The double point a slide cannot keep

A point the machine never reaches found every four-bar’s coupler curve carrying three finite double points, all of them on the circle through the three fixed pivots of Roberts’s cognates. The three are roots of a real polynomial, so the non-real ones pair off and an odd number is real: one or three, never none and never two. The double points that leave for infinity found the slider-crank’s rod curve carrying two, and a whole annulus of coupler points for which both are complex and the curve crosses itself nowhere.

A slider-crank is a four-bar whose rocker is infinitely long. Its slider pin runs along a straight line, and a very long rocker swings its pin along a circle so large that, near the pin, nobody could tell it from the line. So there is a family of four-bars that becomes the slider-crank in the limit, and along that family three double points become two. One of them has to go somewhere.

The essay that found the slider-crank’s pair left the question open with a test attached: the parity. It is the test that decides the whole matter, and it decides it before anything is computed. What the computation adds is where the lost point goes, how fast, and one circle of coupler points on which the answer is different in kind.

A rocker that becomes a slide #

The slider-crank is the same machine throughout: a crank of 1 pivoted at the origin, a rod of 3, and a slide along the x-axis through the crank’s pivot. The four-bar replaces the slide with a rocker of length c, pivoted at (3, −c). Its circle touches the x-axis at (3, 0), halfway along the slider pin’s travel, and as c grows the circle flattens onto the slide there. The coupler point is written as $w = u + iv$ in the coupler’s own units: u along the coupler from the crank pin, v across it, both measured in coupler lengths, so the far pin is at $w = 1$.

The left panel is the four-bar with a rocker of 5. The coupler point (0.4, 0.5) is one of the annulus points, between a crank’s length and a rod’s length from the crank pin, where the slider-crank’s two double points are a complex pair. The four-bar’s curve is two ovals, like the slider-crank’s, and it too crosses itself nowhere. But the four-bar must have a real double point, and it has one: an isolated point far below the machine, on the dashed circle through the pivots, at (0.92, −5.32). Its other two double points are a complex pair near the crank.

The right panel divides everything by the rocker’s length, for rockers from 5 to 100,000. In those units the machine shrinks to the origin, the rocker’s pivot sits at (0, −1), and the circle through the three pivots settles onto a fixed circle through both. The isolated point does not shrink with the machine. It settles onto one point of the fixed circle, marked with a cross. So the lost double point is lost linearly: it stays a fixed fraction of the rocker’s length away, and the rest of the essay is about which fraction and in which direction.

Why the one that leaves is the real one #

The parity argument is short enough to give whole. The four-bar’s three double points are either one real and a complex-conjugate pair, or three real. The slider-crank’s two are either a conjugate pair or two real. Along the family, as c grows, the four-bar’s points move continuously, and whatever arrives in the finite plane in the limit must be the slider-crank’s pair.

A complex double point cannot leave alone. Its conjugate is also a double point of the same real curve, and it is always exactly as far away, so if one runs off to infinity the other does too. But only one point leaves. So the one that leaves is real.

The second half of the argument is a matter of distance. A real configuration of the four-bar puts the coupler point at most $1 + 3|w|$ from the crank’s pivot, since the crank pin is 1 away and the coupler point is $3|w|$ from the crank pin. For (0.4, 0.5) that is 2.92. A real double point further out than that cannot be a crossing, because a crossing is a place the curve actually passes through twice. It has to be an isolated point, reached by two complex-conjugate configurations whose coupler points happen to be real and equal.

The ledger checks it at four coupler points: one inside a crank’s length of the crank pin, where the slider-crank has two crossings; the annulus point; and two beyond one rod, where the slider-crank has two real double points of its own. The four-bar with a rocker of 1,000 has one more real double point than the slider-crank in every row: three against two, one against none. The extra point is the one that leaves, and it is isolated in all four.

That is a satisfying end to the parity question, and it changes what the four-bar’s parity means. Never three circuits and the essay on the isolated point read the odd count as a guarantee that a coupler curve always carries a visible singularity somewhere. The guarantee is real, but it says nothing about where. For a four-bar with a long rocker, the real double point that the odd count guarantees is an isolated point hundreds of units from a curve a few units across; the others, if there are any, pair off as the slider-crank’s do. The curve a designer draws is the slider-crank’s.

