Three discs decide the real double points

A four-bar whose crank and rocker both grow without bound, their pivots receding at right angles, becomes the elliptic trammel, and its coupler curve loses all three of its finite double points on the way. A four-bar becomes a trammel along a family of laws found where they go. With the crank $a = t$ and the rocker $c = \lambda t$, the three leave in proportion to the machine, at the roots of one quartic in a scaled coordinate, and the ratio λ of the two growth rates indexes a whole family of such laws whose two ends are the one-link-at-a-time limits.

That essay also mapped how many of the three are real. Over the coupler’s plane, measured in rods with the crank pin at the origin and the rocker pin at $(1, 0)$, the region where all three are real was the disc one rod from the crank pin when the rocker grew far faster, the disc one rod from the rocker pin when the crank did, and at equal rates an oval about the rod reaching between 0.50 and 0.59 rods from its midpoint. It located that oval by bisection along rays and nothing more. It ended by asking what curve the boundary is, whether its equation is as short as the law’s, and whether every boundary in the family passes through some fixed points, as the two end circles each pass through the other pin.

The boundary turns out to be one polynomial in the coupler point and the ratio together, and the question about fixed points has a better answer than the one it asked for. Every boundary touches one circle and crosses none.

Two double points become real together #

The count changes only one way. A four-bar’s coupler curve always has an odd number of real double points, so as the coupler point moves the count can only jump between one and three, and a jump means two of the three passing from a complex pair to two real points. A complex pair is a point and its mirror image under conjugation, and it can become two real points only by first meeting itself: at the boundary the two coincide.

So the boundary is where two roots of the law coincide, which is the vanishing of a discriminant. The law is a quartic, but one of its four roots is not a double point at all. It is the second place the two pivots’ circles meet, the reflection of the slides’ crossing in the line joining the pivots, which every configuration with both pins there reaches. Dividing that root out leaves a cubic in the scaled isotropic coordinate $Z$, whose three roots are the three departing double points, and whose discriminant is the quantity that vanishes when two of them meet.

Computed symbolically, with $w = u + iv$ and its conjugate treated as independent letters so that nothing is lost to a choice of real and imaginary parts, the discriminant factors. Four of its factors are $(\lambda + i)^2$, $\lambda^2$, $\bar w^2$ and $(\bar w - 1)^2$, which vanish only at a ratio of nought or at one of the two pins, and say nothing about anywhere else. The fifth is a polynomial $R$ in $u$, $v$ and λ of total degree twelve, and the first thing worth noticing about it is that it is real. Its coefficients are complex in $w$ and $\bar w$ separately, and the imaginary part cancels identically once $w$ is written in $u$ and $v$. That had to happen for the boundary to be a curve: a complex function of a real point vanishes, in general, only at isolated points, two real conditions on two real unknowns. Three points that are always closed under conjugation as a set have a discriminant whose complex part can be collected into factors that never vanish off the pins, and those are the four set aside.

The boundary in the pin distances #

$R$ has only even powers of the ratio, up to the fourth, so it is a quadratic in $\lambda^2$. Its three coefficients are polynomials of degree eight in $u$ and $v$, and they become short when written in the squared distances of the coupler point from the two pins, $s = u^2 + v^2$ from the crank pin and $t = (u - 1)^2 + v^2$ from the rocker pin:

$$R = \lambda^4 A + \lambda^2 B + C,\qquad A = s\,(s - 1)^3,\qquad C = t\,(t - 1)^3,$$

$$4B = 8s^2t^2 + 20s^2t + 20st^2 - 20st - (s + t - 1)^2.$$

Nothing else of the coupler point enters. A point’s two distances from the pins determine it up to its mirror image across the rod, and the departing double points of a point and its mirror are mirror images of each other, so their reality could never have depended on anything more.

