A length error is undone by its own size

A parallelogram a micron wrong measured what a change point costs once it is built. A parallelogram four-bar sits on two walls of length space at once, so it cannot be made: change any one of its four lengths by any amount δ and it falls into one of four neighbouring regions. At each of its two flat positions it then either stalls, with the input turning back short of the flat angle, or opens a gap, with the input passing through and the output carried from one motion to the other. Both are square roots of δ. A parallelogram whose input is short by a thousandth has its two circuits 0.1095 radians apart at the flat position — six degrees, from an error of one part in two thousand.

That essay ended by proposing a repair and predicting its size. The square-root separation is what an exact linkage loses; but the exact parallelogram, with its crossing and its choice, is only δ of length away. So a pin clearance of about δ should be enough to reach it, and should therefore restore the change point — a linear cure for a square-root loss.

It is not about δ. It is δ — and the same δ that a third crank charges the same linkage for keeping it out of trouble a different way, which is the coincidence this essay ends on.

Play as a budget on one number #

A revolute joint with radial clearance ρ is a pin of one radius in a hole of a slightly larger one, and what it permits is that the two bodies it joins put their joint centres anywhere within ρ of each other. To find what a linkage with such joints can do, that permission has to be written into the loop.

It goes in additively. The four bars are rigid vectors and the four clearances are four more vectors, each of length at most ρ, and the loop closes when the nine of them sum to nothing. So for an input angle θ and an output angle ψ, define the span residual — how far apart the two coupler pins are, less the coupler’s own length:

$$r(\theta, \psi) \;=\; \bigl|\, (g + c\cos\psi,\; c\sin\psi) - (a\cos\theta,\; a\sin\theta) \,\bigr| \;-\; b .$$

The exact linkage’s configurations are the zeros of that. With play, the configuration is reachable when the residual can be absorbed:

$$|r(\theta, \psi)| \;\le\; \text{the sum of the plays at the pins that have any.}$$

Two things follow from the shape of that statement before anything is measured, and the second is the one worth carrying.

Only the total play enters. Four pins with ρ each do exactly what one pin with 4ρ does, because the four displacement vectors appear in the loop only through their sum.

And displacing any one of the four pins changes the span by that displacement’s component along the line joining the two coupler pins — so the four pins are worth the same. That is not obvious in advance: the ground bearings sit at fixed pivots and the moving pins do not, and it would be reasonable to expect the play at a pivot to be worth more or less than the play at a pin. It is worth the same, and the reason is that a loop equation is a sum and cannot see which of its terms is which.

This is the same object a clearance is a link is about, treated as a member of the chain rather than as an imperfection of one. There the clearance at a joint is given its own two freedoms and counted; here it is given its own displacement and summed, and the two readings agree about what it is — a small, bounded amount of extra reach placed at a named point of the loop.

What is new is that a change point is the one place where a small amount of extra reach does something a small amount of extra reach normally cannot. Away from a flat position the residual’s zeros are simple crossings, and widening the tolerance by ρ widens the reachable band by an amount proportional to ρ: the play buys play. At a flat position the two zeros are about to collide, and the amount of output angle between them is not proportional to anything — it is a square root. The clearance buys the square root’s argument, and the square root is what the mechanism does with it.

The hump, and how tall it is #

At the flat input angle the residual is a function of the output angle alone, and it has a shape that says the whole answer.

The two zeros are the linkage’s two circuits, 0.1095 radians apart. Between them the residual is positive — the two coupler pins are further apart than the coupler can reach — and it has one hump. The play has to lift the budget over that hump, and the hump’s height is 1.000000 × 10⁻³.

That is the length error, to every digit the computation carries. The reason is the one the prediction rested on: at the flat position all four bars lie along one line, so the span at the flat output angle is $g + c - a$, the coupler’s length is $b$, and the difference is exactly whatever was taken off the input. The exact parallelogram is δ of length away, and the residual is the distance to it.

So the threshold is not an estimate with a constant in front of it. It is the error.

Six decades, and a leverage that grows #

Measured over six decades of error, with the threshold found by locating the two zeros and maximising the residual between them rather than by any expansion, the ratio of threshold to error is one to nine figures at every decade.

