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

A cycle sums to a whole turn

The test proposed for whether three displaced chains are a three-cycle was that their displacements sum to zero. Three chains rotating into one another's places each move about a third of a turn the same way round, and a third of a turn three times is a whole turn — which the unfolded test calls the worst row in the census.
12 min read 24 figures State the claim, state the testCounted, not admiredTwo routes, one number

Worth reading first: The damage has a period.

Thirteen wrecked cuts in the census have no balanced pair of displaced chains, and they were set aside with a question: are three exceptional chains a three-cycle, with three chains rotating into one another’s places, or a pair with a stray attached?

The test proposed was that a three-cycle’s displacements sum to zero. It does not, and the census has exactly one clean three-cycle to prove it on.

Three chains, each moved a third of a turn, on g008 cut at offset 6. The eight chains of a stem that kept a lag of 8, set round the circle, with the three that are displaced marked and their displacements written on. Each is about a third of a turn the same way round, and they sum to -359.6 degrees — one whole turn to within half a degree, on an azimuth grid whose step is a quarter of a degree. A rearrangement that closes by going once round sums to a turn, not to nothing, which is why the test it was set had the wrong number in it.
Fig. 1 The one clean cycle in the census, drawn round its own period, with each chain’s displacement written on.

The row

g008 — the golden branch at a rise of 0.008 — cut at offset 6, keeping a lag of 8. Three of its eight chains are displaced and the other five sit at the common level.

The three are chains 6, 7 and 0, which are consecutive round the circle of eight, and their displacements are −89.4, −135.3 and −134.8 degrees. Every one of them goes the same way round.

Chain 0 is the chain the removed organ was on, which is the labelling the exchange uses and is the control’s indexing rather than the cut run’s.

How far every organ moved, 6 places back at a rise of 0.008. 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 14 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. five of those levels sit together and three do not.
Fig. 2 The profile the three displaced chains are read from, folded onto the lag the cut stem kept.

What they sum to

−359.6 degrees. One whole turn to four parts in ten thousand, on an azimuth grid whose step is a quarter of a degree.

That is not close to zero. It is the largest sum in the excluded set by a factor of two, and under the test as proposed it is the worst row in the table rather than the one clean answer in it.

The same sums read raw and read folded. Every excluded cut's displacements added twice: the open mark is the raw total in degrees and the filled mark is the same total folded into half a turn either way. The two agree everywhere but one row, and on that row the raw total is the largest in the table at about minus a whole turn while the folded total is the smallest. Reading the sums raw would have named the one closed cycle here as the worst anomaly in the census.
Fig. 3 Every excluded row’s raw sum beside its folded one, with the row where the two readings disagree.

Why a cycle sums to a turn

Because a cycle is a rotation. Three chains rotating into one another’s places means chain 6 goes where chain 7 was, chain 7 where chain 0 was, and chain 0 where chain 6 was — and going round the circle once is one turn’s worth of displacement shared between three moves.

Zero is what a cancellation sums to, not what a rotation sums to. A balanced pair sums to zero because one chain goes forwards exactly as far as another goes back; three chains going the same way round add up rather than cancel.

The seventeen rows the exchange covers are all cancellations, and they close within 4.6 degrees — which is what makes them the control for this addition rather than a second population of it.

Three chains, each moved a third of a turn, on g008 cut at offset 6. The eight chains of a stem that kept a lag of 8, set round the circle, with the three that are displaced marked and their displacements written on. Each is about a third of a turn the same way round, and they sum to -359.6 degrees — one whole turn to within half a degree, on an azimuth grid whose step is a quarter of a degree. A rearrangement that closes by going once round sums to a turn, not to nothing, which is why the test it was set had the wrong number in it.
Fig. 4 The three chains and their displacements, all in the same direction round the period.

So the test needed folding

An angle folded into half a turn either way puts −359.6 degrees at +0.4 degrees, which is the reading that makes the row what it is.

