A sextic that comes apart

Two earlier essays studied a four-bar’s coupler curve as a single object: a sextic with two ovals or one, and three finite double points on the circle through the cognate pivots, an odd number of them real. Both censuses kept well away from one family of linkages, those at a change point, where the sum of the shortest and longest links equals the sum of the other two.

That exclusion was not caution for its own sake. At a change point the two circuits of a four-bar touch, and on the most familiar change-point chain of all, the parallelogram, the coupler sextic stops being one curve. It factors. This essay divides it, says what each factor is, and finds that the places where the factors meet are the places the essay on Grashof’s condition described as the ones where a parallelogram linkage has to be told which way to go.

A circle that has to be in the equation #

Take a four-bar with its crank equal to its rocker and its coupler equal to its ground: here ground 4, crank 1.5, coupler 4 and rocker 1.5, with the coupler point at u = 0.45 along the coupler and v = 0.5 off it.

One of its motions is obvious from the drawing. Assembled as a parallelogram, the coupler stays parallel to the ground at every crank angle, so the coupler is not rotating at all: it is translating. Every point on it moves exactly as the crank pin moves, displaced by a fixed vector. The crank pin goes round a circle of radius 1.5 about the origin, so the coupler point goes round a circle of radius 1.5 about (ub, vb) = (1.8, 2.0), with no approximation anywhere.

A circle is a curve of degree two, and every point of it is a coupler point of this linkage. The eliminated sextic vanishes on every coupler point the linkage has. A polynomial vanishing on every point of an irreducible curve is divisible by that curve’s equation, which is an algebraic fact rather than a numerical one, so the sextic must be the circle’s equation times something of degree four.

That is a prediction, and it is tested by dividing.

Divided, to rounding #

The circle (x − 1.8)² + (y − 2)² − 1.5² has x² as its highest power of x with coefficient one, so ordinary polynomial long division in x, treating the coefficients as polynomials in y, is exact: no fitting, no tolerance, and a remainder that is identically zero precisely when the circle is a factor.

The remainder’s largest coefficient is 6.3 × 10⁻¹⁶ of the sextic’s largest. That is the rounding of double-precision arithmetic, and the rest of this essay shows it is nothing else.

What comes out of the division is a quartic, and its leading form is exactly a multiple of (x² + y²)². That is a check the division did not have to pass. Every coupler sextic’s leading form is 4b²(x² + y²)³, exactly, and the circle contributes one factor of x² + y² to it. The quartic must supply the other two, which makes it bicircular: it passes through each circular point at infinity twice. It does, with no discrepancy in its leading coefficients.

So the parallelogram’s coupler curve is a circle and a bicircular quartic, and they are two separate curves that happen to share an equation because the product of their equations is what the elimination produces.

Each factor is a motion #

The division is algebra; the question that matters for a machine is what the quartic is.

The linkage is solved at every half degree of crank angle on both solutions of its two-circle problem, 1,442 configurations in all with a worst closure residual of 8.0 × 10⁻¹⁴, and each configuration’s coupler point is tested against both factors. Every configuration lies on one factor or the other. Seven hundred and twenty are on the circle, and they are the parallelogram configurations with the coupler parallel to the ground. The other 720 are on the quartic, and in every one of them the coupler crosses the ground line: the chain assembled as an antiparallelogram, with its two cranks turning in opposite senses.

So the two factors are the chain’s two circuits, one polynomial each. On the crank-rocker the two circuits were two ovals of one irreducible sextic, and the equation could not be separated into a piece for each. On the parallelogram it can, and it separates along exactly the line the mechanism does.

That also accounts for a feature of the factorisation that would otherwise look arbitrary. The circle has the crank’s radius, and its centre depends on the coupler point’s position. Nothing else about the linkage enters. The quartic carries everything else: the crossing, the counter-rotation, and the whole of the dependence on the ground length.

Exactly, or not at all #

A factorisation is an exact statement, so it should fail sharply when the thing it is about is not exactly true. The division is repeated on three other cases.

The kite, with ground 3, crank 3, coupler 4 and rocker 4, has its adjacent links equal in pairs instead of its opposite ones. Its sextic divides by its own circle, of radius 2.6907 about the far pivot, with a remainder of 4.1 × 10⁻¹⁵: rounding again.

