The trammel takes both double points

The double point a slide cannot keep followed a four-bar whose rocker grows without bound into a slide. A four-bar’s coupler curve owes three finite double points and a slider-crank’s owes two, so one had to go, and parity decided which: the real one, because a complex double point cannot leave without its conjugate. It left along a law, $d(|w|^2 - 1)/(\bar w - 1)$ rocker lengths, that knew nothing of the crank or the rod.

The same limit can be taken one step further. Grow the slider-crank’s crank without bound as well, its pivot receding along the slide so that the crank pin’s circle, near the pin, flattens into a straight line square to the slide. The machine becomes one with two slides crossing at right angles: the elliptic trammel, whose rod points draw ellipses. An ellipse is a conic, and a conic has no double points. So both of the slider-crank’s double points have to go, and this time parity gives no guide, because two is even. They could leave as a conjugate pair along one law, as two real points along two, or in some way that depends on the machine.

They leave together, along one law that is simpler than the four-bar’s. And on one circle of rod points, one of them stays.

The machine that becomes a trammel #

The slider-crank is the one the double points that leave for infinity measured, with one number changed. Its rod is 3, its slide runs along the x-axis through the crank’s pivot, and its crank, of length $a$, is pivoted at the origin. The rod point is written $w = u + iv$ in the rod’s own units: $u$ along the rod from the crank pin, $v$ across it, so the slider pin is at $w = 1$. As $a$ grows, the crank pin at the right-hand end of its circle runs up and down a nearly vertical arc through the point $(a, 0)$, and in the limit that arc is a second slide, crossing the first at $(a, 0)$. Near that crossing the machine is a trammel.

The double points are found the way that essay found them, with no reference to the limit. A double point is a point reached by two configurations. Setting the rod point of one copy of the machine equal to the rod point of another gives four polynomial equations in four unknowns, and all their solutions come from following four paths of a homotopy. At a crank of 5 the rod point (0.3, 0.2) has two isolated double points, points of the plane reached by complex configurations that the machine never passes through, at $(-7.90, 2.47)$ and $(-2.10, -2.47)$ from the crossing. At cranks of 10, 100 and 1,000 the nearer is 6.5, 65 and 648 from it, and the farther 16.7, 167 and 1,673.

Where they go #

Divided by the crank, with the crank’s pivot at the origin, the two points settle. At a crank of 1,000 the rod point (0.3, 0.2) has them at $\pm(0.5947, -0.5055)$ crank lengths from the pivot: symmetric through it, and fixed in these units, so leaving in proportion to the crank.

The position can be found without tracking anything, by the same device the four-bar’s law used: asking what a configuration of the machine looks like when its rod point is as far away as the crank. Write each point of the plane as the pair $z = x + iy$ and $\zeta = x - iy$ — isotropic coordinates, in which a real point has ζ the conjugate of $z$ and a complex point has the two independent. Scale everything by the crank, $Z = z/a$ and $Z' = \zeta/a$.

The crank pin satisfies $Z_A Z'_A = 1$ at this scale, and the rod’s two isotropic components have product $9/a^2$, so one of them is small. If the rod’s $z$-component is the large one, the slide condition makes the slider pin’s two components nearly equal, and the rod point is $Z = (1 - w)Z_A + w Z'_A$ with $Z' = Z'_A$. Eliminating the crank pin leaves one condition on the point,

$$ZZ' = (1 - w) + wZ'^2.$$

The mirror-image kind of configuration gives $ZZ' = (1 - \bar w) + \bar w Z^2$. A double point is reached twice, so it needs one configuration of each kind and must satisfy both. Eliminating $Z$ leaves a quadratic in $Z'^2$ with two roots. One is $Z'^2 = 1$, which is the crank pin’s own crossing of the slide at $\pm a$ and not a double point. The other is

$$Z'^2 = \frac{\bar w\,(1 - w)^2}{w\,(|w|^2 - 1)},$$

with $Z = (1 - w)/Z' + wZ'$, and the two signs of $Z'$ give the two double points: always a pair, always symmetric through the crank’s pivot.

For a rod point inside the one-rod circle, $|w| = \rho < 1$ with $w = \rho e^{i\psi}$, the root simplifies to a real point of the plane:

$$q = \pm\, i\,\frac{\rho - e^{i\psi}}{\sqrt{1 - \rho^2}},\qquad |q| = \frac{|1 - w|}{\sqrt{1 - |w|^2}}.$$

For (0.3, 0.2) that is $\pm(0.5947, -0.5055)$, the crosses in the figure. The closed form and the isotropic root agree to $10^{-12}$ at every sample point inside the circle. Nothing in the law involves the rod’s length or the slide’s offset from the crank’s pivot. Both dropped out at the step where the rod’s product shrank to nothing at the crank’s scale.

