What has to be solved together

Every figure on this site is drawn from a solved configuration, and the solve is nearly always Newton’s method against a loop-closure residual. That has been presented as a matter of generality: a solver works on anything, so use one and stop worrying about which mechanisms have closed forms.

This rung asks which ones do, and the answer is a property of the graph.

The arithmetic #

Hold link $g$ still. Turn link $i$, which shares a pin with it, to a chosen angle. Those two links are now known — every point on them has a position.

Take a set $S$ of the remaining links and count. Each unknown link has three coordinates, so $3|S|$ unknowns. Each pin inside $S$ gives two equations. Each pin from $S$ to a link that is already known also gives two equations. A pin from $S$ to a link that is still unknown gives nothing usable, because its other end is not yet placed.

$S$ can be positioned from what is already known exactly when

$$3|S| \;=\; 2\,(\text{pins inside } S) + 2\,(\text{pins from } S \text{ to the known set}).$$

Take the smallest such $S$, place it, add it to the known set, and repeat. What comes out are the Assur groups, and their sizes are what this rung measures.

The four-bar is the smallest case and it does the whole thing in one step. Frame and crank known; the coupler and the rocker remain; one pin between them and two onto known links; $3 \times 2 = 6 = 2 \times (1 + 2)$. One group of two.

Two links is two circles #

A group of two links is called a dyad, and it is the reason the whole classification exists.

Its three pins are: one onto a known link at each end, and one between the two unknown links. The free pin is at a fixed distance from each of the two known ones, so it lies on a circle about each — and two circles meet in two points, in one, or in none.

That is a quadratic. It has a closed form, it has two roots, and the two roots are the two assembly branches every four-bar on this site has. Once the free pin is placed, both links are placed, and the group is done.

So a mechanism whose groups are all dyads can be positioned with a compass, in a fixed order, exactly. A draughtsman in 1890 could do it and so can a formula.

Four links is not two dyads #

A group of four links has six coordinates too many for any pair inside it to be placeable alone — that is what smallest means in the definition — so there is no order in which its links can be taken two at a time.

The arithmetic for it is $3 \times 4 = 12 = 2 \times 6$: four unknown links, six pins between them and onto the known set. It is a system of six equations in twelve unknowns’ worth of structure that does not factor into one circle per unknown.

There is no compass construction. There is no closed form of the four-bar kind. What there is instead is a system to be solved as a whole, and on this site that means the Newton iteration the solver has run since the foundation.

That is the sentence this rung exists for. The solver is a convenience on some mechanisms and the only route on others, and which is which is decided by the graph before any dimension is chosen.

Every group size is even #

A small piece of arithmetic worth pulling out, because it explains the shape of the chart above.

A group of $k$ links needs $3k = 2p$ where $p$ is its total pin count, so $p = 3k/2$ — and that is an integer only when $k$ is even. There are no groups of three links, or five, or seven. The sizes that appear across the whole census are 2, 4, 6 and 8, and nothing else can.

It is the same parity argument that forces the census to have rows only at even link counts, applied one level down — and it is the reason the classical vocabulary talks about classes of Assur group rather than about a continuum. A group’s class in that vocabulary is a statement about its size and its internal arrangement together; here only the size is counted, because the size is what decides whether a closed form exists and the arrangement decides only how unpleasant the system is.

There is one more consequence of the parity, and it is the reason a mechanism cannot be nearly decomposable. A driving choice either yields a set of dyads or it does not; there is no group of three that almost factors, and no continuous parameter that makes a four-link group easier. The classification is discrete all the way down, which is unusual on this site and is what makes it a topology question rather than a numerical one.

Watt against Stephenson, again #

The two six-link chains split on this exactly as they split on everything else, and this is the split that mattered historically.

Watt’s chain comes apart into two dyads from every one of its fourteen frame-and-input pairs. All four of its distinct driving choices are compass constructions.

