Every change point lies flat

A length error is undone by its own size measured what it takes to repair a parallelogram four-bar that has been built slightly wrong. A parallelogram sits at a corner of length space where two walls meet, so any error pushes it off both; at its flat positions its two circuits then either pass each other a small angle apart or stall the input short of the flat angle, both by amounts proportional to the square root of the error. A pin clearance restores the choice between circuits, and the clearance needed turned out to be the length error exactly — not about δ, but δ, at every error size from 10⁻² to 10⁻⁸ — and the same at any of the four bearings.

That essay was careful to say why the parallelogram might be special. At its flat position all four bars lie along one line, so the loop’s shortfall there is a plain difference of lengths and every pin’s displacement lies along the same direction. It predicted that an ordinary change point would be different: a linkage on a single wall, where the circuits meet with the bars not stacked on each other, would need a clearance with a geometric constant in front, and would make the four pins unequal — which would let a designer identify the loose bearing and say which one to open.

Both predictions are testable on a family of linkages, and both fail. The failure has a single cause that is short enough to state first: every change point lies flat.

A family on one wall #

Grashof’s boundary is the set of four-bars whose shortest and longest bars sum to the other two — a test on shape alone, unchanged by any scaling. The family used here puts the crank as the shortest bar and the ground as the longest, $g + a = b + c$, with $g = 4$ and $a = 1$, and lets the coupler and output share their five units in any proportion. At one end, coupler 4 and output 1, the linkage is a parallelogram; at the other, coupler 1 and output 4, a kite. Everything strictly between sits on this wall and no other, with no two bars equal.

Each such linkage has one input angle at which its two assemblies meet, $\theta = \pi$, and there its output angle is $\psi = \pi$ as well. The first figure draws the two circuits of the linkage with coupler and output both 2.5 near that position, with the crank a thousandth short. Built exactly the two curves would cross at the origin; built wrong they pass each other, 0.0400 radians apart at the flat input angle.

Why the flat position is flat #

The collinearity the parallelogram essay treated as a special feature is in fact forced.

Two assemblies of a four-bar meet where they coincide, and that happens where the coupler’s circle about the crank pin and the output’s circle about its pivot touch rather than cross. Touching circles need their centres exactly $b + c$ or $|b - c|$ apart, so at a change point the crank pin is $b + c$ or $|b - c|$ from the output pivot.

How far the crank pin is from the output pivot depends only on the crank angle, and it ranges from $|g - a|$ with the crank along the ground towards the pivot to $g + a$ with the crank pointing away. On a Grashof boundary one of those two extremes equals $b + c$ or $|b - c|$ — that is what the boundary condition says, rearranged — and no intermediate distance can, because the boundary is a single equation in the lengths. So a change point needs the crank along the ground line, and the coupler and output along the same line to touch there.

All four joints are collinear at every change point of every four-bar. The parallelogram is not special in that; it is special only in having two such positions, one at each end of its crank’s travel, because it sits on two walls.

So the threshold is the error, exactly #

The pin clearance that restores a change point is found the same way as on the parallelogram. The loop’s span residual — how far apart the coupler’s two pins are, less the coupler’s length — is computed at the flat input angle as a function of the output angle. Its zeros are the circuits. A total radial clearance at the pins can absorb any residual no larger than itself, so the circuits join when the clearance covers the residual everywhere between them, or, when the crank is stalled and there are no zeros, when it covers the least the residual ever gets to.

At the flat output angle the joints are collinear, so the span is a signed sum of lengths — here $g + a - c$ — and the residual is $g + a - b - c$, which is nought on the wall and exactly $\pm\delta$ after an error of δ in any one length. And because the configuration is collinear, the span is at an extreme there: with the output folded back along the line towards the crank pin, turning it either way can only carry its pin further from the crank pin, so the span is at its least. The residual is $\pm\delta$ at its extreme, and the extreme is the threshold.

Measured on three linkages with couplers of 1.5, 2.5 and 3.5, the residual’s dip is −1.000000 × 10⁻³ in all three, to every digit printed. The dips differ in width — 0.026, 0.040 and 0.061 radians — and the width is not what a clearance has to cover.