Two lines with different slopes is the whole point, and it is worth reading the consequence out rather than leaving it in the picture. At an error of 10⁻², a play of 10⁻² buys back 0.3464 radians of output — 34.6 radians per unit of play. At an error of 10⁻⁸, a play of 10⁻⁸ buys back 3.46 × 10⁻⁴ radians, which is 34,641 radians per unit of play. The leverage is $2\sqrt{3/\delta}$ and it grows without bound as the linkage is made better.

That inverts the intuition a tolerance usually comes with. The finer a parallelogram is made, the less clearance it needs to keep its change point, and the more of the lost choice each unit of clearance returns. A linkage held to a micron needs a micron of play; a linkage held to ten microns needs ten, and gets less for it.

Sharing the play among the four pins divides what each must carry by four and changes no slope, which is the first consequence of the budget rule seen as a design fact: a designer distributing 10⁻³ of total play as 2.5 × 10⁻⁴ at each of four bearings gets exactly what one loose bearing of 10⁻³ would have given.

It is also the reading that explains why the threshold has no constant in front of it while the separation has $2\sqrt{3}$ in front of its square root. The threshold is a statement about lengths and the loop is a statement about lengths, so the two are the same kind of quantity and the answer comes out as a length with no conversion. The separation is a statement about an angle, obtained from a length by a geometry that has the linkage’s own proportions in it, and $2\sqrt{3}$ is where $g$, $a$ and $c$ went.

Just under the threshold #

Below the threshold the two circuits are still two, and how far apart they are is a second square root.

The residual near its top is $\delta - \psi^2/3$, so a budget of ρ reaches out to $\psi = \sqrt{3(\delta - \rho)}$ either side and the gap left is $2\sqrt{3(\delta - \rho)}$ — the same square root as before, taken of what the play is short by rather than of the error. Measured against that law at four fractions of the error the agreement is to within 3 × 10⁻⁵ throughout.

The practical reading is unpleasant and worth stating plainly. The last thousandth of the play closes thirty-two times more of the gap than the first tenth does, so a parallelogram given nine tenths of the clearance it needs is not nine tenths repaired. It still has two circuits, still cannot change between them, and still has a tenth of its original separation left — 0.011 radians on the linkage drawn here, which is most of a degree and is plenty to be visible in a mechanism’s output.

There is no partial credit near a change point. It is restored or it is not, and the approach to restoration is a square root, which means the last approach is the steepest.

Which bearing is the loose one #

The budget rule says the four pins are worth the same, and that follows from an argument about a sum. An argument about a sum is exactly the kind of thing that is right and worth checking anyway, because it has assumed something about what a pin’s displacement can do.

The check does not use the rule. Each pin is given its disc, the loop is re-formed with that pin displaced, and the residual is measured — over every radius and every direction, without the search being told what it should find. The linkage is a thousandth short and the flat position is 1.0000 × 10⁻³ from closing; each of the four pins with 4 × 10⁻⁴ of play leaves 6.0000 × 10⁻⁴, which is the shortfall less the play; and the four agree with each other exactly.

The search is worth having for a second reason. It is the kind of claim two routes to a sensitivity exists to insist on: a derivative computed two ways, one of them not knowing what the other expects. Here the budget rule is an argument about a sum and the disc search is an argument about nothing at all, and the second is capable of disagreeing at any of the four pins and does not.

So a four-bar cannot tell which of its bearings is loose, as far as which configurations it can reach is concerned. That is a narrower statement than it first sounds and the narrowing matters. A loaded mechanism does not take an arbitrary reachable configuration: it takes the one the forces put it in, the pin sits against one side of its hole, and which side depends on the direction of the load at that joint. Two linkages with the same total play distributed differently reach the same set and travel through it differently, and nothing here computes the second. What is settled is the reachability, which is what a change point is a question about.

A gap and a stall want the same play #

A parallelogram has four single-length errors and two flat positions, and the corner it sits on has four regions around it. At a given flat position, two of the four errors open a gap and two stall the input.

The costs are different from each other. A gap separates the circuits by 0.1095 radians; a stall keeps the crank out of 0.0365 radians of its own turn, which is three times smaller and is a loss of a different kind — travel rather than choice. A linkage locked on purpose is a mechanism using a flat position as a feature; this is the same position arriving as a fault.

