Where a slide puts the rest of the degree

A curve of degree four meets every line four times, counting complex meetings and meetings at infinity, and the line at infinity is a line like any other. Nine times through each circular point measured where a linkage’s curves meet it. A point of a machine built only of pins draws a curve that meets the line at infinity at the two circular points and nowhere else, half its degree at each. A slide breaks that. The slider-crank’s connecting rod draws a quartic that passes once through each circular point, and the other two of its four meetings with the line at infinity were left unlocated.

The natural guess, recorded when the question was left open, was the direction of the slide itself. A slide is the one thing in the machine with a direction, and a slide is what broke the rule. It is also the answer the four-bar would suggest: the slider-crank is the four-bar whose output bar has grown without bound along the slide, and whatever the sextic lost in that limit might plausibly have gone the way the bar went.

The guess is wrong, and what replaces it is exact.

Six quartics that look alike #

The machine is the slider-crank in its plainest form: a crank of length 1 on a pivot at the origin and a rod of length 3 whose far end slides along the x-axis, with no offset. A point on the rod is given by two numbers in the rod’s own units: u along the rod from the crank pin, and v across it. The crank pin is (0, 0) in those units and the slider pin (1, 0).

Every one of the six curves is a quartic, and every one is bounded: the crank is short, the rod is finite, and no real point of the machine ever goes far. So nothing on the page meets the line at infinity. All four of each curve’s meetings with it are points no drawing can show, and the only way to find them is to count.

The six are chosen in two groups of three, and nothing in the drawing tells the groups apart. The points on the left are at 0.64, 0.36 and 1.34 rods from the crank pin. The points on the right are all at exactly one rod — as far from the crank pin as the slider pin is.

Where the crank pin goes when the machine goes to infinity #

A point at infinity on the curve is the limit of a complex configuration in which the tracing point grows without bound. The only way for that to happen is for the crank pin to grow without bound too, since everything else is a fixed distance from it. A crank pin a fixed distance a from the origin can be arbitrarily large only as a complex point, and only by running out along one of the two isotropic directions (1, i) and (1, −i), the directions in which $x^2 + y^2$ stays small while x and y do not. Those two directions are the circular points.

Where the crank pin goes, the slider pin must follow at a distance b. It sits on the slide at (B, 0), and the rod’s length gives $B^2 - 2A_x B + a^2 - b^2 = 0$. As the crank pin’s x-coordinate $A_x$ grows, the two roots of that quadratic separate: one grows like $2A_x$ and the other shrinks towards nothing.

On the small root the slider pin stays near the origin, the rod is simply minus the crank pin, and every point of the rod runs out in the same isotropic direction as the crank pin. That is the quartic’s one passage through each circular point, and it is the same for every coupler point.

On the large root the slider pin runs out along the slide at twice the crank pin’s pace, and the rod becomes $A_x$ times (1, ∓i) — the other isotropic direction. A point at (u, v) on the rod is the crank pin, plus u of the rod, plus v of the rod turned a quarter. To leading order it runs out along (1 + u ± iv, v ± i(1 − u)), and a direction is a slope:

$$m_\pm = \frac{2v \pm i\,(1 - u^2 - v^2)}{(1 + u)^2 + v^2}.$$

The same reasoning, applied to the four-bar, accounts for what three linkages, one equation found by working the equation out — a sextic whose top part is a power of $x^2 + y^2$. There, every pin that runs to infinity is tied to a fixed pivot by a distance, every one of them has to go isotropically, and every point of the machine follows. A slide is the one constraint that lets a pin run to infinity in a real direction, and the rod, pinned at one end to a point going one way and at the other to a point going another, runs out along a mixture of the two.

Two things about that expression say most of what this essay finds. The crank length a, the rod length b and the slide’s offset are not in it; they all appeared only at lower order. And it is along the slide, m = 0, only when v = 0 and u = 1 — the slider pin itself, which draws a line.

Counting the meetings #

A prediction about points nobody can draw needs a check that does not use the prediction’s reasoning. The instrument is the one that counts a degree: add a line to the machine’s equations, track every solution path, and count the distinct finite points that arrive. A line through a point at infinity where the curve passes k times meets the curve in the finite plane k fewer times than the degree.

