Phyllotaxis
the spirals are counted, not admired
What a plant might be doing

A step of one organ

The balanced pair inside a wrecked stem's period measures 88.0° to 147.2° against divergences of 99.1° to 138.0° — one organ's step, to within twelve per cent on every row. The residual is not scatter: every stem keeping a 5 or a 7 overshoots and every stem keeping a 4 or an 8 falls short.
13 min read 19 figures Counted, not admiredTwo routes, one number

Worth reading first: The damage has a period · The organ that was taken away · Counting the spirals.

Two adjacent chains, displaced in opposite directions by equal amounts is a picture that makes a prediction. If the two chains have changed places, then each of their organs sits where an organ one place along used to sit, and the displacement is one step of the sequence — which is the stem’s own divergence.

That is a number, it is already measured on every one of these stems, and it can be compared without fitting anything.

The exceptional pair, measured in divergences. One row per wrecked cut whose profile has exactly one pair of exceptional classes, drawn at the size of that pair divided by the stem's own settled divergence. Every row sits between 0.882 and 1.076, so the two chains that came apart moved by one organ's step rather than by two or by half of one. The residual is not scatter: rows are grouped by the lag the stem kept, and every lag sits wholly above the line or wholly below it. Why a surviving 5 or 7 overshoots and a surviving 4 or 8 falls short is not answered here.
Fig. 1 The size of the balanced pair on every row that has one, divided by that stem’s own settled divergence.

The prediction, stated so it can fail

Take the two exceptional classes of a wrecked cut, measure each one’s departure from the level the rest share, and average their sizes. Call that the size of the pair. Take the stem’s settled divergence from the control run. Divide.

If the two chains have exchanged, the ratio should be one. If they have moved by two places it should be about two, and if by half a step it should be about a half — and a divergence near 137° is far enough from 68° and from 275° that the three cases are not near-misses of each other.

A period of 8, and the two classes that are not with the rest. The same wrecked stem, folded on the lag it kept: one row per residue class, each drawn at the mean displacement of its own organs against the level the rest of them share. The bar through each row is the spread inside that class, and the widest of them is 0.87° — so within a class the displacement is a constant. six classes sit at the common level. The two that do not sit at 134.3° and -134.2°, equal and opposite to within 0.0 per cent, and they are neighbouring residues. The stem's own divergence is 137.44°, so an exception is one organ's step.
Fig. 2 One period, with the two exceptional classes drawn against the level the others share.

The answer

Across the seventeen rows that carry a balanced pair, the ratio runs from 0.882 to 1.076.

Sixteen of the seventeen are between 0.96 and 1.08 — within eight per cent of one step. The seventeenth is 0.882, and it is a Lucas 4/7 stem cut four places back whose surviving lag is 4.

So the picture holds. The exchange is by one organ, not two and not a half, on every row that has one.

The exceptional pair, measured in divergences. One row per wrecked cut whose profile has exactly one pair of exceptional classes, drawn at the size of that pair divided by the stem's own settled divergence. Every row sits between 0.882 and 1.076, so the two chains that came apart moved by one organ's step rather than by two or by half of one. The residual is not scatter: rows are grouped by the lag the stem kept, and every lag sits wholly above the line or wholly below it. Why a surviving 5 or 7 overshoots and a surviving 4 or 8 falls short is not answered here.
Fig. 3 The same seventeen rows ordered by the raw size of the pair rather than by the ratio.

The numbers, in degrees

The pairs measure 88.0°, 103.1°, 103.2°, 103.2°, 104.1°, 133.1°, 133.4°, 134.2°, 134.2°, 134.2°, 135.1°, 141.7°, 141.8°, 143.7°, 145.1°, 145.3° and 147.2°.

The divergences those are divided by are 99.1° to 138.0°. The two ranges overlap almost exactly, which is the whole result: a quantity nobody chose the units of comes out the size of a quantity measured somewhere else entirely.

