The clearance that makes the framework generic

Nine bars that ought to be rigid is Dixon’s framework. There are two sets of three joints, every joint of one set is barred to every joint of the other, and the count says nine bars on six joints leave no freedom. It moves anyway, all the way round a loop, when the three joints of each set lie on a line and the two lines are perpendicular. The right angle as a tolerance tilted the lines, took the motion away, and gave it back as a tolerance: the clearance each bar needs grows as the tilt times the square of the push.

That essay held the joints on their lines and asked what clearance lets them move. A clearance also does a second thing, and it works against the first. A pin in a hole can sit anywhere in the hole, so a framework built with clearance has its joints somewhere in a disc about where they were drawn, and six joints anywhere in six discs are not on two lines. The framework moves only because its joints are on two lines. A generic placement of these nine bars has rank nine and no freedom at all.

So clearance absorbs the misfit that stops the framework moving, and it destroys the coincidence that let it move. The question left open was which wins.

The framework as built #

Take the Dixon framework measured in the two essays before this one — joints at −2, 1 and 3 along one line, and at −1.5, 1 and 2.5 along a line square to it — and move each joint to a seeded point inside a disc of radius ρ about where it was drawn. Ten seeds give ten frameworks as built, each with its own nine bar lengths.

Each of the ten has rank nine, so by rank each is a structure. The size of the rigidity is the smallest singular value of the nine-bar Jacobian, $\sigma_{9}$. On the drawn framework it is nought, since there is a direction the joints can move without changing any bar to first order, and that direction extends to a finite motion only on the square framework. On the frameworks as built $\sigma_{9}$ is not nought, and it is proportional to ρ: $\sigma_{9}$/ρ is the same at ρ = 10⁻⁵ and ρ = 10⁻³ to a per cent at every seed.

What it is not is the same from one seed to the next.

Why each one is as rigid as it is #

A nine-bar framework on two sets of three joints has a first-order flex exactly when its six joints lie on one conic. That is Bolker and Roth’s theorem about this graph, and two crossing lines are a conic, a degenerate one. So the size of a framework’s rigidity should be its distance from the nearest conic, and the number that measures that distance is the six-point conic determinant: the determinant of the six rows $[x^2, xy, y^2, x, y, 1]$, which is nought exactly when some conic passes through all six points.

$\sigma_{9}$ runs from 0.0126ρ to 0.2794ρ across the ten seeds, a spread of twenty-two, set by which way each joint happened to move. The ratio of $\sigma_{9}$ to the conic determinant is 3.69 × 10⁻⁴ at every seed, to 0.3%. The singular value comes from the bars’ directions and the determinant from the joints’ coordinates, with no computation in common. Their ratio is a constant of the nominal framework. So the rigidity of the framework as built is its distance from a conic, and a framework whose errors happen to keep the six joints nearly on some conic — not the two lines, but any conic — is nearly as free as the one drawn.

The constant ratio also explains the twenty-two-fold spread. Six joints that went wrong are described by twelve numbers, and lying on some conic is a single equation in them. To first order, then, eleven independent combinations of joint errors leave the six joints on a conic — a slightly different conic from the two lines, but a conic — and keep the first-order flex. Only one combination, the gradient of the determinant, costs rigidity. A random error has a random component along that one direction, and $\sigma_{9}$ is proportional to that component. Seed 6 moved its joints almost entirely along the eleven harmless directions and came out at 0.0126ρ. Seed 10 put a large share into the one that matters and came out at 0.2794ρ. Both were drawn from the same discs.

This is also what separates the random errors from the tilt. A tilt keeps all six joints on two lines, which is still a conic, so it costs no rigidity at all: the tilted framework keeps its first-order flex, as the earlier measurement found, and loses only the finite motion. A random error does the reverse. It usually leaves the finite motion’s geometry almost intact and takes away the first-order flex.

That answers half of the question left open. The clearance does make the framework generic, and by the instrument that question proposed, it becomes a structure with a small compliance: $\sigma_{9}$ is of order ρ, its stiffness against the soft direction is of order $\rho^2$, and there is no rank deficiency left.