Every coupler point gives the same pattern. A general line meets the curve four times; that is the degree. A line through either circular point meets it three times; the curve passes once through each. A line along the slide meets it four times, the full degree, so the curve does not pass through the slide’s point at infinity at all. A line of an arbitrary complex slope also meets it four times. And a line of slope $m_+$ or $m_-$ meets it three times — or, for the three points on the circle, twice.

Four and three and two are small numbers, and the danger with small counts is that a lost path looks like a meeting at infinity. No path stalled in any of the forty-two solves. The unpredicted slope is the control: a line chosen with no reason to pass anywhere special gets the full four.

The equation says the same #

The second check reads the equation of the curve rather than the machine. A quartic has fifteen coefficients, and 240 points of the drawn curve fix them as the null vector of a 240 × 15 matrix, with a gap of at least 10¹³ between the smallest singular value and the next — the equation is determined, not estimated.

A curve’s points at infinity are the roots of the top-degree part of its equation. For these quartics that part is five coefficients. If the curve passes once through each circular point, the top part is divisible by $x^2 + y^2$, and whatever is left is a quadratic whose roots are the slopes of the other two points.

The division is exact to 10⁻¹⁴ on every curve. The slopes that remain match $m_\pm$ to a few parts in 10¹⁵ for the three coupler points off the circle. On the circle they match to a few parts in 10⁷, and that looser agreement is itself part of the answer: on the circle the two roots coincide, and a double root of a polynomial whose coefficients are known to 10⁻¹⁴ is only known to about the square root of that. The prediction says the roots should be one number counted twice, and the equation is behaving exactly as an equation with a double root does.

Changing the machine to a crank of 1.3, a rod of 2.2 and a slide offset by 0.4 moves every curve and leaves the slopes where they were, to 10⁻¹⁵.

Two computations that share neither equations nor method have located the same two points. The slicing never fits anything; the fit never tracks anything. And both of them are possible only because the curve is small: a curve of degree eighteen needs 190 coefficients and gives no singular-value gap to decide by, so on the six-bars the slicing is the only route and the check would have to be a second line rather than a second method.

Why the slide’s direction was the wrong guess #

The slide does reach infinity, in the one complex configuration where the slider pin runs out along it at twice the crank pin’s pace. But the slider pin is only one end of the rod. The other end is the crank pin, and the crank pin cannot follow the slide: a point held a fixed distance from a pivot can only become large isotropically. The rod joining them is pulled one way at one end and another way at the other, and every point on it except the slider pin itself runs out along a direction that is neither the slide’s nor a circular one.

The counts say this directly. A line parallel to the slide meets every rod quartic four times, which means none of them passes through the slide’s point at infinity even once. Only when the coupler point is placed on the slider pin — u = 1 and v = 0 — does the predicted slope become nought, and there the curve is no longer a quartic but the slide itself, a line, which is the one curve that certainly does go to infinity along the slide.

So the slide’s direction is where the prediction starts, not where it ends. Moving the coupler point off the slider pin swings the two directions away from it into the complex plane at once, and they return to the real line only on the circle through the slider pin centred on the crank pin.

A complex pair, except on one circle #

The two slopes are complex conjugates: the same real part, $2v / [(1 + u)^2 + v^2]$, and opposite imaginary parts. A real curve always has its complex points at infinity in conjugate pairs, so that much is forced. What is not forced is where the imaginary part vanishes, and the numerator says exactly where: $1 - u^2 - v^2 = 0$.

Along any ray of coupler points leaving the crank pin, the imaginary part starts positive, passes through nought at a distance of exactly one rod, and goes negative beyond — the two conjugate directions pass through each other and exchange places. The crossing is at one rod whatever the ray’s angle, because the denominator never vanishes and the numerator depends only on the distance.

On that circle both slopes are the same real number, and a curve that passes through two conjugate points at infinity passes through their coincidence twice. That is the two in the ledger: a line of that slope meets the curve twice in the finite plane because it meets it twice at infinity.