It is worth noticing that the seventeen fall into two clusters — around 103° and around 140° — and that those are the two branches. Lucas stems settle near 99° and golden stems near 137°, so the clustering is the divergence’s and not the pair’s.

five limit divergences, all of them 137.5078 over a whole number. The golden angle is the k = 1 member of a family. Real bijugate plants — teasel, Cephalaria — are reported near 68.75°, which is exactly half of it, and the pairs they are counted at are the Fibonacci pairs doubled.
Fig. 4 The two limit angles the branches converge on, which is why the pairs cluster in two groups.

The residual is not noise

Twelve per cent is a loose bound for a quantity read on a grid of 1,536 azimuths, so the natural next question is whether the departures from one are scatter.

They are not. Sorted by the lag the stem kept:

Every row whose surviving lag is 5 has a ratio above one — 1.029, 1.030, 1.050, 1.053, 1.063, 1.076. Every row whose surviving lag is 7 is above one — 1.033, 1.034, 1.034, 1.051. Every row whose surviving lag is 8 is below one — 0.966, 0.968, 0.976, 0.976, 0.976, 0.981. And the single row whose surviving lag is 4 is below one, at 0.882.

Seventeen rows, four values of the lag, and no lag sits on both sides.

The exceptional pair, measured in divergences. One row per wrecked cut whose profile has exactly one pair of exceptional classes, drawn at the size of that pair divided by the stem's own settled divergence. Every row sits between 0.882 and 1.076, so the two chains that came apart moved by one organ's step rather than by two or by half of one. The residual is not scatter: rows are grouped by the lag the stem kept, and every lag sits wholly above the line or wholly below it. Why a surviving 5 or 7 overshoots and a surviving 4 or 8 falls short is not answered here.
Fig. 5 The same rows grouped by surviving lag, on which each group sits wholly above or wholly below one divergence.

Which makes it a leaving rather than a result

A residual that tracks a discrete label is a residual with something in it. What that something is, this essay does not say.

The obvious candidates are all checkable and none of them is checked here. The exchange might be by one step of the surviving family rather than one step of the sequence, which would put a factor involving k in it. The wrecked stem’s own divergence, which differs from the control’s by the slip, might be the right divisor rather than the control’s. Or the two exchanged chains might not land exactly on each other’s positions, relaxing a little afterwards by an amount that depends on how far apart they are.

The third is the one the sign pattern points at, because the spacing between adjacent chains is a function of the arrangement and the arrangement is what k labels. It is left stated rather than tested.

The hops of a 4/7 lattice, shortest first — Lucas, rise 0.020. Every lattice hop at this rise, ordered by how long its step is across the surface of the stem, with the one that a wrecked stem here leaves standing picked out. The two shortest are the contact families the counter returns, 4 and 7, and they differ in length by a factor of 1.082. The lags left standing after a removal are 4, sitting at rank 2 in this order, so the family the rule holds is a short step but not always the shortest one.
Fig. 6 The length ranking of a stem’s lags, on which the surviving lag sits and which any account of the residual would have to use.

What the comparison is not divided by

The obvious alternative divisor is the cut stem’s own divergence rather than the control’s, and it is worth saying why it is not used.

A wrecked stem’s divergence has slipped by a whole number of turns per period, which for a surviving 5 is about 72° and for a surviving 8 about 45°. Folded, its mean divergence is a long way from the control’s — 182° against 137° on one stem, 209° against 137° on another. Dividing by that gives ratios nowhere near one and nowhere near each other.

So the control’s divergence is the right divisor and the agreement is not an artefact of choosing it: the alternative was computed and it does not work.

A wreck is a whole number of extra turns. For each of the 19 stems that never repair, the slip of its settled divergence multiplied by the lag whose hop survived. Every value lands on a whole number of turns — the horizontal lines — with a largest departure of 2.97 degrees, against divergences that have moved between 0 and 103 degrees. 17 of the 19 close on exactly one turn. So a wrecked stem is the stem it was with one extra turn threaded through every period of the family that survived, which is a dislocation with a stated size rather than damage.
Fig. 7 The slip of every wrecked stem, which is what separates the cut stem’s own divergence from its control’s.