The ends of the family can be read straight off it. When the rocker grows a hundred times faster than the crank, $\lambda^4$ is a hundred million and the $\lambda^4$ term rules wherever $A$ is not small, so the boundary closes on $A = 0$. That is the circle $s = 1$, one rod from the crank pin, together with the crank pin itself. The circle appears to the third power, which accounts for six of the octic’s eight degrees; the pin, a circle of radius nought, accounts for the other two. When the crank grows faster the same is true of $C$ and the rocker pin’s circle. The trammel takes both double points found the one-rod circle as the place where a slider-crank’s pair of double points is already at infinity, and here it reappears as the boundary of one end of a family that contains it as a limit.

There is also a symmetry that the form makes obvious. Exchanging the pins exchanges $s$ and $t$, which exchanges $A$ and $C$ and leaves $B$ alone; dividing by $\lambda^4$ then turns λ into 1/λ. So the boundary at a ratio, mirrored across the perpendicular bisector of the rod, is the boundary at the reciprocal ratio — the statement, at the level of the boundary, that growing the crank faster is the same as growing the rocker faster in the machine read from the other end of its rod.

Its sign is the count #

A discriminant says where two roots meet, not which side of the meeting has three real ones. Both ends of the family point the same way: inside the crank pin’s one-rod circle $A$ is negative, because $(s - 1)^3$ is, and the earlier maps put three real there when the rocker grew fast. So the rule to test is that three departing double points are real exactly where $R < 0$.

The test uses the earlier essay’s own grids, which counted the real roots of the quartic numerically at each cell. At ratios of 0.01, 0.3, 1, 3 and 100 the sign of $R$ agrees with that count at every one of the 4,620 cells, 23,100 in all, and the shares of the window with three real come out as before: 23.8% at both ends, 10.6% and 10.2% at 0.3 and 3, 7.7% at equal rates.

Agreement cell by cell says the boundary passes between the same cells; it does not say it passes in the same place. So the boundary was also located along 24 rays from the rod’s midpoint at equal rates, twice: once by bisecting on the law’s count, solving the quartic at every step and asking how many roots are real, and once by bisecting on the sign of $R$, which needs no roots at all. The two disagree by at most $4 \times 10^{-14}$ rods, which is the bisection’s own resolution. The oval of the earlier map runs from exactly half a rod from the midpoint, straight across the rod, to 0.590 rods along it.

The circle every boundary touches #

That 0.500 is the clue to the rest. It is exactly half a rod: the oval comes down to touch the circle that has the rod as its diameter, whose points see the rod at a right angle. The question the earlier essay put was which points every boundary passes through, and the answer the polynomial gives is a circle that every boundary touches.

It follows from reading $R$ the other way round. For a fixed coupler point, $R = 0$ is a quadratic in $\mu = \lambda^2$, and the ratios at which that point changes its count are its positive roots. The quadratic’s own discriminant is again short in the pin distances:

$$B^2 - 4AC = \tfrac{1}{16}\,(s + t - 1)\,(8st + s + t - 1)^3.$$

The first factor vanishes on $s + t = 1$, which by Pythagoras is the set of points from which the rod subtends a right angle — the circle with the rod as its diameter, the circle of Thales. On it the quadratic in μ has a repeated root, so each of its points lies on exactly one boundary of the family, and lies there as a point of tangency. A family of curves meets its own envelope that way. Inside the circle $s + t < 1$ and the discriminant is negative, unless the second factor is too: the quadratic has no real root at all, $R$ never changes sign as the ratio runs from nought to infinity, and the count is three at every ratio, since it is three at the two ends.

Where each boundary touches is the repeated root, $\lambda^2 = -B/2A$, evaluated on the circle. It runs through every positive value. Beside the rocker pin it is enormous, and that is the crank pin’s one-rod circle touching the Thales circle at the rocker pin, where both have the same tangent. At the top of the circle, $(0.5, 0.5)$, it is exactly 1, which is why the oval at equal rates comes down there. Beside the crank pin it is tiny. Every ratio has its pair of touching points, mirror images across the rod: at 0.3 they are at $(0.09, \pm 0.28)$, at 10 at $(0.99, \pm 0.10)$. As the ratio runs from nought to infinity the touching pair travels the whole circle from the crank pin round to the rocker pin, and the boundary rolls round the circle with it, never entering. The hero figure shows the same thing as a slider: the shaded region changes shape from one disc to an oval to the other disc, and the dashed circle is inside it at every stop.

