern at one angle; a different pattern has different gearings and the same rule.

What the table cannot show #

It does not give an actuator’s stiffness. The ratios are relative to the first actuator, and whether a real pair of hinge materials can be a factor of ten apart is a question of materials, not geometry.

It does not price the loss of hold in a real load case. Holding stiffness matters against a load, and which load a folded sheet must resist — its weight in one orientation, a contact force during deployment — decides whether 32 per cent of the hold on a tight pair is enough.

And it treats two actuators, not many. A sheet with three actuators on one freedom stores energy in every pair’s disagreement at once, and the settled position is a weighted average in which every weight is a stiffness times a gearing squared; the rule generalises, but the table does not show it.

Still open: an error in the cutting, as a second driver #

The disagreement between two actuators has a twin that no actuator causes. A panel cut slightly wrong is a sheet whose own geometry disagrees with where a single actuator is trying to put it: the closure the mesh must satisfy and the angle the actuator commands are, in effect, two commands for one freedom.

If that is right, a manufacturing error divides by the same factor. The panel’s own stiffness plays the part of the second actuator and its geometric error plays the part of $\delta$, and the share of the error the sheet absorbs at each crease would be set by the gearing from the actuator to the mis-cut panel, entering squared. Which creases forgive a cutting error would then be computable from the same table as which creases forgive a second actuator, and the actuator would belong where the mis-cut panels it can reach are loosely geared to it.

Sideways from here, the square has a familiar name in machines. A gear train reflects an inertia or a stiffness from its output to its input by the square of its ratio, which is why a small motor through a high reduction can hold a large load. A folded sheet’s gearing between creases is a gear train with no gears, and every result about reflected stiffness in machines should have a counterpart here.

The habit worth carrying is about coupled constraints. When two things are connected through a ratio, ask how the ratio enters each quantity before matching anything across it. Positions pass through a ratio once; forces pass through it once the other way; stiffness and energy pass through it twice, and a rule that matches the wrong power will be wrong by the ratio itself.