The kind is decided before the lengths are

Where three machines keep one area set the standard crank-rocker beside the two four-bars Roberts’s construction gives for its coupler curve, and found the three agreeing about the area the curve encloses to 10⁻¹² while disagreeing about every term of the formula that produces it. The crank-rocker carried the closed-form part in its crank pin’s circle, the double rocker in its coupler’s whole turns, and the third machine in its output pin’s circle; the one term with no closed form was the same number in all three.

It ended by proposing the obvious extension. There are eight kinds of four-bar, not three, and each of them has cognates of some kinds. The table of which, over a census rather than for one machine, would say two things: whether three machines drawing one curve can ever keep their closed form in the same column, and whether a triple rocker, which keeps no closed form at all, has cognates that do.

Both answers are here, and the table turned out to be shorter than expected — short enough to have an algebraic form rather than a measured one.

The census, and what it does not find #

The three signed sums that sort four-bars into eight regions are

$$T_1 = g + c - a - b, \qquad T_2 = g + b - a - c, \qquad T_3 = b + c - a - g$$

with $g$ the ground, $a$ the input, $b$ the coupler and $c$ the output, and each of the eight sign patterns is a region inside which nothing about the motion’s shape can change. The census draws four lengths at random, keeps the chain if it closes and if no sum is within two per cent of nought, runs Roberts’s construction, keeps the result if both cognates close as well, and records which regions the three landed in. Twenty-four thousand chains were kept out of 29,220 drawn, and five tracing points were used in rotation so that a row agreeing with itself is a statement about the point as well as the lengths.

Not one chain produced a second entry in its row. A crank-rocker’s cognates are a double rocker and a rocker-crank, in 3,403 cases out of 3,403. A double crank’s are two double cranks, in 3,459 of 3,459. Every one of the four triple-rocker regions is as fixed as the Grashof ones, and no row crosses the Grashof line in either direction, which is the classical statement that the construction preserves Grashof’s inequality arriving as a by-product rather than as the result.

Every cognate closed. That is worth one sentence because it could have failed: Roberts’s construction is a similarity applied to the triangle on the coupler and it always returns three lengths, and three lengths and a ground are not always a chain that assembles. Over the whole census the count of cognates that could not be built was nought, which is what it must be — the cognate traces the same curve, so it moves, so it closes.

A census with no exception in it is a census asking to be replaced by a reason.

Before that, one thing the census is careful about, because it is where a result of this shape usually goes wrong. A chain within two per cent of a region wall is discarded, and the two per cent is measured against the chain’s own longest bar rather than against a fixed number, so the filter means the same thing for a machine a millimetre across and one a metre across. Without it a chain whose $T_2$ is 10⁻¹⁴ would be classified by its rounding error, and its cognate — whose sums are the original’s multiplied by a factor that can be small — by a different rounding error. The 5,220 chains dropped between the 29,220 drawn and the 24,000 kept are those, plus the ones whose four lengths cannot be assembled at all. Neither kind is an exception being hidden; a chain on a wall has no region to be right or wrong about.

The tracing point enters as a scale #

The reason is in the construction’s own arithmetic. Write the tracing point as the complex number $\lambda$ along the coupler, so that the point sits at $(1-\lambda)A + \lambda B$. Roberts’s first cognate has ground, input, coupler and output

$$|\lambda| \cdot (g,\; b,\; a,\; c)$$

and the second has

$$|1 - \lambda| \cdot (g,\; b,\; c,\; a).$$

The four lengths are the original’s, permuted, and multiplied by one common positive factor. And each of $T_1$, $T_2$ and $T_3$ is homogeneous of degree one in the lengths, so multiplying all four by the same positive number multiplies all three sums by that number and changes no sign.

That is the whole of it. The tracing point enters the construction only through $|\lambda|$ and $|1 - \lambda|$, and those are scales, and a region cannot see a scale. What is left is a relabelling of four lengths, which is a signed permutation of three expressions:

$$(T_1, T_2, T_3) \;\longmapsto\; (T_1,\, -T_3,\, -T_2) \quad\text{and}\quad (T_1, T_2, T_3) \;\longmapsto\; (-T_3,\, T_1,\, -T_2).$$

Two routes with nothing in common past the four lengths. One assembles twenty-four thousand machines, solves each of them and classifies what comes out; the other never builds anything and rewrites three expressions. They agree in all eight regions, which is the only arrangement under which the agreement is evidence rather than a coincidence of method.

It also explains the shape of the answer rather than only its value. A region is a sign pattern, a signed permutation acts on sign patterns, and there are very few of those — which is why a table that could in principle have held sixty-four entries holds eight, and why the entry in each depends on nothing at all except which of the eight rows it is in.