The same parallelogram with its coupler one millionth longer than its ground, divided by the same circle, leaves 2.0 × 10⁻⁷. That is more than eight decades above the exact case, from a change in one length of one part in four million.

The standard crank-rocker, divided by the circle that the parallelogram recipe would assign it (its crank’s radius, about the point the coupler would carry the crank pin to if the coupler translated), leaves 0.687. The recipe describes no motion the crank-rocker has, and the division says so without hesitation.

Lengthening the coupler through six decades, from 10⁻⁶ to 0.1, moves the remainder in exact proportion, a slope of 0.9975 on logarithmic axes with a constant of 0.1938. At equality it drops to 6.3 × 10⁻¹⁶. There is no intermediate regime in which the sextic nearly factors: a first-order remainder appears the moment the lengths stop being equal.

The mechanical reading is the same one the compiled field found for its exact straight lines. A parallelogram chain with its coupler a millionth too long is not a parallelogram with a small error. Its sum of shortest and longest now exceeds the sum of the other two by that millionth, so by Grashof’s condition it is a triple rocker, with one circuit and an irreducible sextic. A drawing of it at the scale of a page cannot be told apart from the parallelogram’s. Its equation is a different kind of object.

A branch of the solve is not a factor #

Every four-bar solver used here finds the rocker pin as an intersection of two circles, and two circles meet at two points. The solver calls the two its branches, and so far in these essays a branch has meant the same thing as a circuit. On the crank-rocker it did: each branch drew one oval all the way round.

On the parallelogram it does not. Each of the two branches spends exactly 360 of its 720 configurations on the circle and 360 on the quartic, and each swaps factor twice a turn, at a crank angle of 0° and of 180°. At those two angles all four pins lie on the ground line, the two circles the rocker pin must lie on are tangent, and the two solutions of the solver are the same configuration.

So the solver’s label changes meaning at the change point without anything in the solver noticing. A branch is a statement about which of two intersection points was taken: above the line from the crank pin to the rocker pivot, or below it. A factor is a statement about which motion the machine is in. On a generic four-bar the two coincide because the intersection points never merge; on a parallelogram they merge twice a turn, and the label that followed the motion until then follows the geometry afterwards.

That matters for anything that uses a branch label as an identity. Branches were components all along made the case that the right object is the connected component of the configuration space, and this is a worked example of why. The component is defined by the motion; the label is defined by a choice of sign, and the two can disagree exactly where it matters most.

Where the machine chooses #

At a crank angle of 0° the parallelogram is flat. The two solutions of its two-circle problem are one configuration, 0 apart. Turn the crank on by 25° and there are two configurations again: one with the coupler still parallel to the ground, whose coupler point lies on the circle to 1.6 × 10⁻¹⁶, and one crossed, whose coupler point lies on the quartic to 7.3 × 10⁻¹⁶. Nothing in the bar lengths decides between them.

This is the event the essay on Grashof’s condition described in words and the three-bar parallelogram solved around: a flat configuration from which the chain can leave as a parallelogram or as an antiparallelogram, which is why locomotive coupling rods are duplicated a quarter turn out of phase and why a real parallelogram linkage needs a third bar, a second input or inertia to stay parallel. What the factorisation adds is that the choice is between two polynomials. The machine leaves the flat configuration on one curve or the other, and the two curves are distinct algebraic objects rather than two halves of one.

Four places the factors meet, and two of them are choices #

A circle and a quartic meet in eight points, counted in the complex projective plane. Substituting the circle’s rational parametrisation into the quartic gives a polynomial of degree eight in the parameter, and its roots can be sorted.

Four of the eight roots sit at the circular points, each twice, since the circle passes through each circular point once and the quartic twice. That leaves four finite meeting points, and on this parallelogram all four are real: at (0.300, 2.000), (3.300, 2.000), (3.240, 1.579) and (0.473, 1.301).

They are not all the same kind of meeting. The first two are the coupler point’s positions at 180° and 0° of crank angle, when every pin is on the ground line. There, one configuration of the machine belongs to both factors at once, and it is the configuration from which the machine chooses. They are the change points.

The other two are places where the circle and the quartic cross without any configuration being on both. The parallelogram machine reaches the point at one crank angle, the crossed machine reaches it at another, and the two never meet there, exactly as the two ovals of a crank-rocker cross without the machines meeting. A drawing of the curve shows four crossings that look alike. Two of them are forks in the motion and two are coincidences of the drawing.