The second factor, $8st + s + t = 1$, is a figure of eight from pin to pin through the rod’s midpoint, lying inside the Thales circle within thirty degrees of the rod. Inside its lobes the discriminant is positive again and the quadratic has real roots, but both are negative values of $\lambda^2$, which no machine has. It is an envelope of the family’s complex members only, and a coupler point inside it behaves like any other point inside the circle: three real, always.

Three discs and four histories #

A coupler point outside the Thales circle has two real roots in $\lambda^2$, and the remaining question is how many are positive. Their product is $C/A$ and their sum is $-B/A$. The signs of $A$ and $C$ say which one-rod discs the point is in, and the answer turns out to need nothing more than that.

Counting positive roots across a grid of 28,500 points over the window, every one obeys

$$\text{switches} = [s < 1] + [t < 1] - 2\,[s + t < 1],$$

with brackets meaning one where the condition holds. A point outside both one-rod discs never changes: one real departing double point at every ratio, 15,852 of the cells. A point in one disc only changes once, from one real at the end where that disc is not the boundary to three at the end where it is, 9,576 cells. A point in the Thales disc never changes either, three at every ratio, 1,976 cells. And a point in the lens where the two one-rod discs overlap, but outside the Thales circle, changes twice, 1,096 cells: three real when either link grows much faster than the other and only one in a window of ratios between.

So the whole history of a coupler point along the family, at every ratio from nought to infinity, is settled by where it sits relative to three circles, two of them drawn about the pins with the rod as radius and one drawn on the rod as diameter. The ends of the family were already known to be decided by the two one-rod circles. The polynomial adds the third circle and the rule for combining them.

A window of one #

The lens is where the ratio matters most, because there neither end describes the middle.

Along a line 0.62 rods off the rod, points from half a rod behind the crank pin to 0.20 rods past it lie in the crank pin’s disc only, and each changes once, from one real to three as the rocker gains on the crank, at a ratio that falls from 27 to about 4 as the point nears the lens. Points past 0.78 lie in the rocker pin’s disc only and change the other way. Between, from 0.22 to 0.78, both changes happen to the same point, and the two lines bound a window in which only one departing double point is real. Directly above the rod’s midpoint the window runs from 0.191 to 5.22. Its two ends multiply to exactly 1, because on the perpendicular bisector $s = t$ and $A = C$; anywhere in the lens the two ends multiply to $\sqrt{C/A}$, since they are the square roots of two numbers whose product is $C/A$.

The width of the window is the practical content. The earlier essay found that at the coupler point $(-0.5, 0.8)$ the direct limit agreed with the trammel’s own picture of three real points only once the rocker grew about a thousand times faster. That point is in the crank pin’s disc only, and the polynomial puts its single change at a ratio of 199.3: below that the one-step picture is wrong about how many of its double points are real, and above it the picture is right. A point in the lens has two such thresholds. Above the rod’s midpoint they are a factor of 3.2 apart just outside the Thales circle, 0.52 rods off the rod, 27 apart at 0.62 and 3,500 apart at 0.80, and the window opens without limit towards the top of the lens, where the two one-rod circles cross. A machine whose two arms differ by less than that factor sits inside the window and has one real departing double point where either one-step limit, taken on its own, would have promised three.

What the octic is #

At any fixed ratio the boundary is a curve of degree eight in the coupler’s plane, and it has two properties that place it beside the curves whose degrees a line through them has counted.

Its highest-degree terms are $(\lambda^2 + 1)^2 (u^2 + v^2)^4$, so far from the pins $R$ divided by the eighth power of the distance tends to $(\lambda^2 + 1)^2$ in every direction, which was confirmed at three ratios to three parts in a hundred thousand at a distance of 100,000 rods. A curve whose top form is a power of $u^2 + v^2$ meets the line at infinity only at the two circular points, and computed symbolically at ratios of 1 and 3 the boundary passes through each circular point four times. Four is half of eight, the most any curve of degree eight can manage, and it is the same maximal circularity that nine times through each circular point found for every curve drawn by a body of a machine built from pins alone, and that the mesh inside keeps the half found for geared five-bars with internal meshes. Here it belongs not to a curve the machine draws but to the curve that sorts the machine’s double points, and the reason is the same: everything was written in distances, and distances are built from $u^2 + v^2$.

