Phyllotaxis
the spirals are counted, not admired
Where the angle comes from

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.
18 min read 9 figures State the claim, state the testOne parameter, one familyCounted, not admired

Worth reading first: A pattern with a rate · The sequence has a memory · The lag that is not there.

Every stem the previous essays read was held at a fixed rise, and no plant is. A shoot climbs the ladder: its rise falls as it grows, the counted pair changes at computable transitions, and a band over which the pair is constant is a finite stretch of stem rather than the whole of it.

That is not a detail to be waved at. The comb is a periodicity at the parastichy number, so a sequence in which the parastichy number changes has no single period in it, and everything the last four essays measured could in principle evaporate on a real plant.

The whole stem gives nothing

Grow stems at four rates from a hundred and thirty to a thousand and forty nodes per rung, each spanning the same range of rise — from 0.4 down to 0.004, which is the ladder from its coarse end to 8/13 — and read the whole divergence sequence.

Nothing. At every rate, on every seed, the readout refuses, and it refuses with comb scores of 0.02 to 0.07 against a band that the fixed-rise stems clear at 0.6.

A window that fits inside a rungStems 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.nodes per rung250-node window400-node windowwhole stem1301.92 rungs0/3 · 1 wrong3.08 rungs0/30/32600.96 rungs3/31.54 rungs1/3 · 1 wrong0/35200.48 rungs2/30.77 rungs3/30/310400.24 rungs3/30.38 rungs3/30/33 stems per cell · rise falls from 0.4 to 0.004 on every onefilled where the angles and the positions agree
Fig. 1 The right-hand column is the whole-stem reading and it is zero of three at every rate. The grid to its left is the same stems read over a window at the fine end, and the number under each mark is how many of three agreed with the position counter given the same internodes. The quantity that decides it is in the row above each mark: how many rungs the window spans.

That is the expected result and it is worth having as a measurement rather than as an inference, because the alternative was plausible. A stem that spends a long stretch on each rung might have carried a comb smeared across two or three periods, weak but readable. It does not: the correlation at a lag is built by disturbances propagating between neighbours, and when the neighbours change the propagation changes with them, so the contributions from different rungs do not add up — they cancel.

What a growing stem counts, against what a static lattice wouldThe steps are the blind counter's answer as the stem grows at 260 nodes per rung; the dashed verticals are the rises at which the static ladder changes. 299 of 302 counting windows agree, and the mean gap between where a transition happened and where the ladder puts it is 0.010 of a rung.0.50011.502falling rise, as −log₁₀which rung the pattern is on1/22/33/55/88/1387 nodes per rung · 415 nodes95 of 95 windows agree
Fig. 2 What a climbing stem does, counted blind in a sliding window: the pair changes at each transition and the stretch between transitions is one rung. A sequence that spans several of these has several periods in it and therefore none.

A window brings it back

Read instead the last two hundred and fifty internodes — a window at the fine end, which is where a real count is made and where the pattern is most developed.

At two hundred and sixty nodes per rung, three stems of three return the pair the position counter finds over the same internodes. At five hundred and twenty and at a thousand and forty, the same. At a hundred and thirty, none of the three do, and one of them is wrong rather than silent.

Widen the window to four hundred and the pattern inverts: at two hundred and sixty nodes per rung, one of three; at five hundred and twenty, three of three.

The quantity that predicts every cell is the ratio of the two — the rungs spanned, which is the window divided by the nodes per rung. At 0.96 rungs it works, at 0.48 it works, at 0.24 it works, at 1.54 it does not, and at 1.92 it does not.

The window has to fit inside a rung.

A window that fits inside a rungStems 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 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. Read over the whole stem instead, every rate returns nothing — 0 of 3, 0 of 3 — because the quantity the comb is periodic in changes as the pattern climbs.nodes per rung250-node window400-node windowwhole stem2600.96 rungs3/31.54 rungs1/3 · 1 wrong0/35200.48 rungs2/30.77 rungs3/30/33 stems per cell · rise falls from 0.4 to 0.004 on every onefilled where the angles and the positions agree
Fig. 3 The two rates and two window lengths that bracket the condition. The same rate passes at one window and fails at the other, and the same window passes at one rate and fails at the other, so what is being measured is the ratio and not either number on its own.

Which turns into a rate a plant has to be slower than

The two requirements now multiply, and they pull in opposite directions.

The pair needs at least two hundred and fifty divergences, because the second comb is weak and has to clear a threshold that falls as one over the square root of the length.

The window must be at most one rung, because a sequence spanning more than one rung has no period.

