The double points that leave for infinity

Where a slide puts the rest of the degree found all four of the places where a slider-crank’s rod curve meets the line at infinity. The curve passes once through each circular point. The other two meetings are a pair of directions, $m_\pm = [2v \pm i(1 - u^2 - v^2)] / [(1+u)^2 + v^2]$ for a coupler point u along the rod and v across it. The pair is complex everywhere except on the circle one rod from the crank pin. On that circle it becomes one real direction that the curve passes through twice.

Passing twice through a point is what a double point does, and that is where the essay stopped. The curve is a quartic. Its configurations are a crank angle and a choice of assembly, and the two assemblies merge at the four complex angles where the rod stands square to the slide. That makes the configuration curve a double cover of a circle branched at four points, which has genus one — the same count never three circuits used for the four-bar. A plane quartic has room for three double points, so a curve of genus one drawn as a quartic owes exactly two. Off the circle those two are somewhere in the finite plane. On the circle something at infinity has swallowed at least one of them.

The question left open was where they are, and whether the thing at infinity takes both.

Four curves, four answers to “where does it cross itself” #

The machine is the one the directions were measured on: a crank of 1 on a pivot at the origin, a rod of 3, and the slider pin running along the x-axis through the pivot with no offset.

The four curves are the whole of this essay in one picture. It helps to say what each one shows before any algebra.

The first point is 0.19 rods from the crank pin, close to it, and its curve is two overlapping loops that cross each other twice. Those two crossings are the quartic’s two double points. Each is a place where the rod’s point arrives in one assembly and, at a different crank angle, in the other.

The second point is 0.64 rods out, and its curve is two separate ovals. They do not touch, so the curve has no crossing at all. This is where the slider-crank parts company with the four-bar. A point the machine never reaches showed that every four-bar coupler sextic has three finite double points with an odd number of them real, so a four-bar’s curve always has at least one real double point. The quartic’s two can both be complex, and here they are.

The third point is 1.20 rods out. Its curve is two thin crescents, and the dots are well away from both of them. They are real points in the plane that satisfy the curve’s equation twice over, and no configuration of the machine ever reaches them. In the four-bar’s vocabulary these are isolated points, or acnodes: two complex-conjugate configurations whose rod point happens to be real.

The fourth point is 1.49 rods out, and it crosses itself again. Its curve is two figure-eights, and the ringed points are real crossings.

So there are two real crossings, then none, then two isolated points, then two crossings again. The rest of this essay is about what decides which, and it turns out to be two circles about the crank pin and one curve that belongs to the rod’s motion rather than to its algebra.

Two ways to find a double point #

With a count this small it is easy to be wrong in a way that looks like a count, so the double points are found twice, by methods that share nothing.

Two copies of the machine. A double point is a place the rod’s point reaches from two different configurations. Write the coupler point as a complex number, $P = (1 - w)A + wB$, where A is the crank pin, B the slider pin and $w = u + iv$. Setting two configurations’ points equal gives $(1 - w)(A_1 - A_2) = w(B_2 - B_1)$. Both slider pins are on the slide, so $B_2 - B_1$ is a real number t times the slide’s direction, and the first crank pin is the second moved by $k\,t$ with $k = w/(1-w)$. That leaves four unknowns: the second crank pin’s two coordinates, the first slider pin’s position, and t. The four conditions are the two cranks’ lengths and the two rods’ lengths. Every solution with t = 0 is a configuration equal to itself, which is not a double point. Dividing that factor out of the two differences of squares leaves a system of degrees 2, 1, 1 and 2, so there are four paths to track by homotopy continuation. Each double point appears twice, once for each order in which its two configurations are listed.

The equation. The second route never touches the machine. As in the equation a four-bar satisfies, the quartic’s fifteen coefficients are the null vector of a matrix built from the drawn curve, with a gap of at least 10¹³ to the next singular value. A double point of an equation F = 0 is a point where F and both of its partial derivatives vanish. The two partial derivatives are cubics, and two cubics have nine common zeros. Most of them are saddles of the surface z = F(x, y), where F is not zero, so they are not on the curve at all. Tracking all nine and keeping the zeros where F vanishes leaves the double points.