Where it goes: two complex configurations, one of each kind #

Divided by the rocker, the leaving point settles onto a fixed point. That point can be found without tracking anything, by asking what a configuration of the four-bar looks like when its coupler point is at a distance comparable to the rocker.

The algebra is cleanest in isotropic coordinates, the device that three linkages, one equation used to read the circular points off the sextic. Write each point of the plane as the pair $z = x + iy$ and $\zeta = x - iy$. For a real point ζ is the conjugate of z. For a complex point the two are independent, and that independence is exactly the freedom an isolated point’s configurations use. It is the same freedom that gave two circles four answers rather than two. A circle of radius r about a centre m becomes $(z - m)(\zeta - \bar m) = r^2$.

The crank pin A satisfies $z_A\zeta_A = 1$. The rod from the crank pin to the far pin is $r = z_B - z_A$ and $\rho = \zeta_B - \zeta_A$, with $r\rho = 9$. The coupler point is $z = z_A + w\,r$ and $\zeta = \zeta_A + \bar w\,\rho$. Now suppose z and ζ are both of the order of the rocker length c. The crank’s product is fixed at 1, so one of $z_A$ and $\zeta_A$ has to be small, of order 1/c, and the other large.

If $z_A$ is the small one, then $z \approx w\,r$: the rod’s z-component is large and its ζ-component $\rho = 9/r$ is small. The far pin is then at $z_B \approx z/w$ and $\zeta_B \approx \zeta$. If $\zeta_A$ is the small one, everything is mirrored: $z_B \approx z$ and $\zeta_B \approx \zeta/\bar w$. These are the only two ways a configuration can put its coupler point far away, and for a real point they are each other’s complex conjugates.

The rocker’s circle is $(z_B - \delta)(\zeta_B - \bar\delta) = c^2$ with its centre $\delta$ at $3 - ic$. Scaling every coordinate by c, writing $Z = z/c$ and $Z' = \zeta/c$, and keeping the leading terms, the circle becomes $Z_B Z'_B + i(Z'_B - Z_B) = 0$: at this scale it is a circle of radius one through the crank’s pivot, and the crank has shrunk to that pivot. Substituting the two kinds of configuration gives

$$ZZ' + i(wZ' - Z) = 0 \qquad\text{and}\qquad ZZ' + i(Z' - \bar w Z) = 0.$$

To leading order each equation gives one configuration of its kind for each point. A double point is a point reached twice, so it needs one configuration of each kind, and it has to satisfy both. Subtracting them gives $Z'(w - 1) = -Z(\bar w - 1)$, and putting that back into either leaves, apart from the origin,

$$q = d\,\frac{|w|^2 - 1}{\bar w - 1},$$

with $d = -i$ the direction from the slide to the rocker’s pivot. Its conjugate comes out as $Z'$, which is the statement that the point is real. The origin, the other common solution, is where the finite double points sit at this scale, because the whole machine has shrunk into it.

Nothing in that law mentions the crank’s length, the rod’s length, where along the slide the rocker touches it, or whether the slide passes through the crank’s pivot. All four dropped out at the step where the crank shrank to a point. The law is a property of the coupler triangle and the direction of the rocker, and nothing else.

Reading the law: a direction and a power #

Multiplying top and bottom by $w - 1$ turns the law into a direction and a size:

$$q = d\,(w - 1)\,\frac{|w|^2 - 1}{|w - 1|^2}.$$

The factor $w - 1$ is the vector from the far pin to the coupler point, in the coupler’s own frame. Multiplying by d turns that frame so the coupler lies along d, pointing from the slide towards the rocker’s pivot. So the point leaves along the line from the far pin to the coupler point, carried round with the coupler laid along the rocker.

