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

The lag that is not there

A pattern built out of its own history should hold its old parastichy pair past the point where a fresh lattice would have changed, and the gap should grow as the shoot is hurried. Over a fifteenfold range of rate it does not — every transition lands within a tenth of a rung of where the static ladder puts it.
16 min read 6 figures Two routes, one numberState the claim, state the test

This essay was written to report a curve and reports a flat line, and the flat line is the result.

The expectation going in was straightforward. A grown pattern is built out of neighbours placed earlier, at coarser rises. The static ladder says which pair of lattice vectors is shortest at each rise, computed fresh, with no history in it at all. A pattern with a history should therefore hold its old pair past the rise at which the new one becomes shorter — it should lag — and the lag should grow as the shoot is hurried, because a hurried shoot has fewer nodes in which to reorganise.

None of that happens.

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 0.087 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 → 0.0012worst mean lag 0.087 rungs
Fig. 1 Each dot is one rate: the mean gap between where the grown pattern changed its count and where the static ladder puts that transition, measured in rungs. A lag of one whole rung would put a dot on the top line.

The measurement

A run covers a fixed range of rise — from 0.4 down to 0.0012, about six rungs — at a stated rate, so a slow shoot is long and a fast one short. A counting window of twenty-two nodes is slid along it, looking only downward, and every place its answer changes is recorded with the rise at which it changed.

Beside each of those sits the rise at which the static ladder puts the same transition: 1/2→2/31/2 \to 2/3 at 0.12472, 2/3→3/52/3 \to 3/5 at 0.04767, 3/5→5/83/5 \to 5/8 at 0.01822, 5/8→8/135/8 \to 8/13 at 0.00696, and downward by a factor of 1/φ21/\varphi^2 each time.

The gap between the two, in rungs, is

ln⁡(hstatic/hmeasured)ln⁡φ2\frac{\ln(h_{\text{static}}/h_{\text{measured}})}{\ln \varphi^2}

Positive means late. Across eight rates from nine nodes per rung to a hundred and thirty-five, the mean gaps are

−0.037, 0.087, 0.024, 0.035, 0.011, 0.013, 0.022, 0.015-0.037,\ 0.087,\ 0.024,\ 0.035,\ 0.011,\ 0.013,\ 0.022,\ 0.015

The worst is under a tenth of a rung. There is no trend with rate, the sign is not consistent, and the whole spread sits inside what one counting window’s resolution can distinguish.

The ladder was computed with no dynamics in it and it survives contact with a rate.

Why this is a result and not an absence

A measurement that finds nothing is worth reporting only if it could have found something, so it is worth being precise about what would have shown up.

A lag would have been large. A pattern one rung behind would be at 5/8 while the equilibrium lattice was at 8/13 — a difference anybody would see, and the sort of thing that would show as a dot near the top line. What is measured is two orders of magnitude smaller than that.

A lag would have had a trend. The mechanism proposed for it — fewer nodes in which to reorganise — is monotone in the rate, so the lag should grow as the shoot is hurried. The measured values do not: the largest is at thirteen nodes per rung and the second largest at eighteen, with the slowest four rates all under 0.023.

And the instrument would have caught it. The same window, on the same runs, resolves the transitions well enough to locate each of them to within a few per cent of a rung. It is not a blunt instrument reporting zero because it cannot see.

So the honest statement is not “no lag was detected”. It is that over the measurable range the grown pattern is at the equilibrium the static calculation predicts, to better than a tenth of a rung, and any lag is smaller than that.

What a growing stem counts, against what a static lattice wouldThe steps are the blind counter's answer as the stem grows at 92 nodes per rung; the dashed verticals are the rises at which the static ladder changes. 82 of 83 counting windows agree, and the mean gap between where a transition happened and where the ladder puts it is 0.012 of a rung.11.502falling rise, as −log₁₀which rung the pattern is on1/22/33/55/88/1392 nodes per rung · 365 nodes82 of 83 windows agree
Fig. 2 One run in detail. The steps are what the grown pattern counts; the dashed verticals are where the static ladder changes. Eighty-two of eighty-three counting windows agree.

