A cut along the zigzag needs a tether

The rows are free found why a Miura’s hinges per layer of compaction climb as the pattern is refined. Along its rows the pattern is an accordion: every band between two straight creases folds exactly onto the band before it, and the folded box does not change as rows are added. Along its columns it walks: every band between two zigzags lands a step of $\sin a$ further along the folded strip, so the box is $\cos a$ wide and $\sec a + c \sin a$ long, and a Miura long in columns spends its hinges on footprint rather than on depth.

It ended on a practical proposal. Splitting a sheet had found reliability improving when a deployable is divided into separately driven modules, since a stuck hinge then costs only its own module. A sheet divided along its zigzags into narrow Miuras would get both benefits at once — few columns each, so little walk, and independent modules — and the open question was whether the joins between modules cost more than the walk they save.

The answer turns on what a join can be, and the first thing to settle is that it cannot be a crease.

Where the walk comes from #

The walk is a fact about reflections. A flat fold places each panel by reflecting it across every crease on a path from a panel held still, and two reflections in parallel lines are a slide square to the lines by twice their spacing. Across the rows the creases are straight and parallel, the slide and the panel’s offset are the same vector, and each second row lands on the first. Across the columns the creases are the zigzag’s slanted segments, the slide is square to them, and the offset is along the row; they differ by $2 \sin a$, pointing along the slanted crease.

So the folded strip’s length has one term for the panel and one for the walk, $\sec a + c \sin a$, and at a slant of 0.35 the walk is 0.343 a column. A Miura sixteen columns wide folds to a strip 6.55 long; four columns, 2.44; one column, 1.41. The walk is the whole of the difference, and it is paid for in columns and in nothing else.

A crease cannot be the join #

The obvious way to divide a Miura into modules is to leave it in one piece and drive each part separately: put an actuator on every fourth column and let each one fold its own stretch. That does not divide anything, and the reason is at the vertices.

A zigzag line of a Miura is crossed by every row crease, and at each crossing four creases meet — the two segments of the zigzag and the two halves of the row crease. A sheet with one freedom is the account of what that does: a degree-four vertex folds rigidly with a single freedom, every one of its four fold angles a function of any one of them. So the row crease’s angle on one side of a zigzag fixes its angle on the other. Two stretches of Miura joined along a zigzag crease are not two mechanisms but one, with the single freedom the whole sheet has.

Driving that one freedom from two places is the situation two drivers and one freedom measured: two commands for one number, which agree or fight. And folding it flat gives the whole Miura’s folded state back, with the whole walk. No arrangement of actuators on a single sheet can divide the walk, because the walk belongs to the flat-folded state and the flat-folded state of a connected Miura is unique up to where it is held.

So a module boundary has to be a cut. The zigzag’s crease is removed — and with it its $R$ hinges, one for each row it crosses — and the modules are separate Miuras, each with its own freedom, its own actuator, and its own folded strip. What holds them together when deployed is something that is not a crease: a tether, a film, a frame. It carries no fold angle and imposes no vertex condition, and a model of paper has nothing to say about it except that it is there and has to work.

What a cut saves #

Once the pieces are separate, each folds to the box of a Miura $C/k$ columns wide, and the stacked pieces share that box: a stack of $k$ strips of equal footprint, their depths adding.

For a Miura sixteen columns by four rows, the uncut sheet folds to a strip 6.551 long holding 11.0 layers on average, with 108 hinges: 9.80 hinges for each layer of compaction, which is the climb the earlier essays found. Cut once, into two modules of eight columns, the strips are 3.808 long and stack 19.7 deep with 104 hinges, 5.29 a layer. Four modules: 2.436 long, 32.6 deep, 96 hinges, 2.95 a layer. Eight: 1.750, 48.3, 80, 1.66. Sixteen single columns: 1.407 long, 64 deep, 48 hinges, 0.75 a layer.

Every row of the table improves every quantity a deployable is priced by. The box shortens because each piece walks over fewer columns; the stack deepens because the same paper lies over a smaller footprint; the hinges fall by four for every cut, one zigzag’s worth; and the hinges a layer fall fastest of all, because both the numerator and the denominator move the right way.