The size is $\left||w|^2 - 1\right| / |w - 1|$ rocker lengths, and its numerator has a name. $|w|^2 - 1$ is the power of the coupler point with respect to the circle one coupler length about the crank pin: positive outside that circle, negative inside, zero on it. When the power changes sign, the direction reverses.

The map shows the law over the coupler’s whole plane. Outside the dashed circle, one coupler length from the crank pin, every arrow points along the line from the far pin to its own position, turned a quarter-turn clockwise. The quarter-turn is d = −i, the rocker hanging below the slide. Inside the circle every arrow is reversed. The arrows shrink to nothing on the circle, where the power vanishes, and grow without bound near the far pin, where the denominator does.

The far pin’s own behaviour is the easiest part of the map to check by hand. The far pin’s curve is the rocker’s circle, which has no double points at all, so a coupler point close to the pin has a curve that is nearly a circle. Its three double points have nowhere finite to be, and the law sends the leaving one off faster the closer the point is to the pin.

The same map answers the question the slider-crank’s essay attached to this one: whether the lost point leaves along the slide. The slide is horizontal in these pictures, and a horizontal arrow needs $d(w - 1)$ to be real. That happens only when $w - 1$ is imaginary, meaning $u = 1$: coupler points straight across the coupler from its far pin. Everywhere else the point leaves in some other direction. Where a slide puts the rest of the degree showed that no rod quartic passes through the slide’s point at infinity, and there was a natural guess that the four-bar might hand its lost point to that direction. It does not, and the law says why: the direction is set by the coupler triangle, and the slide only enters as the direction the rocker hangs from.

The law, against the solve #

A derivation that keeps only leading terms says the leaving point is at $cq$ plus something smaller than c. It does not say how much smaller, and it is exactly the kind of statement worth measuring, since the whole derivation rests on which terms were dropped.

The solid lines are the law’s miss after dividing by the rocker length, and every one of them has slope −1.000 over its last four decades. So the leaving point is at $cq$ plus a fixed offset. Dividing by c turns the offset into the one-over-c miss. The dashed lines are the two double points that stay, measured against the slider-crank’s pair as found by two copies of the machine, and they close in at the same order: slopes between −1.008 and −0.999. At a rocker of 100,000 the two finite points are within 3 × 10⁻⁵ of the slider-crank’s at the ledger’s first point, and within 6 × 10⁻⁴ at the last, whose double points are nine units out.

Four coupler points, one touch point and one slide height could hide a coincidence, and the law claims independence from all of them. So the census varies everything the law says it ignores.

The census draws coupler points over a box four coupler lengths wide, keeping clear of the coupler’s line, where the circle through the pivots degenerates into a line, and of the far pin and the one-coupler circle, where the law’s size is infinite or nought. Each machine’s rocker touches the slide at 0.5, 2, 3 or 5 along it, and the slide runs through the crank’s pivot, 0.3 above it or 0.5 below. At a rocker of 10,000 the worst relative miss is 7.8 × 10⁻³; at 100,000 it is 8.0 × 10⁻⁴; the median machine’s miss falls by a factor of exactly 10.0 for the tenfold rocker, which is what a fixed offset does. The worst rows are the slowest leavers at the left of the plot, where the law’s point is close to the origin and the offset is a larger share of it.

In all 240 machines the leaving point is real, isolated and outside the machine’s reach, and the four-bar has one more real double point than the slider-crank it approaches: 186 with three against the slider-crank’s two, and 54 with one against none. The 54 are annulus points, where the obligatory real double point is the four-bar’s only one.

On the one-coupler circle, all three go #

The law is zero on the circle one coupler length from the crank pin. There the leading term says the leaving point stays at the origin of the scaled picture, so it does not leave linearly, and the slider-crank’s own essay found something special on exactly this circle. For these coupler points both of the slider-crank’s double points are at infinity, together, as a tacnode at the real point at infinity of slope $\tan(\psi/2)$, with ψ the point’s angle from the rod.

So in the limit, a four-bar on this circle has all three of its double points at infinity. None of them can converge on a finite point, and the law gives none of them a linear rate.

