A spare is worth what its pretension leaves

Twelve cables survive a slack one found the fewest cables that keep a three-by-three grid rigid when any one of them goes slack. Twelve are needed and 36 placements of twelve do it. Each gives every row and every column of cells two cables into it and two out, and they fall into three families by how many cells hold a crossed pair: three, four or six. The Latin squares are the three-crossed family, one rising, one falling and one crossed cell in every row and column.

That essay answered a yes-or-no question. For each cable, it asked whether a self-stress exists that leaves that cable out and pulls on all the rest. A builder does not choose a family of self-stresses. They tighten the cables to one pretension, and when a cable later goes slack under load the grid does not jump to whatever stress the census found. It carries the slack cable’s share elastically, through the members that remain, and from then on it has only what that sharing left it. The essay ended by asking which pretension makes a spare worth having, and whether bracings that are equal under the count are equal under load.

They are not, by a factor of four.

The grid as a structure #

To ask what a grid carries, it has to be a structure rather than a graph. Every bar of the frame and every cable is given the same axial stiffness per unit length, so a member’s stiffness is inversely proportional to its length, and a cable on a diagonal is stretched less for the same force than it would be on a side. Loads on the joints are taken in any combination that is in equilibrium on its own, which leaves 29 independent directions of load on a grid of sixteen joints.

A rigid framework under such a load has one set of member forces, found by solving its stiffness equations. The pretension is a self-stress, a set of member forces in equilibrium with no load at all. On these grids there are seven independent self-stresses, since 36 members exceed the 29 a rigid framework needs by seven. While every cable stays taut the framework is linear, so a load $\lambda f$ adds $\lambda$ times its own forces $Bf$ to the pretension, and cable $i$ goes slack when $t_{0i} + \lambda (Bf)_i$ reaches nought.

The load at which that first happens depends on the load’s direction, and the worst direction turns out not to need searching for. Over all loads of unit size, the most that any of them can pull off cable $i$ is the length of cable $i$’s row of $B$, written $\|B_i\|$. So the smallest load in any direction that slackens some cable is

$$\lambda^* = \min_i \frac{t_{0i}}{\|B_i\|},$$

and with the largest cable pretension held at one, the pretension that makes $\lambda^*$ largest is a linear program over the seven self-stresses.

In words, the worst load for a cable is the one that unloads it most per unit of load, and that load is a single, definite pattern of forces on the sixteen joints. Each cable has its own. The grid’s capacity is set by whichever cable is weakest against its own worst pattern, measured against how hard that cable was pretensioned. A pretension that pulls harder on the cables with the largest $\|B_i\|$ raises the capacity, and the linear program finds the balance among them.

Both halves of that were tested. Two thousand random load directions were applied to a sample of the bracings, and none slackened a cable at a load below $\lambda^*$. The load built to be the worst, the limiting cable’s row of $B$ reversed, slackened a cable at exactly $\lambda^*$, to $10^{-9}$. Five hundred random pretensions, normalised the same way, were tried against each sampled bracing’s optimum, and none carried more.

Before a loss, all the same #

Pretensioned at their best for the intact grid, the 36 bracings carry between 1.03 and 1.25 times their largest pretension, in the worst direction, before any cable goes slack. The Latin squares sit at 1.12 to 1.14, the four-crossed family at 1.07 to 1.25 and the six-crossed at 1.03 to 1.09. A designer measuring only the intact grid would find nothing to choose between them, which is what the count also said.

After a loss, nothing alike #

When a cable goes slack it stops pulling on its two joints, and that change is the same as applying its lost pull to them as a load in the other direction. The grid that remains carries that load elastically, and its new member forces are the old pretension, without the slack cable, plus the reduced grid’s response to the lost pull. The capacity left is the same ratio on the reduced grid.

The circles are what a builder gets by tightening each grid to its best intact pretension. After the worst single loss the Latin squares keep between 0.080 and 0.267, the four-crossed family between 0.128 and 0.243, and the six-crossed family between 0.003 and 0.095. The worst six-crossed bracing has a spare in name only. Lose its weakest cable and the remaining eleven are still rigid, as the census promised, but they are pretensioned so unevenly that a load of 0.3% of the largest pretension slackens another.

