A push turns back at the crossing

Dixon’s framework is nine bars joining three joints on one line to three on a line square to it, and nine bars that ought to be rigid found it moving anyway. Its motion runs round a loop through four crossings, each the moment a joint of one line passes through the other line, and at each crossing the framework passes from one family of placements to another. The clearance that makes the framework generic found that a framework built with its joints a little out of place has no motion of its own, so its play has to carry it. The price of the other family measured what the play must be to go on past the first crossing onto either family, and the price is one self-stress found that price in closed form: one inner product of the joints’ errors with the family’s single self-stress.

That essay ended on the question the prices could not answer. With play enough for both families, which one does a framework actually go on to, when it is driven through the crossing by a push on $B_2$, the joint every earlier essay pushed? Perhaps the family whose stress its errors load least, so that a framework goes wherever its play is cheapest. Perhaps something else, in which case the stresses ought to say what.

Something else, and it is decided before any play is added. A push cannot carry the framework through the crossing at all.

Every joint but $W_2$ is the same on both sides #

Dixon’s closed form puts the black joints at $\sqrt{x_i^2 + t}$ along one line and the white joints at $\sqrt{y_j^2 - t}$ along the other, for one parameter $t$. On the site’s framework the white joint nearest the crossing point is $W_2$, at $y = 1$, so the first crossing is $t = 1$: there $W_2$’s distance from the black line, $\sqrt{1 - t}$, reaches nought. The family beyond the crossing is the same formula with $W_2$’s sign reversed.

The formula moves because every bar keeps its length: a black joint at $\sqrt{x_i^2 + t}$ on one line and a white joint at $\sqrt{y_j^2 - t}$ on the line square to it are $\sqrt{x_i^2 + y_j^2}$ apart whatever $t$ is, the $t$ cancelling inside the square root. It is that cancellation, and nothing about the framework’s rigidity count, that lets nine bars on six joints move, and the same square roots that make it move are what make the crossings singular, since each runs out exactly when its joint reaches the other line.

Write $t = 1 - s^2$. Then $W_2$ sits at $s$ on one family and $-s$ on the other, which together are one smooth motion through $s = 0$, with $s$ measuring distance along it. Every other joint sits at $\sqrt{x_i^2 + 1 - s^2}$ or $\sqrt{y_j^2 - 1 + s^2}$, a function of $s^2$ alone: the same at $s$ and at $-s$. Holding the framework by its pins changes none of that, because the pins are $B_1$ and the direction of $B_1W_1$, both even in $s$.

The hero figure draws the framework at $s$ and $-s$ together. Five of the six joints coincide, to the last digit computed, and $W_2$ is mirrored through the line of the black joints. Near its first crossing Dixon’s framework is one joint swinging through a line while the rest of it pauses at the extreme of its own travel.

A push is at a fold #

Any quantity built from the five even joints is itself even in $s$, and smooth, so near the crossing it changes as $s^2$: it has a maximum or a minimum there. The push on $B_2$ is one of them. Measured along the exact motion, it falls from its value at the crossing, 0.5914, by $4.39 \times 10^{-3}$ at $s = \pm 0.1$ and $4.37 \times 10^{-5}$ at $s = \pm 0.01$, identically on both sides, with an exponent of 2.005. $W_2$’s side of the black line is $s$ itself, exponent 1.000.

A coordinate that moves as $s$ passes through its crossing value, and a drive that holds it can carry the framework through. A coordinate that moves as $s^2$ has its largest value at the crossing on both sides, and a drive that holds it meets a fold: it can be pushed up to the crossing and no further, and when it is pulled back the framework has two placements to choose from, one on each family, at every smaller push. That is what the earlier essays meant by “past the crossing every push names two placements”. The push names them, but a push cannot reach them by going forward; it can only come back.

The same thing happens the other way round at the other kind of crossing. The loop’s next crossing is $B_2$ reaching the white line, at $t = -1$, where $B_2$ is the joint at $\pm s$ and every other joint, $W_2$ included, is even. There the push, which moves $B_2$, goes as $s$, exponent 1.021, and changes sign; $W_2$’s side of the black line goes as $s^2$, exponent 2.002. A drive passes a crossing only if it moves the joint that is crossing.

