A stack that has to fit

The networks field’s result about a scissor stack is that its mobility is one whatever the stage count. Add stages and the loop count grows, the joint count grows, the constraint count grows, and the freedom does not: the stages are copies, and a count over copies gives the same answer as a count over one.

That is the same object measured on a different axis, and this time the answer does depend on the count.

The stack, as this site builds it #

A scissor stage here is two bars of length L pinned at their midpoints and at their ends to the stages above and below, with the whole stack standing on a base whose two lower pins are held. That is the networks field’s scissor exactly, at the same proportions, and the only thing added is a half-width of 0.035 with a boss of 1.35 times it.

Three opening angles are used to build the conflict graph — 0.35, 0.8 and 1.25 radians — rather than one, because a pair that clears when the stack is open may not when it is nearly shut, and the graph is meant to hold for the whole motion. Using one angle would give a graph that is right at that angle and possibly wrong at every other.

The fold search then runs on the graph’s own colouring, so the pairs it watches are the ones the assignment put in the same plane. That ordering matters: searching for the flattest angle before deciding the planes would find the angle at which any two bars touch, which is a much larger angle and an answer to a question about a stack built in one plane — a stack that does not exist.

The paper answer #

A scissor stage is two bars of length L crossing at their midpoints, pinned there and pinned at their ends to the stage above and below. At an opening angle φ each stage is L·sin φ high, so a stack of n stages is n·L·sin φ.

As φ → 0 that goes to zero. On paper a scissor stack of any height folds into a flat bundle of lines, which is why the mechanism is the one people use when something has to stow — a lift, a gate, a table, a deployable structure.

The two planes #

Give the bars a width and the first question is the layers question: which bars may share a plane.

Two bars of one stage cross at their midpoints and are pinned there, so they conflict. Two bars of adjacent stages are pinned at their ends, so they conflict. Everything else is measured, over three opening angles, and nothing else conflicts.

The conflict graph that comes out is two-colourable at every stage count tested, so a scissor stack needs exactly two planes whatever its height — one for the bars leaning one way and one for the bars leaning the other. Add stages and the graph grows and the answer does not.

That is not obvious in advance and it is the structural reason a scissor folds at all. A four-bar needs three planes and therefore three plate thicknesses of depth; a scissor of six stages needs two, the same as a scissor of one. The mechanism people use for stowing is the one whose plane count does not grow, and until a link had a width there was no way to say that.

What stops the fold #

With two planes, bars leaning opposite ways pass each other freely — that is what the planes are for. What cannot pass is two bars in the same plane, and in an alternating assignment those are the bars two stages apart.

So the fold is bisected for rather than computed: reduce φ until the first same-plane pair touches. For bars of half-width 0.035 with bosses of 1.35 times that, the answer is

φ = 0.09508 radians,

and the height it leaves is 0.09493 per stage.

Which is 2 × 0.04725 = 0.0945, the two boss radii, to within the bisection’s own tolerance. What stops a scissor folding is two bosses meeting, and nothing about the stack, the stage count or the bar length enters it.

Why bisection is the right instrument here #

The fold angle is found by bisection on φ, and the reason it is sound is the same monotonicity the width search uses, read on a different parameter.

As φ decreases every bar rotates towards the horizontal and every same-plane pair comes closer. There is no φ at which a closing pair opens again, so the predicate does every same-plane pair clear is monotone in φ, its crossing is unique, and bisection converges to it rather than to whichever crossing it happened to bracket.

That is worth checking rather than assuming, because a scissor’s geometry is not obviously monotone: the bars swing, the joints translate, and the stack shortens all at once. What makes it monotone is that the whole stack is a similarity as φ changes in one respect — every stage does the same thing — so a pair that is closing at one stage is closing at every stage, and there is one crossing rather than n.

The bisection runs to 10⁻⁵ in φ, which is a height of 10⁻⁵ per stage: far finer than the quantity is meaningful to, and cheap enough not to bother tightening.