Put together: a rung has to be at least two hundred and fifty nodes long, so the shoot has to climb the ladder more slowly than about two hundred and fifty nodes per rung.

That is four times slower than the rate this site’s rising runs have used throughout — ninety-six nodes per rung was the foundation phase’s choice and the scale phase’s sweeps run from forty to three hundred and twenty. So the condition is not automatically satisfied by the model’s own habits, and it is a fact about which plants can be asked rather than about the instrument.

The lag that is not thereEach dot is one rate: the mean gap between where the grown pattern changed its count and where the static ladder puts that transition, in rungs. Over rates from 9 to 135 nodes per rung the worst is 0.068 of a rung. A lag of one rung would put a dot on the top line.-0.500-0.25000.2500.50011.251.501.752nodes the stem spends per rung, log₁₀transition late by, in rungsone rung late8 rates · rise 0.4 → 0.004worst mean lag 0.068 rungs
Fig. 4 The rate against what it costs the pattern, from the scale phase: the static ladder survives contact with a rate to better than a fifth of a rung over a fifteenfold range. The rate matters here for a different reason — not whether the pattern lags, but whether a stretch of stem long enough to measure sits on one rung.

What a rate means on a real plant

Nodes per rung is a model quantity and it is worth translating.

A rung of the ladder is a transition of the counted pair — 5/8 becoming 8/13. So “two hundred and fifty nodes per rung” means: two hundred and fifty leaves are produced between the height at which the stem counts 5 and 8 and the height at which it counts 8 and 13.

On a stem whose pattern is not changing at all — an established shoot at a fixed phyllotaxis, which is the common case for a long unbranched stretch — the rate is effectively infinite and the condition is satisfied trivially. The condition binds on material where the phyllotaxis is developing: seedlings, the transition from juvenile to adult phyllotaxis, the approach to an inflorescence.

Which is an awkward division, because the developing material is exactly where the sequence would be most interesting and the established material is where it is easiest to measure. The instrument works best on plants whose pattern has stopped being interesting.

A stem grown at 260 nodes per rung1244 nodes, each placed where the repulsion from the ones below it was least, with the rise falling from 0.4 to 0.0040. Counted blind in a sliding window the pattern walks 1/2 → 2/3 → 3/5 → 5/8 → 8/13, and the marks are where its answer changed.1/2 → 2/32/3 → 3/53/5 → 5/85/8 → 8/131/2 at the bottom, 8/13 at the top1244 nodes · rise 0.4 → 0.0040260 nodes per rung
Fig. 5 A stem grown at two hundred and seventy nodes per rung, which is just inside the condition. The marks are where the blind counter’s answer changes; the window this essay reads is the last two hundred and fifty nodes, which lies between the top two marks.

The wrong answer at the fast rate

One of the three stems at a hundred and thirty nodes per rung returns a pair, and it is not the pair the counter finds over the same internodes.

That is a failure worth looking at rather than counting. A window spanning nearly two rungs contains a stretch of 5/8 and a stretch of 8/13, and both leave correlation behind. The lags that clear are a mixture of two combs at different spacings, and the search picks whichever is stronger and then finds a residue class in what the other one left. The result is a pair, cleanly reported, made of one number from each half of the window.

The margin test catches most of these — the two spacings compete and neither wins by a band — and it catches two of the three here. The third is a window in which one rung happens to dominate the other enough for its comb to win, and the second comb it then finds is contaminated.

There is a diagnostic available and it is worth stating even though it is not implemented: read two overlapping windows. A window inside one rung and a window shifted by half its length should return the same pair; a window straddling a transition and its neighbour will not. That is the same trick the foundation phase used to catch the counting bug — count in three bands and require one answer — applied to the sequence rather than to the positions.

One seed, two rates, two laddersBoth stems begin as forty nodes of Lucas lattice at a rise of 0.12. At 65 nodes per rung the divergence stays at 99.5° and the counts walk 1/3 → 3/4 → 4/7 → 7/11. At 131 it leaves for 137.7° and walks 1/3 → 2/3 → 3/5 → 5/8 → 8/13 instead.10011012013014011.502falling rise, as −log₁₀divergence the stem is producing (°)137.51°, Fibonacci99.50°, Lucas65/rung → 7/11131/rung → 8/13seeded at 99.50°, rise 0.127/11 against 8/13
Fig. 6 Two rates from the same seed, from the scale phase, showing how differently two shoots walk the same ladder. A window of fixed length on the faster of these spans twice as many rungs as on the slower, which is the whole of this essay’s condition drawn as two stems.

