Where three machines keep one area

Three linkages, one curve drew Roberts’s theorem the way it is usually drawn: a four-bar with a point on its coupler, two other four-bars built from it, and one curve traced by all three. Three linkages, one equation then checked it a second way, by fitting the sextic that vanishes on each trace and finding the same coefficients to 5 × 10⁻⁷.

The area a coupler point encloses ended by proposing a third check. The area inside a coupler curve can be written as a sum of four terms — the areas the two pins enclose, the coupler’s turning, and one integral with no closed form — and three machines that draw one curve would have to agree about the total while, in general, disagreeing about every term. That is the test run here. It passes. What it finds on the way is more interesting than the pass.

Three machines from one #

The original is the crank-rocker used throughout these essays: ground 4, crank 1, coupler 3.5, rocker 3, with the tracing point written as one complex number λ = 0.45 + 0.5i. That means it sits 0.45 of the way along the line from the crank pin to the rocker pin and half a coupler length to one side.

Roberts’s construction puts a third fixed pivot at λ times the ground, and hangs from it two new four-bars whose bars are the original’s bars, rescaled and with their roles exchanged. That third pivot is not an arbitrary point either: the circle through it and the original’s two pivots is the circle every double point of the curve lies on. The first cognate is the original scaled by |λ| = 0.673 with its crank and coupler swapped; the second is scaled by |1 − λ| = 0.743 with its bars permuted the other way. Each carries its own tracing point, and all three points draw one curve.

The three machines are not three of a kind. Sorted by the signs of their three length sums, the original is a crank-rocker: its input turns and its output swings. The first cognate’s shortest bar is its coupler, which makes it a double rocker whose coupler turns fully while both grounded links only swing. The second cognate’s shortest bar is its output link, so its output turns and its input swings; in the eight kinds it is a rocker-crank.

The curve they share also has two pieces. A crank-rocker’s two assemblies draw two separate ovals, and the cognates must draw both. Each machine has two circuits, and each circuit is matched to an oval by where its points are, never by its area, since area is what is being compared. Placed back in the original’s frame, every circuit lies on one oval to within 4 × 10⁻⁷ and at least 2.9 from the other.

The formula, three times #

The area formula from the essay on enclosed areas is

$$S \;=\; (1-u)\,S_A \;+\; u\,S_B \;-\; \bigl(u - u^2 - v^2\bigr)\,\pi b^2 N \;-\; v \oint B\cdot dA,$$

where $S_A$ and $S_B$ are the areas the crank pin and the rocker pin enclose, N is the number of turns the coupler makes, b is the coupler’s length and (u, v) is the tracing point in the coupler’s own coordinates. A pin that goes round encloses its whole circle; a pin that swings back and forth along an arc encloses nothing. So which terms a machine keeps is decided by which of its parts go round.

The crank-rocker’s crank pin goes round and nothing else does, so its area is its crank term plus the integral. The double rocker’s pins both swing and its coupler turns once, so its area is the coupler term plus the integral: Holditch’s term, a chord turning once. The rocker-crank’s output pin goes round and nothing else does, so its area is its rocker term plus the integral.

The ledger shows three things at once. First, the totals agree, which is Roberts’s theorem passing its area test: the first oval encloses 2.116353777 and the second 1.339398142, for all three machines, to 10⁻¹². Second, the closed-form part of the area is in a different term for each machine: the crank term for the crank-rocker, the coupler term for the double rocker, the rocker term for the rocker-crank. Third, and not required by anything said so far, those three different terms are the same number, 1.727876. And the mixed integral, which has no closed form, is the same number too: 0.388478 on the first oval and −0.388478 on the second.

The two ovals together say something the single oval does not. On each machine the closed-form term is the same for both ovals and the mixed term changes sign between them, so the mean of the two ovals’ areas is the closed-form term alone: (2.116353777 + 1.339398142)/2 = 1.727876. The area essay found that the crank-rocker’s two assemblies straddle the area a point on the coupler line would enclose, because their rocker pins are mirror images across the line from the crank pin to the rocker pivot. The cognates straddle the same number, each for its own reason: the double rocker because its coupler turns the same amount on both circuits, the rocker-crank because its output pin’s circle is the same circle on both. Three machines with three different mechanisms for it agree about where the middle is.

One number in three closed forms #

The third observation has a one-line explanation, and it is worth having because it turns a coincidence of these lengths into a property of the construction.

The crank-rocker’s closed-form term is $(1 - u)\pi a^2$, the crank pin’s circle scaled by how far along the coupler the point sits: 0.55π = 1.727876.