What the same clearance lets it do #

The other half is what the framework as built can do with the clearance it was built with. A clearance of ρ at each joint lets each bar’s effective length be out by up to 2ρ, one ρ at each end, and the question is how far that lets the framework be pushed.

The measurement is the one the tilt essay used: bisect on an allowance until a placement exists that pushes $B_{2}$ a stated distance along its line with every one of the nine bars inside it. The only difference is that the framework’s own lengths are now those of the framework as built, not the drawing. The descent starts from the drawn framework’s own placement at that push, with the same joint errors added, so it is aimed at the branch the framework is on.

The picture is almost empty, and that is the answer. The worst of the ten frameworks, pushed all the way to the first crossing, needs 0.14ρ on its worst bar. The clearance gives 2ρ. The framework as built uses at most 7% of the clearance its own errors came with, so the damage the clearance does is paid for by about a fourteenth of it, at every seed.

In lengths a designer would recognise: put the framework’s joints on a frame a metre across, so that the unit is about a fifth of a metre, and give every joint a clearance of 50 microns with holes drilled to the same accuracy. That is ρ = 2.5 × 10⁻⁴ in the framework’s own units. The worst framework as built then needs 7 microns of play on its worst bar to reach the crossing, where 100 microns are available. The holes drilled wrongly by up to 50 microns have made the framework rigid by rank, and have cost it 7 of the 100 microns that let it move.

The crossing is the limit, and for a reason that is not about clearance at all. It is where $W_{2}$ arrives at the other line, and the push, measured along the line $B_{1}B_{2}$ had as built, is at its largest there along the whole motion, 0.591. A larger push is not a placement of the moving framework, square or not. The tilt essay originally read the jump in allowance there as the cost of reaching the crossing off square. That was wrong, and it is corrected there: the square framework, which moves exactly, needs the same allowance past 0.591 as the tilted one.

One curve per framework, at any size #

The second thing the curves say is in their units. Each is drawn in units of ρ, and the curve does not depend on ρ.

For one of the seeds, measured at ρ = 10⁻⁴ and ρ = 10⁻², the allowance per unit ρ agrees to between 0.6% and 2.3% at every push up to 0.55. So a framework ten times more precisely made is ten times stiffer against its soft direction and has ten times less clearance to pay with, and the two cancel exactly. There is no size of joint error at which the framework as built crosses over from mechanism to structure. With the lines square, it is a mechanism with a tolerance at every ρ.

The shapes of the curves are informative too. Each starts in proportion to the push. Most then bend upward, but two turn back down before the crossing, so the order of the seeds at the crossing is not their order at the start. The first-order behaviour belongs to $\sigma_{9}$. The second-order behaviour depends on which way each joint moved, not only on how rigid the framework became.

The first push is the rigidity #

The first-order part can be checked against the singular value directly. If $\sigma_{9}$ is the rate at which the soft direction changes the bars, the allowance needed for a small push should be $\sigma_{9}$ times the push times a constant of the framework.

The ten points lie on one line through the origin, at 0.401 to 0.421 times $\sigma_{9}$ — the same to 5% — across a twenty-two-fold spread in $\sigma_{9}$ itself. The bisection that finds an allowance never computes a singular value, and the decomposition that finds $\sigma_{9}$ never asks about an allowance. Their agreement says that the framework’s rank-nine rigidity is exactly what a small push has to overcome, and that 0.41 $\sigma_{9}$ h is its price in allowance.

That makes the comparison exact at first order. The price is 0.41 $\sigma_{9}$ h, $\sigma_{9}$ is at most 0.28ρ for any of these frameworks, and the clearance is 2ρ. At first order, a push would have to be about seventeen units long to use up the clearance. The loop is fifteen units long, and the first crossing arrives at 0.591.

What the tilt does, and what the errors add #

So with square lines, the clearance wins outright at every size. The question that remains is what happens when the lines are not square, which is where a real framework starts. There a tilt charges an allowance of its own that grows as the square of the push. The errors add their $\sigma_{9}$ term, and both come out of the same 2ρ.