Over all eight single-length errors, at errors of 10⁻³ and 10⁻⁶, on the same three linkages, the least clearance that rejoins the circuits or releases the stall is the error to within $1.4 \times 10^{-10}$ of itself. There is no geometric constant in the threshold anywhere on the wall.

The four bearings are still alike #

The second prediction was that the four pins would stop being worth the same once they were no longer symmetrically placed.

They are placed asymmetrically: the output pivot is 4 from the crank pivot, the crank pin 1 from it on the other side, the coupler pin 1.5 from the output pivot. Each is given a play of $4 \times 10^{-4}$ and its disc searched, with no rule telling the search what to expect, and each leaves the loop exactly $6.0000 \times 10^{-4}$ from closing: the shortfall less the play.

The reason is the same collinearity. A pin’s displacement changes the span by its component along the line joining the coupler’s two pins; at the flat configuration all four joints are on that line, so a displacement of a given size along it moves the span by the same amount at any of them. Away from the flat configuration the pins would differ, but that is not where a clearance is needed.

So the conclusion the parallelogram essay reached with a caveat holds without one: as far as which configurations it can reach, a four-bar at a change point cannot tell which of its bearings is loose.

The constant moved to the angle #

Something does change along the wall, and the first figure’s dial shows it: the circuits’ separation at the flat angle is 0.026 radians on one linkage and 0.061 on another, with the same error and the same threshold.

The laws are a law of cosines each. For a gap — an error that lowers $g + a - b - c$ to $-\delta$ — the crank pin at the flat angle is $b + c - \delta$ from the output pivot, and the output’s angle away from the ground line satisfies

$$\cos\varphi = \frac{c^2 + (b + c - \delta)^2 - b^2}{2c\,(b + c - \delta)} \approx 1 - \frac{b\,\delta}{c\,(b + c)},$$

so $\varphi^2 = 2b\delta / c(b + c)$ and the two circuits, one either side of the line, are $2\varphi$ apart:

$$\text{gap} = 2\sqrt{\frac{2b\,\delta}{c\,(g + a)}}.$$

For a stall — an error that raises it to $+\delta$ — the crank pin is too far away at the flat angle, and turning the crank back by $x$ brings it nearer by $ga\,x^2/2(g + a)$. The crank reaches the first assembly when that covers δ:

$$\text{stall} = \sqrt{\frac{2(g + a)\,\delta}{g\,a}}.$$

Measured from the exact closure at an error of $10^{-8}$ across eight linkages on the wall, both constants lie on their closed forms to $10^{-7}$.

The stall’s constant has no coupler or output in it. It is $\sqrt{2(g + a)/ga} = 1.5811$ at every point of this wall, because a stall is the crank failing to bring its pin close enough, and how fast the pin approaches as the crank turns back is a property of the crank and the ground alone.

The gap’s constant grows as $\sqrt{b/c}$, from 0.73 near the kite to 2.38 near the parallelogram, and at the parallelogram itself it would be $2\sqrt{8/5} = 2.53$. The two cross where $b/c = (g + a)^2/4ga$, which is a coupler share of 0.610 here: short of that, a crank a little short costs less output angle than a crank a little long costs crank travel; past it, the other way.

How far the square root can be trusted #

A leading-order law is only as useful as the range of errors over which it holds, and for a change point that range is the whole of practice.

On the linkage with coupler and output both 2.5, the gap measured from the exact closure — the difference of the two circle-intersection angles at the flat input — against the closed form differs by $1.7 \times 10^{-6}$ of itself at an error of $10^{-4}$, by $1.7 \times 10^{-5}$ at $10^{-3}$, and by $1.7 \times 10^{-4}$ at $10^{-2}$. The relative departure is almost exactly δ/60, growing in proportion to the error, which is what the next term of the expansion predicts. A four-bar a hundredth of its crank length wrong — a gross manufacturing error on a crank of any real size — still has its gap given by the square-root law to two parts in ten thousand.

That matters for the argument rather than for the arithmetic. The claim that the constant has moved from the clearance to the angle could have been an artefact of looking only at tiny errors. It is not: at every error a machine could plausibly be built with, the clearance is δ exactly and the angle is the square root with its geometric constant.

The other two walls #

The family above sits on one of the three walls of length space — the one where the crank and the ground together equal the coupler and the output. A four-bar can also be on a boundary where the crank and the coupler balance the ground and the output, or the crank and the output balance the ground and the coupler, and the collinearity argument claims to cover all three.