Real inside the one-rod circle, complex outside #

The law answers the question the four-bar’s essay attached to this one: whether the pair leaves as two real points or as a conjugate pair. It depends on the rod point. Inside the one-rod circle the square root is of a positive number and the pair is two real points, which are isolated: they are thousands of units from a curve a few units across, far beyond any configuration’s reach. Outside it the pair is complex.

The ledger shows something the law implies without stating. On the site’s own slider-crank, with a crank of 1, the three rod points inside the one-rod circle all have complex pairs and the two outside have real ones. At a crank of 1,000 it is the other way round. The double points that leave for infinity found the pair complex for rod points between a crank’s length and a rod’s length from the crank pin. In rod units that annulus runs from $a/3$ to 1, and as the crank grows past the rod the annulus turns inside out: it runs from 1 to $a/3$. Measured at a crank of 5, where $a/3$ is 1.67, a rod point at 1.62 has a complex pair and one at 1.83 a real one; at a crank of 2, where $a/3$ is 0.67, a rod point at 0.81 has a complex pair and one at 0.36 a real one. At the limit its inner edge is the one-rod circle and its outer edge has gone to infinity, so everything outside the one-rod circle is complex and everything inside real. The annulus is not a feature of short cranks. It is the same region seen from the other side.

Checked against the solve, and against what the law ignores #

A law derived from leading terms says the double points are at $a\,q$ plus something smaller than $a$. How much smaller is a measurement, and it was worth making, because the four-bar’s law carried a remainder of order one.

For three rod points inside the one-rod circle and two outside, and cranks from 10 to 1,000, the gap between the solved double points and the law, in crank lengths, falls with slope −2.000 on log axes. That is one power better than the four-bar’s law: each double point sits at the crank times the law plus a remainder that shrinks like one over the crank, not one that stays of order one. At a crank of 1,000 the rod point (0.3, 0.2) is on the law to $5 \times 10^{-7}$ crank lengths.

The census varies what the law claims to ignore. A hundred and twenty machines each get a crank of 1,000, a random rod from 2 to 5, a random offset of the slide from −1.5 to 1.5 and a random rod point away from the two special circles below.

Inside the one-rod circle the pair is real every time and within $3 \times 10^{-3}$ crank lengths of the law; outside it the pair is complex every time and within $7 \times 10^{-3}$. The median gap is $9 \times 10^{-4}$. An offset breaks the machine’s symmetry through the crank’s pivot, and it shows: the remainder is larger than on the symmetric machine and no longer falls as fast. But it stays a remainder. The rod point decides where the pair goes, and nothing else does.

Two circles #

The law has two places where something happens, and they are different circles with different meanings.

The one-rod circle. At $|w| = 1$ the denominator $|w|^2 - 1$ vanishes and the root runs off. The pair is at infinity for every crank, not only in the limit. Asked for double points at three rod points on this circle — at 40°, 90° and 140° — with cranks of 10, 100 and 1,000, the solver finds none that are finite, at every one. The double points that leave for infinity found the pair meeting at infinity on this circle for the site’s short-cranked slider-crank, and it stays there at every crank length. The lead that asked for this measurement suspected this circle might become the one on which the trammel’s own geometry changes. It does not.

The circle with the rod as its diameter. The trammel’s own geometry changes on a different circle. A trammel moves its rod so that every point on the circle through the crank pin and slider pin with the rod as diameter runs along a straight line through the crossing of the two slides. This is the property of the Cardan circle, the rolling circle of the trammel’s motion. In rod units that circle is $\rho = \cos\psi$, and on it the law gives

$$\rho - e^{i\psi} = -i\sin\psi,\qquad q = \pm 1$$

exactly. One of the pair sits at $+1$: one crank length from the pivot along the slide, which is the crossing of the two slides. The other sits at $-1$, on the far side of the pivot.

So on this circle one double point does not leave. It settles on the crossing. Measured at three rod points round the circle, at 60°, 90° and 120°, its distance from the crossing falls as one over the crank, with slopes of −1.000, and it is a real crossing of the curve every time. The curve the rod point draws crosses itself there, and in the limit the curve itself becomes the straight line the trammel draws, traced out and back, with the crossing on it. Its partner, a crank’s length on the far side of the pivot, is two cranks from the crossing and leaves.