Two routes to one number

The size of the pair comes from the displacement profile — a comparison between two runs, organ by organ, in the top hundred and twenty organs of each.

The divergence comes from the control run alone, as a mean over its last stretch, and is the same number the ladder sweep reads to find a rung.

Nothing computes one from the other. They agree to eight per cent on sixteen rows and twelve on the seventeenth, which is the site’s standard check — two independent routes to a quantity — arriving without having been arranged.

How far every organ moved, 4 places back at a rise of 0.013. One mark per organ above the hole, at the angle it sits from where the same organ sits in a control that shares its history. The collection has read two numbers out of profiles like this one — the largest displacement anywhere, and the first organ's — and never the profile. It is not a bump that decays. After about 3 organs it settles into a repeating pattern of five levels, one per residue class modulo 5, which is the lag whose hop this stem kept. three of those levels sit together and two do not.
Fig. 8 The profile the first number is read from, which shares nothing with the run the second is read from except a history.

What the picture now is

A wrecked stem is the control’s arrangement with two things done to it.

The whole stem has slipped: every organ displaced by one common amount, which is the whole-turn slip already measured seen organ by organ.

And two adjacent chains have changed places: displaced from the common level by plus and minus one divergence step, to within eight per cent.

Three numbers describe the difference between a wrecked stem and its control — the slip, the exchange, and which two chains — where before this round there were three hundred.

A period of 5, and the two classes that are not with the rest. The same wrecked stem, folded on the lag it kept: one row per residue class, each drawn at the mean displacement of its own organs against the level the rest of them share. The bar through each row is the spread inside that class, and the widest of them is 1.62° — so within a class the displacement is a constant. three classes sit at the common level. The two that do not sit at 141.1° and -142.3°, equal and opposite to within 0.8 per cent, and they are neighbouring residues. The stem's own divergence is 137.77°, so an exception is one organ's step.
Fig. 9 A period with the two parts of the description marked: the level the slip sets, and the pair the exchange sets.

Why an exchange and not a shift

A shift would displace every chain by one step, not two of them, and the profile would have no common level at all. That is not what is measured.

An exchange of two adjacent chains is the smallest rearrangement that leaves the arrangement’s own regularity intact, which is presumably why the rule ends up there: the placement rule minimises a sum over neighbours, and an arrangement with two chains swapped is still a lattice by every measure a counter applies.

That last part is testable and is worth stating as a prediction rather than a conclusion. If two adjacent chains are exchanged, the counted pair should be unchanged — and it is, on every wrecked stem in the census, which is a fact that has been sitting there without an account of why.

The two spiral families a counter finds between 0.43 and 0.67 of the radius. 21 spirals one way and 34 the other, found from the point positions alone — the counter is never told the divergence angle.
Fig. 10 What a counter reads off an arrangement, which two chains changing places leaves alone.

The thirteen rows without a pair

Thirteen of the thirty wrecked cuts have three or more exceptional classes and are not scored here. A stem whose difference from its control is more than one exchange has no single “size of the pair” to measure.

They are not evidence against the picture; they are cuts where the picture needs more than one term. Two of them have six exceptional classes out of eight, which is a stem whose arrangement has been rearranged rather than nudged.

Whether those decompose into several exchanges is a question this measurement can be pointed at and has not been. It would need the exceptions sorted into pairs, and with six of them there are fifteen ways to do it.

How constant the displacement is inside one residue class. One row per wrecked cut in the census, drawn at the widest spread found inside any one residue class when the profile is folded on the lag that stem kept. 25 of 30 rows sit between 0.12 and 6.09 degrees, which on a quantity whose between-class differences run past a hundred and fifty degrees is a constant. The five that do not sit from 10.3° up. There is nothing in between, so the line drawn at 10° could have been drawn anywhere in a wide interval.
Fig. 11 Every wrecked cut, on which the rows with several exceptions can be found.

