A cable is half a brace

Which diagonal rigidifies a grid answered a question anyone who has braced a fence panel has asked. A grid of square cells, pin-jointed, is a mechanism: every column of cells can lean by an angle and every row by another, and a cell shears by the difference. A rigid diagonal in a cell forces its column’s angle to equal its row’s. So a set of braces is a set of edges joining columns to rows, and the grid is rigid exactly when those edges connect every column and row, Bolker and Crapo’s theorem. A three-by-three grid needs five braces, a spanning tree on six nodes, and eighty-one of the hundred and twenty-six ways of placing five will do.

That essay assumed every diagonal was a strut, as stiff in compression as in tension. Many are not. The diagonals of a braced tower, a shed wall or a stage truss are often rods or wires that can pull and cannot push: pushed, they bow or slacken and carry nothing. A cable is a joint of a particular kind, like a ratchet with no teeth: it acts in one direction and is free in the other. The question is what that does to the rule.

A cable is one direction of an edge #

A cell of the grid shears when its column’s lean and its row’s differ, and one of its diagonals lengthens while the other shortens. Which one lengthens depends on the sign of the difference. A cable on the rising diagonal stops the shear that would stretch it and allows the one that would shorten it. A cable on the falling diagonal does the opposite. So a cable does not identify its column’s angle with its row’s. It forbids one of them to exceed the other, and allows the reverse.

That is a directed edge. Draw it from the column to the row for a rising cable and from the row to the column for a falling one. Following the arrows then goes from angles that may not exceed to angles that may not be exceeded. The grid’s angles can all differ from one another in some allowed way unless every node can be reached from every other along the arrows. So a grid braced with cables is rigid exactly when its directed graph is strongly connected. A strut forbids both senses at once and is an edge in both directions, which is why the undirected rule held for struts.

The hero figure shows a rigid placement and its graph. The six arrows run round a single cycle, column to row to column, through all six nodes, so every node reaches every other. The last two placements the slider offers are not rigid, and they fail differently. The fourth puts every cable on a rising diagonal, so every arrow runs from a column to a row, and no row can get back. The fifth uses the same six cells as a cycle, which would be rigid with struts, but two of its cables are on the other diagonal, and the arrows no longer run round.

Two routes that share only a list of cables #

The directed rule is a claim, and it can be tested against a computation that knows nothing of graphs. A framework with cables is rigid to first order exactly when two things hold, by a theorem of Roth and Whiteley. It must be rigid with every cable read as a bar, which is a rank. And it must carry a self-stress that pulls on every cable: bar forces that balance at every joint, with every cable in tension. The second is a small linear program over the framework’s self-stresses, asking whether some combination puts every cable above a positive floor.

That route builds the grid from coordinates, takes a singular value decomposition, and solves a linear program. The graph route runs a reachability search on six labels. They share the list of cables and nothing else.

There are 4,032 ways to place five cables, 5,376 to place six, 4,608 to place seven, 2,304 to place eight and 512 to place nine. The two routes agree on every one of the 16,832. No placement of five is rigid. Five edges cannot make a strongly connected graph on six nodes, since every node needs an arrow in and an arrow out. Twelve placements of six are rigid, 108 of seven, 216 of eight and 102 of nine. With struts, five were enough and 64% of the sets of five worked. With cables, six are needed and 0.22% of the sets of six work.

What the count sees and what it does not #

Maxwell’s count has something to say here, and it is not nothing. A three-by-three grid has sixteen joints and so twenty-nine freedoms once its three rigid motions are removed, and twenty-four bars of its own. Five braces make twenty-nine, a count of nought: exactly enough, if every brace is independent. Six make thirty, a count of minus one: one member more than rigidity needs, and so at least one self-stress.

For struts the extra member is waste. For cables it is required. A cable framework is rigid only if some self-stress pulls on every cable, and a framework with no self-stress at all has nothing to pull with. So a cable grid needs at least one member more than a strut grid, and the count, which cannot see a cable’s direction, does see why six is the least. What it cannot see is which six. It gives the same count, minus one, for all 5,376 placements, and twelve of them work. A constraint that has been said already found the count right about the difference between freedoms and self-stresses and wrong about both. Here it is right about the self-stress and blind to where it pulls.

Twelve cycles #

Why twelve is visible once the rule is directed. A strongly connected graph on six nodes with six edges is a single cycle through all six. In the complete bipartite graph between three columns and three rows, a cycle through all six must alternate column, row, column, row, column, row. There are six such cycles, and each can be run in two directions.

Running a cycle the other way swaps every cable to the other diagonal of its cell, so the twelve grids come in six pairs, each pair using the same six cells. Every one uses six different cells, two in each column and two in each row, which is what a cycle alternating between columns and rows must do. The scattering rule the struts’ essay gave — never two braces in a row unless their columns are already joined — becomes stricter for cables: two per row, two per column, and the diagonals arranged so that the arrows chase each other round.

Twice the longer side #

The count of cables a grid needs follows from the same observation that ruled out five. Every node of the directed graph needs an arrow in and an arrow out, and every arrow has one end at a column and one at a row. So the side of the grid with more nodes, columns or rows, needs at least two arrow ends per node, and each cable supplies one end there. A grid needs at least twice its longer side in cables. Struts need columns plus rows less one.

The enumeration bears it out on three sizes. A two-by-two grid needs three struts or four cables, and two of the sixteen placements of four work. A three-by-three grid needs five struts or six cables. A two-by-three grid is where the difference is sharpest: four struts, but six cables, one in every one of its six cells, and six of the sixty-four ways of orienting them work. For a square grid the cables need only one more than the struts. For a long, thin one they need nearly twice as many, and the grid has to be braced in every cell.

