The area a coupler point encloses

The four curves at the top of this page are drawn by four different crank-rockers. Their grounds run from 2.6 to 5, their couplers from 2.5 to 3.5, their rockers from 1.9 to 3.8, and the curves they draw are four different shapes in four different places. What the machines share is a crank of length 1 and a tracing point three tenths of the way along the line from the crank pin to the rocker pin.

Each curve encloses an area of 2.199115. That is 0.7π, and it is not a coincidence of the four sets of lengths chosen. It holds for every crank-rocker with that crank and that tracing point, on both of its assemblies, and the reason fits in a line of algebra that turns a curve’s area into the areas its pins enclose.

What a coupler point draws is a sextic whose shape is violently sensitive to every length in the machine. The area inside it is, in the commonest case, sensitive to one of them. This essay is about how that can be, where it stops being true, and the one term that has no closed form here.

Half a cross product, and what falls out of it #

The area inside a closed curve is half the integral of the cross product of position with its own increment, taken once round. It is a signed quantity: positive when the curve runs anticlockwise, negative when it runs clockwise, and for a curve that crosses itself it is the sum of the loops with their own signs. Every area in this essay is that signed quantity, because it is the one that has a formula.

A coupler point is a fixed combination of the two moving pins. Written with one complex number ℓ = u + iv, where u is the fraction along the line from the crank pin A to the rocker pin B and v is the fraction perpendicular to it, the point is P = (1 − ℓ)A + ℓB. That is the same attachment the solver uses, with the offset measured in coupler lengths.

Put that combination into the area integral and it expands into four pieces. Two of them are the areas the pins themselves enclose, weighted by the squared sizes of their coefficients. One is the area the coupler’s own direction sweeps, which is the square of the coupler length times the angle it turns through. The fourth is a mixed integral of one pin against the other, and it splits once more, because the increment of the product of the two positions integrates to nothing round a closed circuit.

Collected, the result is

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

where $S_A$ and $S_B$ are the signed areas the crank pin and the rocker pin enclose, N is the number of whole turns the coupler makes in one circuit, b is the coupler length, and the last term is the integral of the rocker pin’s position against the crank pin’s motion.

Nothing in that derivation is about four-bars. It holds for any rigid bar whose two ends move on closed paths, which is why it is an old result in a new setting: it is Holditch’s theorem of 1858, generalised past the single convex track his chord slid round, and it is the identity a polar planimeter’s arm integrates when it reads an area as a rolled distance.

On the coupler line of a crank-rocker, only the crank is left #

Now take the commonest case drawn in these essays, a crank-rocker, and put the tracing point on the coupler line, so that v = 0.

The crank pin goes round its circle once, so $S_A = \pi a^2$.

The rocker pin never goes round. It swings out along an arc of its circle and comes back along the same arc, so the region it encloses has no inside, and $S_B = 0$ — exactly, whatever the rocker length and whatever the swing.

The coupler rocks as well. It tilts one way and back again during a turn of the crank, and returns to its starting direction without having gone round, so N = 0.

And the mixed integral is multiplied by v, which is zero on the line.

Three of the four terms vanish identically, and what remains is $S_P = (1 - u)\pi a^2$. The ground length, the coupler length and the rocker length have all dropped out, along with the shape of the curve, its position in the plane and the angle the rocker swings through.

The straight line in that figure has a reading at each end worth making explicit. At u = 0 the tracing point is the crank pin, and the area is the crank circle’s. At u = 1 it is the rocker pin, which only swings, and the area is nought. Past the rocker pin the area goes negative, and a negative area is not a malfunction: the curve is being traversed clockwise, and a curve traced by a point beyond B turns the other way round from one traced by a point short of it.

Two crank-rockers with nothing in common but the crank lie on the same line, dot for dot. The shapes of their curves differ throughout the figure, and the areas do not differ at all.

Two routes, and how they approach each other #

The number 2.199115 has so far been produced by an argument. The site’s habit is to produce it twice, by routes that share no inputs past the four lengths, and to watch how the two approach each other.

The traced route positions the linkage at a list of crank angles, records the tracing point at each, and adds up the shoelace sum of the polygon through those points. It knows nothing about pins, turns or Holditch. It is the area of a polygon.

The formula route never looks at the tracing point. It reads πa² off the crank length, reads nought for a pin that retraces an arc, counts the coupler’s turns and rounds the count to the integer it must be, and evaluates the one integral that remains only when v is not zero.

A polygon inscribed in a smooth closed curve misses the thin slivers between its chords and the curve, and those slivers shrink as the square of the spacing. So the traced route should close on the formula at second order in the number of positions, and one Richardson step — combining the sums at two spacings so the second-order error cancels — should leave a fourth-order error behind.

