The third pole goes anywhere but the circle

A moving plane has, at each instant, one point that is not moving: the velocity pole, where the coupler is turning. It has one point that is not accelerating, the acceleration pole, and the other pole is on the circle found that point confined to the inflection circle, whatever the drive. The drive’s schedule — how the crank’s speed is changing — chooses where on the circle it sits, through one number, the crank’s angular acceleration over its speed squared. The geometry supplies the circle.

The next derivative gives a third point, the one whose jerk is nought: the rate of change of its acceleration. That essay ended by asking whether this pole of the third order is confined too. Its position depends on two numbers of the drive rather than one — the crank’s acceleration and its jerk, each over the right power of its speed — and the natural guess was that the geometry would again supply a locus, with the cubic of stationary curvature the obvious candidate.

It supplies none. Two numbers are exactly enough to put the pole anywhere, and the geometry’s only contribution is one curve the pole cannot reach.

Why two numbers are enough #

A point of the moving plane has a position $z$, and with the crank turning at a steady unit speed its first three derivatives with respect to the crank angle are $z_1$, $z_2$ and $z_3$, each affine in the point: every point has a centre, and every derivative is a fixed part of the body’s motion plus a rotation of the point about the crank pin. Drive the crank with an angular acceleration κ and jerk λ, per unit of its speed squared and cubed. The chain rule gives the point’s jerk in time as

$$z''' = z_3 + 3\kappa\,z_2 + \lambda\,z_1.$$

That is affine in κ and λ, and it is the whole argument. Asking for $z''' = 0$ at a chosen point is two real equations in the two unknowns κ and λ. They have exactly one solution whenever $z_1$ and $z_2$ point in different directions. They point in the same direction exactly where $\mathrm{Im}(\bar z_1 z_2) = 0$, which is the definition of the inflection circle: the points whose acceleration is along their velocity, so that their paths are momentarily straight. So every point of the plane off the inflection circle is the third-order pole of exactly one drive, and the drive can be written down.

The acceleration pole’s confinement came from the same kind of arithmetic with one fewer unknown. Its condition, $z_2 + \kappa z_1 = 0$, is two equations in one unknown, solvable only where $z_1$ and $z_2$ are parallel — on the circle. One more derivative brings one more free number, and the confinement turns into its opposite.

Checked by solving both ways #

The argument gives two computations that should agree. One takes a point, solves the two equations for its drive, and computes the jerk at the point under that drive. The other takes a drive and solves for the point whose jerk is nought: the jerk is also affine in the point, so that point comes from one complex division.

Four hundred points scattered over an eight-unit square round the four-bar each have their drive found, and that drive’s pole comes back to the point to $3 \times 10^{-15}$ of its distance from the crank pin. The jerk at each point under its own drive is nought to $3 \times 10^{-16}$. Neither computation knows about the inflection circle except as the place where the first one’s two equations stop being independent.

Run the other way, over a grid of 625 drives, the poles spread across the plane rather than along any curve. That is what two free numbers mapped into a plane do unless something collapses them, and nothing does. The inflection circle is conspicuous in the picture only by what is not on it.

The circle is where the drive has to be enormous #

The circle is not a place the pole avoids by luck. As a point approaches it, the two equations its drive must satisfy become nearly the same equation, and the drive that solves them grows without bound.

More than a tenth of a circle radius from the circle, a drive of at most 1.6 — acceleration and jerk together — will put the pole there. Between a thousandth and a hundredth of a radius from it, the drive needs 9 to 47. Within a thousandth, 112 to 820. The drive runs off as the point runs onto the circle, and the product of the two stays bounded: the drive needed is roughly one over the distance. A real crank cannot be driven with a jerk eight hundred times its speed cubed without the linkage noticing, so for practical purposes the pole’s reach stops a little short of the circle, by an amount the drive’s limits decide.

A coupler point over a turn #

The same fact read along a single point’s path says when a drive can make that point jerk-free. A coupler point stays fixed on the coupler while the inflection circle moves with the linkage, and the point is on the circle at exactly the instants its own path is momentarily straight: the circle of points going straight is the set of points whose paths have an inflection there. So a point whose path has inflections cannot be made jerk-free at those instants by any drive, and needs an enormous one near them.