Every row agrees. The two copies arrive four times and give two points. The equation gives the same two points to at worst 6 × 10⁻¹⁴, together with seven saddles off the curve. The routes also agree on the type of each point, although they decide it in different ways. The copies route calls a real point a crossing when both of its configurations are real. The equation route calls it a crossing when the Hessian of F is negative there, meaning the curve has two real tangents.

The plain machine has a symmetry that shows up in the answers. Turning the crank half a turn and reflecting the slider pin through the pivot reflects every point of the curve through the origin. So the double points come in pairs ±p, and a complex pair on this machine is always $\pm i\,p$ for some real p. In the ledger that is the ±i entry: points with a real part of exactly nought.

Walking a point out along a ray #

The ledger gives snapshots. The figure below joins them up: the coupler point moves out from the crank pin along a ray at 52° to the rod, and at each stop the curve it draws is shown with any real double points.

At 0.2 and 0.3 rods the curve crosses itself twice. The two crossings sit on opposite sides of the crank’s pivot and close in on it as the point moves out. Just past the inner circle they are gone. They have not left the page. They met at the pivot and turned into a complex pair.

The inner circle is one crank length from the crank pin, which is a third of a rod on this machine. The reason is short. A point of the rod at distance ρ from the crank pin can sit on the pivot only if ρ equals the crank’s length, because the crank pin is exactly that far from the pivot. When it does, it does so in two configurations that are each other’s half-turn about the pivot, so the curve passes through the pivot twice. On that circle the two crossings have become one point. Inside it they are the two real points either side of the pivot. Outside it they are the complex pair $\pm i\,p$, the same approach continued along the imaginary axis.

From 0.45 rods to 0.9 the curve has no real double point, and the complex pair grows: 0.81 from the pivot at 0.45 rods, 2.04 at 0.7, 4.78 at 0.9. This is the annulus where the curve is two separate ovals. It is also where the open question about the double points applies. The pair has to be somewhere and is not in the real plane, and as the point nears one rod it is heading off to infinity.

Just past one rod the pair is real again, as two isolated points almost seven units from the pivot at 1.08 rods. They come in closer as the point keeps moving: 5.08 at 1.2 rods. At 1.5 rods they are crossings. What changed between 1.2 and 1.5 is the subject of the section after next.

How fast the pair runs off, and in which direction #

The essay that located the directions predicted one real direction at infinity for points on the circle: tan(ψ/2), where ψ is the coupler point’s angle from the rod. If the double points really are what goes to infinity, they should head off in that direction and at a rate that says what kind of singular point they build there.

Both approaches are straight lines on this plot, with slope −0.505 from inside and −0.495 from outside, fitted over the last decade before the circle. The pair’s distance from the pivot grows as one over the square root of the gap to the circle. The direction settles as well. A thousandth of a rod from the circle, the pair points within 0.06° of half the ray’s angle. Inside the circle that is the direction of the imaginary part, and outside it is the direction of the real point itself.

The two sides join up once the complex pair is written in projective coordinates. A point $i\,p$ is the same projective point as $(p : -i)$, and as p grows without bound that tends to $(\hat p : 0)$, the real point at infinity in the direction of p. The same happens to $-i\,p$. So as the coupler point reaches the circle, both members of the complex pair arrive at the same real point at infinity. As it crosses, they leave that point as two real points on opposite sides of the pivot. They pass through each other at infinity, the way two real roots of a quadratic pass through each other when its discriminant changes sign. The difference is that here the meeting place is at infinity rather than on the page.

The exponent −½ matters for what comes next. Two double points that merge like $\pm\sqrt{s}$ as a parameter s goes to nought are what a tacnode does when it is perturbed. A tacnode is a point where two branches of a curve touch with a common tangent, and it counts as two double points. So the square root is the first sign that on the circle the two double points are not simply absorbed into the double passage through infinity. Together they are that passage.