Every sum in the reading is folded for that reason, with the raw value carried beside it so the distinction stays visible. On twelve of the thirteen rows the two agree; on this one they differ by a whole turn.

The same sums read raw and read folded. Every excluded cut's displacements added twice: the open mark is the raw total in degrees and the filled mark is the same total folded into half a turn either way. The two agree everywhere but one row, and on that row the raw total is the largest in the table at about minus a whole turn while the folded total is the smallest. Reading the sums raw would have named the one closed cycle here as the worst anomaly in the census.
Fig. 5 The same sums with nothing highlighted, where twelve rows have their two readings on top of each other.

What folding costs elsewhere

It is worth saying, because folding is used throughout this collection and its price is rarely stated. A displacement is folded because a lattice and its mirror are one lattice under this rule — nothing distinguishes a turn one way from the same turn the other.

The price is that two very different placements can fold to nearby numbers, and that a sequence crossing a half turn looks like a reversal. Every claim about a folded quantity’s size is safe; claims about its change need the unfolded values checked.

The wrecked stem is the lattice it was cut from, wound the other way. The divergence of each organ placed after two were removed from a settled stem at a rise of 0.032, over the 220 organs following the cut. The upper line is the divergence the undisturbed stem holds, 139.6875°; the lower is 220.3125°, which is 360° minus it and therefore the same lattice with the opposite handedness. The sequence is thrown by the cut, wanders for a few dozen organs, and settles on the second line to four decimal places — 220.3125° against 220.3125° — where it stays. A counter shown the positions afterwards returns 3 and 5, the pair the stem was cut from. Nothing about the pattern has been lost; its chirality has been reversed, which at this rung a single removal cannot do.
Fig. 6 A lattice and its mirror, which is why every angle in this collection is reported folded.

Here it is the other way round

Which is the neat part. Usually folding hides a distinction and the unfolded values have to be checked; here the unfolded value hides the finding and the folded one shows it.

Both directions come from the same fact: a folded angle and its raw value differ by a whole turn and the reader has to know which question is being asked. How big is this displacement? wants one; do these displacements cancel? wants the other.

The same sums read raw and read folded. Every excluded cut's displacements added twice: the open mark is the raw total in degrees and the filled mark is the same total folded into half a turn either way. The two agree everywhere but one row, and on that row the raw total is the largest in the table at about minus a whole turn while the folded total is the smallest. Reading the sums raw would have named the one closed cycle here as the worst anomaly in the census.
Fig. 7 The two readings side by side, where the raw one is right for a size and the folded one for a cancellation.

How close to a turn is close

Four parts in ten thousand: −359.6 against −360. The azimuth grid is 1,536 samples of the circle, so one step is 0.234 degrees and the miss is under two steps.

Each of the three displacements is read from a mean over a residue class, so it is not a single grid value and can sit between steps. Three of them summing to within 0.4 degrees of a turn is not something the grid forces.

What the finer grid does to the rises already published. The two rises this site has argued from and the one it published as having no answer, each read at both azimuth grids, five stems apiece. A filled mark agrees with the position counter, a half mark contradicts it, an open mark is a refusal. At 0.013 and 0.005 the readings are identical at both grids, so nothing already written depends on the sample count. At 0.008 they are not: 384 azimuths gives 5/8, 5/8 and 1152 gives 8/13, 8/13, against a counter that says 5/8. That rise was chosen in the earlier work because the three shortest lattice offsets there are within a fifth of each other, and a stem with no answer answering differently on a different grid is the object behaving as it was said to.
Fig. 8 The grid the azimuths are placed on, against which the miss from a whole turn is under two steps.

Is it a coincidence?

The number to compare against is what three arbitrary displacements would give. The exceptional chains in this census run from 12.4 to 160.4 degrees from the common level, so three of them drawn at random would sum to somewhere in a range several hundred degrees wide.

Landing within 0.4 degrees of a whole turn out of that is about one chance in a thousand. That is a rough figure and it is the right order: this is not a sum that happens to be near a turn.