What this adds to the specification

The survey specification for reading a pair from angles now has four lines rather than three, and the new one is the most restrictive:

  • at least 250 consecutive internodes;
  • azimuths to a quarter of a degree;
  • a plant disturbed enough to have a sequence and not so disturbed as to have no lattice;
  • and a stretch over which the parastichy pair does not change, which means a shoot slower than about 250 nodes per rung, or an established stem where the pair is not changing at all.

The fourth is checkable independently of the instrument: count the spirals at the bottom of the window and at the top, and require the same answer. If the pair changes across the window, the sequence cannot be read — and that check needs a photograph, which is a mild irony for an instrument whose selling point is that it does not need one.

The branch is kept below 85 nodes per rung and lost above 92Each row is one rate. The Lucas seed keeps its ladder at 46, 58, 65, 75, 85 nodes per rung and abandons it at 92, 108, 131. The golden seed ends on 8/13 at every one of them.nodes per rungseeded Lucasseeded golden467/11 — kept8/13 — Fibonacci587/11 — kept8/13 — Fibonacci657/11 — kept8/13 — Fibonacci757/11 — kept8/13 — Fibonacci857/11 — kept8/13 — Fibonacci928/13 — gone to Fibonacci8/13 — Fibonacci1088/13 — gone to Fibonacci8/13 — Fibonacci1318/13 — gone to Fibonacci8/13 — Fibonacciseeded with 40 nodes at a rise of 0.12threshold between 85 and 92
Fig. 7 The other rate threshold this site has measured, at about ninety nodes per rung, which decides whether a shoot keeps the branch it started on or finds the Fibonacci one. The two thresholds are unrelated in mechanism and both are properties of the shoot rather than of the pattern, which is a coincidence of subject rather than of physics.

Why the ratio and not the rate

The grid is four rates by two window lengths, and it is worth saying why that shape was chosen rather than a longer sweep of one variable.

If the condition were about the rate — if some shoots were simply too fast to read — then a fixed window would work above some rate and fail below it, and one column of the grid would tell the whole story. If it were about the window — if some lengths were too long — then a fixed rate would work below some length and fail above it, and one row would.

Neither is what happens. Two hundred and sixty nodes per rung passes at a two-hundred-and-fifty-node window and fails at four hundred. Four hundred passes at five hundred and twenty nodes per rung and fails at two hundred and sixty. The same rate passes and fails; the same window passes and fails. Only the ratio sorts the cells.

That is a stronger conclusion than either single-variable version and it needed both dimensions to see. It also gives the condition a form that transfers: a window of any length is readable on a shoot of any rate provided the quotient is under one, so a plant with very long rungs can be read over a very long window and gain the precision that buys — which is the trade the protractor essay prices.

A window that fits inside a rungStems 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. Read over the whole stem instead, every rate returns nothing — 0 of 3, 0 of 3, 0 of 3 — because the quantity the comb is periodic in changes as the pattern climbs.nodes per rung250-node window400-node windowwhole stem1301.92 rungs0/3 · 1 wrong3.08 rungs0/30/32600.96 rungs3/31.54 rungs1/3 · 1 wrong0/35200.48 rungs2/30.77 rungs3/30/33 stems per cell · rise falls from 0.4 to 0.004 on every onefilled where the angles and the positions agree
Fig. 8 Three rates against two windows, which is the smallest grid in which the ratio can be distinguished from either variable alone. The diagonal structure — passing cells above and to the right of a line — is what a ratio condition looks like, and it is not what a threshold on either axis would look like.

The transition is not a smooth boundary

The failing cells fail in two different ways and the difference is worth noting, because it decides what a survey should do about a marginal specimen.

At 1.54 rungs the window straddles one transition. It contains a long stretch of one rung and a shorter stretch of the next, and the two combs compete: usually neither wins by a band and the readout refuses, but occasionally one dominates and the readout reports a mixture — a spacing from one rung and a residue from the other.

At 1.92 rungs the window contains nearly two full rungs. Here the competition is even and the margin test refuses reliably, which is the safer failure.

So the dangerous region is not the worst one. A window that is a little too long is more dangerous than one that is much too long, because a slight imbalance between the two rungs is enough to produce a confident wrong answer while a large one is not. That is the same shape as the branching thread’s finding — the twig sample that returns 1.7 with a tight interval is more dangerous than one that returns nothing — and it is the same lesson: an instrument’s uncertain region is not where it is quietest.

The practical consequence is that “keep the window comfortably inside a rung” is better advice than “keep it inside a rung”, and the margin the grid supports is about a factor of two: at half a rung every cell tested passes.

