No drive holds the third pole

The other pole is on the circle found that a drive’s schedule decides where the coupler’s acceleration pole sits, through one number, the crank’s acceleration over its speed squared, $\kappa = \ddot\theta/\dot\theta^2$. The pole stays on the inflection circle and the number slides it round. The third pole goes anywhere but the circle went one derivative further. A coupler point’s jerk under a schedule is

$$\dddot z = \dot\theta^3\,\big(z_3 + 3\kappa z_2 + \lambda z_1\big),$$

with $z_1$, $z_2$, $z_3$ the point’s derivatives in the crank angle at a steady crank and $\lambda = \dddot\theta/\dot\theta^3$ a second schedule number. That is two real equations in two unknowns, so at one crank angle every point off the inflection circle is the third-order pole, the point of zero jerk, of exactly one drive.

That essay ended by asking the obvious next question. A machine designer does not want a coupler point jerk-free for an instant. A pick-and-place gripper or a film claw wants it smooth over a stretch of the stroke. Is there a crank schedule that holds a chosen point’s jerk at nought throughout?

There is not, and the reason is that the two numbers the instant needs are not independent over a stretch.

One function, two numbers #

A crank’s schedule is its speed as a function of its own angle, $\omega(\theta)$. Everything the drive does follows from that one function. The crank’s acceleration is $\ddot\theta = \omega\omega'$, so the first number is

$$\kappa = \frac{\ddot\theta}{\dot\theta^2} = \frac{\omega'}{\omega},$$

the slope of the speed’s logarithm. The crank’s jerk is $\dddot\theta = \omega(\omega'^2 + \omega\omega'')$, so the second number is

$$\lambda = \frac{\dddot\theta}{\dot\theta^3} = \Big(\frac{\omega'}{\omega}\Big)^2 + \frac{\omega''}{\omega} = \kappa' + 2\kappa^2.$$

At one instant both numbers can be anything, because $\omega'$ and $\omega''$ can be anything at a point. Over a stretch they are tied. Choose $\kappa$ along the stroke and $\lambda$ is determined: it is the slope of $\kappa$ plus twice its square.

So to hold a point’s jerk at nought over a stretch, the schedule has to deliver at every angle the pair $(\kappa^*, \lambda^*)$ that the earlier essay’s two equations demand. The first fixes $\kappa = \kappa^*$ along the whole stretch, which fixes $\kappa'$, which fixes $\lambda$. The second equation then holds only if

$$\lambda^* = \kappa^{*\prime} + 2\kappa^{*2}.$$

That is a condition on the point and the linkage, not on the drive. Its failure, the mismatch $m(\theta) = \lambda^* - \kappa^{*\prime} - 2\kappa^{*2}$, measures how far the point is from being holdable.

What the relation means for an ordinary crank #

The relation $\lambda = \kappa' + 2\kappa^2$ is easy to see on schedules a machine actually uses. A crank turning steadily has $\kappa = 0$ and $\lambda = 0$: neither number can be moved without the other moving with it. A crank whose speed grows by the same proportion for every degree, $\omega = e^{c\theta}$, has $\kappa = c$, constant, and $\lambda = 2c^2$: its jerk number is fixed at twice the square of its acceleration number and cannot be chosen at all. A crank driven through a gear pair whose ratio wobbles sinusoidally about one, $\omega = 1 + \epsilon\sin\theta$, has $\kappa \approx \epsilon\cos\theta$ and $\lambda \approx -\epsilon\sin\theta$ to first order in the wobble. The jerk number then runs a quarter-turn behind the acceleration number, at the same size.

In every case the second number follows from the first. The earlier essay’s single-angle result imagined a drive that could set both at will, and such a drive exists only for an instant, or over a stretch only when the point’s own needs happen to obey the same relation.

The mismatch on the standard four-bar #

The hero figure draws it for the coupler point at $(0.45, 0.5)$ in the coupler’s own frame, the point the curvature essays have used throughout, over the crank’s stroke from 30° to 150°. The $\lambda$ the point needs climbs to about 1.5 through the middle of the stroke. The $\lambda$ that the needed $\kappa$ forces stays near nought, because the needed $\kappa$ is small and slowly varying. Their difference reaches 1.77 and crosses nought once, at 39.8°.

The other points tell the same story. The point at $(0.3, 0.8)$ has a mismatch of up to 1.61, nought only at 46.8°. The point at $(0.5, 0)$, on the line of the coupler’s two pins, has a mismatch up to 2.10 that is nought nowhere on the stroke. The point at $(1, 0.3)$, which stays nearest the inflection circle, needs schedule numbers that run large, and its mismatch reaches 9.37 and is nought at two isolated angles. Since $\kappa^*$ and $\lambda^*$ are smooth functions of the crank angle away from the inflection circle, the mismatch is too, and a smooth function that is not identically nought is nought only at isolated angles. No point tried can be held jerk-free over any stretch, however short.