Rigidity is the wrong measure of what is left because it only asks whether a stress pulling on every remaining cable exists. The loss does not choose that stress. The grid is left with whichever stress the elastic sharing produces, and in some bracings that stress leaves one cable barely taut.

A pretension chosen for the loss #

The sharing is linear, so the stress left after losing cable $j$ is a fixed linear map of the pretension. The pretension that keeps the most after the worst single loss is then also a linear program. Maximise the smallest capacity over all twelve losses, every cable’s pretension at most one, over the seven self-stresses. The squares in the scatter are its answers. Each was checked by recomputing every loss with the redistribution and confirming that the smallest capacity left is the program’s optimum.

It lifts every family. The Latin squares now keep 0.322 to 0.404 after any loss, the four-crossed family 0.215 to 0.266, and the six-crossed family 0.095 to 0.096. The cost is paid before the loss. The best Latin square carries 1.124 intact when pretensioned for the intact grid and 0.998 when pretensioned for the loss, giving up 11% of its intact capacity to keep half as much again afterwards. On one four-crossed bracing the price is steep, 1.25 intact falling to 0.65.

The losses show what the program did. For the best Latin square, five of the twelve losses leave exactly the worst case, 0.404, and the others leave more, up to 0.561. The program has raised the worst losses until several meet, the usual signature of a max-min. For the four-crossed family’s best, all twelve losses leave exactly 0.266, a bracing so symmetric that every cable is equally critical. For the six-crossed family’s best, six losses leave 0.096 and the program can do nothing about them. Whatever pretension is put into those grids, losing one of six particular cables leaves them nearly unloaded in some direction.

That is a property of where the crossings are rather than of the pretension. A six-crossed bracing puts two cables in each of six cells and leaves three cells with none, a pattern the earlier essay drew as a cycle of crossed pairs, so every one of its twelve cables is half of a crossed pair. Half of those twelve are the critical ones, each leaving 0.096. Why those six and not the others is not derived here; the measurement shows only that no pretension in the seven-dimensional space can reach past them.

What choosing for the loss costs before it #

A pretension chosen for the loss is a compromise, and the intact capacity it gives up varies more than the margin it buys. Across the twelve Latin squares, pretensioned for the loss, the intact capacity runs from 0.990 to 1.100, against 1.117 to 1.138 for the intact-best pretension. Each gives up between 2% and 11% to keep 0.32 to 0.40 after a loss instead of 0.08 to 0.27. Across the six-crossed bracings it runs from 0.946 to 0.978 against 1.026 to 1.093, a similar cost for almost nothing in return, since their worst loss is pinned at 0.096 whatever is done. The four-crossed family is the one where the trade can be expensive: its pretensions chosen for the loss carry anywhere from 0.65 to 1.25 intact. On the bracing whose twelve losses are all equally critical, keeping 0.266 after any of them, against 0.243 with the intact-best pretension, costs nearly half the intact capacity, 1.242 falling to 0.653.

For a designer the reading is simple. On a Latin square, pretension for the loss: it costs at most a tenth of the capacity the grid would rarely use and raises what it keeps when it most needs it by half again or more, fourfold on the worst of them. On a six-crossed bracing no pretension rescues the spare, so the choice of placement has to be made before any cable is tightened.

The seventh contact found a redundant contact that constrains nothing until something deforms. A spare cable is the opposite case, a redundant member that carries pretension all the time and whose redundancy is used only when another member fails. How much of the pretension survives that failure is what decides its worth.

Twelve Latin squares are three #

The count cannot tell any two Latin-square bracings apart, and neither could the intact capacity, which varies by under 2% among them. After a loss they separate cleanly.

They fall into three classes of four. Pretensioned for the loss, one class keeps 0.404 after its worst loss, one 0.330 and one 0.322. The classes are what is left of the twelve once the grid’s own symmetries are taken out. A quarter turn or a reflection of the square grid swaps rising and falling cables, since it turns one diagonal of each cell into the other, and keeps a crossed cell crossed, so it maps a Latin-square bracing to another. Applying all eight of the square’s symmetries to the twelve gives exactly three sets of four, and each set is one of the three classes, every member at the same capacity to $10^{-4}$. The best class puts its three crossed cells on one of the grid’s two diagonals: RFX / FXR / XRF reading the rows upwards, crossed cells running from the bottom right to the top left. The other two classes put theirs on broken diagonals, which wrap round the grid’s edge. A builder told only to use a Latin square has a one-in-three chance of the best kind.