Square, the framework as built reaches the crossing at every clearance from 3 × 10⁻⁶ to 10⁻³. A degree off square, the travel is the tilt essay’s: 0.085 at 3 × 10⁻⁶, 0.150 at 10⁻⁵ and 0.406 at 10⁻⁴, growing as the square root of ρ until the square law bends, and reaching the crossing at 10⁻³. Adding the worst seed’s joint errors to the tilted framework changes its travel by at most 1.2% anywhere on the curve.

So the answer to the question as posed — at what clearance, relative to the tilt, does the mechanism with a tolerance give way to the structure with a compliance — is that the joint errors never decide it. The random errors cost about a tenth of what they are given, at every size. The tilt is systematic and costs in proportion to the square of the push, so it is the tilt alone that sets the travel. The square law on its own would say that a framework a degree off square reaches its crossing once the clearance at each joint is about $1.5 \times 10^{-4}$ of the unit its joints are placed in. The law bends upward before the crossing, and the measured threshold lies between $3 \times 10^{-4}$, which reaches 0.568, and $10^{-3}$, which reaches the crossing. With less clearance the framework goes a shorter way, as the square root of the clearance. The same errors that took away its rank deficiency are not what stops it.

What the rank was measuring #

This is the second time these measurements have found rank and motion saying different things. It moves to first order and not at all found a tilted framework with a rank deficiency and no motion. Here the frameworks as built have no rank deficiency, and on their own clearance they move all the way to the crossing.

The rank describes the framework with every joint a perfect point and every bar a perfect length. That is exactly the framework that cannot exist once clearance is admitted, because clearance is what put the joints where they are. Asking whether the framework as built is rigid, and answering with its rank, asks about a zero-clearance version of a framework that owes its shape to clearance. The honest question is what the audit of what decides whether it moves always had to add to the count: how much motion, for how much play. Measured that way, the framework as built is a mechanism with a tolerance, and its tolerance is set by the tilt of the lines.

The rank reading still has a use. $\sigma_{9}$ is the framework’s first-order stiffness against its soft direction, and if the joints are preloaded — held against one side of their holes by a spring, as precision fixtures are — the play is taken out and $\sigma_{9}$ becomes the whole story. A framework with preloaded joints and errors of ρ has a soft mode whose stiffness grows as $\rho^2$. That is a structure with a compliance, and choosing to remove the play is how a designer chooses it. The choice is real in both directions. A fixture that must not wander wants the preload and gets a very soft structure, whose softness is set by how far its holes are from a conic. A mechanism that must move wants the play, and it gets a tolerance set by the squareness of its two lines, whatever its holes did.

What this does not settle #

The joints are displaced but not free. The joint errors here are fixed displacements, like holes drilled in the wrong place. A real clearance also lets each pin wander within its hole as the framework moves, and that freedom is what the allowance of 2ρ stands for. Treating the two as independent draws of the same size is a model. A framework whose holes are drilled accurately and whose pins are loose is the tilt essay’s case, and one whose holes are drilled wrongly and whose pins are tight is the rank’s.

Ten seeds. The ratio of $\sigma_{9}$ to the conic determinant is constant across ten seeds and is not derived. A first-order expansion of both about the two lines would give it in closed form, and it should be a property of where the six joints sit along their lines.

The loop has four crossings. Everything stops at the first, where the push is at its largest. What happens beyond it — whether a framework as built can be taken round the loop at all on its own clearance, and whether it chooses the same branch at the crossing as the square one — is not measured.

Still open: a framework that chooses its branch at the crossing #

At the crossing, $W_{2}$ lies on the other line and three bars lie along it. There the square framework’s motion passes from one family of placements to the next, and a framework as built, having no motion, has nothing to follow. The same situation was measured in a four-bar: a parallelogram a thousandth wrong loses its change point, and the radial play that joins its two motions again is exactly the error.

The distinct argument there would be that question for nine bars. At the crossing, how much allowance lets a framework as built pass from one family to the other, and is it proportional to ρ, as the four-bar’s was, or to a different power because nine bars share it? Is the family it arrives in decided by the signs of its joint errors, as a parallelogram’s is decided by which bar is long? The measurement would be the allowance to push past the crossing and then back again, to a push of 0.5 on the far branch, for the ten seeds and three sizes of ρ.