Lattice offset — where it appears
Named by 67 essays across 5 fields — each of them below, with the objects they name alongside it.
Counting the spirals
Almost every claim about phyllotaxis is a claim about how many spirals run through a pattern, and the count is almost never done. It can be done from the points alone, by a count that is never told what angle built them — and then a count of 34 is evidence rather than a restatement.
The counts change with radius
The same head gives 13 and 21 near the centre, 21 and 34 further out, 34 and 55 beyond that, and 55 and 89 at the rim. The transitions are at computable radii, and a photograph captioned with one pair is a statement about one annulus rather than about a flower.
The six are the spirals
Label every contact between two cells in a seed head with the difference between the two nodes' placement indices. The labels are the parastichy numbers — 34, 55, 21, 89 — and the six sides Euler forces turn out to be about two from one family, one and a half from the next, and one each from two more.
A counter that sees no positions
This site has counted spirals two ways, and both were handed coordinates. A third counter is handed a list of angles and nothing else. It returns one number instead of two, it refuses more often, and where it refuses it would have been wrong every time.
Every family but two is a sum
A seed head has six spiral families and everybody reports two. That looks like a convention hiding information and it is the opposite — every family but the two smallest is the sum of two others, so a third count is a prediction rather than a measurement, and a check that catches a wrong pair.
The angles name the branch
Seed the same rule at the Lucas angle and the readout returns 4 and 7, then 7 and 11 — the pairs the position counter finds, and not Fibonacci numbers. So a list of divergence angles carries not only how many spirals there are but which family of ladders the plant is on.
What a forgery has to know
A lattice with transported errors reproduces the comb and the pair, so one quantity is left: the two combs' relative strength. Weighted by distance the forgery puts more in the second comb than the first; the rule does the opposite. It matches only if the coupling is turned three to one towards the further neighbour, which no falloff supplies.
A periodicity is not a lattice
Give a lattice's errors a period of eight and a comb appears at spacing eight, on an arrangement with no rule in it. But the partner it names is 10, then 12, then 11, then nothing — an accident of the disturbance rather than a measurement of the pattern. The forgery is caught by reading a second stem, and by nothing else.
A period that is not a count
Eighteen wrecked stems settle into a block whose period is one of their own spiral counts, and one settles into a block of four on a lattice counted 8 and 13. The odd one is not noise. It is the case that shows what the rule is actually conserving, and it is the reason this thread is about lattice steps rather than about spirals.
A disturbance that is not passed on
The disturbance that forges every observable measured here does two things at once — it correlates an organ's error with its contact neighbours', and it hands that error on to be handed on again. Every result about it has been unable to say which half did the work. This is the control that takes the second half away and keeps the first.
The sequence has a memory
Every measurement this collection has made of a stem's divergence angles throws the order away. A spread is invariant to shuffling. Put the angles back in order and there is a large correlation between one and the next — 0.54 with no noise at all — which is the rule correcting itself, and which nothing had looked at.
The forgery needs a history
A disturbance passed between touching organs manufactures the comb, the second comb and the parastichy pair on an arrangement with no rule in it — which is why the comb stopped being evidence. Give the organs the same correlation with no accumulation in it and the forgery collapses: one seed in eight returns a pair, and the comb is the noise floor.
The order carries the count
Take the divergence angles off a stem, throw away every coordinate, and autocorrelate what is left. The result is periodic at the smaller parastichy number — peaks at it and at every multiple of it. A list of angles, with no picture and no position in it, carries the spiral count.
A neighbourhood is a hypothesis
Every simulation of this kind stops summing somewhere. The earlier work found that where it stops decides what pattern comes out — so the stopping place is not a detail of the program but a claim about how far a primordium's influence reaches, and it should be written down as one.
A hard edge is not a falloff
The prediction was that cutting the neighbourhood at three spacings would reproduce the pattern truncation had manufactured. It does — if the cut is smooth. A hard cut at the same distance produces no pattern at any width, and the reason is that it is the only one of the three whose neighbour set depends on where the candidate is.
The second comb
The autocorrelation of a divergence sequence has peaks at the smaller parastichy number and at every multiple of it. It also has a second set of peaks, at the same spacing, offset by the difference of the pair — so a list of angles with no coordinate in it returns both numbers rather than one.