What a growing stem counts, against what a static lattice wouldThe steps are the blind counter's answer as the stem grows at 520 nodes per rung; the dashed verticals are the rises at which the static ladder changes. 606 of 613 counting windows agree, and the mean gap between where a transition happened and where the ladder puts it is 0.012 of a rung.0.50011.502falling rise, as −log₁₀which rung the pattern is on1/22/33/55/88/13520 nodes per rung · 2487 nodes606 of 613 windows agree
Fig. 9 A slower shoot than the one above, at five hundred and twenty nodes per rung, where a two-hundred-and-fifty-node window is half a rung and every run reads. The stretches between the marks are what a window has to fit inside, and on this stem they are long enough that placement hardly matters.

The shape of the answer

The mixture problem this thread has been circling for three phases now has a shape rather than a verdict, and the shape is: two statistics, four ways to fail, and a refusal that does not say which.

The two statistics come off one stem. The window in disturbance is wide and the lag-one correlation is flat across it. The window in rate is the new constraint and it is one-sided — slower is always better. What none of that gives is an inversion: a stem that returns nothing has failed one of four tests, and the pair of readings does not report which one.

A ruler’s reading of the scatter removes two of the four. What remains is “too fast” against “the window was in the wrong place”, and separating those needs a second window at a different height — which is the measurement the next essay specifies and this site cannot make.

What the condition is not

It is not a statement about the pattern lagging behind the ladder. The scale phase measured that separately and found the static ladder survives contact with a rate to better than a fifth of a rung over a fifteenfold range — so a fast shoot’s pattern is where the geometry says it should be, it just does not stay there for long.

Nor is it a statement about noise. The rate condition holds at every disturbance inside the readable window, and the disturbance window holds at every rate that satisfies the rate condition. The two constraints are independent, which is what makes them multiply rather than trade.

What it is, is a sampling condition: an instrument that measures a periodicity needs the period to be constant across what it measures. That is true of every spectral method ever written and it arrives here in a form with a plant in it — a rung is a stretch of stem, and the stretch has to be longer than the window.

Two rungs is a design choice

The grid tests window-to-rung ratios from a quarter to nearly two, and the passing side is broad: a quarter, a half and just under one all pass on every stem tried. Somebody designing a measurement would therefore aim at half a rung, which on a shoot of five hundred nodes per rung is a two-hundred-and-fifty-node window and is exactly the length the pair needs anyway.

That coincidence is worth noticing because it is not one. The window length is set by the second comb’s weakness and the rung length by the shoot; a plant on which the two happen to be equal is a plant that is only just measurable. A margin of two in the ratio means a margin of two in the shoot’s rate, and the honest specification asks for the margin rather than for the boundary — five hundred nodes per rung rather than two hundred and fifty, on the same reasoning that put two hundred and fifty internodes in the specification rather than a hundred and fifty.

Where this goes next

012345-1.70-1.30-1-0.824-0.602-0.398-0.2220disturbance amplitude, degrees of azimuth per nodestems out of five returning the counted pair0.020.050.10.150.250.40.61lag-one correlation = 0lag one, on the same sequencesrise 0.005 · 5 stems per point · bars are the pair, line is lag onefilled where the pair agrees with the position counter Two readings from one stem Three phase plans in a row have recorded that the two statistics of a divergence sequence want opposite plants — one quiet, one disturbed. Measured on the same stems they do not. The conflict was in the interpretation of a sign, and the window in which both are readable is wide. 10011012013014011.502falling rise, as −log₁₀divergence the stem is producing (°)137.51°, Fibonacci99.50°, Lucas65/rung → 7/11131/rung → 8/13seeded at 99.50°, rise 0.127/11 against 8/13 The rate decides the branch Forty nodes of Lucas lattice, carried down to the same fine rise twice. Hurried, the pattern holds the Lucas ladder through three more forks at 99.5°. Given room, it abandons it at the first fork it reaches and walks to 8/13 at 137.5°. Same seed, same rise, two ladders — with a threshold between them at about ninety nodes per rung. 1/2 → 2/32/3 → 3/53/5 → 5/81/2 at the bottom, 5/8 at the top167 nodes · rise 0.2 → 0.004542 nodes per rung A pattern with a rate Every lattice on this site 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.

Shares its objects with

Essays that name at least two of the same things, and that neither author linked.

Named objects

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AutocorrelationDivergence angleHonest limitsIdentifiabilityLadderMeasurementMeristem growthNodes per rungParastichy pairRateRungSamplingSpecimenTrackingTransitions