What twelve per cent is worth

The bound in the assertion is twelve per cent below and twelve above, and sixteen of seventeen rows are inside eight. That gap between the bound and the data is deliberate: the assertion is written to fail if the picture is wrong, not to be tight around the numbers that happen to have come out.

A tight bound here would be a bound fitted to seventeen rows, and the next stem grown would fall outside it for no reason worth reporting. A loose bound that would still reject an exchange of two steps, or of half a step, is a bound that tests the claim rather than the sample.

The exceptional pair, measured in divergences. One row per wrecked cut whose profile has exactly one pair of exceptional classes, drawn at the size of that pair divided by the stem's own settled divergence. Every row sits between 0.882 and 1.076, so the two chains that came apart moved by one organ's step rather than by two or by half of one. The residual is not scatter: rows are grouped by the lag the stem kept, and every lag sits wholly above the line or wholly below it. Why a surviving 5 or 7 overshoots and a surviving 4 or 8 falls short is not answered here.
Fig. 12 The seventeen ratios against the bound, with the space between them being what makes it a test rather than a fit.

What the seventeen rows are

Six on golden stems whose surviving lag is 5, six on golden stems whose surviving lag is 8, four on Lucas stems whose lag is 7, and one Lucas stem whose lag is 4.

They come from nine of the census’s twelve lattices, spanning rises from 0.026 down to 0.005 on the golden branch and 0.013 to 0.008 on the Lucas one, at offsets from three to nine places back. So the seventeen are not one lattice’s cuts or one offset’s; they are a spread across the census, selected only by having exactly one balanced pair of exceptions.

That selection is the one place a bias could enter. Rows with three or more exceptions are excluded, and if the excluded rows were systematically the ones whose exchange is a different size, the seventeen would be a filtered sample. There is no way to tell from here, because a row with six exceptions has no single size to measure.

How constant the displacement is inside one residue class. One row per wrecked cut in the census, drawn at the widest spread found inside any one residue class when the profile is folded on the lag that stem kept. 25 of 30 rows sit between 0.12 and 6.09 degrees, which on a quantity whose between-class differences run past a hundred and fifty degrees is a constant. The five that do not sit from 10.3° up. There is nothing in between, so the line drawn at 10° could have been drawn anywhere in a wide interval.
Fig. 13 The whole census, of which seventeen rows carry exactly one balanced pair and thirteen do not.

What an exchange predicts about the counted pair

Nothing changes. Two adjacent chains swapping members leaves every chain intact — the same number of organs at the same spacing, in a different order — so a counter run over the wrecked stem’s points returns the same pair.

It does, on every wrecked stem in the census, and it has been reported for two rounds as a fact with no account attached. An exchange is an account of it: the counted pair is a property of the chains, the exchange preserves the chains, so the pair is preserved.

The prediction runs the other way too and is stronger for it. If a wrecked stem’s difference from its control were anything that moved organs between chains, the counted pair would move — and no wrecked stem in the census has a different pair from its control.

The two spiral families a counter finds between 0.68 and 0.92 of the radius. 34 spirals one way and 55 the other, found from the point positions alone — the counter is never told the divergence angle.
Fig. 14 A counter reading an arrangement’s chains, which an exchange within a chain leaves alone.

Why the divisor is not the surviving family’s step

The obvious alternative to dividing by the divergence is dividing by the hop of the surviving family, since that is the lag the profile is folded on and the one quantity the wrecked stem demonstrably keeps.

It does not work and the reason is arithmetic rather than empirical. The surviving family’s hop is the azimuth from an organ to the one k places above it, which on a 5/8 stem at 137° is 5 × 137° folded, or about −33°. The pairs measure 134° to 147°. They are not that number at any of the four values of k.

The exchange is between adjacent chains, not between an organ and its own family member, so the step involved is the sequence’s step and not the family’s. That is the same fact the adjacency measurement reports, arriving through the size.