It was checked on one linkage from each of the other two, with nothing borrowed from the family. A linkage with ground 3, crank 1, coupler 4 and output 2 satisfies $g + c = a + b$; its circuits meet with the crank pointing at the output pivot, at $\theta = 0$ rather than $\pi$, where the crank pin is $g - a = b - c$ from the pivot and the coupler folds back over the output. All eight of its single-length errors of $10^{-5}$ give a threshold of 1.0000 δ there. A linkage with ground 2, crank 1, coupler 3.5 and output 2.5 is on the same wall with different proportions and gives the same. And a linkage with ground 3.5, crank 1, coupler 2 and output 2.5 is on the first wall with its change point at π, and gives the same again.

At each of those linkages’ other flat angle the loop is nowhere near closing: the residual there is tens of thousands of times the error, because it is a position the linkage simply does not pass through. That is the single-wall linkage’s other difference from the parallelogram, seen from the clearance’s side: one flat angle is a change point and costs δ, the other is not one at all.

The parallelogram is the worst place on its wall #

That answers the question the parallelogram essay posed about a worst case, though not in the form it expected. A given manufacturing error does not demand more clearance anywhere on the wall — the clearance is the error everywhere. What varies is how much choice the error takes away before the clearance is supplied, and it takes the most where $b/c$ is largest, which on this wall is its end: coupler equal to the ground and output equal to the crank.

Of all the linkages on one Grashof boundary, the one that loses the most output angle to a given length error is the parallelogram at its end. A designer who wants a change point that degrades gently as it is built wrong should move away from the corner, towards a longer output and a shorter coupler — and it costs nothing in the clearance needed to restore it.

The leverage the parallelogram essay described, the output angle a unit of clearance buys back, is the gap divided by δ, $2\sqrt{2b/c(g + a)\delta}$. It grows as the linkage is made more precisely, as it did there, and it is largest at the corner for the same reason.

Eight errors, one change point #

A parallelogram has two change points, and each of its eight single-length errors produces a gap at one and a stall at the other. A linkage on one wall has one, and the table of errors is correspondingly simpler.

The four errors that lower $g + a - b - c$ — ground short, crank short, coupler long, output long — open a gap of 0.012649 radians; the four that raise it stall the crank by 0.015811. Every measured value matches the law to the six figures printed, and every threshold is 1.000000 δ. At the other flat input angle, $\theta = 0$, nothing happens: the crank pin is at its nearest to the output pivot, far from any tangency, and the linkage passes through an ordinary position.

That is also the sense in which branches are components on this family. The exact linkage’s configuration curve has one crossing, not two; an error either separates it into two circuits, never three, or leaves one circuit with a crank that cannot turn fully — and a clearance of δ undoes either.

What this does not settle #

Nothing is loaded. As in the parallelogram essay, what is measured is which configurations are reachable; a mechanism under load occupies one of them, and which one depends on the forces at each pin.

Only single-length errors are measured. An error in two lengths that keeps $g + a - b - c$ at nought leaves the linkage on the wall, and its change point survives with no clearance at all. A general error is a sum of single ones as far as $g + a - b - c$ is concerned, so the threshold is the net change in that sum, which can be much smaller than any individual error.

The joints are ideal discs. A worn or out-of-round bearing is not a disc and does not sum like one.

Still open: the clearance that makes a framework generic #

A four-bar has one loop and one residual, which is why the collinear argument settles everything. Nine bars that ought to be rigid is a framework with a dependency among many loops, which moves only because its joints lie on two perpendicular lines, and the right angle as a tolerance measured how much clearance lets a tilted version move.

Its distinct argument would be the opposite effect of clearance on that framework. Clearance absorbs the misfit a tilt produces, but it also moves the joints off their lines into general position, where a generic placement has no freedom at all. Two things would come out of it. Whether, for joints placed randomly within their clearance discs, the smallest singular value of the constraint matrix stays small enough to call the framework a mechanism with a tolerance or grows large enough to call it a structure with a compliance; and at what clearance, relative to the tilt, one description gives way to the other — which is the multi-loop version of the question answered here in one line, and one where no single collinearity is available to settle it.