The last row is worth pausing on. A Miura one column wide has no interior zigzag, only its four rows’ straight creases, and its folded state is a pleated strip: the limit of dividing a Miura along its zigzags is a stack of accordions. Three hinges a strip and four layers, three quarters of a hinge a layer, better than the preliminary base’s one — and not a Miura at all, since the slant that gives the pattern its single freedom and its walk has been cut away with the zigzags.

Counting what a cut removes #

The hinge count is simple enough to write down, and writing it down shows why the saving per cut is fixed. A Miura $C$ columns by $R$ rows has $R - 1$ interior straight creases, each crossing all $C$ columns, and $C - 1$ interior zigzags, each crossing all $R$ rows. Every crossing splits a crease into a separate hinge, so the sheet has

$$H = C(R - 1) + R(C - 1)$$

hinges: 108 for sixteen by four, 112 for eight by eight. Cutting along one zigzag deletes that zigzag’s $R$ segments and leaves every straight crease’s segments where they were, now belonging to one module or the other. So each cut removes exactly $R$ hinges, whichever zigzag it follows and however many cuts have already been made.

That is why the break-even below depends on the rows and not on the columns. The columns decide how much walk a cut saves — the box shortens by $\sin a$ for every column a module no longer has to carry — and the rows decide how many hinges it removes. A sheet long in columns and short in rows is the one where cutting saves most footprint and removes fewest hinges for each actuator it adds.

The same cells, a squarer sheet #

The saving depends on how long the sheet was in columns, because the walk is proportional to the columns.

The same sixty-four cells drawn eight by eight fold uncut to a box 3.808 long at 5.69 hinges a layer — already much better than the sixteen-by-four sheet’s 9.80, because half as many columns walk half as far. Cut into eight single columns they reach 1.407 and 0.88. The eight-by-eight sheet has further to go in hinges per layer at the end than the long one did, because each cut on it removes eight hinges rather than four and its accordions are eight rows deep.

So cutting along the zigzags is worth most on a sheet long in columns, which is exactly the sheet the walk hurt most. The rows are free made the same point from the other side: a Miura long in rows was already as cheap to trust per layer as a waterbomb. A cut turns a sheet long in columns into several sheets long in rows.

What a cut costs #

The price of all that is not in the table. Every module is a separate mechanism with its own freedom, so every module needs its own actuator, and every join between modules is a part that is not a hinge. The earlier essay’s model of reliability puts a number on both.

Let every hinge work with probability $p$ and every actuator with $q$, independently. A module of $h$ hinges opens with probability $p^h q$, which is also the share of the area expected to open. Everything opens with probability $p^H q^k$, where $H$ is all the hinges left after $k - 1$ cuts. Each cut changes that product by a factor $q/p^R$: it removes $R$ hinges and adds one actuator.

The share expected to open rises with every cut, whatever the actuators. With hinges at 0.999 and actuators as reliable as hinges, it goes from 89.7 per cent for the uncut sheet to 97.5 with four modules and 99.6 with sixteen: a stuck hinge now strands a quarter of the sheet, or a sixteenth, instead of all of it.

The chance that everything opens rises only if an actuator fails less often than $R$ hinges together. With actuators at 0.9999, each cut is a clear gain and complete success rises from 0.898 to 0.952 at sixteen modules. With actuators exactly as reliable as a hinge, each cut trades four hinges for one part and still gains, from 0.897 to 0.938. With actuators at 0.99 — a motor, a spring release, a shape-memory element, each much likelier to fail than a crease — an actuator fails as often as ten hinges, each cut trades four for ten, and complete success falls from 0.889 to 0.812.

That is the answer to the question as it was put. The joins cost nothing in hinges — they remove them — and the modules pay for themselves in footprint and in expected coverage at any actuator reliability. What the joins and their actuators cost is certainty: whether the deployment is complete, rather than how much of it is. The break-even is a single comparison, $q$ against $p^R$, and it depends on the rows, since a cut along a zigzag through more rows removes more hinges for the one actuator it adds.

