The price of the other family

Dixon’s framework is nine bars on six joints that ought to be rigid and is not. Nine bars that ought to be rigid put three joints on one line and three on a line square to it, barred every joint of one set to every joint of the other, and found the whole thing moving. The motion runs round a loop and crosses itself four times, at the placements where a joint of one set reaches the other line.

The clearance that makes the framework generic built the framework the way it would be built in a shop: every joint somewhere inside a disc of radius ρ about where it was drawn. The joints then lie on no conic, and by rank the framework is a structure with no motion at all. But the same clearance lets each bar be as much as 2ρ out, and the framework as built could be pushed all the way to its first crossing using at most a fourteenth of that. It stopped there, because the crossing is where something else begins.

At the crossing the framework’s motion passes from one family of placements to another. Up to it, pushing one joint along its line makes the push grow; beyond it, on the next family, the push falls again. So past the crossing every push names two placements. A framework with no errors follows its motion from one to the other. A framework as built has no motion to follow, and something has to carry it. The question left open was what that costs, whether the cost goes as the error or as some other power of it, and whether the framework’s own errors decide which family it ends up in.

One push, two placements #

The framework is the one measured before: joints at −2, 1 and 3 along one line, at −1.5, 1 and 2.5 along a line square to it, held by the first joint $B_1$ and the direction of the bar $B_1W_1$. The push is how far $B_2$ has been moved along its line from where it was built. It grows to 0.591, where $W_2$, the joint of the second set nearest the crossing point, arrives on the first line.

The two placements at a push of 0.5 differ in one joint. Every bar has its drawn length in both, and five of the six joints sit in the same places; only $W_2$ is on the other side of the first line. That is what a family is, in Dixon’s own construction: the six joints’ positions along their lines are fixed up to a sign for each, and the families are the sign choices. Passing through the crossing is $W_2$ changing sign by going through nought, which is exactly the placement where it lies on the other line and three of the bars lie along that line with it.

Dixon’s own construction makes the families concrete. Every placement of the moving framework has the first set’s joints at $\pm\sqrt{x_i^2 + t}$ along their line and the second set’s at $\pm\sqrt{y_j^2 - t}$ along theirs, for one parameter t, with the drawn framework at t = 0. As t grows the first set spreads out and the second closes in, and at t = 1 the second set’s joint at 1, $W_2$, has closed in all the way to the point where the lines cross. It cannot go further on its own side, since its coordinate would have to be the square root of a negative number. The motion carries on by changing that joint’s sign: t starts to fall again, and $W_2$ comes back out on the other side of the crossing point. Every placement on the far family is a placement of the near one with $W_2$ reflected through the crossing point, which is what the dashed framework shows.

It moves to first order and not at all found that the same construction breaks the moment the lines are not square: the square roots no longer describe a motion. Here the lines stay square and it is the joints that are off them, which leaves the construction intact for the drawn framework and gives the framework as built nothing to follow.

For the exact framework the two placements are joined by the motion. For a framework as built, the question is how much play it takes to push it to 0.5 on each side.

The price along the whole passage #

The measurement is the one the approach to the crossing used: bisect on an allowance, the same on every bar, until a placement exists that pushes $B_2$ the stated distance with every bar within it. The descent starts from the exact framework’s placement on the family being asked about, with the framework’s own joint errors added, so that it is aimed at that family and not at the nearest placement of any kind.

Left of the crossing are the curves the essay on the framework as built measured. Right of it is the new part, and most of the curves keep rising: going on past the crossing costs more than arriving did. The dearest is seed 2, which needed 0.133ρ to reach the crossing and needs 0.204ρ to be pushed back down to 0.5 on the other side. Two curves do the opposite. Seeds 4 and 7 are cheaper past the crossing than before it, and at a push of 0.5 each is cheaper on the far family than on its own.

The whole picture fits under a quarter of ρ. A clearance of ρ at every joint gives 2ρ on every bar, ten times the dearest price here, so every one of the ten frameworks can be carried through the crossing by the clearance its own errors came with. The approach cost at most a fourteenth of the clearance. The passage costs at most a tenth.

Proportional to the error #

The four-bar version of this question was settled in a length error is undone by its own size. A parallelogram built a thousandth wrong loses its change point: its two motions end up a tenth of a radian apart, which is the square root of the error, but the radial play that joins them again is a thousandth exactly, the error itself. It was possible that nine bars sharing the misfit would change that power, since a crossing of this framework is a more complicated event than a four-bar’s change point, with three bars lining up at once.

It does not change the power. Every line in the figure is flat across two decades of error, to 4.2% at worst, so the price of the far family is proportional to ρ with a coefficient that belongs to each framework: 0.0023 for seed 7, 0.204 for seed 2. Dividing by ρ puts all three sizes of error on one number. The near family’s prices at the same push are flat as well, to 2% at nine of the seeds and 5% at the cheapest, where the price is 0.002ρ, which is the approach curves’ collapse onto one per framework, seen again at a single push.

There is a short reason, and it is the same reason the approach was proportional. A framework as built differs from the drawn one by joint errors of size ρ, so its nine bar lengths differ from the drawn ones by amounts of size ρ. The drawn framework’s far placement satisfies the drawn lengths exactly. To first order, then, the far placement of the framework as built is the drawn one plus a correction of size ρ, and the least allowance that makes it feasible is the least largest bar error over all such corrections. That is a linear problem in the errors, and its answer scales with them. The coefficient is the one thing that needs the framework’s geometry, and it comes out small because nine bars and twelve coordinates give the correction a lot of room.