The number, and what it is not #

A six-stage stack of unit bars stows at 0.570 rather than at nothing. Five stages at 0.475, four at 0.380, three at 0.285, two at 0.190.

That is linear in the stage count, with a slope of one boss diameter, and it is a floor rather than an estimate: it is what the geometry allows, with no allowance for plate thickness, running clearance, pin heads or anything else that a real stack has. The honest reading is that a scissor stows to at least a boss diameter per stage and probably rather more.

The single stage is the exception in the table, at φ ≈ 0, and it is an exception for an uninteresting reason: one stage has two bars, they are in different planes, and there is no same-plane pair to touch. A one-stage scissor really does fold flat, and the floor arrives with the second stage.

Why the floor is per stage #

This is the part that matters for a design, and it is worth separating from the arithmetic.

The obstruction is local: two bosses, two stages apart, on the same side. It does not involve the whole stack, it does not get worse with height, and it does not depend on the bar length. So it contributes a fixed height per stage, and the stowed height is that times the count.

The consequence is that stowing does not improve with scale in the way the paper formula suggests. Doubling the bar length doubles the deployed height and leaves the stowed height alone, which is good — the ratio improves. Doubling the stage count doubles both, which is neutral. And making the bars thinner improves the stowed height directly, which is the only lever that touches the floor and is the one a designer of a deployable structure spends their effort on.

Which pair touches, and where #

The pair that stops the fold is worth naming precisely, because two bars sharing a plane covers several candidates and only one of them is the one that meets.

In the alternating assignment, the bars leaning one way are in plane 1 and the bars leaning the other are in plane 2. Within plane 1 the candidates are stage k and stage k+2, stage k and stage k+4, and so on. It is the nearest pair — two stages apart — that closes first, and the contact is between the boss at the top end of the lower bar and the boss at the bottom end of the upper one, which are two bosses that sit at nearly the same place when the stack is flat.

That is why the answer is two boss radii and not, say, two half-widths: the contact is boss-to-boss rather than side-to-side, because the ends of the bars are the parts that come together as the stack shortens.

It also says which change would help. Thinning the bars in the middle does nothing at all — the middles never touch. Reducing the boss does everything, and reducing the boss means reducing the pin, which is where a deployable structure’s designer ends up: the stowed height of the whole assembly is set by the pins.

What the network field’s result becomes #

Put the two measurements of the same object side by side.

Mobility: one, at every stage count. A statement about the rank of a constraint Jacobian, decided by the equations, indifferent to the count because the stages are copies.

Planes: two, at every stage count. A statement about a conflict graph, decided by the shapes, indifferent to the count for the same reason — the conflicts are between neighbours, and adding a neighbour adds the same conflicts again.

Stowed height: a boss diameter per stage. A statement about material, decided by a bisection, and proportional to the count, because the obstruction is local and each stage contributes its own.

Three quantities, one object, and two of them are invariant under repetition while the third is additive under it. That is a compact summary of what repetition does and does not preserve, and it is the networks field’s founding question asked of a quantity that field could not compute.

The measurement without a mechanism #

There is a small piece of machinery worth mentioning because it is what let this essay exist at all.

Everything else in the bodies field is built on a dressed mechanism: a Mechanism with joints, constraints and a solver, with bodies hung on its links. A scissor stack is not one of those here, and it does not need to be — its geometry is explicit, every joint’s position is a formula in φ rather than the output of a Newton solve, and the networks field has already established the freedom count that a solver would confirm.

So the conflict graph is built from frames of bare segments: a list of named line segments per configuration, with a rule saying which pairs are joined. The colouring, the census and the fold search all take that, unchanged, because none of them ever looked at the mechanism — they look at a graph and at placed bodies.