What a trammel’s line is, algebraically #

The staying double point says something about the trammel that is easy to miss. A trammel’s rod point on the Cardan circle draws a straight line segment. As an algebraic curve, that is not a line. The ellipse the other rod points draw has degenerated, and a degenerate conic that looks like a segment is a line counted twice: its equation is the square of a line’s.

A line counted twice is singular everywhere along it. Every one of its points is a double point in the algebraic sense, because the curve passes through each of them twice, once going out and once coming back. So the slider-crank’s curve near the Cardan circle, which has two double points, is approaching a curve that is all double points. One of the two stays on it. The other leaves, because the limit has room for only one where the machine actually is.

The four-bar’s essay found the one-coupler circle as the place where all three of its double points left together, like the cube root of the rocker. Here the one-rod circle is the place where both have already gone, at every crank, and a different circle is the place where one never goes. The two limits have their special circles in different places, and for different reasons.

Why the four-bar’s law does not carry over #

It was reasonable to expect the four-bar’s derivation to carry over, with the crank’s circle playing the part the rocker’s did. It does not, and the reason is in the scaling. In the four-bar’s limit the crank stays small while the rocker grows, so at the rocker’s scale the crank shrinks to a point and the coupler’s own triangle is all the law can see. Here the growing link is the crank itself, which carries the rod. At the crank’s scale the crank pin’s circle does not shrink; it is the unit circle, and the law comes from how the rod’s two isotropic directions meet that circle and the slide.

The four-bar’s law, applied here, misses the pair at (0.3, 0.2) by 1.13 crank lengths at a crank of 1,000. The two limits are different kinds of limit. One removes a pivot from the finite plane, the other turns a pivot’s circle into a line through the machine. The paths that leave found algebra going to infinity in limits like both of these, and how it goes is decided by which link grows.

What a long crank does to the curve a designer draws #

A slider-crank with a crank much longer than its rod is not a machine anyone builds on purpose, but a trammel with a slightly curved guide is. A second slide is often made as a pivoted arm, and the slider-crank with a very long crank is exactly a trammel whose second guide has been bent into a large circle. Four kinds of slider-crank sorted slider-cranks by how their links compare; this one sits at the far end of that list, where the crank is so long it can only rock through a small arc and its pin barely leaves a straight line.

For the designer the measurement is reassuring in the same way the four-bar’s was. The curve the rod point draws converges on the trammel’s ellipse, and nothing in the finite plane has to be unlearned. The two double points that make the difference are, over most of the rod’s plane, isolated points or a complex pair: points the machine never reaches, hundreds of crank lengths away when the crank is long. The algebra keeps its count of two exactly, at every finite crank, by putting both where nobody looks.

The exception is the Cardan circle. There a real crossing of the curve stays on the machine, at the slides’ crossing, for every crank length. A designer using a rod point on that circle to get a nearly straight path from a trammel with a curved guide has a curve that crosses itself close to the centre of the stroke — which is the straight-line motion folding back on itself, seen through a guide that is not quite straight.

What the pair does not settle #

The offset machine’s special circles. Everything on the two circles was measured with the slide through the crank’s pivot. With an offset, the symmetry through the pivot is gone, the pair is no longer exactly symmetric, and whether one point still stays on the Cardan circle, or whether the one-rod circle still holds both at infinity, has not been measured.

The next term. On the symmetric machine the remainder falls as one over the crank, but its coefficient is not derived. With an offset the remainder is larger and its law is not measured.

The degree. The slider-crank’s quartic becomes the trammel’s conic, so two degrees of the curve are lost, as well as two double points. How the quartic’s equation, divided by the right power of the crank, comes apart into a conic and what the other factor is has not been computed.

Still open: the four-bar that becomes a trammel in one step #

Both limits so far were taken one link at a time: a rocker into a slide, then a crank into a slide. A four-bar can become a trammel directly, with its crank and its rocker growing together — two pivots receding at right angles to each other. The four-bar’s three double points must then all leave, one more than either step took.

The distinct argument would be the order of that limit. If crank and rocker grow at the same rate, the three points might leave along a law that is neither of the two here, or the real one might leave first on the four-bar’s law and the pair follow on the trammel’s. Growing the rocker a hundred times faster than the crank, or the crank a hundred times faster than the rocker, should recover the two one-step laws in sequence. The measurement would say whether the direct limit is the composition of the two steps or something the steps cannot see.