The mismatch was computed from the earlier essay’s formula for $(\kappa^*, \lambda^*)$ at each of 241 angles, with $\kappa^{*\prime}$ by central differences. It does not depend on any schedule, so it is a property of where the point sits on the coupler.

Where on the coupler the points sit #

The four points were chosen to span the coupler. The point at $(0.45, 0.5)$ is the curvature essays’ usual coupler point, well inside the coupler’s triangle. The point at $(0.3, 0.8)$ is higher above the line of the pins. The point at $(0.5, 0)$ lies on that line, halfway between the pins. The point at $(1, 0.3)$ sits just above the rocker’s pin, and of the four it stays closest to the inflection circle, 0.40 inside it at 30° of crank, where the others are 1.5 to 3.2 away.

That last point matters because of what the earlier essay found about the inflection circle. A point on the circle of points going straight cannot be made the third-order pole at all, since there its first two derivatives are parallel and the two equations collapse into one. As a coupler point approaches the inflection circle the schedule numbers it needs grow without bound, and its mismatch grows with them. That is why the point nearest the circle has a mismatch five times the others’, largest at the start of the stroke where it is nearest. The points that can come closest to being held are the ones well away from the circle throughout the stroke, and even they cannot be held.

The best a schedule can do #

If the jerk cannot be held at nought, the useful question is how small a schedule can make it. With $\kappa(\theta)$ free along the stroke and $\lambda$ tied to it, the point’s jerk per crank speed cubed is

$$J(\theta) = z_3 + 3\kappa z_2 + (\kappa' + 2\kappa^2)\, z_1,$$

and the best schedule minimises its root mean square over the stroke. That is a least-squares problem in one function. Sampled at 61 angles, with $\kappa'$ by differences, it was solved by Gauss–Newton from a steady crank.

For the point at $(0.45, 0.5)$ the steady crank’s root-mean-square jerk is 0.947 and the best schedule’s 0.647, a third less. The largest jerk falls from 1.14 to 0.82. The best schedule’s jerk dips to 0.020 at 32°, close to 39.8° where the mismatch vanishes, and there a schedule can briefly do what the earlier essay’s instant allowed. Everywhere else it stays well above nought. The other points do similarly or worse: 64% of the steady jerk remains at $(0.3, 0.8)$, and 80% at $(0.5, 0)$, whose mismatch has no zero.

What the schedule asks of the crank #

The best schedule is a crank speed that rises across the stroke.

For the three points it runs from about 1.0 at 30° to 1.5, 1.4 and 1.6 at the end of the stroke, a smooth acceleration of the crank by half across 120°. A crank driven that way, by a servo or by a non-circular gear pair that turns a steady input into this output, takes a third off the jerk at its best point and much less at others. That is a real gain for a machine limited by vibration, and it is the most any drive can give. The gear pair would be cut from the schedule the way that essay cut its pairs from a demanded ratio: the ratio is the speed $\omega(\theta)$, rising smoothly from one to one and a half across the stroke. Over the rest of the turn they would have to bring the ratio back down, which is where the full-turn question below begins. The rest of the jerk belongs to the linkage, and only a different linkage can remove it.

Checking the formula in time #

Everything above rests on the jerk formula and on the relation $\lambda = \kappa' + 2\kappa^2$. Both can be checked by doing what a machine does. Run the crank under the best schedule in time, integrating its angle from its speed with a fourth-order Runge–Kutta step. Solve the four-bar at each instant, place the coupler point, and difference its position three times. That route uses neither formula. It needs the schedule, the linkage’s own loop closure and arithmetic.

At mid-stroke, for each of the three points with a best schedule, the jerk differenced in time and the formula’s agree to about $10^{-6}$: 1.10011 against 1.10011 for the first point, 0.94246 against 0.94246 for the second, 1.41843 against 1.41842 for the third. The relation that ties $\lambda$ to $\kappa$ is inside that agreement, because the time-domain route has only the crank’s speed to work from.

The instant’s answer, read along the stroke #

The earlier essay already drew the instant’s answer as a function of the crank angle, without asking whether the answers could be strung together.

