A drive that goes round is an angle

Dixon’s framework is nine bars joining three joints on one line to three on a line square to it, which Grübler’s count calls a structure, and nine bars that ought to be rigid found that it moves. Its motion runs round a loop, and four times on the way round a joint of one line passes through the other line: a crossing, where the framework passes from one family of Dixon’s placements to the next. A push turns back at the crossing found what each kind of crossing does to a drive. Where $W_2$ reaches the black line, every joint but $W_2$ moves as the square of the distance along the motion, so a push on $B_2$ is at a fold and cannot carry the framework through, while $W_2$’s own side of the black line passes straight through. Where $B_2$ reaches the white line the roles reverse.

So neither of those drives takes the framework all the way round. That essay proposed one that would seem to: a lever moving $B_2$ and $W_2$ together, $B_2$’s push plus some multiple of $W_2$’s side, which has a nonzero rate at every one of the four crossings. It asked what play such a drive would need to take a framework as built all the way round, and whether going round costs more than its hardest crossing.

The lever cannot go round, and the reason is not about crossings at all. A drive that does go round exists, but it is not a push of any kind. It is an angle.

The loop is a circle #

Dixon’s placements move each black joint from $x_i$ to $\sqrt{x_i^2 + t}$ along its line and each white joint from $y_j$ to $\sqrt{y_j^2 - t}$ along its own, and every bar keeps its length because each bar’s length squared is $x_i^2 + y_j^2$. In this framework $B_2$ starts at 1 and $W_2$ at 1, the nearest joints to the crossing point on each line, so the motion stops being given by the formula when $t$ reaches 1, where $W_2$ arrives at the black line, or −1, where $B_2$ arrives at the white. Beyond each of those the next family takes over, with that joint’s sign flipped.

Joined up, the four families are one loop, and it has a simple coordinate. Put $t = \cos 2\psi$. Then

$$B_2 = \sqrt{1 + t} = \sqrt2\cos\psi,\qquad W_2 = \sqrt{1 - t} = \sqrt2\sin\psi,$$

with the signs that the four families supply, and as ψ runs from 0 to 360° the framework goes once round, through $W_2$’s crossings at 0° and 180° and $B_2$’s at 90° and 270°. Every other joint is a smooth function of $\cos 2\psi$. And $B_2$ and $W_2$ are the two ends of the bar between them, of length $\sqrt2$, with each end on one of two perpendicular lines. That is the elliptic trammel, and the bar turns exactly once as the framework goes once round. Checked against Dixon’s own placements at 64 points round the loop, the trammel’s formulas hold to $10^{-12}$ and no bar’s length moves by more than $2 \times 10^{-15}$.

This is what the earlier essays saw piecemeal. A crossing is the trammel bar lying along one of its two lines. The four families are its four quadrants. And the loop that nine bars that ought to be rigid found by following the bars alone until they came back, with no formula at all, is in this coordinate simply a circle, which is why following it eventually returned to the start.

Every drive that is a number folds twice #

A push on $B_2$, $W_2$’s side of the black line, and any lever between them each reduce the framework’s whole configuration to a single number. Round the loop each is a smooth function of ψ, and a smooth function on a circle has a largest value and a smallest, at least one of each. At those two places the drive is stationary: moving the framework a little either way changes it only to second order. That is a fold, and it is the same thing that stopped the push at $W_2$’s crossing. A drive that can only increase stops at its top.

The push has its top and bottom at $W_2$’s two crossings, 0° and 180°, which is the fold the earlier essay found, seen from the whole loop. The side has its top and bottom at $B_2$’s crossings, 90° and 270°. A lever has both its folds somewhere between: push plus side folds at 34.1° and 239.3°, push plus four times the side at 72.4° and 258.6°, push plus a quarter of it at 11.0° and 208.9°, push minus the side at 120.7° and 325.9°. In every case there are exactly two, one top and one bottom.

If the push were simply $B_2$’s coordinate on its line, $\sqrt2\cos\psi$, a lever would be $\sqrt2(\cos\psi + k\sin\psi)$, a sinusoid whose top is at $\arctan k$: 45° for push plus side, 76° for four times the side, 14° for a quarter of it. The measured folds sit near those, a few degrees short, because a push measured from the pinned $B_1$ carries more than $B_2$’s coordinate. $B_1$ moves along the black line too, as $\sqrt{4 + t}$, and with $B_1$ pinned and its bar to $W_1$ held in direction, the black line itself turns as the framework moves. Those are second-order corrections to where the folds are, and nothing at all to whether they are there.