A harmonic is a step taken twice
The spectrum contains the larger parastichy number, their sum, and echoes of the smaller one, and no ranking of peak heights separates them. What separates them is arithmetic: a harmonic is a multiple of the spacing and a family is not, and the two kinds sit in different residue classes.
Two shapes, one threshold
Read in the same unit, an exponential falloff and a gaussian one disagree about where the lattice ends by half. The quantity they agree on turns out to be one the earlier work measured for an unrelated reason — and it agrees with a bracket left by a sweep of a completely different parameter.
The neighbourhood was already settled
The earlier work explained a small difference between two kinds of noise by saying a jostle is diluted among some thirty neighbours. Sweep the neighbourhood sixfold and the difference does not move — because past four spacings the rule builds the identical lattice, internode for internode. There was nothing to dilute.
The boundary belongs to the pattern
Three kinds of noise, in three incommensurable units, destroy a lattice at the same place — about a degree and a half of divergence scatter. The earlier work measured that of two kinds and called it a scale rather than a constant. With a third it looks less like a coincidence and more like a property of what a lattice is.
The rung was not the instrument
The earlier work said the pair readout has a ceiling one rung above where it works, that this is arithmetic rather than statistics, and that no amount of stem fixes it. The arithmetic is right and gives a band of lag windows that is never empty; what was actually stopping the reading was an eight-node seed and a grid of 384 azimuths.
Two thirds of a cell
The founding claim of this field is that the six sides Euler forces are the spiral families. Measured against the tessellation it names two thirds of a cell's walls exactly, in every band of a head and at every rise of a stem, and the missing third is the same third everywhere.
Two rankings, one list
An essay in this collection claimed that the four shortest index hops on a seed head and the four largest shares of its cell walls are the same four numbers in the same order, and called the correspondence exact. Measured again from the same points, the two lists hold the same four families and order them differently, and they order them differently in five of the six bands the head can be read in.
Removing a neighbour costs least
Take away an organ that is a direct chain-neighbour of the growing tip and the next organ moves by under thirty-one degrees. Take away anything else inside the front and it moves by at least sixty-three. Thirty cuts, two groups, a factor of two between them and nothing in the gap.
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°.
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.
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.
Both walls of the slot
The growing tip sits between its two chain-neighbours. Removing either alone is a cheap removal on all six lattices — 2.3° to 41.7°. Removing both together throws the next organ past the expensive line on three of them, and the interaction runs from −25.8° to +132.9°.
The hole on the other branch
Near a transition, the run of offsets a stem notices stops being a run: there is quiet past the front and then one isolated offset, felt as hard as anything inside it. Where that offset sits was pinned down on Fibonacci lattices, where the numbers to check it against are 5, 8 and 13. On the Lucas branch they are 4, 7 and 11 — and the rule holds there too.
The response with a hole in it
Removing an organ is felt out to the larger parastichy number and no further — that is the intervention's headline, and it holds in the middle of a rung. Swept towards a transition the run of felt offsets stops early and one lone offset past it comes alive, with three quiet organs in between. The lone offset is one place inside the count the stem is about to have.
The organ that guards the second slot
An organ twelve places back is the furthest of any from where the next one goes, and removing it moves the next one by a whole divergence. The reason is that the rule's profile has two low points rather than one, the second is the slot after next, and that organ is holding it up. The comparison between what it holds up and how far behind it is decides the whole thing.
The rung that two organs wreck
On the coarse 3/5 stem both walls of the slot heal when removed alone and wreck when removed together — and the wreck keeps no rigid hop at all. Two of the six pairs in the design end at a destination single removals almost never reach.
Which chains changed places
A wrecked stem's displacement profile is a set of levels, one per chain, with two of them out of line — equal and opposite, on neighbouring chains. Nothing said which two. They are the hole's own chain and the one below it, on ten of the seventeen cuts that carry a pair.
One way round, seventeen times
The two chains that change places in a wrecked stem are adjacent, which is symmetric and says nothing about direction. Label them by lag from the hole and the one displaced forwards is always the lower of the two — on every row of the census, without an exception.