The framework as built, pushed #

A framework as built has its joints out of place and no motion of its own, and the fold might not survive that. So the push was simulated. Ten frameworks, the ten the earlier essays measured, each joint displaced by up to $\rho = 10^{-3}$, every bar free to take up an allowance of half that. The push on $B_2$ was raised in small steps, and at each step the joints were moved as little as possible — the least total motion, by linear program on the bars’ linearised lengths — to reach the new push with every bar within its allowance. When no step could reach a larger push, the push was lowered again to 0.45.

Every framework turns at a push between 0.5916 and 0.5930, just past the exact framework’s 0.5914, with $W_2$ still on its own side of the black line. Pulled back, every one comes down the family it went up. With four times the play, two thousandths on every bar, the turning points move out to between 0.5946 and 0.5960 and the outcome is the same. With the largest single joint motion made least instead of the total, the outcome is the same again, framework by framework. Play lets a framework as built be pushed a little further than the exact one, by an amount that grows with the play, but it does not let $W_2$ cross. The fold is still there, blurred by the allowance.

How much further is close to a fixed multiple of the play. Averaged over the ten frameworks, the turning point lies $0.82 \times 10^{-3}$ past the exact crossing with an allowance of $5 \times 10^{-4}$ on every bar, $1.8 \times 10^{-3}$ with $10^{-3}$ and $3.7 \times 10^{-3}$ with $2 \times 10^{-3}$: between 1.6 and 1.9 times the allowance, the steps of the drive near the turn being $2 \times 10^{-4}$ long. At a quarter of ρ some frameworks turn slightly before the exact crossing instead of after it, because that little play is not enough to keep their errors absorbed all the way up. The overshoot is the play being spent on reach, and it is spent in the direction the fold allows, along the push, not across the black line.

Which stress the push loads #

So the question the earlier essay put has a plain answer. A push on $B_2$ loads the self-stress of the family the framework arrived on, all the way to its turning point and all the way back, because the framework never leaves that family. The play it uses is that family’s: the price is one self-stress found it to be the errors’ inner product with the near family’s stress, and nothing in the drive ever asks for the other.

Cost does not enter. Frameworks 4 and 7 are the two of ten that are cheaper beyond the crossing, 0.0327 against 0.0336 per unit ρ for framework 4 at a push of 0.58 and 0.0080 against 0.0122 for framework 7. Both retrace exactly as the eight that are cheaper where they started. The driver is choosing something other than cost, as the earlier essay suspected it might, and what it chooses by is parity: whether the coordinate it holds is even or odd in the motion at the crossing. The stresses do not say what. The closed form does.

Driving $W_2$ instead #

The coordinate that is odd at this crossing is $W_2$’s side of the black line, so the same ten frameworks were driven by that instead: from its value at a push of 0.45 on the starting family, through nought, to the opposite value, in steps of least motion with the same play.

All ten go through. The push rises as $W_2$ approaches the line, reaches its largest, between 0.5903 and 0.5917, as $W_2$ reaches it, and falls again as the framework goes on along the family beyond. Driven by the coordinate that passes, a framework as built crosses with the play it has, onto the family beyond, whether or not that family is the cheaper.

What going through costs #

The least play that lets the side drive through can be found by bisection on the drive itself, and it can be predicted without driving anything.

The prediction is the first-order price, computed as the earlier essay computed it but with $W_2$’s side holding the framework rather than the push: the least largest bar misfit, over first-order adjustments of the exact placement, that meets the joints’ errors at that point of the motion. Each framework’s price is a smooth curve along the motion, and the side drive needs exactly its largest: 0.0088 per unit ρ for framework 1 by bisection against 0.0087 predicted, 0.2172 against 0.2173 for framework 2, and within the bisection’s resolution for all ten. For eight of them the worst point is the far end of the path, a push of 0.45 on the family beyond; for frameworks 4 and 7 it comes before the crossing.