So the lever does exactly what was claimed for it at the crossings. Its rate is nonzero at all four. It has simply moved its folds into the open stretches between them, where nothing about the framework is singular and the fold is a property of the drive alone. At such a fold the framework is at the drive’s dead centre, as a slider-crank driven from its slider is at the top of the stroke and a toggle locked on purpose is at its own, and pushing harder accomplishes nothing. No choice of lever ratio removes the folds, because the reason for them is that the loop is closed: a number that went up all the way round would have to come back to where it started higher than it began.

Four angles turn round #

What goes round a circle without folding is an angle. The framework has eighteen of them that a motor could drive: at each of its six joints three bars meet, and any two of them make an angle that a rotary actuator mounted at the joint could turn.

Fourteen of the eighteen rock. Each swings back and forth and comes back to where it started, its rate passing through nought twice, so each is a number in the sense above and folds twice. Four turn a whole revolution: at $B_2$, the angle from $B_2W_1$ to $B_2W_2$ and the angle from $B_2W_2$ to $B_2W_3$, and at $W_2$, from $W_2B_1$ to $W_2B_2$ and from $W_2B_2$ to $W_2B_3$. Every one of the four has the trammel bar $B_2W_2$ for one of its sides, which is why it turns: the bar turns once and the other side, a bar that only rocks, cannot undo it. And none of the four ever stops. Their rates stay between 0.22 and 1.78 in size, never crossing nought.

The steadiest of the four is the angle at $W_2$ from $W_2B_2$ to $W_2B_3$. Its rate is least, 0.553, at 0° and greatest, 1.447, at 180°, both at $W_2$’s crossings, and it passes through the two $B_2$ crossings at exactly 1. Those extremes are $1 - 1/\sqrt5$ and $1 + 1/\sqrt5$. The bar $W_2B_2$ turns at exactly one radian per unit of ψ all the way round, since its direction is ψ itself; the bar $W_2B_3$ only rocks, and it rocks fastest when $W_2$ is crossing the black line, where $W_2$ moves along its line at $\sqrt2$ and the bar to $B_3$ is $\sqrt{10}$ long, turning it at $\sqrt2/\sqrt{10} = 1/\sqrt5$, against the trammel bar at one crossing and with it at the other. A motor turning that joint at a steady rate carries the framework round at a speed that varies by a factor of 2.62 over a turn, through all four crossings, without a fold anywhere. That is the drive the earlier essay asked for, and it is a rotary motor at one joint, not a lever between two.

What going round costs #

A framework as built, with its joints a little out of place, has no motion of its own: the clearance that makes the framework generic found that its play has to carry it, and the price is one self-stress measured that play as a price. With a driver held, the bars’ constraints at each point of the motion have a single self-stress, and the least play that lets the framework be at that point, per unit of the joint errors ρ, is one inner product of the errors with it.

With the motor’s angle as the driver, the price is a smooth curve round the whole loop for every one of the ten frameworks, finite everywhere including the four crossings, where the angle passes through as the side drive did. Every curve is nought at 45°, where the framework was built: there the exact placement is the as-built one and nothing needs to give. At $W_2$’s first crossing the price agrees with the side drive’s crossing price from the earlier essay to the digit, 0.0013 per unit ρ for framework 1 and 0.1403 for framework 2, which is a check on both: two different drivers, held at the one place where both pass, ask the same of the bars.

Every price curve starts from nought at 45° and climbs as the motor turns the framework away from where it was built, in either direction, and how fast it climbs is set by the joint errors. The ten curves are ten different shapes because the ten frameworks’ errors load the self-stress differently at each point of the loop: the stress itself changes as the framework moves, and so does which of the errors it weighs most. Framework 7 climbs steadily from 0.002 at 48° to 0.126 at 128° and peaks at 0.142 at 163°; framework 1 never needs more than 0.087 anywhere on the loop, and framework 2 needs five times that.