They leave together, and they leave like the cube root of the rocker. At four angles from the coupler, from 34° to 149°, the three distances grow with slopes between 0.333 and 0.334 over the last two decades, and at a rocker of a million they are equal to within 0.02%. A rocker a thousand times longer puts all three ten times further out.

One of the three is real, and it points along a direction the slider-crank already knew: within 0.01° of the angle ψ/2 from the slide, the real direction at infinity where the slider-crank’s tacnode sits. The two complex ones are the tacnode’s two branches arriving from the four-bar’s side. So on this circle the four-bar sends all three of its double points to the one real point at infinity that the slider-crank’s curve passes through twice.

It fits the linear law from both sides. Just off the circle the law’s direction $d(w - 1)$, with w near $e^{i\psi}$, is $2\sin(\psi/2)\,e^{i\psi/2}$ times a positive number: the direction ψ/2 again, reversed inside the circle and not reversed outside. The linear departure’s direction is continuous across the circle up to its sign, and its size falls to nought there. The cube root is what takes over when the linear term has nothing left to say. The crossover comes when the linear term, of order c times the distance to the circle, falls to the cube root’s $c^{1/3}$, which is at a rocker near the distance to the circle to the power −3/2.

What a long rocker does to a designer’s curve #

In a machine, a four-bar with a long rocker is the usual way to replace a slide that is hard to build or lubricate, and four kinds of slider-crank is the family it approximates. Everything the designer looks at is in the finite plane: the curve, its crossings and cusps, the places where a coupler point hesitates.

The measurements here say that part converges quietly. The two double points that stay approach the slider-crank’s like one over the rocker, and a double point that is a crossing on the slider-crank is a crossing on the four-bar close to it. The only thing a long rocker adds is a single isolated point far out, which no configuration reaches and no tolerance in the machine can make reachable. The parity of the four-bar’s double points is a fact about the whole complex curve, and a long rocker spends it where nobody will ever build.

The same reasoning explains why the paths that leave found so much of a machine’s algebra at infinity in limits like this one. The algebra keeps its counts exactly, and a limit that changes the real machine a little can only keep them by moving solutions a long way. Here one double point moves linearly, the circle through the pivots grows with it, and the count of three survives every finite rocker.

What the derivation leaves unexplained #

The cube root is measured, not derived. On the one-coupler circle the leading term of the derivation vanishes, and the next terms decide the rate. The slope is 0.333 to 0.334 at four angles and the three distances agree to 0.02%, which is strong evidence for a cube root and for three points leaving as the three cube roots of one quantity. It is not a proof, and nothing here names the quantity.

The offset machine’s circle is not examined. With the slide offset, the slider-crank’s two double points no longer reach infinity together: one of them turns real just inside one rod and passes through infinity alone. The linear law holds for offset machines — a third of the census had an offset — but what happens on the offset machine’s special circle, whether two leave together or one, and at what rate, has not been measured.

Where the sextic’s lost degree goes. The four-bar’s sextic, the equation a four-bar satisfies, passes three times through each circular point and the slider-crank’s quartic once, so two passes through each circular point and two degrees of the curve are lost in the limit as well as one double point. The measurements here follow the double points only. How the curve’s equation, divided by the right power of c, comes apart in the limit has not been computed.

Still open: the two double points a slider-crank loses on the way to a trammel #

A slider-crank whose crank grows without bound, with the crank’s pivot receding at right angles to its slide, becomes a machine with two slides: the elliptic trammel, whose rod points draw ellipses. An ellipse is a conic and has no double points at all. So the same limit, taken one step further, has to lose both of the slider-crank’s double points, and the parity gives no guide, because two is even.

The distinct argument there would follow the slider-crank’s two double points as its crank lengthens into a second slide. It would ask whether both leave, which is forced, and whether they leave as a conjugate pair along one law or as two real points along two. It would also ask whether the one-rod circle, where the pair is already at infinity, becomes the circle on which the trammel’s own geometry changes. The derivation above should carry over with the crank’s circle playing the part the rocker’s did here. The test is the same pair of checks: a law from the leading terms, and a census that varies what the law claims to ignore.