That is a small refactoring with a general point in it. A field’s machinery is worth building against the narrowest input that carries the question, and here the question is about bodies and conflicts rather than about solving. Anything whose positions are known in closed form — a pantograph, a folded stack of anything, a linkage somebody has already solved by hand — can be asked the same questions without being rebuilt as a mechanism first.

The other deployables, and why they are not here #

A scissor is one of several mechanisms whose whole point is to be small and then large, and the measurement above applies to all of them in principle and to none of them here.

A pantograph is a scissor’s near relative with the pins off the midpoints, and it has the same two-plane structure and the same boss-to-boss floor. Nothing about the arithmetic changes; only the geometry does.

A folded sheet — the origami-like structures the networks field builds from spherical vertices — has a floor of a completely different kind, because sheets stack rather than pass each other and the flat state is a question about layer ordering through the whole sheet rather than about two bosses. That is a genuinely different problem and it belongs to a different subject.

And a telescoping mast has no folding at all: its floor is a section thickness times a stage count, which is the same linear form arrived at without any mechanism.

The common structure is that every deployable has a floor set by its smallest feature and a reach set by its largest, and the ratio between them is the product. What differs is which feature and which obstruction, and the scissor’s is the simplest of the three: one contact, between two bosses, two stages apart.

What is not modelled #

Plate thickness. Every plane here is an index, not a height. A real two-plane stack is two plates plus a running clearance thick, and that adds a constant to the whole stack rather than to each stage.

Pin heads. Whatever retains a pin sticks out past the outermost plane, and it is the same constant again.

The end fittings. A real scissor lift has a base, a platform, and usually a hydraulic ram between two of the pins — which this site does model elsewhere — and all of them are larger than the bars.

And the bars’ own ends. The bars here are straight, of uniform width, with bosses at their ends and at their midpoints. A real scissor bar is often waisted or shaped precisely to get the fold flatter, which is a designer buying stowed height with machining time.

Each of those makes a real stack thicker than the number here. The number is a lower bound derived from the one obstruction the geometry forces, and its value is that the obstruction turns out to be two bosses rather than anything about the stack.

What it says about a lift #

The catalogue this site keeps of machines a reader has met includes a scissor lift, whose selling number is a stowed height and a raised height and the ratio between them. That ratio is what the whole product is.

The floor here says where the ratio comes from. Deployed height is n·L·sin φ at the working angle, which is set by the bar length and the stage count. Stowed height is n × a boss diameter, which is set by the pin. So the ratio is

$$\frac{L \sin\varphi_{\text{max}}}{2 r_{\text{boss}}},$$

with the stage count cancelling entirely. A stack of three stages and a stack of ten have the same extension ratio, and adding stages buys height at constant ratio rather than improving it. That is a fact about the mechanism a catalogue does not state and a designer knows: stages are for reach, and the ratio is bought with slender bars and small pins.

It is also a fact this site could not have derived four essays ago. The stowed height needed a body; the ratio needed the stowed height; and the cancellation needed both to be linear in the same count.

The general shape #

A quantity that goes to zero in a model where the members are lines does not go to zero when the members are made of something, and what stops it is usually the smallest feature in the model rather than the largest.

Here the largest thing in the picture is a stack of six bars a unit long, and the answer is set by a disc 0.047 across. That is a general property of fold and stow problems rather than a fact about scissors: a folded state is where the material is closest together, so it is where the smallest dimension in the design decides everything, and every larger dimension has stopped mattering.

This site has met the same phenomenon once before, in the field where a member has no length of its own: a strand wrapped round a pulley has a bend radius that no model of it as a line contains, and the radius is set by the strand’s own thickness. Same structure, different object, and in both cases the quantity the line model reports is zero.

Which is also why the paper formula is not a bad model that gets a wrong answer. It is an exact model of a scissor made of lines, and lines have no smallest feature. The failure is at the modelling step and not in the arithmetic, and it produces a prediction of zero — the one answer that is wrong by an infinite factor.