The hop that survived
A stem that never repairs after an organ is removed settles into an exactly repeating block of angles, and the period of that block is a spiral count of the lattice it was cut from. Nobody could say why. Read the wrecked stem by lags rather than by neighbours and the answer is one line: one family of the original lattice is still standing, organ by organ, and the block is its period.
One turn per survivor
If a wrecked stem keeps one family of its old lattice exactly, then the angle it settles at is not free. Over the period of the family that survived, the pattern has to come back to where that family left it — which means the whole change in the divergence is a whole number of turns spread over a small whole number of organs. Measured, it is one turn, at seventeen of nineteen.
Six lattices were not enough
The interaction between the two walls of a slot came back at −25.8° to +132.9° on six lattices, three above zero and three below, with no ordering by rise, by counted pair or by branch. A quantity that looks free on six rows is usually a quantity that has been sampled at six rows.
When the second wall is free
On six of thirty lattices, removing both walls of the slot costs exactly what removing the larger one alone costs — 35.9° and 35.9°, 12.0° and 12.0°, agreeing to the last digit of the grid the azimuths sit on. The smaller wall is not a wall on those rows.
The rung decides the sign
Twenty-four lattices where both walls of the slot are really there. Thirteen give a strongly positive interaction, at 85° to 135°; eleven give a negative or null one, at −25° to −0.5°. Nothing lies between. Every rung's lattices fall on the same side as each other.
The exception was already labelled
The larger counted number sorts twenty-two of twenty-four lattices by the sign of their slot interaction. Both misses are on the Lucas 3/4 rung — the one rung a different measurement had already singled out, for reasons with nothing to do with this one.
A survivor has to be a neighbour
A stem that never repairs after a removal keeps exactly one lattice hop rigid, and nothing predicted which one. Sweep every offset at twelve lattices and the answer narrows sharply: at twenty-nine of thirty the surviving hop is one of the two families a counter returns, and the one exception is a step six times too long to be one.
The angle is not the actor
Cut an organ out of two stems that settled on the same divergence and return different counted pairs, and the family left standing is different at every one of the four pairs where both stems wreck. The angle is held to a hundredth of a degree underneath.
Where the survivors meet
At a matched pair the two stems keep exactly the counted numbers their two pairs have in common — the 5 where 3/5 meets 5/8, the 8 where 5/8 meets 8/13, the 7 where 4/7 meets 7/11, and nothing at all where 3/5 meets 8/13. Four rows, including the empty one.
One offset, two answers
Which contact family a wrecked stem keeps is decided by where the cut landed, at twenty-five of thirty offsets, by the simplest rule anybody would write down. It is refuted by two runs: the same counted pair, the same offset, two different rises, and two different surviving families.
One rung, two answers
The offset accounts for twenty-five wrecked stems of thirty and is refuted by a single pair of runs that differ in nothing but the rise. Sweep one rung at a thousandth and the refutation stops being an anomaly: the same offset on the same lattice keeps one family at the coarse end and the other at the fine one.
The family that lost a member
The offset rule restated in the arrangement predicts that the chain whose organ was taken is the chain that breaks. Scored on the nine offsets where the question can be asked, it is right none of the time and its opposite is right all nine.
The front deepens down a rung
The offsets that never repair grow from one to five across a single rung, while a counter returns the same pair at every rise. The extra offsets are not a random extension of the ones already there: they are the ones past the smaller counted number, and they are the ones that keep the larger family.
The ordering on six bands
A hundred and fourteen wrecked cuts across four bands, and at every offset of every one of them the family left standing is the same immediately above the handover and immediately below it. Where the answer does change — on the widest band, at three offsets — it changes somewhere else.
The shortest hop was a coin flip
The reading that a wrecked stem keeps its shortest hop was refuted at twelve of twenty-nine across the census. Re-scored along a single rung, where the counted pair is held and the step ordering reverses, it is right at sixteen of thirty-one — which is not a refutation but an absence of information.
The organ that was nobody's neighbour
Twenty-one of the thirty wrecked offsets remove an organ that lies on neither contact chain through the tip. The reading that explains the other nine has nothing to say about them, and the honest thing is to say so rather than to widen the definition until it does.
Every rise of a band
A band is cut at nine rises because the quantity it was built to test is a constant, and a constant is checked at the ends and at the crossing. On the widest band that quantity turned out not to be constant, which makes nine the wrong number. This is all hundred and twenty-six.