The hops of a 5/8 lattice, shortest first — golden, rise 0.010Every lattice hop at this rise, ordered by how long its step is across the surface of the stem, with the two that a wrecked stem here leaves standing picked out. The two shortest are the contact families the counter returns, 5 and 8, and they differ in length by a factor of 1.076. The lags left standing after a removal are 5 and 8, sitting at rank 2 and 1 in this order, so the family the rule holds is a short step but not always the shortest one.85133161021181122624629lag, in organs — ordered by the length of its stepstep length, in turns of the stemkept: lag 5, lag 8golden, rise 0.010 · pair 5/8 · offsets that wreck: 4, 5, 6, 7, 8generated from a stated rule, not drawn to look right
Fig. 15 The lengths of a stem’s lags, on which the surviving family’s step is not the size the exchange comes out at.

The one row at 0.882

A Lucas 4/7 stem, cut four places back, surviving lag 4, pair size 88.0° against a divergence of 99.79°.

It is the only row whose surviving lag is 4, and it is the only row outside eight per cent. Those two facts are the same fact if the residual tracks the lag, and one row is not enough to say whether it does — a second stem keeping a 4 would settle it, and there is not one in the census.

Reported rather than dropped, because it is the row that sets the twelve per cent in the assertion, and a bound set by a row the reader has not been told about is a bound that looks tighter than it is.

The exceptional pair, measured in divergences. One row per wrecked cut whose profile has exactly one pair of exceptional classes, drawn at the size of that pair divided by the stem's own settled divergence. Every row sits between 0.882 and 1.076, so the two chains that came apart moved by one organ's step rather than by two or by half of one. The residual is not scatter: rows are grouped by the lag the stem kept, and every lag sits wholly above the line or wholly below it. Why a surviving 5 or 7 overshoots and a surviving 4 or 8 falls short is not answered here.
Fig. 16 The seventeen ratios with the four groups by surviving lag, of which one group has a single member.

What would make this an account

A reason the exchange should happen at all, and the design has none.

The picture is descriptive: a wrecked stem is its control with a slip and an exchange in it. Why the placement rule, having lost an organ, ends up putting two neighbouring chains the other way round rather than doing something else, is not addressed by measuring the size of what it did.

The route to an account runs through the transient, which is where the exchange happens. Above the onset nothing changes; below it the stem is moving. So the question “why an exchange” is a question about seven to three hundred organs of placement immediately above a hole, and those runs exist.

How far every organ moved, 6 places back at a rise of 0.01. One mark per organ above the hole, at the angle it sits from where the same organ sits in a control that shares its history. The collection has read two numbers out of profiles like this one — the largest displacement anywhere, and the first organ's — and never the profile. It is not a bump that decays. After about 18 organs it settles into a repeating pattern of eight levels, one per residue class modulo 8, which is the lag whose hop this stem kept. six of those levels sit together and two do not.
Fig. 17 The transient, drawn from the hole, which is the stretch an account of the exchange would have to read.

Why the two routes matter here in particular

Because the claim is a coincidence of two numbers, and a coincidence of two numbers computed the same way is not a coincidence.

The size comes from comparing a cut run against a control run, organ by organ, across their top hundred and twenty organs. The divergence comes from the control run alone, as a mean over its last stretch — the same value a ladder sweep reads to find a rung.

They share the control run and nothing else: one is a difference between two runs and the other is a property of one of them. There is no arithmetic path from either to the other, which is what makes them agreeing to eight per cent worth reporting rather than a restatement.

The divergence slides along the 5/8 rung. Measured at every rise on one rung, where a counter returns 5 and 8 spirals throughout. The settled divergence slides from 136.6406 to 137.8672 degrees. The ratio of the two contact steps falls to 1.0131 at a rise of 0.016 and the ordering changes hands at 0.015: above it the shorter step belongs to the 5-family and below it to the 8-family. Neither quantity is available to a counter, which is shown positions and reports a pair, and both of them move while that pair does not.
Fig. 18 The settled divergence read from a control run, which is one of the two independent routes.