The comparison with the four-bar is the refutation of an intuition. A parallelogram’s coefficient is one: its play has to be the whole of its error. Dixon’s frameworks need between a five-hundredth and a fifth of theirs. More bars sharing the misfit makes it cheaper, because more of the error can be absorbed by moving joints the bars do not care about.

The signs of the errors do not choose #

The four-bar result had a second half. Which of a parallelogram’s two motions a framework built slightly wrong ends up on is decided by which bar is long, so the sign of the error chooses the branch. The open question asked whether the same is true here.

Reversing every one of the six joint errors, so that each joint sits exactly opposite where it was, changes no price by more than 1.3%. The largest change is on one of the cheapest seeds, where the price is 0.002ρ and 1.3% of it is at the edge of what the bisection resolves. So the signs of the errors do not choose the family.

The linear argument says why. The least largest bar error is a norm, and a norm does not change when its argument changes sign: if a correction makes the framework with errors d feasible, the opposite correction does the same for errors −d, to first order. The two frameworks that prefer the far family prefer it by very different margins. Seed 4 is barely cheaper there, 0.027ρ against 0.029ρ at a push of 0.5. Seed 7 is cheaper by a factor of nearly six, 0.0023ρ against 0.0131ρ: its errors happen to lie almost entirely in directions the far family can absorb and its own family cannot. Nothing about the size of either framework’s errors says this; both are drawn from the same discs.

What survives the reversal is the pattern of the errors, the direction they point in the twelve-dimensional space of joint displacements, and that is what separates the eight seeds that prefer their own family from the two that prefer the other. The same thing was true of rigidity: the size of a framework’s smallest singular value was set by the component of its errors along one direction, the gradient of the conic determinant, and not by the errors’ size alone. Here there is no single direction to name, since each family’s price is a norm of its own linear map.

What a given play buys #

A framework built with less play than its errors, pins fitted snugly into holes that were drilled slightly wrong, is the case in which the framework itself can choose. A clearance is a link treated the play at a pin as a short extra link with a free direction, and that is the right picture here: the play is what the framework moves on, and the allowance measured is how long those extra links have to be for a given placement to be reachable. The ledger turns that into a picture.

A framework whose play reaches the end of its bar can be pushed to the crossing. If the mark lies beyond the bar, as it does for eight of the ten, then with play between the two the framework can reach the crossing and can only come back the way it came. It has in effect chosen its own family, and the choice was made by its error pattern, not by anything done at the crossing. For seed 2 the gap is 0.071ρ: a framework built with joint errors of 50 microns, allowed 7 microns on every bar where 10 would be needed, would reach its crossing and return. For seeds 4 and 7 the mark is inside the bar, so anything that reaches the crossing can leave it on either family.

With the play a clearance of ρ actually gives, 2ρ on every bar, every one of the ten can take either family, and nothing in the framework chooses. The choice is then made by whatever is pushing it: the direction it is driven in at the crossing, the friction in its joints, which way it is loaded. A designer who wants the framework to stay on its own family, as a fixture would, has a narrow window: enough play to allow the approach and not enough to allow the far family, a band of 0.071ρ at the widest seed and nothing at all at seeds 4 and 7. In practice that window is too narrow to machine to, and backlash is an allowance is the better model of what a designer does with play: size it for the motion wanted and add a guide for the rest.

That is the position a parallelogram with slack in its pins is in, and it is why a parallelogram a micron wrong needs a third crank or a guide to be driven through its change point reliably. The kinematics has said everything it can: the passage is open, and its price is small.

What the crossing adds to the audit #

What decides whether it moves was about counting freedoms and finding the count unreliable. The essays on this framework have been adding what the count leaves out, one measurement at a time: the right angle as a tolerance found that a framework off square moves only as far as its clearance lets it, with the travel growing as the square root of the clearance. The essay on the framework as built found the random errors paying for their own damage fourteen times over on the way to the crossing.

The crossing completes the first quarter of the loop. A framework as built, square, reaches its first crossing and passes it, on its own clearance, at a price proportional to its error and independent of the errors’ signs, and with any real clearance it can leave on either family. The other three crossings are the same kind of event at other joints, and nothing here measures them. If they behave like this one, a square Dixon framework built with ordinary clearance is a mechanism around its whole loop — a result the rank said nothing about and the count of bars said the opposite of.

What the passage leaves open #

The other three crossings. Each is a joint of one set passing through the other line, but the joints are at different distances from the crossing point, and the price depends on which bars line up. Only the first crossing is measured here.

The tilted framework’s crossing. The frameworks here are square with random joint errors. The tilt essay found a framework a degree off square unable to reach its crossing below a clearance of between 3 × 10⁻⁴ and 10⁻³ of the unit its joints are placed in, and one that does reach it has spent most of its clearance on the approach. Whether the passage then costs it anything more is not measured.

Why seeds 4 and 7. Two of the ten are cheaper on the far family, and nothing here predicts which. A first-order expansion of both prices about the crossing would write each as a norm of a linear map of the errors, and the preference as a comparison of two norms. That expansion has not been done.

Still open: a framework driven through its crossing #

With play enough for both families, the kinematics leaves the choice to whatever drives the framework, and the natural driver is a push along a fixed line with friction in the joints. The distinct argument there would be the framework as built driven quasi-statically through its crossing by a push on $B_2$, with each joint resisting motion within its clearance by a small friction. It would ask which family the framework leaves on, and whether that depends on the direction of the push, on the ratio of friction to play, or on the error pattern that decided the prices here. The measurement would be the same ten frameworks, each pushed through the crossing in a sequence of small steps with a minimum-effort rule for how the play is used, and the family each ends in.