That is what the measurement shows, on both linkages, to two decimal places in the slope. The value of the figure is not that the gap is small at 2,048 positions — any error is small somewhere — but that it falls at exactly the rate a sampling error falls. A term missing from the formula would not fall at all. It would sit at a fixed height while the polygon’s error slid down past it, and the curve would flatten out onto it.

The triple rocker needed one repair before it behaved. Its crank does not turn, so its circuit runs out along one assembly and back along the other, and at each limit the pin positions have a square-root corner in the crank angle. Parametrising the crank angle as the midpoint plus half the range times the sine of a new angle turns each corner into a smooth reversal, and the circuit becomes a periodic function the Richardson step can work on. The first version also handed back a failed position at the limit itself, where rounding made a zero slightly negative, and placed the pin at the rocker pivot. That cost the fourth decimal of the area and made the convergence first-order, which is exactly how it was found.

When the rocker pin also goes round #

Ground the shortest link and both frame-adjacent links turn fully. The inversion is a double crank, and the formula changes character, because now every term is alive.

The crank pin encloses πa² = 28.274. The rocker pin encloses πc² = 32.170, because it goes all the way round too. And the coupler turns once in each circuit, carried round with the two cranks, so N = 1 and the third term is −u(1 − u)πb².

Put together for u = 0.3, that is 0.7 × 28.274 + 0.3 × 32.170 − 0.21 × π × 12.25, which comes to 21.361. The traced polygon agrees to 3.1 × 10⁻¹³. The tracing point’s curve is smaller than a straight average of the two pin circles would suggest, and the deficit is precisely the area the coupler sweeps as it turns once, shared out between the two ends in the proportion u(1 − u).

The same machine with the tracing point off the line, at (0.45, 0.5), encloses 56.949, and there every one of the four terms contributes: 15.551 from the crank pin, 14.476 from the rocker pin, 0.096 from the coupler’s turn — small because u − u² − v² is nearly zero there — and 26.826 from the mixed integral. No term in the formula is decorative in general. The crank-rocker’s case is special because three of them happen to vanish.

Holditch’s chord, with both ends swinging #

The third Grashof class is the one that makes the connection to Holditch exact.

Holditch’s theorem, as he stated it, concerns a chord of fixed length whose two ends slide round a closed convex curve. A point dividing the chord into lengths p and q traces a second closed curve inside the first, and the area between the two is πpq — independent of the curve.

A double rocker with the coupler as its shortest link is that chord with the curve taken away. Both rockers only swing, so neither pin encloses any area. But the coupler, being the shortest link, turns fully relative to its neighbours and to the ground, so N = ±1. The formula on the coupler line leaves a single term, u(1 − u)πb², and with p = ub and q = (1 − u)b that is πpq.

Three machines with a coupler of 1 and nothing else in common land on one parabola; the fourth, with a coupler of 1.5, lands on a parabola 2.25 times taller. Neither the crank nor the rocker nor the ground has a say.

In Holditch’s version the ends travel round a curve that encloses an area, and the chord’s point encloses that area less πpq, so the theorem is stated as a difference. Here the ends travel on arcs that enclose nothing, and the difference is the whole of it. The four-bar supplies what the theorem needs — two ends kept a fixed distance apart, each returning to where it started, and a chord that turns once — without anything convex anywhere.

A figure-eight that encloses nothing #

The fourth class is the non-Grashof chain, where no link turns fully relative to any other. Nothing goes round: the crank pin retraces an arc, the rocker pin retraces an arc, and the coupler rocks and returns. On the coupler line every term is zero.

The curve in that panel is not small. It crosses itself once, and each of its two lobes encloses 6.790 — one anticlockwise and one clockwise — so the signed total is 4 × 10⁻¹⁵. The formula predicts nought and the tracing polygon returns nought to rounding, and the lobes are the reason nought is not the same statement as encloses nothing.

A second argument gives the same zero, sharing nothing with the formula. Every four-bar’s configurations are carried onto configurations by reflection in the ground line: the crank angle changes sign and the rocker pin moves to the other side of the line from the crank pin to the rocker pivot. A triple rocker has a single circuit, which runs out on one side and back on the other, and the reflection carries that circuit onto itself in the same direction of travel. A point on the coupler line reflects to the same point of the reflected configuration. So the reflection carries the curve onto itself without reversing it, while reversing every signed area — and the only signed area equal to its own negative is nought.

The same reflection applied to a crank-rocker carries one assembly’s circuit onto the other’s, which is the second reason the two assemblies of a crank-rocker enclose the same area on the coupler line. It says nothing about a point off the line, because a reflection turns a point at +v into a point at −v.