The 13 cuts the exchange sets aside, chain by chain. One row per wrecked cut with no balanced pair, one cell per chain of the lag it kept, and the chains displaced away from the common level filled. The lag is printed at the left and the sum of the displacements at the right, folded into half a turn either way. seven of the 13 sum to within 12 degrees of nothing and six do not, and every row that closes kept a lag of 7 or 8 while every row that does not kept 4 or 5. The rows are stacked by that lag.
Fig. 9 The excluded rows whose sums do not close, whose values are spread across hundreds of degrees.

And the chains are consecutive

The second piece of evidence. The three displaced chains are 6, 7 and 0 out of eight, which are adjacent round the circle — a rotation among three chains that are not neighbours would be a stranger object.

Adjacency is not proof: several of the other excluded rows have consecutive exceptional chains and do not close. What it does is make the reading coherent, which a sum near a turn among three scattered chains would not have been.

It also fits what the balanced rows do. The exchange runs one way round on every row it covers — the chain displaced forwards immediately precedes the chain displaced backwards — so adjacency is what a rearrangement in this census looks like.

The 13 cuts the exchange sets aside, chain by chain. One row per wrecked cut with no balanced pair, one cell per chain of the lag it kept, and the chains displaced away from the common level filled. The lag is printed at the left and the sum of the displacements at the right, folded into half a turn either way. seven of the 13 sum to within 12 degrees of nothing and six do not, and every row that closes kept a lag of 7 or 8 while every row that does not kept 4 or 5. The rows are stacked by that lag.
Fig. 10 The excluded rows with their chains, where consecutive runs are common and closing is not.

What the other three-exception rows do

Four rows in the census have exactly three exceptional chains. This one closes; the other three do not.

Their folded sums are −16.9, +41.6 and −173.6 degrees, which are not near zero and not near a turn. So three exceptions is not a category with one answer, and the question as posed — are three exceptions a cycle or a pair plus a stray? — has the answer sometimes one and sometimes neither.

The 13 cuts the exchange sets aside, chain by chain. One row per wrecked cut with no balanced pair, one cell per chain of the lag it kept, and the chains displaced away from the common level filled. The lag is printed at the left and the sum of the displacements at the right, folded into half a turn either way. seven of the 13 sum to within 12 degrees of nothing and six do not, and every row that closes kept a lag of 7 or 8 while every row that does not kept 4 or 5. The rows are stacked by that lag.
Fig. 11 The excluded rows that do not close, three of which carry exactly three exceptional chains.

Which is a real answer

Not a satisfying one, and it is the shape the census supports. Of four rows with three exceptions, one is a clean three-cycle with each chain moving about a third of a turn, and three are rearrangements whose displacements do not cancel and do not close.

The three that do not are not a pair plus a stray either. A pair plus a stray would have two displacements equal and opposite with a third left over, and none of the three has that.

The 13 cuts the exchange sets aside, chain by chain. One row per wrecked cut with no balanced pair, one cell per chain of the lag it kept, and the chains displaced away from the common level filled. The lag is printed at the left and the sum of the displacements at the right, folded into half a turn either way. seven of the 13 sum to within 12 degrees of nothing and six do not, and every row that closes kept a lag of 7 or 8 while every row that does not kept 4 or 5. The rows are stacked by that lag.
Fig. 12 The excluded rows that do close, of which only one has three exceptions.

What closes besides this one

Six more rows close, and none of them is a three-cycle: they carry four, five and six exceptional chains, and their folded sums are −0.2, +2.6, +3.0, −6.2, −8.0 and −10.5 degrees.

Those are cancellations rather than rotations — several displacements in both directions adding to nothing. So closed covers two different objects and only one row in the census is the rotating kind.

How far each rearrangement is from closing, against the lag it kept. One bar per excluded cut, its length the distance from the displacements summing to nothing, ordered smallest first, and each labelled with the lag the cut left standing. The seven shortest bars are the cuts that kept a lag of 7 or 8 and the six longest are the cuts that kept 4 or 5, with a gap of 6.3 degrees between the two groups and no row in it. The rule sorts every one of them.
Fig. 13 How far each excluded row is from closing, of which seven are inside the line.