The double rocker’s coupler is the original’s crank scaled by |λ|, and its tracing point is 1/λ in its own coupler’s coordinates. The real part of 1/λ is $u/|\lambda|^2$ and its squared size is $1/|\lambda|^2$, so $u - u^2 - v^2$ for that point is $(u - 1)/|\lambda|^2$. Multiplied by π times the coupler’s squared length, $|\lambda|^2 a^2$, the $|\lambda|^2$ cancels and what is left is $(1 - u)\pi a^2$, turned once.

The rocker-crank’s output link is the original’s crank scaled by |1 − λ|, and its tracing point is 1/(1 − λ), whose real part is $(1 - u)/|1 - \lambda|^2$. Times π times the output link’s squared length, $|1 - \lambda|^2 a^2$, the scale cancels again and leaves $(1 - u)\pi a^2$.

The double rocker’s version has a classical name, and a sign that looks wrong until the name is examined. A chord of fixed length whose two ends travel on closed paths, with a point dividing it into lengths p and q, is the setting of Holditch’s theorem, and the area term it contributes is πpq. For a point between the ends p and q are both positive. The double rocker’s tracing point is not between the ends of its coupler. For a tracing point on the original’s coupler line at u, the cognate’s point is at 1/u along its own coupler: at u = 0.3 that is 3.33 coupler lengths along a chord of length 0.3, far beyond its other end. Then q is negative and so is πpq: with its coupler turning the same way, a point beyond the end of a chord goes round its curve in the opposite sense to a point between the ends, and the chord term carries that sign. Measured on that line, all three machines enclose 2.199115 = 0.7π, and for the double rocker all of it is the chord term, with p = 1 and q = −0.7, so that π|pq| is 0.7π.

So the construction does not merely preserve the total. It carries the crank’s circle of the original into a coupler’s turn in one cognate and an output pin’s circle in the other, rescaled so exactly that each is the same number. And once the totals agree and the closed-form parts agree, the mixed integrals have to agree as well, since they are what is left. That is a derivation, and the ledger measures what it predicts, to 10⁻¹².

A machine called by the wrong name #

The formula needs to know which pins go round, and one of the three machines is exactly where a classification gives the wrong answer to that question.

The shortest bar of the second cognate is its output link, and its output link is next to the frame. Grashof’s rule, as it is usually applied, names a linkage whose shortest bar is next to the frame a crank-rocker, whichever end of the frame that bar is at. So the rule calls this machine a crank-rocker, which is the name of a machine whose input turns and whose output swings. This one does the opposite. The eight kinds of four-bar found 530 linkages in a census of 4,000 that the usual classifier names this way.

An area formula that decides from a name whether the rocker pin goes round — for instance by counting the rocker’s circle only for a double crank, the one Grashof name that means both ends turn — is right for the crank-rocker and the double rocker and wrong for the second cognate. It counts that machine’s area as 0.388478, the mixed integral alone, and leaves out $u\pi c^2$ = 1.727876 of the 2.116354 the machine encloses.

The traced polygon, which knows nothing about names, is not fooled. Computed both ways, a name-based formula and the trace disagree by exactly the missing term, on exactly the machine whose name is wrong, and on nothing else, which is the whole case for computing an area two ways. Asking the pin instead of the name settles it: a pin’s traced path encloses either nothing or its whole circle, and the second cognate’s output pin encloses 1.7357 against a circle of 1.7357, which is not a close decision. Read that way, the formula and the trace agree on all three machines and both ovals, and that agreement is the ledger above.

That the misnamed machine turned up here is not an accident. Roberts’s construction exchanges the roles of the bars, so it is exactly the operation that turns a crank-rocker into a machine with its turning link at the other end of the frame. A classification that cannot tell the two ends apart is guaranteed to meet the difference in the cognates of any crank-rocker.

The part with no closed form #

The mixed integral is the term the area essay could evaluate and not write down: the rocker pin’s position integrated against the crank pin’s motion, over a circuit. Its integrand carries the square root of a quartic in the crank angle’s cosine, and nothing here claims a closed form for it.

It is also not a property of the tracing point. The mixed term is −v times the integral, and the integral itself depends only on the four lengths. For the original it is −0.776955634 wherever on the coupler the tracing point is put.

The cognates each have their own integral, over their own circuit, with their own lengths. Their mixed terms equal the original’s. Their tracing points’ offsets from their coupler lines are the imaginary parts of 1/λ and 1/(1 − λ), which are $-v/|\lambda|^2$ and $+v/|1 - \lambda|^2$. So the three integrals must stand in fixed ratios: the first cognate’s is $-|\lambda|^2$ times the original’s and the second’s is $+|1 - \lambda|^2$ times it.