What sits at infinity on the circle #

The rate suggests a tacnode, and the fitted equation can confirm it. Write the quartic in homogeneous coordinates about its real point at infinity. Take Y as the coordinate across the line of slope tan(ψ/2) through the pivot and Z as the coordinate off the line at infinity. For a tacnode tangent to that line, every term of weighted degree below four has to vanish, counting Y with weight two and Z with weight one. What remains, $\alpha Y^2 + \beta YZ^2 + \gamma Z^4$, factors into two branches $Y = \lambda Z^2$. They are real or complex depending on whether $\beta^2 / 4\alpha\gamma$ is at least one or below it.

The low-order terms vanish to between 10⁻¹⁶ and 10⁻¹⁵ of the leading ones at all five angles, so the point is a tacnode. The tangent is the line of slope tan(ψ/2) through the crank’s pivot, to rounding. The ratio that decides the branches comes out as a closed form that nothing in the derivation asked for: it is $\cos^2(\psi/2)$ at every angle, to 3 × 10⁻¹⁴. The ratio is below one everywhere on the circle except at ψ = 0, the slider pin, whose curve is the slide itself. So the tacnode’s two branches are complex conjugates. No real branch of the curve comes near the point.

That settles both questions left open about the circle. The count of finite double points at exactly one rod is nought, because all four paths of the two-copy system leave for infinity at every angle tried. The point at infinity is not a node with complex branches and not a point where the curve touches the line at infinity. It is a tacnode that holds both of the quartic’s double points, and it is tangent to a real line through the crank’s pivot, which the real curve never reaches.

The curve that decides crossing or isolated #

Past one rod the pair is real, and the walk showed it arriving as isolated points and later becoming crossings. For a real double point to change type, its two tangents must first coincide. At that instant it is a cusp, a point where the rod’s point comes to rest for an instant before reversing.

A point of a moving body is at rest exactly when it is the body’s instant centre, as every point has a centre put it. So the coupler points whose curves have a cusp are the points the rod ever turns about, written in the rod’s own frame. That set is the rod’s moving centrode. On a slider-crank the rod’s instant centre is where the line of the crank meets the perpendicular to the slide through the slider pin. Following it round a turn of the crank and writing it in the rod’s frame draws the centrode directly. The frame seen from the coupler is the same exchange of viewpoints worked on a four-bar.

The map puts the three pieces together. Inside the crank-length circle both double points are crossings. Between the two circles they are a complex pair. Outside the one-rod circle they are real, and the centrode, drawn solid, bounds the region where they are crossings: an hourglass standing on the slider pin, with its waist tangent to the one-rod circle there. Everywhere else outside the circle they are isolated points.

The boundary is not fitted to the colours. The centrode is computed from the instant centre alone, and the map from the two-copy system alone. As a check, points were taken just either side of the centrode along its normal, at 44 places along it. The type changes across it at all 44.

The centrode touches the one-rod circle only at the slider pin, and it can never cross inside. The instant centre lies on the crank’s line, directly above or below the slider pin, so its distance from the crank pin is the horizontal run of the rod divided by the cosine of the crank angle: $\sqrt{b^2 - a^2\sin^2\theta}\,/\,|\cos\theta|$. That is never less than b, because the rod is longer than the crank. So every cusp a point of this rod can draw belongs to a point at least one rod from the crank pin. That is why the annulus, where the pair is complex, has no cusps to cross: a complex pair can become real only by colliding with itself or by passing through infinity, and the pair here does the second.

An offset changes the circles and splits the pair #

Everything so far has used the plain machine, with the slide passing through the crank’s pivot. That machine has the half-turn symmetry that made every complex pair $\pm i\,p$. An offset slide breaks the symmetry. Four kinds of slider-crank treats the offset as a design choice, and here it tests how much of the picture above was the symmetry.