Three, four and five double points #

The essay on double points counted three finite double points on every coupler curve, all on the circle through the cognate pivots, an odd number real. That count assumed a generic linkage, and change points are where it stops holding.

The crank-rocker has three, all on the circle. A change-point chain that is neither a parallelogram nor a kite, ground 4, crank 1, coupler 3 and rocker 2, has four: the three on the circle, and one more at (0.35, 1.5), which is where its coupler point is when every pin is on the ground line. Its sextic does not factor. It is a single curve whose two circuits touch at that one point, and the touching point is a double point the construction on the circle cannot see. The circle finds places where two different orientations of the coupler carry the point to the same spot. Here there is only one configuration, with its orientation fixed along the ground line, and what is undecided is not where the coupler points but which way the motion leaves.

The parallelogram and the kite have five each: the three on the circle and two change points off it. On the parallelogram, two of the four meeting points found above are on the orientation circle and the two change points are off it, so the circle’s three are the two plain crossings of the factors and one singular point of the quartic itself.

So the generic count of three, and the parity argument that went with it, are properties of linkages away from change points. At a change point the curve acquires one extra double point where its circuits touch, or two where the curve comes apart, and those points are exactly the configurations the machine can leave in more than one way. That essay’s census excluded these linkages, and this is the reason it had to.

The kite’s circle is a motion the crank does not make #

The kite’s factorisation holds to the same standard, and its circle is stranger than the parallelogram’s.

All 1,438 configurations that a sweep of the crank solves lie on the quartic. Not one is on the circle. The circle has radius 2.6907, which is the coupler length times the distance of the coupler point along it, 4 × |0.45 + 0.5i|, and its centre is the far fixed pivot, (3, 0). It is a genuine factor of the sextic, and at first sight no motion of the machine draws it.

It is drawn by a motion in which the crank does not move. The crank and the ground are both 3 long, so the crank pin can sit exactly on the far pivot. When it does, the coupler (from the crank pin) and the rocker (from the far pivot) are two bars of length 4 hinged at the same point and meeting at the same rocker pin. They lie on top of each other and can turn together about the pivot as a single body, carrying the coupler point round a circle. Every position of that turn satisfies every bar exactly.

A sweep that drives the crank and solves for everything else cannot find that motion, because on it the crank angle is constant and everything else moves. The equation found it anyway. The kite has a circuit that no crank sweep visits, and its sextic contains it as a factor. It is the clearest case among coupler curves of the equation knowing more about a machine than any single way of driving the machine reveals.

What this essay does not establish #

Two reducible families are divided here, the parallelogram and the kite, each at one coupler point. The division is an exact test and would detect any other circle factor, but no search was made for other reducible coupler sextics, nor for factors that are not circles. Degenerate placements of the coupler point, such as a point on one of the pins, whose curve is a circle traced by the pin itself, were not examined either.

The meeting points are counted on one parallelogram. On other parallelograms some of the four finite meetings may be complex, and the split into two change points and two plain crossings is established for this linkage only; the two change points, being the flat configurations, are real on every parallelogram.

The kite’s circle is a motion of the chain only if the crank pin can be placed on the far pivot and held there, which a real kite’s joint at that pivot may or may not allow. The equation does not know about pins, washers or how a bar would pass over another.

What comes next #

A symmetric curve from an asymmetric machine. Some coupler curves have a mirror line although the linkage that draws them has none, and the condition for it is a pair of equal lengths of a different kind from the parallelogram’s. A symmetric curve from a lopsided machine measures the symmetry exactly and finds the axis.

What a fit makes of a reducible curve. The degree-six fit of the equation a four-bar satisfies was decided by one clear gap. On a parallelogram’s circle factor alone, every product of the circle with any quartic vanishes on the traced points, so the fit should find a null space of dimension fifteen rather than one. Whether the singular values show that as a clean gap, and whether a fit to both factors recovers the product uniquely, is a measurement not yet made.

The change point as a design tolerance. A coupler a millionth too long destroys the factorisation and turns the chain into a triple rocker. What it does to the motion is a separate question with a size: how far past the flat configuration the lengthened chain can go before it stops assembling, as a function of the error. That would say what a parallelogram linkage’s change point costs in a machine that is built rather than drawn.