At twelve tracing points spread round a loop in the coupler plane, the twenty-four ratios match their predictions to 2.3 × 10⁻¹² of themselves.

That test is harder than it looks, because the original is the only one of the three machines that stays the same as the tracing point moves. Its lengths do not depend on where the point is, so its integral does not either. The cognates’ lengths are the original’s scaled by |λ| and |1 − λ|, and their pivots sit at λ times the ground, so every tracing point round the loop hands the construction a different pair of machines. Their integrals range accordingly: the first cognate’s from 0.008 to 0.861 of the original’s size, the second’s from 0.012 to 1.013. Twenty-four different machines, each with an integral nobody can write down, and each integral lands on the value the construction predicted for it. So Roberts’s theorem makes a statement about three integrals none of which has a closed form: they are proportional, with constants the construction states in advance. The totals are the theorem; the proportionality is its consequence for the one term nobody can write down.

The signs are the part of that prediction easiest to get wrong. One cognate’s tracing point is 1/λ and the other’s 1/(1 − λ), and they put the offsets on opposite sides of their coupler lines, so one ratio is negative and the other positive. The measurement settles that independently of the algebra: the first ratio is negative and the second positive at every one of the twelve points, and both agree in size as well.

How finely an area can tell #

An equality is only as strong as the smallest difference that would have been noticed. So one cognate is built wrong on purpose: its tracing point is moved along its coupler by a small amount ε, and both comparisons — the area and the curve — are asked whether anything has changed.

Both respond in proportion. The area changes by 1.7357 times the error, which is the output pin’s circle, $\pi c^2$, because the only term with u in it for that machine is $u\pi c^2$. The curve moves by about 2.23 times the error, which is modest for a coupler curve: a coupler point’s curve is sensitive to every dimension, and a tracing point is one of the gentler ones. Neither comparison is more sensitive than the other in any useful sense.

What differs is the floor. Three exact machines’ areas agree to 1.9 × 10⁻¹³, the rounding left after the trace’s correction, while an exact cognate’s traced points already sit 4.0 × 10⁻⁷ from the original’s polyline, because a curve can only be compared at the points it was sampled at. The original’s oval is represented by 4,096 positions joined by straight chords, and a chord of an oval about eight units round cuts inside the curve by a few parts in ten million. A point exactly on the curve therefore sits that far from the polyline, and no comparison of positions against positions can resolve a difference smaller than the sag of its own chords. The area sums the same positions and has no such floor, because the correction that removes the chords’ error from a sum has nothing to correct in a distance. So the area can tell a misplaced tracing point of about 10⁻¹³ from a correct one, and the curve one of about 2 × 10⁻⁷. Six decades separate them, and they are decades of resolution, not of sensitivity.

That puts the three checks of Roberts’s theorem in order. Overlaid traces, the first check, agree to their sampling: 1.3 × 10⁻⁵ for the overlay drawn above. Fitted equations agree to 5 × 10⁻⁷. Areas agree to 10⁻¹², and they could detect a construction error a million times smaller than the curves could.

What one original and a sampled curve cannot show #

One original. Every number here belongs to the cognates of the standard crank-rocker, at three tracing points and a loop of twelve. The cognates of a double crank, a double rocker or a triple rocker are different machines, and whether their closed-form terms move between the same columns is not measured.

The mixed integral itself. Its proportionality across the three machines is derived from the equal totals and the equal closed-form parts, and measured. A closed form for any one of the three integrals is not found, and the claim that none exists is about the form of the integrand, not a proof.

The curve between samples. The area test compares numbers, and an area is blind to changes that enclose nothing: a curve deformed so that a gain on one side exactly cancels a loss on the other would pass it. The overlay and the fitted equation are not redundant with it, only coarser.

Branch points. Where a cognate’s two circuits would meet, at a change point, the circuits are not separate and matching them to ovals has no meaning. No change-point linkage is run.

What comes next: where the other kinds keep their area #

The crank-rocker’s area moved from the crank’s circle to a coupler’s turn and an output pin’s circle because its cognates are a double rocker and a rocker-crank. Every one of the eight kinds of four-bar has cognates of some kinds, and in each the closed-form part of the area must land in whichever terms those kinds keep.

Its distinct argument would be that table: for each of the eight regions, the regions of its two cognates, found over a census rather than for one machine, and the columns the closed-form term occupies in each. Two questions have answers waiting in it. Whether the three machines’ closed-form terms can ever share a column, and whether a triple rocker, which keeps no closed-form term at all, has cognates that do.