The second property answers the question about fixed points directly. Both pins lie on every boundary, since $A$, $B$ and $C$ all vanish there. But they lie on it as isolated points. Near the crank pin the lowest terms of $R$ are $-\lambda^2\left[(\lambda^2 + 1)u^2 + \lambda^2 v^2\right]$, a quadratic form that is negative in every direction, so $R$ is zero at the pin and negative on every small circle about it: the pin is an isolated real point of the boundary, sitting inside the three-real region, with no branch of the boundary passing through it. Checked at three ratios on a circle of a thousandth of a rod about each pin, $R$ is negative all the way round. The points where the two one-rod circles cross, $(0.5, \pm 0.866)$, are on no boundary at any finite ratio: there $R = 27\lambda^2/4$, positive, so they have one real departing double point. They belong to the family’s two ends and to nothing in between. The earlier essay suspected that the boundaries might form a pencil through fixed points; they form instead a quadratic family whose real envelope is one circle, and whose only common points are two isolated ones.

A point the machine never reaches found isolated points of a coupler curve that no configuration visits. The pins are the same kind of point, one level up, on the curve that bounds where such points are real.

What a designer reads from it #

Every trammel built with two long pivoted arms in place of its slides is one of these four-bars, and its departing double points are the places where its coupler curve, seen at the scale of the arms, crosses itself or has a point no configuration reaches. A designer choosing a coupler point does not need the law to know how many of them are real. The three discs say it at a glance: a point within half a rod of the rod’s midpoint, seeing the rod at more than a right angle, has three at every proportion of the arms; a point outside both one-rod discs has one; and a point in the lens has three only if one arm is much longer than the other, with the threshold given by one square root.

The circle with the rod as diameter is also the circle that a trammel moves along straight lines through the slides’ crossing, the circle on which the trammel’s own double points settled at the crossing rather than leaving. That the same circle is the envelope of the reality boundaries is not obviously the same fact, and nothing here proves that it is. It is the circle on which the rod is seen at a right angle, which is the geometric content of both.

What the boundary does not settle #

Which real points are crossings. A real double point of a coupler curve is either a crossing, reached by two configurations of the machine, or an isolated point, reached by none. The boundary separates one real from three real; it says nothing about how the three divide between the two kinds. The count that does not move showed how much of a mechanism’s character lives in exactly that kind of division.

Oblique slides and unequal powers. The pivots recede at right angles and both links grow linearly. Neither the Thales circle’s role nor the three-disc rule has been tested for pivots receding at another angle, where the Thales circle would presumably be replaced by the circle on which the rod subtends that angle, or for a rocker growing as a power of the crank.

Why the figure of eight. The second factor of the discriminant is a real curve with no real role, an envelope of members of the family at negative $\lambda^2$. Its shape, from pin to pin within thirty degrees of the rod, is measured and not explained.

Still open: which of the three real points the machine reaches #

Inside the Thales circle all three departing double points are real at every ratio, and the earlier essays have shown that the four-bar’s real double points can be crossings or isolated points. At the family’s ends the answer is known: the far point that leaves along the four-bar’s law is always isolated, and the trammel’s pair is real and isolated inside the one-rod disc. In the middle of the family nothing has said.

The distinct argument would be the second boundary, the one inside the three-real region where a real double point changes from isolated to a crossing. At a crossing two real configurations meet; at an isolated point two complex conjugate ones do. So the boundary is where those two configurations themselves pass from a complex pair to two real ones, which is again a coincidence and again a discriminant, this time of the configurations rather than of the points. The measurement would compute it, test it against curves traced at large crank lengths, and ask whether it too is short in the pin distances and whether the Thales circle is its envelope as well.