The one line

The balanced pair inside a wrecked stem’s period measures 88.0° to 147.2°, against control divergences of 99.1° to 138.0°: one organ’s step, within eight per cent on sixteen of the seventeen rows and twelve on the last. The residual is not scatter — every stem keeping a 5 or a 7 sits above one divergence and every stem keeping a 4 or an 8 sits below — and what that tracks is not answered here.

The exceptional pair, measured in divergences. One row per wrecked cut whose profile has exactly one pair of exceptional classes, drawn at the size of that pair divided by the stem's own settled divergence. Every row sits between 0.882 and 1.076, so the two chains that came apart moved by one organ's step rather than by two or by half of one. The residual is not scatter: rows are grouped by the lag the stem kept, and every lag sits wholly above the line or wholly below it. Why a surviving 5 or 7 overshoots and a surviving 4 or 8 falls short is not answered here.
Fig. 19 The seventeen pairs, each drawn against the divergence of the stem it came from.

Where this goes next

A period of 8, and the two classes that are not with the rest. The same wrecked stem, folded on the lag it kept: one row per residue class, each drawn at the mean displacement of its own organs against the level the rest of them share. The bar through each row is the spread inside that class, and the widest of them is 0.87° — so within a class the displacement is a constant. six classes sit at the common level. The two that do not sit at 134.3° and -134.2°, equal and opposite to within 0.0 per cent, and they are neighbouring residues. The stem's own divergence is 137.44°, so an exception is one organ's step. One level and two exceptions Inside a wrecked stem's period most residue classes sit at one level and a couple do not. On seventeen of the thirty cuts the exceptions are exactly two, equal and opposite to within five per cent — and on all seventeen they are neighbouring residues, which was not looked for. How far every organ moved, 6 places back at a rise of 0.01. One mark per organ above the hole, at the angle it sits from where the same organ sits in a control that shares its history. The collection has read two numbers out of profiles like this one — the largest displacement anywhere, and the first organ's — and never the profile. It is not a bump that decays. After about 18 organs it settles into a repeating pattern of eight levels, one per residue class modulo 8, which is the lag whose hop this stem kept. six of those levels sit together and two do not. The damage has a period Every wrecked stem in the census has had two numbers read out of its displacement profile and the profile itself read out of none of them. Folded on the lag the stem kept, twenty-five of the thirty are constant inside each residue class to between 0.12° and 6.09°. How constant the displacement is inside one residue class. One row per wrecked cut in the census, drawn at the widest spread found inside any one residue class when the profile is folded on the lag that stem kept. 25 of 30 rows sit between 0.12 and 6.09 degrees, which on a quantity whose between-class differences run past a hundred and fifty degrees is a constant. The five that do not sit from 10.3° up. There is nothing in between, so the line drawn at 10° could have been drawn anywhere in a wide interval. The plateau was a prediction Last round's search for a reference organ found that the largest displacement above a hole is a plateau rather than a peak, and reported it as a failure. A profile constant on each of k residue classes has exactly k levels, so its maximum is attained by a whole class — a ninth to a quarter of every window, forever.

What links here

Computed from the collection, not written here: the essays that point at this one.

Shares its objects with

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

  • How wide a band should be — both name claim testing, divergence angle, measurement, negative result, prediction, residual, resolution, tolerance
  • The organ that moved furthest — both name ablation, claim testing, divergence angle, honest limits, lattice offset, measurement, negative result, rigid hop
  • The rung that two organs wreck — both name ablation, claim testing, lattice offset, measurement, mechanism, negative result, prediction, rigid hop
  • The shallower front turns over — both name ablation, claim testing, divergence angle, honest limits, measurement, negative result, prediction, tolerance
  • Where the survivors meet — both name ablation, claim testing, divergence angle, lattice offset, mechanism, negative result, prediction, rigid hop
  • A count with a factor in it — both name ablation, claim testing, divergence angle, honest limits, measurement, rigid hop, slip

Named objects

A flat tag is an object no other essay names yet.

AblationClaim testingDivergence angleHonest limitsLattice offsetMeasurementMechanismNegative resultPredictionResidualResolutionRigid hopSlipTolerance