Away from the crossings the curves rise and fall with no regard for them. The worst point on the loop is within six degrees of a crossing for none of the ten frameworks: framework 1’s worst is 0.087 at 213°, framework 2’s 0.440 at 233°, framework 8’s 0.163 at 323°. The worst over the whole loop is larger than the worst of the four crossings by a factor between 1.002, for framework 9, and 1.459, for framework 8. So the answer to the earlier essay’s second question is yes, going round costs more than its hardest crossing, by up to half as much again, and the reason is that the hardest place on the loop is usually not a crossing at all. A crossing is where the framework’s family changes; it is not where its errors are loaded hardest.

Driven round, it settles in one turn #

The price is a first-order statement about where the framework can be. Whether a drive actually gets round needs the framework driven, quasi-statically, the way the earlier essays drove it: each step advances the motor’s angle a little and moves the joints by the least total amount that keeps the pins, reaches the new angle, and leaves every bar within its allowance.

Bisected on the allowance, the drive gets round frameworks 1, 2 and 3 with 0.0879, 0.4395 and 0.1563 per unit ρ, against loop worsts of 0.0869, 0.4398 and 0.1555: the least play that takes a framework round is its worst first-order price round the loop, to within one per cent, which is the step size’s resolution. At three per cent less than that worst, none of the three gets round.

The more surprising number is where the framework ends up. After one turn the motor’s angle is back where it started, but the joints are not: they end 0.21, 2.19 and 0.45 ρ from where they were built. The drive has used its play, and the least-motion rule, having pushed the framework into one corner of what the play allows at the hardest point of the loop, has no reason to bring it back to the as-built placement afterwards. Driven round a second and a third time, though, each framework ends where the first turn left it, to $2 \times 10^{-6}$ ρ or better. The first turn settles the framework onto a cycle, and every turn after repeats it. A framework as built and driven round is in a different place from the one it was assembled in, and stays there.

What a builder can take from it #

Put the motor at $B_2$ or $W_2$, on the trammel bar. Of the eighteen joints a rotary drive could turn, only the four angles that include the bar $B_2W_2$ take the framework round, and the steadiest of them varies its speed by 2.6 to 1. Nothing that pushes, pulls or levers can do it.

Budget the play for the loop, not for the crossings. The crossings are where the framework changes family, and the earlier essays measured them because that is where the mathematics is singular. The play a framework as built needs is set somewhere else on the loop, and is up to half as much again.

Expect the first turn to move it. A framework assembled with errors and then driven round finds its own cycle in one turn, a few ρ from where it was built. A measurement taken before the first turn and after it will disagree by that much, and one taken after any later turn will not.

What the motor does not settle #

Inertia and friction. The drive is quasi-static and the least-motion rule stands in for whatever a real framework’s joints do with their play. A motor driving a real framework fast would carry it through a lever’s fold by momentum, as a flywheel carries a slider-crank through its dead centres; the claim here is only that a quasi-static drive cannot.

Other frameworks. The trammel is this framework’s: $B_2$ and $W_2$ are the joints nearest the crossing point on their lines and both start at 1, so the loop is a circle of radius $\sqrt2$ in their coordinates. A Dixon framework whose nearest joints start at different distances has the same construction with an ellipse in place of the circle, and nothing here has checked that the same four angles turn round on it.

Where the settled cycle lies. That every turn after the first repeats it is measured; what the cycle is, which corner of the play it sits in and whether a different least-motion rule would choose a different one, is not.

Still open: a framework whose loop is not a circle #

On this framework the two nearest joints start at equal distances from the crossing point, which makes the trammel bar’s ends move on a circle in ψ and puts the motor’s speed ratio at 2.62 with its slowest and fastest points at the crossings. If $B_2$ started at 1 and $W_2$ at 1.5, $t$ would run from −1 to 2.25, and $B_2$ and $W_2$ would be $\sqrt{1 + t}$ and $\sqrt{2.25 - t}$. Their squares would still add to a constant, so the two would still go round a circle, of radius $\sqrt{3.25}$, but the other joints would no longer sit symmetrically about it.

The distinct argument would be what that does to the drive. The bar $B_2W_2$ still turns once round, so the four angles that include it still turn; but their rates are no longer symmetric between the two kinds of crossing, and one of the four may come close to stopping where its other side is shortest. The measurement would take a family of Dixon frameworks with the ratio of the two nearest distances running from one to three, find the least rate of each of the four angles round the loop, and ask whether some ratio leaves no angle that a motor can drive at a speed ratio below a given value — the point at which a framework that moves stops being a framework a motor can usefully drive.