Half the angle #

A coupler point one rod from the crank pin is u = cos ψ, v = sin ψ, where ψ is the angle between the rod and the line from the crank pin to the point. Its real slope is 2 sin ψ / (2 + 2 cos ψ), and that is tan(ψ/2).

The figure draws it. The rod lies horizontal in its own frame; the circle is every point a rod from the crank pin; and through each point on the circle a short line shows the direction in which that point’s quartic passes twice through infinity, measured from the slide. The slider pin, at ψ = 0, points along the slide. A point a quarter-turn round, directly across the rod from the crank pin, points at 45°. A point almost opposite the slider pin points almost straight across the slide.

Nothing about the drawn curves shows this. The quartic for (0, 1) and the quartic for (0.4, 0.5) are both bounded loops. One passes twice through a real point at 45° on the line at infinity and the other passes through two complex points there, and the difference is exactly as invisible as the circular points are.

What a slide does to the accounting #

The rule for pins was that all of a curve’s degree is at the circular points. The rule a slide leaves in its place can now be written as a ledger.

The four-bar’s sextic puts three meetings at each circular point and none anywhere else. The slider-crank’s crank draws a circle, and a circle is always once through each. The slider-crank’s rod quartic puts one at each circular point and two elsewhere: a complex pair off the circle, one real point twice on it. The elliptic trammel, with both ends of its rod on slides, draws an ellipse that passes through neither circular point and puts both of its meetings elsewhere.

The reason for the pattern is the one that explained half. Written in z = x + iy and z̄ = x − iy, a pin’s distance condition is a product of something in z and something in z̄, and nothing a pin does mixes them otherwise. A slide along the x-axis says y = 0, which is z = z̄, and ties the two together. Each slide gives the machine a way to reach infinity that is not isotropic, and each such way is a meeting with the line at infinity somewhere other than a circular point. The rod quartic has one slide and one extra pair; the trammel has two slides and no circular meetings left at all.

What this does not decide #

The finite double points. A quartic of genus one has two double points, counted with multiplicity. A passage once through each circular point uses none of them, so off the circle both are finite. On the circle the curve passes twice through one real point at infinity, and that double point has to come from somewhere. The natural reading is that the two finite double points run off to infinity as the coupler point approaches the circle and meet there. It is not measured here.

What the real point at infinity does to the real curve. Every drawn curve is bounded, so no real branch approaches the real point at infinity; it is reached only by complex configurations. Whether it is a node whose two branches are a complex conjugate pair, or a point where the curve is tangent to the line at infinity, is a question about its local structure that the counts here do not separate.

How many answers the machine has. The curve’s meetings with a line are configurations of the machine, and a count of them is a count of solutions like any other. Nothing here changes how many configurations a slider-crank has for a given crank angle; it changes only where the curve’s own degree sits.

A general slider machine. The derivation used the slide passing along the x-axis and a single slide in the machine. A six-bar with a slide in one of its loops, or a slide placed so that it locks a pin, has more ways to reach infinity, and each would have to be followed separately.

Why the circle is the rod’s length. The condition $1 - u^2 - v^2 = 0$ fell out of the algebra. A geometric reading exists — the coupler point, the crank pin and the slider pin form an isosceles triangle, and the real direction at infinity is the direction the bisector of its apex angle has when the rod lies along the slide — but it has not been turned into an argument that would predict the same condition on a machine whose rod is not the only link joining a pin to a slide.

Still open: the double points that leave for infinity #

The quartic’s genus leaves room for two double points. Off the circle they are somewhere in the complex plane; on it, the curve’s double passage through a real point at infinity accounts for them. The distinct argument there would follow them.

A double point is a place the tracing point reaches from two different configurations, which is a polynomial system in two copies of the slider-crank with their tracing points set equal and their configurations required to differ. Tracked for coupler points along a ray from the crank pin, its finite solutions would show whether two double points exist inside the circle, whether they are real or complex there, whether they run to infinity in the direction tan(ψ/2) as the ray reaches one rod, and what they become beyond it. The test of the account would be the count at exactly one rod: none finite, if the double point at infinity has absorbed both.