The map does not close, and the reason is the motor #

There is something wrong with reading the table as a partition, and it shows up the moment the construction is applied twice.

Start with the standard crank-rocker. Its cognates are a double rocker and a rocker-crank, so the three machines that draw its curve occupy three different regions. Now start with that double rocker and apply the construction to it, with its own tracing point. Its cognates are two crank-rockers. The set has changed: the first reading says the triple contains a rocker-crank and the second says it contains a second crank-rocker.

Both readings are correct measurements and the disagreement is not in the geometry. A region is a sign pattern of three sums in which $a$ and $c$ appear differently, so it says which of the two ground pivots carries the motor as well as what the four bars are. Swapping the input and output lengths is a relabelling — the same four bars in the same order, read from the other end — and under it

$$(T_1, T_2, T_3) \;\longmapsto\; (-T_3,\, T_2,\, -T_1),$$

which fixes four of the eight regions and moves the other four in two pairs. A crank-rocker driven from its other pivot is a rocker-crank. Grashof’s rule gives those two the same name for exactly this reason and is criticised for it; the eight regions separate them, which is usually the improvement, and here it is the thing that has to be taken back out.

Roberts’s construction does not only produce a chain. It produces a chain with its ground pivots already chosen — the first cognate stands on the line from the original’s input pivot to the tracing point, the second on the line from its output pivot — so it makes the labelling decision on the reader’s behalf, and a map between regions reports that decision alongside the geometry. The check that this is the whole discrepancy is one line of the same algebra: composing the relabelling with the first cognate’s substitution gives the second cognate’s substitution exactly. The second cognate is the first one driven from the other end.

Four sets of three #

Quotienting the eight regions by that relabelling leaves six classes of chain, and on those the construction closes. Four sets of three, and starting from any member of a set returns that set.

The Grashof half holds the crank-and-rocker set — two chains with one member turning and one swinging, and a double rocker — and the double crank’s set, which is three double cranks. The non-Grashof half has the same shape one step over: the both-inner triple rocker is a set of three alike, and the remaining three triple-rocker regions collapse into a set of two alike and one other. That the two halves should have the same shape is not something the census was looking for and is not explained here; it is what the signed permutations happen to do, and whether it survives the same question asked of a five-bar or a six-bar is open.

The second of the two questions is already answered by this figure and needs no measurement. A triple rocker’s cognates are triple rockers. The four triple-rocker regions form two sets between them and neither set has a member outside, so there is no machine anywhere in the construction that draws a triple rocker’s curve and is not itself a triple rocker.

What a designer cannot buy #

Roberts’s construction is not usually met as a classification result. It is met as the answer to a practical difficulty: the linkage draws the right curve and its ground pivots are in the wrong place, or one of its bars fouls something, and the construction produces two other machines that draw the same curve with their pivots somewhere else. What a coupler point draws is the problem and the three machines are the room to move in it.

The table says what that room is, and it is smaller than it looks. The two alternatives are not two arbitrary four-bars. Their kind is fixed before any length is chosen, so a designer who has a crank-rocker and needs a machine that can be driven continuously has exactly one alternative of that kind — the rocker-crank — and the other is a double rocker, which has no member that turns and therefore no place to put a motor at all. The room is in where the pivots are, never in what the machine is.

At the other end the constraint is absolute. A triple rocker’s curve can be drawn by three machines and all three are triple rockers. There is no arrangement of the construction, at any tracing point, that produces a chain with a continuously turning input for such a curve. If a curve is wanted from a motor and the machine that draws it is a triple rocker, cognation is not the way out, and the way out has to be a different curve or a different chain — the six-bar route rather than the cognate route.

The double crank is the opposite case and is the most generous. All three of its machines turn at both ends, so all three can be driven from either pivot, which is six places to put a motor for one curve.

Where the area goes #

The area formula has four terms. Three are closed forms — the crank pin’s circle $\pi a^2$, the rocker pin’s circle $\pi c^2$, and the coupler’s whole turns — and each is exactly nought unless that member goes all the way round. The fourth is an integral of the rocker pin against the crank pin’s motion and has no closed form. A machine’s entry in the table below is therefore the set of closed-form columns it occupies, and because a pin either retraces an arc or goes round its whole circle that set has nought, one or three members and never two.

The crank-rocker’s set behaves as the essay below it described: three machines, three different columns, and the mixed integral the same number in all three. It is the only one of the four that does.

In the double crank’s set every member of every chain goes round — that is what a double crank is — so all three machines have all three closed-form terms at once. They are not three different decompositions of one number into three parts. Sorted, the three machines’ sets of three are the same three values to 1.1 × 10⁻¹⁴.

