Phyllotaxis
the spirals are counted, not admired
Series

Rising — the series

3 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. A stem grown at 42 nodes per rung. 167 nodes, each placed where the repulsion from the ones below it was least, with the rise falling from 0.2 to 0.0045. Counted blind in a sliding window the pattern walks 1/2 → 2/3 → 3/5 → 5/8, and the marks are where its answer changed.

    A pattern with a rate

    Every lattice in the essays before this one is a static object indexed by a parameter, and a plant is not. Put the rise on a clock, place each node where the repulsion from the ones below it is least, and the object that comes out has a history — which is the first thing here that could disagree with the ladder.

    part 3 · emergence
  2. The memory belongs to the rise, not to the lattice. The lag-one correlation of a noiseless rising stem, against how fast it climbs the ladder. Below about sixty nodes per rung it is negative; above it, 0.54, 0.74, 0.58, 0.58, 0.55 — flat across a fivefold change in rate. The horizontal line is the same rule with the rise held FIXED, where the correlation is -0.68. So the +0.74 the earlier work called the sequence's own memory is the pattern chasing an equilibrium that is moving under it.

    A shoot too fast to remember

    Sweep the rate at which a stem climbs the ladder and the correlation between one divergence and the next changes sign — negative below about fifty-five nodes per rung, positive above it, with the flip inside one step of the grid. The instrument the earlier work proposed is unavailable on a fast shoot, and nothing said so.

    part 4 · emergence
  3. A window that fits inside a rung. Stems that climb the ladder at four rates, read over a window at the fine end. The condition is a ratio: the window has to be shorter than a rung. 250 internodes at 130 per rung is 1.92 rungs and agrees on 0 of 3; 400 internodes at 130 per rung is 3.08 rungs and agrees on 0 of 3; 250 internodes at 260 per rung is 0.96 rungs and agrees on 3 of 3; 400 internodes at 260 per rung is 1.54 rungs and agrees on 1 of 3; 250 internodes at 520 per rung is 0.48 rungs and agrees on 2 of 3; 400 internodes at 520 per rung is 0.77 rungs and agrees on 3 of 3; 250 internodes at 1040 per rung is 0.24 rungs and agrees on 3 of 3; 400 internodes at 1040 per rung is 0.38 rungs and agrees on 3 of 3. Read over the whole stem instead, every rate returns nothing — 0 of 3, 0 of 3, 0 of 3, 0 of 3 — because the quantity the comb is periodic in changes as the pattern climbs.

    A window inside a rung

    A stem that climbs the ladder has no comb in it at any rate, because the quantity the comb is periodic in changes as it goes. Read a window instead and it comes back, on one condition: the window has to be shorter than a rung — which makes the shoot's rate the thing that decides whether a plant can be asked.

    part 5 · emergence

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