Stephenson’s comes apart that way from eight of its fourteen, and the other six leave all four remaining links as one group.

The reason is the loop structure. Watt’s two loops share exactly one pin, so once the first loop is closed the second has two unknown links and three usable pins — a dyad, exactly. Stephenson’s two loops share a whole binary link, so closing either one requires the other, and the two must be solved together.

That is a genuine engineering difference and it is the reason a draughtsman had to know which six-bar was on the drawing board. It is also why the two names survived when most of the nineteenth-century vocabulary did not.

A group is not a structure #

Two zero-mobility objects have now appeared in this field and they are different objects. It is worth separating them explicitly, because both are computed by a count coming out at nought and the counts look alike.

A rigid subchain — the thing that makes a graph not a chain — has zero mobility on its own, with one of its own links held. Its arithmetic is $3(k-1) - 2p = 0$: hold any member and nothing inside moves. A triangle is the smallest.

An Assur group has zero mobility relative to what it is attached to. Its arithmetic is $3k - 2p = 0$, with no link of its own held, because the links it is held by are outside it. A dyad is the smallest, and a dyad on its own is emphatically not rigid — two links joined by one pin flap about freely.

The two arithmetics differ by exactly the three coordinates of the held link, which is why they are so easy to confuse. The consequences are opposite: a rigid subchain means the mechanism has fewer links than it looks like, and an Assur group means the position problem comes apart. One is a defect and the other is the structure of a solution.

Reading a decomposition off the picture #

The routine is a search over subsets, but on the mechanisms a reader is likely to meet, the answer can be seen.

A dyad is two links in a row with a free pin between them and both outer pins on links already placed. In a drawing that is a chain of two bars whose ends are both attached to something that has stopped moving in the argument, and it is exactly what a designer means by “add a dyad” when extending a four-bar into a six-bar.

A four-link group is what appears when two loops share more than a pin, which is the difference the two six-bars turn on. It looks like a small braced quadrilateral whose corners all reach outward, and the recognisable sign is that no two of its links have both of their outer pins on placed parts.

The practical rule is the one that falls out of the arithmetic: count the pins a candidate pair has onto already-placed links. A pair needs two of them, one at each end, and it needs exactly one pin between its own two links. If either count is wrong, the pair is not a group and the search moves on.

Which mechanisms this makes easy #

It is worth listing what the dyadic case covers, because it covers most of what anybody draws.

Every four-bar, of course: one dyad. Every slider-crank, which is a four-bar with a pin sent to infinity. Every Watt six-bar, from every driving choice. Eight of Stephenson’s fourteen. Every mechanism built by the standard move of take a coupler point and hang a dyad from it, because that move adds a dyad by construction and a dyad added to a dyadic mechanism keeps it dyadic.

That is why the closed-form case feels universal to anyone who has met linkages through four-bars and six-bars. The whole of the classical repertoire is dyadic, the exceptions start at eight links, and eight-link mechanisms are where nobody has an intuition anyway.

Where it comes from, and what the word means #

The decomposition is Assur’s, from 1914, and the groups carry his name. The idea he had is the one above: a mechanism is a frame and an input with zero-mobility groups hung off them, and the groups are the units of the position problem.

The arithmetic of a group is worth restating in his terms. A group has zero mobility relative to the links it attaches to. Attach a dyad to two known links and nothing moves; attach it to two links that are themselves moving and it follows them exactly. That is why a mechanism can be built up group by group: each one adds parts and adds no freedom.

It also explains why the mobility of the whole is one however many groups there are: the frame contributes nothing, the input contributes one, and every group contributes nought. A mechanism’s degree of freedom is entirely in its input, which is a way of saying what the mobility count says from the other end.

The decomposition is not unique and the sizes are #

One caution. When several sets satisfy the arithmetic at the same step, the routine takes the smallest, and among equally small ones it takes whichever it finds first. So the order of the groups can depend on the search, and on a chain with symmetry the particular links in a group can too.