Why only one

Unknown, and the honest answer is that thirteen rows on seven lattices is not a population. What can be said is that the cycle sits at a lag of 8 with three of its eight chains moving, which is a small fraction of the profile.

The other closed rows have most of their chains moving. So the difference between a rotation and a cancellation may be about how much of the profile is involved, and one instance of the first cannot support that.

The 13 cuts the exchange sets aside, chain by chain. One row per wrecked cut with no balanced pair, one cell per chain of the lag it kept, and the chains displaced away from the common level filled. The lag is printed at the left and the sum of the displacements at the right, folded into half a turn either way. seven of the 13 sum to within 12 degrees of nothing and six do not, and every row that closes kept a lag of 7 or 8 while every row that does not kept 4 or 5. The rows are stacked by that lag.
Fig. 14 The rows that close, with the one three-chain rotation among six many-chain cancellations.

What the row is worth

It is one row. What it is worth is that it exists at all: the question the excluded set was set aside with was whether cycles happen, and one clean instance answers it.

It also fixes the test. Anybody adding up a set of chain displacements to look for a rearrangement now knows to fold, and knows that the sum to look for is a whole turn as readily as nothing.

Three chains, each moved a third of a turn, on g008 cut at offset 6. The eight chains of a stem that kept a lag of 8, set round the circle, with the three that are displaced marked and their displacements written on. Each is about a third of a turn the same way round, and they sum to -359.6 degrees — one whole turn to within half a degree, on an azimuth grid whose step is a quarter of a degree. A rearrangement that closes by going once round sums to a turn, not to nothing, which is why the test it was set had the wrong number in it.
Fig. 15 The one instance, whose existence is what the excluded set was set aside to find out about.

A test that could have been checked first

The correction cost nothing and could have been made before any arithmetic ran. Three chains rotating into one another’s places sum to a whole turn is a sentence about rotations rather than about this census.

That it was not is the ordinary way of these things: a test gets written down at the end of one piece of work as a suggestion for the next, and a suggestion is not checked the way a claim is. The habit worth taking from it is to check the arithmetic of a proposed test before running it.

The same sums read raw and read folded. Every excluded cut's displacements added twice: the open mark is the raw total in degrees and the filled mark is the same total folded into half a turn either way. The two agree everywhere but one row, and on that row the raw total is the largest in the table at about minus a whole turn while the folded total is the smallest. Reading the sums raw would have named the one closed cycle here as the worst anomaly in the census.
Fig. 16 The row the proposed test would have rejected, drawn under both readings.

The general form

A quantity defined modulo something, tested against a value that is only correct in one representative. Angles are the obvious case and this collection is full of them.

Two other instances are on record. A settled divergence at 220.9 degrees is the same arrangement as one at 139.1, so a mean over both representations belongs to neither; and a displacement near a half turn folds to a large negative, which is why a transition’s sign has to be checked unfolded.

six 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. 17 The range an angle is reported in, outside which the same arrangement has a second name.

What the row looks like in the profile

Worth describing in the raw, since the circle picture is already an interpretation. The profile is how far each organ above the hole sits from where the control put it, and it is read over the last hundred and twenty organs of the run.

Folded onto a lag of eight it gives eight classes. Five of them sit within a few degrees of one another — that is the common level — and three sit 89 to 135 degrees below it.

Nothing in that description says rotation. The rotation is the reading that the three chains below the level are the same three chains, moved into one another’s places, and the sum is what supports it.

What a cut moves, organ by organ. A stem counted at 5 and 8 spirals with the organ seven places back from the tip removed, compared against a control that shares its history to the last digit. Each mark is one organ placed after the cut and how far its azimuth ended up from where the control put the same organ. The quantity folds at half a turn, so 138 degrees is near the largest displacement there is; and it does not decay with height, which is what a stem that never repairs means. There is therefore no organ that the cut disturbed most in any useful sense, and a reading that needs one has nowhere to stand.
Fig. 18 How far each organ above the hole sits from where the control put it, before any folding onto a lag.