That also explains where the earlier essay’s passage figure stopped. It priced each family with the push holding the framework, and stopped at 0.58 on both sides because the single-family formula broke down near the crossing. It broke down because of the fold: near it, a small change in the push needs a large movement along the motion, of the order of the square root of the change, and first-order adjustments that large are no longer small. Holding $W_2$’s side instead keeps the adjustments small all the way through, which is why the price can be followed into the crossing and out the other side.

The curves are continuous through the crossing. Framework 1’s price is 0.0013 per unit ρ a thousandth of the motion either side of it; framework 2’s is 0.1403 on one side and 0.1404 on the other. The price is one self-stress read the crossing as a singular placement where a second stress appears and supposed the price must jump there for a framework that switches family. The second stress is real, but it belongs to the push. At the crossing the push is at its fold and no longer fixes where the framework is along its motion, so with the push held the bars have one more free combination and the framework one more self-stress — one more of the dependencies among its bars that a constraint that has been said already counted as constraints repeating each other. Held by $W_2$’s side, the framework at the crossing has one stress like everywhere else, and its price passes straight through: at the crossing it is 0.0013, 0.1403, 0.0675 and so on, and for nine of the ten it lies between the prices at a push of 0.58 either side. Framework 4, whose two families cost almost the same, has its crossing price, 0.0339, a little above both.

Two crossings, two drivers #

The loop has four crossings, two where $W_2$ passes the black line and two where $B_2$ passes the white one, and a drive passes a crossing only if it moves the crossing joint. A push on $B_2$ therefore carries the framework through the $B_2$ crossings and folds at the $W_2$ crossings; a drive on $W_2$’s side does the reverse. Neither alone drives the framework round its loop. A drive that moves both joints, in some combination, has a nonzero rate at every crossing and could, which is the shape of the question below.

A length error is undone by its own size met a branch point of the same kind in a parallelogram’s change point, where the choice between two motions is lost once the linkage is built a little wrong and returns when its pins are given play equal to the error. There the question was whether the choice could be had at all. Here the play is ample for both families, and the question is who makes the choice: the drive does, by which coordinate it holds, and the errors only decide what each choice costs.

What a builder can do with it #

A framework, or any overconstrained linkage, that has to be taken from one branch of its motion to another through a branch point is not helped by more play or by better-placed errors if it is driven by the wrong joint. The joint to drive is the one whose motion is odd at the branch point: the one that passes through something there, as $W_2$ passes through the black line. A drive on any joint that merely pauses at the branch point will bring the linkage up to it and back down the way it came, however the play is shared out.

And the price of going through is easy to know in advance. Held by the right joint, the first-order price is a smooth curve along the motion, computable from one linear program per point, and the play the drive needs is that curve’s largest value on the path, not something special at the branch point.

What the drive does not settle #

One rule for friction. The least-motion steps stand for a framework whose joints resist motion more than its bars resist taking up play. Two different rules, least total motion and least largest motion, gave the same outcome at every framework and every play tried, which says the fold dominates the rule, but a real joint’s friction is neither.

Quasi-static. Nothing here has mass. A framework pushed hard towards its fold and released might carry through on its own momentum, which a quasi-static drive cannot see.

The first crossing. The parity argument covers all four crossings, and the $B_2$ crossing’s exponents were measured, but the simulations ran only at the first.

The play’s own pattern. Under least motion every bar’s play is taken up to its limit at once, which is how the drive reaches its turning point. Whether the pattern the play settles in along the way follows the family’s stress, bar by bar, was not measured.

Still open: a drive that goes all the way round #

A push folds at two of the loop’s four crossings and a drive on $W_2$’s side at the other two, so neither carries Dixon’s framework round. A drive that moves $B_2$ and $W_2$ together — $B_2$’s push plus some multiple of $W_2$’s side, say, which a lever between the two joints would supply — has a nonzero rate at every crossing.

The distinct argument would be the least play such a drive needs to take a framework as built all the way round. At each crossing the drive is held by a different mix of the two coordinates, so the stress it loads changes from crossing to crossing, and the price along the whole loop is a curve with four crossings on it. The measurement would choose the mix, drive the ten frameworks round, and compare the play the full loop needs with the worst of the four crossings’ own prices, to say whether going round costs more than its hardest crossing.