That is a stronger statement than sharing a column. The construction here moves no area at all between the closed form and the integral; it permutes the closed form’s three numbers among its three columns, so the sum of the three is the same in each machine and the term-by-term disagreement that the crank-rocker’s set shows is reduced to a change of which pin is responsible for which piece.

The reading that makes this less surprising is about what the closed-form terms are. Each of them is the area a pin sweeps in going once round its own circle, and a pin’s circle is a property of the chain’s lengths rather than of the curve. The construction permutes the lengths, so it permutes the circles; what it cannot do, in a region where every pin goes round, is turn one of those circles off. The crank-rocker’s set is the interesting one precisely because there the permutation does turn terms off and on, since which pin goes round changes from machine to machine.

The double rocker’s set does share a column, in the plainest sense. Its two cognates are both crank-rockers, both have their crank pins going round and neither has anything else, so both keep their closed form in the crank column and both keep the same value, −0.007854. The original keeps its own in the coupler column. Two machines, one column, one number.

And in both triple-rocker sets there is no closed form to place. No member of any of the six chains goes round, every closed-form term is exactly nought, and the whole enclosed area is the integral.

That case is worth a second look, because the two cognates are not the same machine. Their lengths differ — one is the original’s scaled by $|\lambda|$ and the other by $|1 - \lambda|$, and those are different numbers — and their ground pivots stand in different places. They are two distinct crank-rockers that happen to enclose the same crank-circle term, and they do so because the term is $(1 - u)\pi a^2$ evaluated in each machine’s own coupler coordinates, where both the scale and the tracing point have moved in step. Two different machines arriving at one number by two different routes is the usual arrangement in this field, and here it arrives without anybody having set it up.

A rule that was a fact about one machine #

The essay below this one left behind a working rule: that the three machines drawing a curve carry their closed form in three different columns. It was written on the standard crank-rocker, where it is true, and it reads as a statement about Roberts’s theorem.

It is not one. Fed a double crank it is refused, because all three machines are in all three columns. Fed a double rocker it is refused, because two of the three are in the same column. Fed a triple rocker it is refused, because there are no columns occupied at all.

This is the standing hazard a rule calibrated on one machine always carries, in its usual form: a claim with a number or a count in it is often a fact about the first case rather than about the family, and it holds there and fails on the next. The repair is not to weaken the rule. It is to say which of the four sets it is about, which is a sharper statement than the one it replaced — the three-different-columns behaviour belongs to the crank-and-rocker set, and belongs to it because that set is the only one whose three machines differ in which members turn.

That is also the reading that makes the other three cases obvious in hindsight. The closed-form column a machine occupies is decided by which of its members goes round; the set’s three machines occupy three columns exactly when they disagree about that in all three ways. A double crank’s set cannot disagree, because every member of every one of them turns. A triple rocker’s cannot, because none does. A double rocker’s set has two machines that agree.

What this does not settle #

The regions are a planar four-bar’s. Nothing here is about the slider-crank, whose kinds are four rather than eight because one of the three sums is lost in the limit, and nothing is about a six-bar. Roberts’s construction has analogues for both and the substitutions would be different ones.

The areas are compared at one tracing point per region. The column table and the permuted values are computed at $(0.45, 0.5)$ of each machine’s own coupler. The columns cannot depend on the point, since which members turn does not, but the values in them do, and no sweep of the point is reported.

The two halves having the same shape is a coincidence until it is not. The Grashof and non-Grashof sets come out as one set of three alike and one set of two-and-one in each half. That is a property of two signed permutations acting on eight sign patterns and it is checked rather than derived, and no argument here says it had to happen.

Nothing is said about the curve’s pieces. A coupler curve can have two ovals or one, and the column table is read off the first oval. A set’s three machines draw every oval of the curve, and whether a machine’s column can differ between a curve’s two ovals is a question the ledger could answer and this essay does not ask it.

Still open: whether the areas are permuted anywhere else #

The double crank’s set does something none of the other three does: its three machines carry the same three closed-form values in different columns, so the multiset of closed forms is an invariant of the set and not only the total. The obvious question is whether that is special to the double crank or a general fact hiding under the other sets’ zeroes — in the crank-and-rocker set the three values are $\pi a^2$ scaled, nought and nought for one machine and a different arrangement for the next, and it is not clear whether those agree as multisets too or merely happen to sum alike.

Its distinct argument would be that comparison run over a census rather than a representative: for every set, the three machines’ multisets of closed-form terms, compared entry by entry after sorting, with the tolerance stated and the tracing point swept. If they agree everywhere, the area’s decomposition has a stronger invariant in it than the total, and Roberts’s construction is a permutation of that invariant rather than a rearrangement of it. If they agree only for the double crank, the reason must be that the double crank’s set is the one whose three machines are all in the same region, and the invariance is a statement about that coincidence instead.