The play that undoes them is 1.000 × 10⁻³ in all four cases. The reason is the same sentence read twice. At a gap, the flat input angle has two output angles and the residual between them peaks at δ. At a stall, the flat input angle has no output angle at all and the residual never gets closer to nought than δ. Either way the loop is δ from closing at the flat position, because either way the exact parallelogram is δ of length away, and the play does not care which side of the wall the linkage fell.

The distinction between the two losses is the one the branches essay draws. A stall does not change how many components the configuration space has; it shortens the one the linkage is on. A gap is about whether two components are one, and the change point is exactly the configuration at which they would have touched. So the play is buying two different things for the same price — a piece of a circuit in one case, and a junction between circuits in the other — and it is only by measuring both that they turn out to cost the same.

The number a third crank charges #

There is one more mechanism that meets this number, and it arrives from somewhere else entirely.

The mechanism Grübler says cannot move is the three-crank parallelogram, where a third crank is added to carry the coupler past the flat positions. A three-crank chain cannot take the crossed motion — the crossed configuration is not available to all three cranks at once — so a length error there cannot be spent on a gap or a stall. It has to be spent on a misfit: the largest amount some bar would have to stretch to keep the chain assembled over a turn.

For the linkage here the misfit is $\sqrt{g^2 + 2g\delta + \delta^2} - g$, which is $\delta$ exactly, at every crank angle where it is largest. So the three-crank chain’s misfit and the two-crank chain’s clearance threshold are the same quantity, and not merely the same order.

That is a sharper statement than either essay could make alone, and it says something about what a redundant constraint costs. A designer facing a parallelogram that has to pass its flat positions has two ways out. Add a third crank, and the price is δ of misfit taken up somewhere in the chain’s own elasticity or clearance. Leave two cranks and open the bearings, and the price is δ of clearance. The same length error is paid for with the same length either way, and what differs is where it is paid: as a fit that is wrong all the way round in one case, and as a looseness that is only needed at two positions in the other.

What this does not settle #

Nothing here is loaded. Every configuration counted is one the linkage can reach, and a mechanism under load occupies one of them rather than all. Which one, and therefore what the output actually does as the input passes the flat position, depends on the direction of the force at each pin. The play that makes the change point available is what is measured; the play that makes it happen the way a designer wants is not.

The linkage is one loop. A four-bar has one loop and one residual, which is why a play budget is a single number. A framework with a dependency among nine bars has nine residuals and no single scalar to compare a budget against, and how a clearance shared among several joints behaves there is a different question with a different shape.

The bearings are perfect except for their size. A pin in a hole is treated as a joint centre free within a disc, with no friction, no tilt, no out-of-round and no wear. A worn bearing’s clearance is not a disc, and a bearing carrying a steady load has its own contact geometry.

One parallelogram. The threshold is the length error for the linkage drawn here and for the four single-length errors at both of its flat positions, over six decades. That the constant in front is exactly one is a fact about the parallelogram’s flat position, where all four bars are collinear and the residual reduces to a difference of lengths. A general change point, on a linkage that is not a parallelogram, sits on one wall rather than two and its residual has a different shape; what the constant becomes there is not computed.

The play is radial and equal. Four pins with the same clearance, or one pin with all of it. A real linkage has different fits at different joints, and the budget rule says the total is what counts — but it says so under the same reachability reading, and the case where one bearing is orders of magnitude looser than the others is where the load argument above would bite hardest.

Still open: the same measurement on a general change point #

A parallelogram is the extreme case: two walls at once, four regions around it, and every one of its bars along one line at the flat position. That last is what makes the threshold exactly the error rather than some multiple of it, because a collinear loop’s residual is a plain difference of lengths and every pin’s displacement lies along the same line.

Its distinct argument would be the same measurement made on an ordinary change point — a linkage on a single wall of length space, where the four bars are not collinear when the two circuits meet. Two things would come out of it. The threshold would acquire a constant, and that constant would be a function of the angle between the bars at the change point, so there would be a worst case: a geometry at which a given manufacturing error demands the most play. And the four pins would stop being alike, because a displacement’s usefulness is its component along the coupler’s own direction and the four pins no longer share a line — which would make the loose bearing identifiable after all, and would say which bearing of a mechanism is the one to open.