What does not depend on the search is the multiset of sizes. A driving choice that yields two dyads yields two dyads however the search is written, because a group of four containing a placeable pair would not have been minimal. That is the quantity every count in this field uses, and it is the one worth trusting.

What it buys downstream #

Two things, and one of them is a prediction the graph makes about a count of configurations.

The number of assemblies. A dyad has two solutions. A chain solved as $k$ dyads in sequence therefore assembles in $2^k$ ways at a given input angle — a prediction made from the graph alone, checked against a count of distinct converged solutions from random seeds, and correct on every case tested. That is the next essay in the algebra ladder.

And a warning about the solver. A group of four or more has to be solved as a system, which means a seed, which means branch selection is decided by where the seed was — and there is no compass construction to say what the branches even are. On a dyadic mechanism the branches are enumerable and a solver can be told which one to take. On a non-dyadic one they are not, and following one continuously is the only reliable way to stay on the branch a sweep started on.

That is why the sweeps in this field are continuations from the previous frame rather than cold solves at each angle, and why a frame that fails to converge is reported rather than drawn.

The last figure is the next rung. Four of the sixteen eight-link chains have no all-dyadic driving choice at all — not a bad one, none — and on those the compass has run out, permanently and for reasons that no choice of dimensions can repair.

Why a group always has an even number of links #

The group sizes that turn up are two, four, six and eight, and never three, five or seven. That is not an accident of the census and it does not have to be observed — it follows from the arithmetic that defines a group.

An Assur group is a set of links that, when attached to parts already placed, adds no freedom: it must be exactly rigid against what is known. For $g$ links and $p$ pins joining them to each other and to the placed parts, that is $3g - 2p = 0$, so $p = 3g/2$ — and a pin count is an integer, so $g$ must be even.

The pin counts come with it. A group of two links has three pins, which is the dyad: two outer pins onto placed parts and one between the pair, the two circles that meet. A group of four has six, a group of six has nine, and so on up. Every one of those is a set with exactly enough pins to be rigid and not one more, which is what makes it the smallest thing that can be placed as a unit.

So the decomposition can only ever return even sizes, and the interesting boundary is between the first of them and all the rest. Two links and three pins is a quadratic — two circles, two intersections, a compass. Four links and six pins is a system with no such factorisation, and it is where this site’s solve stops being a convenience. There is nothing in between, and the reason there is nothing in between is this line of arithmetic rather than any fact about the sixteen chains.

That also explains a shape in the census that would otherwise want explaining. A chain with an odd number of movable links after the frame and the input are removed cannot come apart into dyads at all, whatever its wiring — the sizes must sum to an even number and every size is even, so the parity has to work out before any structure is considered. The chains where the decomposition is interesting are therefore the ones where the parity permits an all-dyadic answer and the wiring refuses it, which is exactly the set four of the sixteen belong to.

The parity argument has a second use, and it is a cheap check rather than a result. A decomposition that returns an odd group has a bug in it, and so does one whose group sizes do not sum to the number of links left after the frame and the input are set aside. Neither of those requires knowing anything about the chain, so both can be asserted on every decomposition the census runs — which on sixteen chains at twenty driving choices each is three hundred and twenty opportunities for an off-by-one to announce itself. A check that costs an addition and fires on a whole class of errors is the sort this site collects, and this one came free with the arithmetic that says the sizes are even.

There is a limit to how far the parity argument reaches, and it is the same limit the whole decomposition has. It says the sizes are even and it says nothing about which even sizes occur, so a chain whose leftover links number six may come apart as three dyads, as a dyad and a four, or as a single six — three outcomes the arithmetic cannot separate. Deciding between them is the structural question, and it is the one that has to be computed per chain and per driving choice. What the parity buys is a floor under the bookkeeping rather than an answer, which is the ordinary relationship between a counting argument and the thing it counts.