Why the ladder is so hard to leave

The result is surprising for about a minute and then it is not, and the reason is worth having because it says something about what kind of object a parastichy pair is.

A parastichy pair is not a position, a phase or an amplitude. It is an ordering: which two index offsets have the shortest hops. Orderings are discrete, and a discrete quantity does not lag continuously — it either changes or it does not.

For the pattern to be a rung behind, the arrangement it is holding would have to remain the local optimum well past the point where a different arrangement became the global one. But the rule places one node at a time against its immediate neighbours, and the immediate neighbours are exactly what a parastichy pair is a statement about. There is no long relaxation time here for a lag to accumulate in: the pattern’s memory is a couple of rows deep, and a couple of rows is a fraction of a rung at every rate tested.

Put differently: the lattice adjusts locally and the ladder is a local statement. A quantity with a long memory — a phase, an overall rotation, an accumulated error — could lag. This one has nowhere to store the lag.

Six stems built, forgotten and recoveredEach row is a lattice built from a divergence and a rise, counted by machinery shown only the coordinates, and reconstructed from the counts and the two hop lengths. The worst error in the recovered angle is 3.0e-13°.137.51°, rise 0.092.8e-14°counted 2/3137.51°, rise 0.031.7e-13°counted 3/5137.51°, rise 0.0122.8e-14°counted 5/899.50°, rise 0.083.0e-13°counted 1/3151.14°, rise 0.071.1e-13°counted 2/399.50°, rise 0.027.1e-14°counted 4/7error in the recovered divergence anglecounts and hop lengths onlyworst 3.0e-13°
Fig. 3 What the static side of the comparison rests on. A stem built from two numbers, counted blind, and reconstructed from the counts and the hop lengths to the last digit — so the rises the transitions are compared against are measured rather than labelled.

What a rung is worth, in nodes

There is a way of restating the result that makes it look almost inevitable, and it is worth doing because it also bounds how far the result can be pushed.

A rung is 2ln⁡φ≈0.9622\ln\varphi \approx 0.962 in the log of the rise. At a rate of TT, the pattern spends 0.962 T0.962\,T nodes crossing it. The pattern’s memory — the neighbourhood the placement rule actually sees — is about 6/h6/\sqrt h nodes, which at a rise of 0.01 is sixty and at 0.001 is a hundred and ninety.

So at the slow end the memory is a fraction of a rung and at the fast end it is several rungs. That last part is the surprise: at nine nodes per rung the placement rule is looking at nodes placed five or six rungs earlier, and still the transitions land on schedule.

The reason is that a neighbourhood spanning several rungs is not a neighbourhood spanning several arrangements. The nodes at the far end of the window are far away in distance too, and the repulsion falls off as the inverse cube — so they contribute almost nothing to where the next node goes. The effective memory is set by the distance falloff rather than by the window, and the distance falloff is local at every rate.

Which means the result would not survive a rule with a slower falloff. That is a prediction rather than a finding, and it is the obvious thing for a next phase to test.

Where the expectation came from

It is worth saying where the expected curve came from, because it was not idle: it came from a real phenomenon in a neighbouring subject, and the reason it does not transfer is the interesting part.

In a system driven slowly through a bifurcation, the state lags the equilibrium by an amount that grows with the driving rate. That is standard — it is why a slowly cooled magnet has a different domain structure from a quenched one, why a swept oscillator’s response peaks late, and why hysteresis loops open up as the sweep is hurried. The general shape is a relaxation time competing with a driving time, and the ratio of the two is the lag.

Phyllotaxis has a bifurcation structure — the van Iterson tree forks, and the forks are where the pattern has to choose — so the analogy is not far-fetched. The expectation was a relaxation time somewhere in the arrangement, and a lag proportional to the rate divided by it.

What the analogy misses is that the quantity being tracked is not continuous. A magnet’s magnetisation can be part-way; a swept oscillator’s phase can be behind. A parastichy pair cannot be between 5/8 and 8/13. There is no partial state for a lag to occupy, and the transition happens when the ordering flips, which is a local comparison made afresh by every node.

So the correct analogy is not a driven oscillator. It is a system whose order parameter is discrete and whose correlation length is short — and such a system does not lag, it just changes.