A strip needs crossed cables #

The limit of a long, thin grid is a strip one cell high. Its graph has one row node and a column node for each cell, and the row node must reach every column and be reached from every column. So every cell needs an arrow each way, which one cable cannot give.

Of the sixteen ways to put one cable in each of four cells, none is rigid. With both diagonals crossed in every cell, the strip is rigid, and the linear program finds a self-stress pulling on all eight cables. The struts’ essay noted that a cell’s second diagonal adds nothing, since the first already triangulates it; the cell that repeats for ever counted that second diagonal as a pure redundancy. For cables the second diagonal is the other half of the brace. This is why a braced bay of a steel shed, a tower leg or a long fence carries an X of rods or wires where one strut would have done. The X is the minimum.

What holds a cable grid #

A cable carries nothing until it is pulled, so a grid held by cables alone has to be pulled before it can resist anything. The self-stress that the linear program finds is that pull, and it is not arbitrary.

On the first rigid six-cable grid the six cables carry equal tension, and the bars they pull against carry compression of 0.71 or 1.41 times it: the components along the bars of a unit pull on a diagonal of a unit square, added where two cables meet a bar. Twelve of the twenty-four grid bars carry nothing. A constraint that has been said already read a self-stress as the combination of constraints that comes to nothing. Here that combination is the one the grid must be put into before it is built, as pretension. Every cable must be tightened against the frame in these proportions, or some cable is slack at the first load that tries to shorten it, and the grid moves.

That is also what the linear program’s floor means physically. It asked for every cable’s tension to be at least one, and a stress that exists at a floor of one exists at any floor, scaled. The grid can be pretensioned as hard as its bars allow, and the harder it is pulled, the larger a load in either direction it can take before a cable goes slack. The price is one self-stress found a single self-stress deciding what a built framework’s play has to absorb. Here a single self-stress is what makes a cable framework a structure at all.

No grid of single cables survives a slack cable #

A grid held by the least number of cables has no spare: every arrow of its cycle is needed, and a cable that stretches under load or loses its pretension takes its arrow away. More cables might be expected to buy a margin, and the census can say whether they do. For every rigid placement, it removes each cable in turn and asks whether the grid is still rigid.

None is. Of the 108 rigid grids with seven cables, 72 contain one of the twelve six-cable cycles with a seventh cable added, and 36 are strongly connected without containing one; not one of either kind stays rigid when every single cable is tried. The same holds for all 216 with eight cables and all 102 with nine, a cable in every cell. A three-by-three grid braced with one cable per cell cannot be made to survive the loss of any one of them.

The reason is the same count that set the minimum, one level up. A directed graph stays strongly connected after losing any one edge only if every node keeps an arrow in and an arrow out after the loss, so every node needs two of each. Six nodes with two arrows out each is twelve arrows, and a grid of nine cells with one cable per cell has at most nine. So a cable-braced grid that must tolerate one slack cable needs at least twelve cables. Twelve cables on nine cells means at least three cells crossed. The seventh contact found a contact added to an exactly constrained part constraining nothing and able to fail to touch. A cable is that kind of constraint, and a slack one is a contact that has stopped touching.

The rule a builder can carry #

The struts’ rule was a drawing: columns on one line, rows on another, a line for each brace, rigid if connected. The cables’ rule is the same drawing with arrows.

Draw each cable as an arrow, from its column to its row if it rises and from its row to its column if it falls. The grid is rigid when every column and row can be reached from every other along the arrows. A cycle of arrows through every node is the least that will do, and a strut is an arrow each way.

Three corollaries follow without drawing anything. A grid needs at least twice its longer side in cables, so a grid braced with cables in fewer cells than that cannot be rigid, however the cables are placed. A pair of crossed cables in one cell is exactly a strut, so any bracing rule for struts holds for crossed cables, at twice the count. And a grid that must stay rigid with any one cable slack needs two arrows in and two out at every node, twice the least count again, which a grid of single cables can never reach.

What it does not settle #

Pretension is required, not measured. The analysis is first-order, and this field knows that a first-order answer need not be the last word; it also assumes every cable is taut. How much pretension a grid needs to carry a stated load without a cable going slack is a question about the size of the self-stress against the load, and it is not computed.

One cable per cell. The census puts at most one cable in a cell, except in the strip. Mixing crossed cells, single cables and struts in one grid follows the same directed rule, since a crossed cell or a strut is an arrow each way, but the enumeration here does not include mixtures.

Square cells. As for struts, the rule depends on every cell being a parallelogram, so that a brace compares a column’s lean with a row’s. A grid of general quadrilaterals has no such rule, braced either way.

Still open: the fewest crossed cells that survive a slack cable #

A grid of single cables cannot survive the loss of any one of them, and the degree count says a grid that does needs at least twelve cables, three cells crossed at the least. Whether twelve is enough is not settled by the count: two arrows in and out at every node is necessary for surviving any single loss and is not sufficient.

The distinct argument there would be the smallest cable bracing of a three-by-three grid that stays rigid with any one cable slack, and where its crossed cells go. The census would allow each cell to hold nothing, one cable either way or two crossed. It would search from twelve cables upward for placements whose directed graph stays strongly connected after the loss of any one edge, checking each against a self-stress as here. That would give a designer the least redundant cable bracing and, as the struts’ essay gave for struts, a drawing rule for placing it. The loops are in the graph began this field with a count read off a drawing. This would be the first question in it whose answer is a count read off a directed one.