Measured over a turn at a quarter of a degree, the coupler point at (1.3, 0.2) — in coupler lengths, along the coupler and across it — has a path with inflections at 89.6° and 152.1° of crank. The drive it needs runs off at 89.75° and 152.25°, the samples either side of the inflections, and is of order one everywhere else. The same holds for a point at (0.8, −0.4), whose path inflects at 57° and 281°. The point at (0.45, 0.5), which this field has used throughout and whose path is convex all the way round, never needs a drive larger than 1.73.

That gives the third-order pole a practical reading. A designer who wants a tool or a load on the coupler to be momentarily free of jerk at some position of the stroke can have it, by shaping the crank’s speed, anywhere except near the places the tool’s path inflects. There, the tool’s acceleration is along its direction of travel and the jerk has a component across the path that no timing of the crank can remove.

On the circle, two points #

On the circle itself $z_1$ and $z_2$ are parallel, so $z_3 + 3\kappa z_2 + \lambda z_1$ can vanish only if $z_3$ lies along the same line. That is one real condition on a point of the circle, so it picks out isolated points, and they have a meaning. A point whose first three derivatives are all along one line is a point whose path is straight to third order.

Measured round the circle, the sine of the angle between $z_1$ and $z_3$ vanishes at exactly two points. One is the velocity pole, where $z_1$ is itself nought and the condition is empty. The other is Ball’s point: the straightest point there is, where the inflection circle meets the cubic of stationary curvature, whose path has zero curvature and zero rate of change of curvature. Found this way it sits at (1.2362, 6.7434) from the crank pin, and found as the circle’s meeting with the cubic — a separate construction — at the same place to four decimals.

At Ball’s point all three derivatives are along one line, so the pole condition is one equation in two unknowns, and a whole line of drives puts the pole there. Every other point of the plane has exactly one drive; Ball’s point has a one-parameter family; the rest of the inflection circle has none.

The drives come in pencils of circles #

The pole’s dependence on the drive has a shape worth drawing. For a fixed crank acceleration, the pole is a ratio of two expressions each affine in the jerk — a Möbius function of one real number — and a Möbius function carries a real line to a circle.

So as the jerk alone varies, the pole runs round a circle, and as the jerk grows without bound it tends to the point that makes the jerk’s coefficient vanish, which is the point whose velocity is nought. Every constant-acceleration circle passes through the velocity pole. Likewise, as the acceleration alone varies the pole runs round a circle through the point whose $z_2$ vanishes — the acceleration pole of the steadily driven crank, which the other pole is on the circle placed on the inflection circle at an angle set by the coupler’s own acceleration. The poles lie on their circles to $10^{-14}$ of the radius, and the circles pass through their points to $3 \times 10^{-16}$.

The pencils also give a way to read a drive off the drawing without solving anything. To put the pole at a chosen point, draw the circle through that point and the velocity pole that belongs to the right acceleration, and the circle through the point and the steady acceleration pole that belongs to the right jerk; the two families are coordinates on the plane, as a pair of pencils of circles always are, and each point has one circle of each. The velocity pole and the steady acceleration pole are the coordinate system’s two foci, and the inflection circle, which passes through both, is the one curve on which no circle of one family meets a circle of the other, apart from Ball’s point.

Both centres of the pencils are on the inflection circle, and that is the circle’s other role. It is not a locus the pole is confined to, but it carries the two points from which every family of drives fans out: the velocity pole, which the jerk alone pushes the pole towards, and the steady acceleration pole, which the acceleration alone pushes it towards.

Not on the cubic #

The obvious candidate for a locus was the cubic of stationary curvature, whose defining quantity is built from $z_1$, $z_2$ and $z_3$. The measurement rules it out plainly: the cubic’s own residual at the third-order pole, as a share of its size near the machine, is 7.3 under a steady crank, 9.2 under a drive of acceleration 1 and jerk 2, 0.006 under acceleration −1 and jerk 5, and 0.05 under acceleration 0.5 and jerk −3. The pole wanders on and off the cubic as the drive changes, the way any point in the plane does.