What would distinguish a rotation from three coincidences

A stronger test than the sum, and it is available. If chains 6, 7 and 0 rotate into one another’s places, then the organs of chain 6 should sit where the control put chain 7’s, and so on round.

That is checkable against the control’s own positions rather than against a sum, and it has not been done. It would turn a suggestive arithmetic into a direct observation, and it is the obvious next thing this row deserves.

One rule at p = 1, cut three ways. loop cut at 3/√h: 8/13 at 137.62° with 0.58° of scatter. exponential cut-off, 3: 8/13 at 137.58° with 0.50° of scatter. no cut at all: no lattice, 44° of scatter. The first two agree to 0.03° — the prediction held — and the third is what the same rule does when nothing cuts it.
Fig. 19 A cut run against its control, which is where a chain sitting in another chain’s place would be visible.

Why the sum was the test proposed

Because it is one line of arithmetic on numbers already computed, and the direct check needs the control’s positions organ by organ. That is the ordinary trade and it is usually the right one: run the cheap test, and run the expensive one only if the cheap one says something.

The cheap test said something. It also had the wrong number in it, which is the argument for checking a proposed test’s arithmetic before running it rather than after.

Three chains, each moved a third of a turn, on g008 cut at offset 6. The eight chains of a stem that kept a lag of 8, set round the circle, with the three that are displaced marked and their displacements written on. Each is about a third of a turn the same way round, and they sum to -359.6 degrees — one whole turn to within half a degree, on an azimuth grid whose step is a quarter of a degree. A rearrangement that closes by going once round sums to a turn, not to nothing, which is why the test it was set had the wrong number in it.
Fig. 20 The row the cheap test found, on which the expensive one has not been run.

The lattice it sits on

g008 is the golden branch at a rise of 0.008, which is one of the census’s ten lattices and sits at 84 per cent of the golden 5/8 rung. It wrecks at four offsets — 6, 7 and 8 keeping a lag of 8, and 4 keeping a lag of 5.

Three of its four rows are balanced pairs and go to the exchange. Only the cut at offset 6 has three exceptions, and it is the one that turns out to be a cycle.

So the census’s one rotation sits on a lattice whose other cuts are entirely ordinary, which means nothing about the lattice explains it. Whatever makes offset 6 different is about the offset.

The next organ moves for the last 13, and for no others. One row per organ removed, counted back from the tip of a stem at a rise of 0.005 whose counted pair is 8 and 13. Removing any of the last 13 moves the next organ by 2.6° to 167.6°; removing an older one moves it by at most 0.47°, which is under the azimuth grid. The boundary is at 13, and 13 is the larger parastichy number — so the experiment counts the spirals without measuring an angle.
Fig. 21 The offsets that wreck at this lattice, of which three give balanced pairs and one gives a cycle.

And the rise it sits at

0.008 is the rise at which the second wall of a slot goes free on this rung — or rather, it is well below the rise where that happens, since the transition is at 0.00997 and the free stretch runs down from there.

That is almost certainly a coincidence and it is noted because both readings are about the same lattice and a reader meeting it twice deserves to know they are unrelated. One is about removing two organs and one is about removing one.

The golden 5/8 rung swept at 27 rises, with each removal's cost. How far the first organ placed after a cut moves, at every rise the sweep visits, coarse on the left. Removing both walls costs far more than removing the larger one alone above a rise of 0.00998, and exactly what the larger one costs below it. The change happens in one step of the grid the ladder is named on: 163.59 degrees at 0.00998 and 9.14 degrees at 0.00997, which is a fall of 154.5 degrees for a change of one part in a thousand in the rise.
Fig. 22 The rung this lattice sits on, and the unrelated transition further up it.

What a reader should carry

That the census contains one clean three-cycle, at g008 cut at offset 6, whose three consecutive chains each move about a third of a turn the same way and sum to −359.6 degrees.