Which to buy #

The two quantities answer different missions, and splitting a sheet drew the same line for modules cut any way. A solar array that loses power in proportion to the panel that fails to open wants the expected share, and every cut along the zigzags improves it. A reflector or an antenna whose function needs the whole surface wants complete success, and there cuts are worth making only with actuators more reliable than a zigzag’s hinges.

What is specific to the zigzag cut is the footprint and the hinges a layer, which improve on both missions. The pattern cheapest to trust priced a deployable by the tests needed to demonstrate it opens, in proportion to its hinges per layer. By that price the sixteen-by-four sheet cut into four modules is more than three times cheaper to demonstrate than the uncut one, 2.95 hinges a layer against 9.80 — before the actuators, which each need testing too and which a gearing reflects stiffness squared is a reminder are never as simple as a hinge.

What this cannot show #

What a tether is. The model counts a join as a part that must work and nothing more. A real tether must let two folded strips stack while tied at their edges, must not tangle, and must pull the deployed modules into a flat sheet with their edges aligned. Whether that is possible at all for strips that lie at their own slant — the folded strip lies at its own slant — is a mechanical question the geometry of paper does not reach.

How the stacked strips lie. Each module’s box is measured on its own folded state and the stack is taken to share it. Strips folded at one slant lie at one angle to their own drawing, so stacking them in one box needs every module turned the same way, which a tether along a zigzag may or may not arrange.

And whether any Miura-like mesh cancels the walk without a cut. The other open direction the earlier essay named — a quadrilateral mesh whose columns alternate their slide, keeping one freedom — would divide the walk with no joins and no actuators at all. Nothing here searches the family the Miura belongs to for one; the cut is the answer available without one.

The idealisations underneath #

Every cell is a unit square sheared to the slant, every hinge is a line of zero width, and the folded state is the flat-folding model’s, composed from reflections. The mean depth is the stack’s average over its own footprint, measured on a grid of the folded state. Reliability treats every hinge and actuator as failing independently with a fixed probability, which is the simplest model with a number in it and ignores every failure that takes out neighbours together — a jammed stack, a torn tether.

How the numbers were checked #

Every module’s box is measured, not assumed. Each Miura is folded flat, the smallest rectangle round its folded panels is found exactly from their hull, and it is required to be $\cos a$ by $\sec a + c \sin a$ to six figures; the table would stop on a module whose folded state disagreed.

Every cut is required to remove exactly one zigzag’s hinges, $R$ of them, counted on the crease patterns rather than computed.

And the two reliability claims are required together: the expected share rising with every cut at all three actuator reliabilities, and complete success rising with the cuts for the best actuators and falling for the worst.

Still open: a join that folds #

The tether is a concession, and the cleanest way to remove it would be a join that is a crease after all. A straight cut across the zigzags is not the only cut. A join along a line that the folded state maps to a single straight edge of each strip could be a single straight hinge folded through half a turn, stacking one module’s strip onto the next like pages of a book. Whether a Miura has such lines — whether, in the folded state, some line of the sheet lands on a straight edge of the folded strip at both of its sides — is a question about the folded state that a direct search could answer, and a yes would give a divided Miura with no parts but paper.

The other open measurement is the one the reliability model leaves out. Correlated failures are the real ones: a stack that jams takes several hinges together, and a module whose neighbour sticks may be pulled off its own path by the tether between them. A simulation that let one module’s failure load its neighbour would say whether the expected-share gain survives when failures are not independent — which is the assumption that makes every modular argument look better than it may be.

Sideways from here, the deciding set does not move found the creases where an actuator on a whole Miura decides its folded state. A module is a small Miura with one actuator, and where on a narrow module the actuator belongs is the same question on a smaller sheet — with the difference that a four-column module has fewer creases to choose from and fewer states to choose between.

The habit worth carrying is about dividing a mechanism. Before splitting a system into independent parts, check that the boundary can be independent. A boundary drawn through a coupled element is not a boundary but a second input to the same element; here the coupling is the degree-four vertex, and the only way to divide the Miura was to cut through it and pay for what holds the pieces instead.