The reason the guess was reasonable, and wrong, is the one the earlier essay gave in advance. Stationary curvature is a condition on the path’s shape, and shape does not depend on timing: every drive gives the same cubic. The third-order pole is a condition on a derivative in time, and it moves with the drive. Only where the two coincide — at Ball’s point, whose path is straight whatever the timing — does the drive stop mattering.

One point, one drive, in numbers #

For the coupler point at (0.45, 0.5) and the crank at 60°, the drive comes out as an angular acceleration of 0.144 and a jerk of 1.252, each per the right power of the crank’s speed. With the crank turning at one radian a second that is an acceleration of 0.144 rad/s² and a jerk of 1.25 rad/s³ at that instant; at ten radians a second, 14.4 and 1,252. The point’s own derivatives with respect to the crank angle at a steady crank are of order one — 0.58, 1.39 and 0.93 in coupler lengths — so the drive is modest: a small speed-up of the crank, rising quickly, cancels the point’s jerk.

That is also the size of a drive a real input can deliver. A crank driven through a ratio that is a function of the angle — non-circular gears, or a linkage ahead of it — has an acceleration and jerk fixed by that ratio’s first and second derivatives, and a ratio that varies by tens of per cent over a turn supplies drives of order one. So jerk-free instants at a chosen coupler point are within reach of an ordinary variable-ratio drive, anywhere off the inflection circle, and a drive’s two numbers are exactly the freedom such a ratio has at one instant: its slope and its curvature.

What the drive cannot do at one instant is choose which point: it has two numbers, and the point takes both. A designer wanting two coupler points jerk-free at once asks four numbers of two, and gets them only where the two points’ requirements happen to agree.

What this means for the order of the poles #

The field’s objects sorted themselves once into the geometric, unchanged by how fast the input is driven, and the kinematic. The velocity pole is geometric. The acceleration pole is kinematic but tied to a geometric circle: the drive chooses one number, and the circle takes up the other. The third-order pole is kinematic and tied to nothing but the plane, except for the one circle it cannot enter.

That sequence stops here in an informative way. A fourth-order pole — the point whose snap is nought — would depend on three drive numbers, and three numbers mapped into a two-dimensional plane leave a free line of drives for every point. Only the counting changes: zero free numbers fixes a point, one fixes a curve, two fix the plane, and three leave the drive underdetermined. The velocity pole’s rigidity, the acceleration pole’s circle and the third-order pole’s freedom are the same fact, counted three times. Four positions brought together met the same counting from the synthesis side, where each extra precision position is one more condition and each extra derivative of the motion one more number.

What the third pole does not settle #

The fourth order. The counting above is an argument, not a measurement. That the snap’s pole is underdetermined follows if the snap is affine in the three drive numbers, which the chain rule suggests and nothing here has checked.

Other mechanisms. Everything was measured on one four-bar at one crank angle. The argument uses nothing about the four-bar except that its derivatives are affine in the point, which is true of every planar motion, but the positions of Ball’s point and the circle are this machine’s.

Where the pole is useful. A point with no jerk is a point whose acceleration is momentarily steady, which matters for a tool or a load that should not be shaken. Whether a drive chosen to put the third-order pole at a working point does anything useful over a stroke, rather than at one instant, is not examined.

Still open: a drive that holds the pole still #

At one instant the drive’s two numbers put the third-order pole anywhere off the circle. Over a stretch of motion the pole moves with the linkage, and the drive’s acceleration and jerk change with the crank angle.

The distinct argument would be the inverse problem over a stroke: choose a body point, and ask whether there is a crank schedule — the crank’s speed as a function of its angle — that keeps that point’s jerk nought throughout. At each crank angle the argument above gives one acceleration and one jerk, and a schedule has only one function’s freedom, whose acceleration and jerk are not independent: the jerk is the acceleration’s own derivative. So the two numbers the pole needs cannot in general both be met, and the measurement would be how far a best schedule falls short — the smallest achievable jerk at the chosen point over a stroke — and whether some points of the coupler, perhaps near Ball’s point, can be held jerk-free for a stretch after all.