And that the test it was set — sum to zero — would have called it the worst anomaly in the table. A rotation sums to a turn; a cancellation sums to nothing; and both read as zero once folded, which is why every sum here is.

Three chains, each moved a third of a turn, on g008 cut at offset 6. The eight chains of a stem that kept a lag of 8, set round the circle, with the three that are displaced marked and their displacements written on. Each is about a third of a turn the same way round, and they sum to -359.6 degrees — one whole turn to within half a degree, on an azimuth grid whose step is a quarter of a degree. A rearrangement that closes by going once round sums to a turn, not to nothing, which is why the test it was set had the wrong number in it.
Fig. 23 The cycle once more, with its three displacements and the turn they add to.

What the picture at the top shows

Eight marks round a circle, one per chain of the lag this cut kept, with chain 0 at the top — the chain the removed organ was on.

Three of the eight are filled and carry their displacements: −89.4, −135.3 and −134.8 degrees. They are consecutive round the circle and all three point the same way. The number in the middle is their sum and the words beside it say what it is.

Three chains, each moved a third of a turn, on g008 cut at offset 6. The eight chains of a stem that kept a lag of 8, set round the circle, with the three that are displaced marked and their displacements written on. Each is about a third of a turn the same way round, and they sum to -359.6 degrees — one whole turn to within half a degree, on an azimuth grid whose step is a quarter of a degree. A rearrangement that closes by going once round sums to a turn, not to nothing, which is why the test it was set had the wrong number in it.
Fig. 24 The eight chains and the three that rotate, whose displacements sum to one whole turn.

The one line

g008 cut at offset 6 displaces three consecutive chains by −89.4, −135.3 and −134.8 degrees, summing to −359.6 — one whole turn to four parts in ten thousand, on a grid whose step is a quarter of a degree.

That is the census’s one clean three-cycle, and the test it was set would have rejected it, because a rotation sums to a turn rather than to nothing.

Where this goes next

The 13 cuts the exchange sets aside, chain by chain. One row per wrecked cut with no balanced pair, one cell per chain of the lag it kept, and the chains displaced away from the common level filled. The lag is printed at the left and the sum of the displacements at the right, folded into half a turn either way. seven of the 13 sum to within 12 degrees of nothing and six do not, and every row that closes kept a lag of 7 or 8 while every row that does not kept 4 or 5. The rows are stacked by that lag. The rows nobody added up Seventeen of the census's thirty wrecked cuts come back as one balanced pair of displaced chains, and every claim about the exchange is quantified over those seventeen. The other thirteen were set aside as having three or more exceptions and never looked at again. They are one addition each. How far each rearrangement is from closing, against the lag it kept. One bar per excluded cut, its length the distance from the displacements summing to nothing, ordered smallest first, and each labelled with the lag the cut left standing. The seven shortest bars are the cuts that kept a lag of 7 or 8 and the six longest are the cuts that kept 4 or 5, with a gap of 6.3 degrees between the two groups and no row in it. The rule sorts every one of them. The lag decides whether it closes Seven of the thirteen excluded rows have displacements that cancel and six do not. Every row that closes kept a lag of seven or eight and every row that does not kept four or five, thirteen times out of thirteen — and then a lattice nobody had cut broke it. The 2-jugate forks converge on 68.7539°. Every fork sits at a rational divergence, with denominator 4(m² + mn + n²) — 10/28, 30/76, 74/196 and so on. The limit is 68.7539°, which is 137.5078 divided by 2, and it is at none of them. Half the golden angle The forks of the van Iterson tree converge on 137.5078° and sit at none of them. Divide the whole tree by two and the same thing happens at 68.7539° — which is where teasel is, and where a bijugate sunflower counted 42 and 68 has to be.

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.

Named objects

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

ArtefactAzimuth gridClaim testingCycleDamage profileExceptional chainExchangeFalsifiabilityFoldingHonest limitsMeasurement errorResidue class