One run, in detail

The aggregate is a flat line, and it is worth looking at a single run to see what the flatness is made of.

At ninety-two nodes per rung, a stem covering the rise from 0.2 to 0.0045 is three hundred and sixty-five nodes long, and the sliding window gives eighty-three counts along it. Eighty-two of the eighty-three agree with the static ladder’s pair at that window’s rise.

The one that disagrees is at a transition, and it disagrees by being a window early. That is not the pattern being wrong; it is the window straddling a change and reporting one of the two answers. A window of thirty-six nodes at that rise spans roughly a fifth of a rung, so it will contain a transition for about a fifth of a rung’s worth of positions, and during that stretch the answer it gives is decided by where the majority of its nodes are.

Which is the sense in which the resolution and the result are the same size. The pattern’s transitions are at the ladder’s rises to better than the width of the instrument, and making the instrument narrower makes it noisier rather than sharper — a window of a dozen nodes cannot resolve a pair of thirteen and twenty-one at all.

There is a version of this measurement that would be sharper, and it is worth naming as work not done: track the individual lattice vectors rather than the counted pair, and record the rise at which two of them cross in length. That is a continuous quantity and it would give a lag with an error bar rather than a lag inside a window width. It would also stop being a blind measurement, since the vectors are what the ladder is computed from — which is why it was not done.

Where a rate does matter

None of this means the rate is irrelevant, and the next essay is the case where it decides everything.

The distinction is between which rung and which ladder. The rungs are local, discrete and adjusted locally, and the pattern hits them on schedule at every rate. The branch — Fibonacci or Lucas or something else — is a global property of the whole arrangement, it is not re-derived at each node, and it turns out to be metastable: a pattern can hold a branch that is not the lowest one available, and whether it holds it through a fork depends on how many nodes it spends there.

So the two results together say something sharper than either alone. The static ladder describes a growing stem exactly, provided the branch it is on is already known — and knowing that is a question about history, not about rise.

What the model’s limits do to this

Three bounds, and the first is the one that matters most.

Nine nodes per rung is the fast end, and it is set by the measurement rather than by the model. A counting window of twenty-two nodes has to contain enough of a rung to resolve where a change happened; below about six nodes per rung it does not, and the window is straddling two transitions at once. So the claim is about rates from nine upward. Whether something happens at two nodes per rung is not known here, and a shoot at two nodes per rung would be passing a transition every other node, which is not obviously a thing a plant does.

The rule is a sampled circle with a truncated neighbourhood. The azimuth is chosen from 512 sample points, quantising the divergence to about 0.7°, and the window is capped. Both are the same class of limitation the disc implementation records for its own sweep, and both are stated rather than worked around.

And this is one rule. Inverse-cube repulsion against a finite neighbourhood is a model of form, not a mechanism, and a different rule with a genuinely long memory — a mechanical one with elastic stresses, say, or a chemical one with slow diffusion — could behave differently. What has been shown is that this rule tracks, and that the tracking has a reason which is about the locality of the quantity rather than about the details of the rule.

Every transition as the rise fallsThe pair climbs 1/2 → 2/3 → 3/5 → 5/8 → 8/13 → 13/21. Consecutive transitions are 0.382, 0.382, 0.383, 0.382 of the previous rise — 1/φ² is 0.3820.0.2500.5000.75011.25-2.50-2-1.50-1-0.500log₁₀ of the rise between nodes (falling to the right is the plant growing)log₁₀ of the larger parastichy number2/33/55/88/1313/21500 rises, shortest vectors recomputed at eachratio 0.3820 against 1/φ² = 0.3820
Fig. 4 The rungs the grown pattern is being measured against. They are computed from a periodic lattice with no dynamics in it, and a stem grown at any rate here lands on them.

What the tracking buys the rest of the site

Three phases of this site have computed properties of static lattices and applied them to plants. The application has always carried an implicit assumption, never stated and never tested: that a growing organ is, at each moment, approximately the equilibrium lattice at its current rise.

That assumption is now measured, and it holds to a tenth of a rung over a fifteenfold range of rate.

Which retroactively licenses a good deal.