Each point of those curves is a pair $(\kappa^*, \lambda^*)$ valid at its own angle. Read as a schedule, the curve of $\kappa^*$ against the angle is a perfectly good crank speed, the logarithm of $\omega$ integrated from it. The curve of $\lambda^*$ is a second, different demand on the same speed. A drive can follow either curve, and it cannot follow both, because following the first already fixes what the second must be.

The figure’s large values make the same point from the other side. Where a point’s path is near an inflection the instant’s drive runs off to infinity, as every point has a centre of curvature that runs off to infinity there. A schedule following those numbers would have to accelerate the crank without limit, and the mismatch measures something different: how badly the two demands disagree even where both are modest. For the usual coupler point they never exceed about 1.7 and still disagree by as much as they are large.

Why the instant misleads #

The earlier essay’s result was correct and is still true: at any one angle, any point off the inflection circle can be made the third-order pole. What it did not say, because at one angle it does not arise, is that the two numbers it used are a function’s value and its first two derivatives. At an instant, a function’s slope and curvature are free. Over an interval they are the slope and curvature of one curve, and a curve cannot have arbitrary slope and curvature at every point of an interval independently.

Where the curvature stands still found the coupler points whose path curvature has momentarily stopped changing, another condition met at an instant, by a locus of points that moves as the crank turns. Conditions at an instant are cheap in this field. A single condition at a single point is cheap. The same condition held along a stretch is a differential equation, and a differential equation with two unknown functions and one free function has no solution in general. The third-order pole goes anywhere at an instant. Over a stretch it goes where the drive’s one function lets it, which is almost nowhere a designer would choose.

What a designer can take from it #

Schedule for the jerk a schedule can remove, not the jerk a designer would like gone. A crank schedule reaches a coupler point’s jerk through one function. The best one takes a fifth to a third off the root-mean-square jerk at the points tried, and the steady crank’s jerk sets the scale of what remains.

Choose the point where the mismatch has a zero. The best schedule’s jerk dips almost to nought near angles where the mismatch vanishes. A point whose mismatch is nought where the motion’s critical stretch lies, at the moment a film claw enters the film or a gripper meets a part, can be made nearly jerk-free there, briefly. That is a point-placement question, answered by the mismatch curve rather than the earlier essay’s instant. On this four-bar the usual coupler point’s mismatch vanishes at 39.8°, the higher point’s at 46.8°, and the point nearest the inflection circle’s at 44.8° and 133.8°; the point on the line of the pins has no such angle on the stroke. At the usual point the best schedule’s jerk falls to 0.020 at 32°, against the steady crank’s 0.56 there, a twenty-eighth, at the stroke’s start where the mismatch is passing through nought.

To do better, change the linkage. What remains after the best schedule is the linkage’s own. The straightest point there is found the coupler point whose path is straight to the highest order at an instant. The corresponding stretch question, which point’s jerk the linkage itself keeps small, is a synthesis problem rather than a scheduling one.

What the schedule does not settle #

A whole turn. The stroke here is 120° of a turn. A crank that repeats needs a periodic schedule, and the speed that rises by half across the stroke has to come back over the rest of the turn, putting jerk there instead. Over a full turn the best periodic schedule would do less. A speed that rises by half across 120° must fall by the same factor over the remaining 240°, at about half the rate of change, and every coupler point’s jerk on that part of the turn pays for it, whether or not the machine is doing useful work there.

Inertia and the drive’s own jerk. A crank whose speed rises by half needs a torque to accelerate it, and the drive that supplies the schedule has its own jerk. Neither is weighed.

Size. The jerks here are per crank speed cubed on the standard four-bar, whose crank is one unit long. A curvature is a size with a minus sign found that every curvature in this field scales with the linkage, and a jerk scales with it too, in proportion to its size. The shares the best schedule removes do not depend on size, since every term of the jerk scales alike, so they transfer to a machine of any size. The jerks themselves do not.

Other measures. The schedule minimises the root mean square of the jerk. A designer limited by the largest jerk, or by its rate of change, would choose a different schedule, and the share it removes would differ.

Still open: the coupler point a linkage keeps smoothest #

The schedule’s limit puts the question back on the linkage. Every coupler point has a steady-crank jerk along the stroke and a mismatch, and between them they say how smooth a scheduled drive can make it. A designer free to choose the point would pick the smoothest.

The distinct argument would be the map of that choice over the coupler plane. For each point of a grid on the coupler, it would record the best schedule’s root-mean-square jerk over the stroke. It would find the minimum and ask whether it lies near Ball’s point, where the path is straightest at an instant, or somewhere the instant gives no hint of. The measurement would also compare the best point’s scheduled jerk with its steady jerk. It would say whether choosing the point and then scheduling the crank is better than either alone, and by how much.