The alternation is not a period
Nine sampled rises gave 8, 4, 8, 4 at one offset of one band, and a period was the obvious thing to look for. At full resolution it is thirteen islands one to three rises wide, with gaps of 1, 2, 3, 6, 7, 8, 9, 16, 31, 44 and 48 — and a fitted period buys exactly nothing.
Three offsets, three crossings
The claim the band design rests on is that the survivor does not change where the two contact steps change places. It holds at full resolution: nineteen changes and not one at the handover. Where they are is three different rises, eight, nineteen and twenty-nine below it.
The offsets that never change
Three of the six offsets that wreck anywhere on the band keep the same family at every rise they wreck at — 98, 22 and 81 rises of the 126. And which offsets wreck at all is a function of the rise, which no reading of a band had drawn.
A band with nothing inside it
Five offsets wreck on the Lucas 7/11 band and every one of them keeps the same family at every rise it wrecks at. There are no islands, no transition region and no period to look for, which is what makes the picture from the other band a picture of that band.
One rise below the census
Ten lattices were cut at every offset and their surviving lags came back as four numbers. One rise further down a rung the census already sweeps, three cuts keep a lag of eleven — which is a fifth number, on a lattice nothing about was unusual except that nobody had cut it.
The side the census sat on
Eight of the ten lattices the ablation census wrecks at were grown past their rung's handover, one before it, and one so close that the ordering it quotes differs by parts in a thousand. A reading scored over the step ordering was therefore scored against a quantity the census was nearly holding fixed.
The lag that never survives
A correction to the exchange's size rests on four hop clusters, and a fifth would be the first real test of it. The prediction was written for a golden lattice at a lag of eleven. No golden lattice on this ladder reaches one, and the reason is a fact about the rule rather than about the search.
The ordering was not the actor
Cut an organ out of every rise of a band where the counted pair and the settled divergence are held and the two contact steps change places, and the family left standing does not change. Fifty-five wrecked cuts on two branches, and the ordering reverses underneath every one of them.
The second band, cut whole
One band was cut at every one of its rises and came back with a transition region — a stretch where three offsets change their answer, in short islands with uneven gaps. The obvious question is whether that is a picture of bands or a picture of that band. The other wide band answers it.
The organ that moved furthest
A reading that works on nine of thirty rows needs a reference organ, and the obvious repair is to measure one rather than to choose it. Measured, the disturbance turns out to have no far edge at all, so there is no organ that moved furthest in any sense the reading can use — and the generalisation that does work needs no reference organ.
The wrecking set moves again
Which offsets wreck a stem was assumed to be a property of the lattice. On one band it turned out to be a property of the lattice and the rise, changing on nearly a fifth of that band's steps. On the second band it changes more, and one offset's wrecking is broken into five separate stretches.
An offset that arrives
The rise where a lattice first keeps a new family is located to one step of the grid, and what happens there is not what the question assumed. No cut changes its mind: a cut that was not wrecking starts, and what it keeps is the new family.
The fourth band, cut whole
Three bands cut at every rise left one account of which bands change their answer standing, and the account was the one nobody had a reason to prefer. The band that would have killed it has now been cut, and it did not kill it.
Two bands that wreck nothing
The census that reads a band refuses two of the six, and the refusal is correct: a band with no wrecked cut has no surviving family, so it has no answer to change. Cutting them anyway turns a refusal into a measurement, and the measurement has a third tone in it that the census cannot see.
A wrecking set with a range
Which offsets wreck a stem was taken to be a property of the lattice, and every band cut whole has found it to be a property of the rise instead. Six bands turn that replication into a measured range, and the range is a factor of fifty-one.
New islands or old edges
Halving a band sweep's step found two more changes of surviving family, and there are two quite different things they could have been. Every coarse change and every coarse island turns out to be carried by exactly one fine one, so the extra pair is a rise the coarse grid stepped over rather than a boundary it misplaced.
Named alongside it
The objects these essays reach for when they reach for this one.
MeasurementAblationParastichy pairHonest limitsRiseClaim testingNegative resultRungControlRigid hopThe placement ruleDivergence angle