The ladder is a statement about growing stems and not only about periodic lattices. The disc’s transition radii, predicted from the cylinder’s ladder, are predictions about real heads to the extent that a head is a slow enough shoot — and a head at a thousand elements is passing rungs at hundreds of elements each, which is the slow end of this sweep. The cone’s φ² spacing is likewise a statement about growing cones.

None of those was in doubt exactly. All of them rested on an assumption that was easier to make than to check, and this is the check.

The plane of stems: divergence across, rise upEach shade is one parastichy pair. The marked points are the lattices where three families are equally short — the forks — and the Fibonacci ones run up the middle towards 137.51°.-2.50-2-1.50-1-0.500100120140160180divergence angle (°)log₁₀ of the rise between nodes1,2,32,3,548 × 150 lattices, each solved548 runs drawn
Fig. 5 The plane a growing stem is travelling through. A rate turns a point in this diagram into a path, and the result of this essay is that the path crosses the region boundaries where the diagram says it should.
The same counter, on a stem and on a discThe stem returns 2 and 3 in all three bands. The disc returns 21/34, 34/55, 55/89 — three answers to one question, which is why a published count needs to say where it was taken.stem, lower third2 and 3stem, middle third2 and 3stem, upper third2 and 3disc, r = 0.18–0.3221 and 34disc, r = 0.45–0.6234 and 55disc, r = 0.82–0.9955 and 89stem at rise 0.060, disc of 1600 points, both at 137.51°counted from coordinates onlyone answer against three
Fig. 6 The two static geometries this result licenses. Both are statements about equilibrium lattices, and both are now statements about growing organs to within a tenth of a rung.

Two things this does not license

The tracking result is broad and it is worth fencing it, because it is the kind of finding that invites being stretched.

It does not say a static lattice is a plant. What it says is that the sequence of parastichy pairs a growing rule produces matches the sequence a static calculation predicts. That is one quantity. The divergence wanders by a couple of degrees where a static lattice has none; the spacing between neighbours is not uniform where a static lattice’s is; the arrangement near a transition is genuinely intermediate for a stretch of nodes where a static lattice is never intermediate at all. Every one of those is a real difference and none of them shows up in a counted pair.

And it does not say the history is irrelevant. It says the history does not shift the rungs. Whether the history decides which ladder the pattern is on is a completely separate question with a completely different answer, and it is the next essay. A reader who takes “the static ladder describes a growing stem” away from this page and forgets the qualifier will get the following one exactly backwards.

The distinction is worth one more sentence because it is the useful one. A parastichy pair is a local fact — which two offsets have the shortest hops right here — and local facts equilibrate fast. A branch is a global fact about the whole arrangement, it is never re-derived from scratch, and it does not equilibrate at all unless something forces it to.

The result, stated as it is

A grown pattern’s transitions land where a periodic lattice’s do, to better than a tenth of a rung, at every rate between nine and a hundred and thirty-five nodes per rung.

That is a negative result about lag and a positive one about the ladder, and the second is the more useful. It is also the kind of finding this site exists to produce: the claim was plausible, the expected answer was a curve, the machinery was built to measure it, and what came out was a flat line that licenses three phases of earlier work.

An assertion that has never rejected anything proves nothing. This one rejected the essay it was written for.

Where this goes next

1/2 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. 0.2500.5000.75011.25-2.50-2-1.50-1-0.500log₁₀ of the rise between nodes (falling to the right is the plant growing)log₁₀ of the larger parastichy number2/33/55/88/1313/21500 rises, shortest vectors recomputed at eachratio 0.3820 against 1/φ² = 0.3820 The Fibonacci ladder Lower the rise on a cylinder and the parastichy pair climbs — 1 and 2, then 2 and 3, then 3 and 5 — each rung the sum of the two before it. Nothing in the arithmetic mentions Fibonacci, the transitions sit at computable rises, and consecutive ones stand in the ratio 1/φ². 0501000.2500.5000.7501growth parameter Gspread of the last 30 steps (°) — 0 means settledfilled dark: convergedthe usable range is stated, not implied Where the model stops Below a growth parameter of about 0.18 this implementation does not converge — the settled angle wanders over a hundred degrees however long the run. That is the range where the literature says the interesting behaviour lives, and it is worth a figure rather than a quietly chosen axis.