Rigidity — the series · Crystallography & Symmetry

11 frameworks, 3 where the count is wrong. Every net in this collection read as a framework of rigid bars and free joints, with the cell free to change shape. Maxwell's count and the number of mechanisms agree on most of them and not on all: a framework with a state of self-stress has a bar the count treats as removing a freedom that the others had already removed, and it has a mechanism the count cannot see. Here that is fes, snb, ring5, where the count says 2, -1, -4 and the rank says 3, 0, 9. The identity Maxwell is always right about — count equals mechanisms minus self-stresses — holds on every row.

The count that promises a mechanism

Count the joints, count the bars, subtract. The number that comes out promises rigidity when it is small and a mechanism when it is large, and it is wrong in both directions — because it assumes every bar removes a freedom the others have not already removed.

part 1 · applied

the kagome net: one of 1 mechanism. An infinitesimal mechanism of the kagome net, drawn as a velocity at every joint. The vector is an exact solution of the rigidity matrix — a set of joint velocities and a rate of change of the cell's metric under which no bar's length changes to first order — with the two rigid translations projected out so that what is left is a motion rather than a shift. Whether it continues into a finite motion is a separate question that a first-order calculation cannot answer, and this collection answers it for one framework by constructing the motion explicitly.

A fold that keeps its symmetry

The kagome framework has exactly one mechanism, and it does not stop at first order. Every triangle turns, alternate ones the other way, the cell shrinks to half its size, and not one bar changes length — and the count that found the mechanism cannot see how many there really are.

part 2 · applied

the kagome net: 34 of 144 wavevectors carry a mechanism. The zone of the kagome net, with a mark at every wavevector whose rigidity matrix drops rank — which is to say at every wavevector that carries a motion of the bars. There are few of them and they are isolated, so enlarging the cell adds mechanisms slowly. The ranks at the half-integer wavevectors are exact; the others are computed with a stated tolerance, because the matrix there has genuinely complex entries.

A mechanism that is a wave

The framework essays found the kagome net's mechanism count growing with the cell it was looked for in, and recorded it as a finding without an explanation. Here is the explanation: the motions lie along lines in reciprocal space, and a larger cell samples a line at more places.

part 3 · applied

One framework has a count of zero, one mechanism and one self-stress. Every net this collection has a placement for, as a periodic bar-and-joint framework in a fixed cell: its point group, the joints and bars of one cell, the scalar Maxwell count 2n − e − 2, and the mechanisms and self-stresses found exactly from the rank of the rigidity matrix. The scalar count is always the difference of the last two, which is Maxwell's identity — and the bathroom net is the row that shows what the identity costs: nought equals one minus one, and a framework that reads isostatic moves.

The mechanisms a count cannot see

Maxwell's count subtracts constraints from freedoms, and a mechanism and a state of self-stress cancel in the subtraction — so a framework with one of each reports the same number as a rigid one. The bathroom net reports nought and moves. Doing the same subtraction with representations instead of numbers separates them, because a mechanism and a self-stress cancel only when they belong to the same representation.

part 4 · applied

Five solids, twice each. Each Platonic solid as a framework of rods hinged at the corners, and again with its faces made rigid by adding their diagonals. The rank of the rigidity matrix reaches 3V − 6 exactly when the framework cannot move; the shortfall counts the ways it can. Three of the five are rigid as rods and all five are rigid as plates, which is Cauchy's theorem in the form a rank computation can see.

The polyhedra that can flex

A cube of rods folds and a cube of cardboard does not, and the difference is a rank. Cauchy proved in 1813 that a convex polyhedron with rigid faces is rigid; the rank of a rigidity matrix sees it directly, and it also sees where the hypothesis is doing the work. Drop convexity and an octahedron flexes — followed here for forty steps with every edge length held to five parts in a thousand million million.

part 5 · applied

One determinant, three dimensions. The Cayley–Menger determinant of a set of squared distances, at three sizes. Its value is the squared content of the simplex those distances describe, times a factor that alternates in sign with the dimension. At three points it is Heron's formula rewritten; at four it gives a tetrahedron's volume from its six edge lengths with no coordinates anywhere. The alternating sign is not a convention — a value of the wrong sign means the distances belong to no set of points at all.

What six lengths decide and nine do not

A tetrahedron's volume is a determinant in its six edge lengths, with no coordinates anywhere. Add a fifth vertex and the lengths stop deciding: two shapes with identical edges and identical faces have volumes in the ratio 2.6. What survives is that the possibilities are finite — which is the whole reason a flexing polyhedron cannot change its volume.

part 6 · applied

Nine frameworks, counted and then decided. Maxwell's count subtracts bars from twice the joints; the pebble game inserts the bars one at a time and discards any that cannot be paid for. The two agree on most of these frameworks and not on all, and where they differ the count is the one that is wrong — it assumes every bar is an independent constraint, and a bar added to a part that is already rigid is not. The redundant column is how many bars the game refused.

A game that decides what counting only bounds

Maxwell's count subtracts bars from twice the joints and is a bound, not an answer, because it assumes every bar constrains something new. In the plane there is an exact repair: Laman's condition, run as a game in which each joint holds two pebbles and a bar is admitted only if four can be gathered at its ends. Two rigid bodies sharing a joint are what the count gets backwards.

part 7 · applied