The two boundary circles survive. For each of five machines the crossings stop being real at the same radius on every ray, and the complex pair turns real again at a second radius that is also the same on every ray. So both boundaries are still circles about the crank pin. What changes is their size. Measured in lengths rather than rods, the two radii R are the roots of

$$(R^2 - a^2)(R^2 - b^2) + e^2 R^2 = 0,$$

where a is the crank, b the rod and e the offset. The measured radii match this to about 10⁻⁸ on every machine tried. With no offset the roots are the crank’s length and the rod’s length, which are the two circles above. With an offset the two roots move towards each other: their product is always ab, so in rod units it is the crank over the rod, while their sum of squares falls by $e^2$. The annulus closes completely when e = b − a, which is exactly the offset at which the crank can no longer turn all the way round.

At an offset of 0.4 the inner circle is at 0.3367 rods instead of a third, and the outer is at 0.9899 instead of one. Between the outer circle and one rod the pair is real — two isolated points on most rays. They are therefore no longer a symmetric pair $\pm i\,p$ heading off together. The complex pair met itself at a real point, split into two real points, and then only one of them goes to infinity at one rod. On the circle itself the two-copy system reaches one finite double point at every angle tried. The singular point at infinity is correspondingly an ordinary node: after shifting onto its tangent, its degree-two part keeps a $Z^2$ term of the same sign as the $Y^2$ term, so its two tangents are complex conjugates. The real tangent line of the plain machine has not disappeared. It has moved to cross the vertical through the pivot at the height of the slide, to 10⁻⁹.

So the plain machine’s tacnode is a coincidence of its symmetry, and a clean one. The general slider-crank sends one double point through infinity on the one-rod circle, not two. The test proposed when the question was left open — “none finite, if the double point at infinity has absorbed both” — holds for the plain machine and fails with any offset.

The centrode shows the same split. With no offset its two branches through the slider pin are traced by both assemblies alike. With an offset each assembly traces its own pair, and the band between them is where one double point is a crossing and the other is isolated, because each cusp belongs to one of the two points only.

Where the slider-crank sits against the four-bar #

The four-bar’s coupler sextic has three finite double points, and an odd number of them are always real. Its curve cannot avoid crossing itself or having an isolated point. The slider-crank’s quartic has two, so the parity argument gives nothing, and the map shows the consequence: a whole annulus of coupler points whose curves are clean ovals.

The accounting behind this is the same kind of ledger as in nine times through each circular point. A quartic has room for three double points and its genus of one uses up one, leaving two. The circular points take none of them, because the quartic passes through each only once. Every double point the curve has is therefore available to the finite plane or to the one real point at infinity that the slide provides. On the one-rod circle, the plain machine sends both there and the offset machine sends one.

What this does not settle #

Why the ratio is $\cos^2(\psi/2)$. It is exact to 10⁻¹⁴ at every angle tried and on three different pairs of crank and rod lengths, and nothing here explains it. It says the tacnode is flattest on the rod and becomes a pair of real branches only in the limit of the slider pin itself, where the curve degenerates into the slide.

Why the two circles are circles. The closed form was found by measuring the radii on five machines and recognising a pattern, and it was checked on each of them to 10⁻⁸. It has not been derived. The measurements say the type of a coupler point’s double points depends only on its distance from the crank pin inside the one-rod circle, and outside that circle it plainly does not, since the centrode is not a circle. A derivation would have to explain both.

The offset machine’s escape rate. Only one double point passes through infinity there, and nothing here measures how quickly. A single real point crossing the line at infinity would be expected to grow as one over the distance, not one over its square root, but that has not been measured.

Still open: the third double point a four-bar loses on the way to a slide #

A slider-crank is the limit of a four-bar whose rocker has grown without bound, and the paths that leave showed how much of a machine’s algebra goes to infinity in limits like this. The four-bar’s sextic has three finite double points. The slider-crank’s quartic has two, and the rest of the sextic’s degree is absorbed into the circular points and the slide’s point at infinity.

The distinct argument there would follow the four-bar’s three finite double points as its rocker lengthens. It would find which one goes to infinity and in which direction. It would check whether that direction is the slide’s, which the count of meetings at infinity showed is where no rod quartic passes. It would also check whether the other two arrive where this essay found them. The test is parity. A four-bar always has an odd number of real double points and the slider-crank can have none, so somewhere along the way a real double point has to leave. Finding where it goes is the whole argument.