Tuesday, March 23, 2021
Oblique locked crossings
Thursday, March 18, 2021
Sinusoidal notches for locked crossings
The fourth notch is engaged by bending the sides of both weavers upward, as in the photo below:
The square pencil packing in Diamond Weft
Wednesday, March 17, 2021
The NbO net and 3D weaving
Denser than Diamond Weft
Tuesday, March 16, 2021
Weaving with locked crossings
Thin, flat weavers can be locked together at a fixed position and angle of crossing with four side notches in each weaver. In an x-ray view of the completed locked crossing, the respective notches just barely overlap. For a 90-degree crossing the overlap areas form a square (the notches on each weaver may not appear to be arranged in a square because the locus of overlap is eccentrically located on each notch.)
These photos are of Diamond Weft. I used an earlier version of locked crossings with just two notches per weaver in They Urned It (a data sculpture based on the expansion of the Fed balance sheet), but using just two notches relies on a certain interplay between the surface curvature and the notch location to keep the crossing locked. While it may seem it would be difficult to engage all four pairs of notches at a crossing, if the material is thin and flexible enough, engaging the fourth pair of notches is a move similar to getting the last corner down in the familiar weave method of closing the flaps on a cardboard box. Here are some accurately cut paper weavers with a 1/8" punch used to shape the bottom of the notch, along with an "X-ray" view of the interlocked crossing.Thursday, October 29, 2020
Periodic, isotropic weaves from overlays of three quasi-cartesian grids
Three copies of the cartesian grid cannot be overlaid into a pattern that is both periodic and isotropic. This is a familiar problem in designing halftone patterns for printing. The traditional arrangement of three square-grid halftones avoids distracting moire patterns with a 120° rotation dispersal that results in a pattern of dot 'rosettes' that is isotropic but not periodic. Wang and Loce show the way towards periodicity. If the orthogonal grids are not square but 1:0.866 rectangles (i.e, based on the base-to-altitude ratio of equilateral triangles), and still arranged 120° to each other, isotropic and periodic patterns are possible.
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| Three copies of a rectangular grid with aspect ratio 0.866 can be rotated 120° to each other and still form periodic patterns. |
We may prefer, instead of sacrificing the equilateral property of the cartesian grid, to sacrifice its orthogonality. Three copies of a slightly sheared (skew) cartesian plane can be overlaid with 120° rotation dispersion to form periodic patterns. (The sheared grid is formed by rotating two copies of a parallel ruling ± 43.898 degrees.)
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| Three copies of a slightly sheared cartesian grid can be rotated 120° to each other and form periodic patterns. |
A cartesian grid has two line directions, after three copies are dispersed 120° from each other, there are six line directions (see figure below.) It is somehow easier to perceive these six directions as three narrow-angle 'flashlight beams' with (half-angle) beam-spread of 15°.
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| Three right-angles rotated 120° to each other tend to be perceived as three narrow beams with half-angle 15°. |
In Panda, Maulik, Chakraborty, and Khastgir there a several periodic solutions for billiards played on an equilateral triangular table. Any of these solutions that preserve the full symmetry of the triangle could decorate the equilateral triangles of a deltahedron resulting in a surface wrapped by geodesic lines.
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| Periodic solutions to billiards on an equilateral triangle table (from Panda, Maulik, Chakraborty, and Khastgir.) |
The solution labelled (m=2, n=5) caught my eye, me thinking that the lines were orthogonal (in fact, they are just slightly oblique.)
Flat weaving in this pattern would look something like this:
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| Flat weaving the (m=2, n=5) periodic solution to billiards-on-a-triangle. Heavy lines emphasize the kagome-like organization. |
Extended in this way, it becomes easier to see that the lines are not quite orthogonal. This geometric construction shows that an angle that needs to be 15° for orthogonality is actually atan(sqrt(3)/7) = 13.8979°.


