Re-tensioning, and how tight a cable gets #

A grid that loses a cable can be re-tensioned rather than left as it is, and that sets a ceiling on what any pretension chosen beforehand could keep. Re-tensioning means choosing a new self-stress of the eleven cables that remain, six-dimensional now, with the largest at one again, so it is the same linear program on the reduced grid.

Re-tensioned after the worst loss, the Latin squares carry 0.487 to 0.516, the four-crossed family 0.387 to 0.502 and the six-crossed 0.405 to 0.422. Re-tensioning nearly equalises the families, because the eleven cables left in each are rigid and can be given any of their own self-stresses. What separates the families is the loss that happens under load with no one there to re-tension, and there the Latin squares keep four fifths of their re-tensioned capacity and the six-crossed family a quarter.

For the best Latin square that ceiling is 0.516, and the pretension chosen for the loss keeps 0.404 without anyone touching the grid, four fifths of what re-tensioning would recover. So the choice of pretension made at assembly does most of the work that maintenance after a failure would do. The four-crossed family’s best keeps 0.266 against a ceiling of 0.502, about half, and the six-crossed family’s 0.096 against 0.42, under a quarter.

The last column is a warning for the cables themselves. Sharing out a lost pull raises some of the other cables’ tension above the largest pretension: to 1.50 times it on a Latin square, 1.42 on the four-crossed family and 1.23 on the six-crossed. A cable rated just above its pretension will be overloaded by its neighbour’s failure. A slack-proof bracing needs cables with half as much strength again as their pretension, or it survives the first loss by inviting the second.

What the counting essays could not see #

A cable is half a brace and which diagonal rigidifies a grid answered their questions with graphs, and graphs know nothing of force. A cable there is an edge that knows which side it is on, an arrow that forbids one sign of a shear, and an arrow carries no number. The price is one self-stress found a mechanism’s cost of passing a crossing in one self-stress, and here the grid’s capacity is likewise a fact about its self-stresses. They are seven-dimensional, and the question is which point in that space the pretension occupies.

A constraint that has been said already called a redundant constraint one that the others imply. A spare cable is a redundant constraint by design, and this is what redundancy costs and buys once forces enter. It costs nothing in the intact capacity, which is the same with or without the choice of family. It buys a margin after a loss whose size depends on where the redundancy is placed, which no count of constraints can say. The count says how many and not where found the same distinction in a lattice’s mechanisms, and it holds here for a grid’s spare strength.

What the elastic grid does not settle #

Equal stiffness. Every member has the same axial stiffness per unit length. Real cables are far more compliant than the frame’s bars, and a stiffer frame would carry more of a lost pull through the bars and less through the other cables, which changes both the capacities and the tightest cable. The ranking of the families under realistic stiffness ratios is not computed.

Small displacements. The analysis is linear. A cable gone slack under a large load, or a grid deflecting enough for its geometry to change, is outside it. So is the sudden release of a cable that snaps rather than slackening slowly, which loads the rest dynamically, beyond what the static sharing gives.

The frame’s bars. The bars are taken to carry any force in tension or compression. Under the redistribution some carry more than before, and a slender bar might buckle. The bars’ forces are computed in the model but not checked against any limit.

One loss at a time. Every measurement takes away one cable. A grid that loses one and is then loaded until a second goes slack is the next event, and the census at twelve says the grid is no longer guaranteed rigid after two.

Still open: the grid that loses two #

The survivors were defined by one loss, and a builder who rates the cables at half as much again as their pretension has bought a margin for one loss and not two.

The distinct argument would be the fewest cables that survive any two going slack, and the pretension that keeps the most after the worst pair. The count extends: three arrows in and out at every node, eighteen cables at least on a three-by-three grid. The measurement would enumerate placements with up to three cables a cell or more cables than cells, test each pair of losses for rigidity, and price the survivors under load as here. It would ask whether the Latin square’s advantage persists at the next level, or whether a bracing that survives two losses has to spread its cables so evenly that every placement is as good as every other.