Off the line, one integral stays unevaluated #

Move the tracing point off the coupler line of the crank-rocker and the last term wakes up. The coupler still does not turn, so the term in v² stays asleep, and the area becomes (1 − u)πa² − v ∮B·dA: a straight line in v whose slope is the mixed integral.

Both lines are exactly straight, because nothing in the formula is quadratic in v while the coupler does not turn. Their slopes are equal and opposite, and they cross where the point is on the line.

The integral ∮B·dA is the part of this essay that is computed rather than derived. Written out, it is an integral over the crank angle of the rocker pin’s offset from the line to the rocker pivot, times a factor linear in the cosine of the crank angle. The offset involves the square root of a quadratic in that cosine, so after the natural substitution the integrand carries the square root of a quartic. Integrals of that kind are elliptic in general, and no closed form is claimed here. The site evaluates it by the same corrected quadrature as the area, and the straightness of the two lines to 10⁻¹⁰ is the check that the quadrature and the formula agree about what it is.

So the claim that an area depends on the crank alone is a claim about the coupler line. Off it, the area depends on the whole machine through one number, and that number has no expression in the four lengths that can be written down here.

Why the two assemblies straddle the line #

The two slopes being equal and opposite is not a property of the one crank-rocker in the figure, and it has a reason.

At each crank angle, the rocker pins of the two assemblies are mirror images of each other across the line from the crank pin to the rocker pivot, because they are the two intersections of the same pair of circles. Their average therefore lies on that line. The mixed integral of that average against the crank pin’s motion splits into a part along the crank pin’s own position, which integrates to nought because the crank pin moves at right angles to it, and a part along the ground, whose integrand is odd in the crank angle and cancels over the turn. So the two assemblies’ integrals sum to nought exactly.

The two curves in that figure look nothing alike — one a long leaning oval above the ground line, the other a flat lens below the coupler — and between them they enclose exactly twice what a point on the line would. What one assembly gains by the tracing point sitting off the line, the other assembly loses.

That is also a caution about reading the hero figure. Its four areas are equal on both assemblies because the tracing point is on the line. A point a coupler plate puts anywhere else gives two different areas for one machine, depending on which way it was assembled.

Which length an area answers to #

The survey of quantities by scale fitted the power an enclosed area follows when every length is multiplied together, and found two. The formula agrees and says more.

Every term is a length squared times a pure number: πa², πc², πb² times a count, and the mixed integral, which is a position times a displacement. So multiplying every length by k multiplies the area by k², as the survey found. But on the coupler line of a crank-rocker the only length in the area is the crank, which means the exponent two belongs to the crank alone, and a machine whose crank is exactly right and whose other three lengths are all 1% long encloses exactly the area it was drawn to enclose.

That is the sense in which the area is a narrower measurement than the curve. The curve’s shape is sensitive to every dimension, with the coupler hole the worst of them. Its area, for this one class and this one line, is a reading of one hole spacing on one link, and the survey’s rule that a size error propagates with a factor equal to the exponent applies to that link and no other.

What the formula does not reach #

Four things are outside what this essay establishes, and each is a different kind of gap.

The closed form of the mixed integral. It is evaluated, not derived, and the claim that it is elliptic in general is a statement about the form of its integrand rather than a proof that no elementary expression exists for four-bars in particular.

Change-point linkages. When the shortest and longest lengths sum exactly to the other two, two circuits touch at a configuration where the coupler can go either way. The count N is then not a property of the linkage but of which way it went, and the circuit parametrisation assumes simple limits. Nothing here was run at the boundary.

Holditch’s theorem as stated. The case tested is the degenerate one where both ends retrace arcs. A chord with its ends on one convex curve is a different mechanism — two sliders on one closed track — and it has not been built here.

The area the machine sweeps. The region a tracing point’s curve encloses is not the region the machine occupies. The room a machine sweeps is a union over every link’s body at every position, computed on a grid and bracketed by a certificate, and it depends on everything. The two areas answer different questions and neither bounds the other.

Where this goes from here #

The next step is a test of Roberts’s theorem by area, and it is a sharper test than any run so far.

Three four-bars draw one coupler curve, and that has been checked by overlaying traces and by fitting one equation to all three. Each cognate has its own crank, its own coupler and its own tracing point, and generally its own class: the site’s standard crank-rocker has a double-rocker cognate. The formula therefore evaluates one area by three different sums — a crank circle and a mixed integral for one, Holditch’s chord and a mixed integral for another — and Roberts’s theorem says the three must agree.

That is a claim with content, because the terms that carry the area differ between the machines while the curve does not. It would compute the three decompositions side by side, report which terms each cognate’s area lives in, and use the agreement of the totals as an identity between three mixed integrals that have no closed form here.