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.
4 essays call
frame-flex. The drawing above is what it returns with no arguments at all; every
call below passes it something, because a placement that passes nothing draws whichever member
of the family the generator happens to default to rather than the one its essay argues about.
Every one of this site's 393 essays names its parameters at the
call site, which the standard pass of 2026-08-09 established and param-floor
holds.
Where it is called
Changing this generator changes every one of these figures.
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.
The level that does not move
Three levels cross the kagome net's zone and one of them is a horizontal line. The reason is a state that alternates in sign round a single hexagon and is exactly zero everywhere else — a solution with no wavevector in it at all, which is why no wavevector can move it.
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.
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.