Field

Stems and cones

A stem is a cylinder, and on a cylinder the spiral counts are the same the whole way up — which on a disc they never are. Two numbers fix the lattice, both come back out of the points exactly, and lowering one of them climbs the Fibonacci ladder.

A stem unrolled: 28 nodes at 137.51° with a rise of 0.090 circumferences. The counter is shown these coordinates and the circumference, and finds 2 parastichies one way and 3 the other. The faint strips left and right are the same stem: a family leaving one edge re-enters at the other.

A stem is a cylinder

The sunflower is the photograph, and it is the hard case. Nearly all real phyllotaxis happens on a stem, where the geometry is a lattice on a cylinder with two parameters — and where the spiral counts, which on a disc change with radius, are the same the whole way up.

The same counter, on a stem and on a disc. The 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.

Counting up the stem

The same counting machinery, pointed at a stem instead of a seed head, returns one answer three times where the head returned three answers. That contrast is a measurement rather than a preference, and it is the one the whole cylindrical argument rests on.

Six stems built, forgotten and recovered. Each 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°.

Two numbers out of the points

A seed head's divergence angle can be recovered from its spiral counts only to within an interval, because a range of angles gives the same counts. On a stem the counts come with lengths attached, two measurements pin two unknowns, and the lattice comes back to the last digit it was built with.

Every transition as the rise falls. The 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.

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.

Where the disc's counts change, predicted from a cylinder. The dashed lines are the transition radii the cylinder's ladder gives through h = c²/4πr², with nothing fitted. The dots are what the blind counter returns from the disc: 15 of 16 bands agree, and the ones that do not sit on a transition.

A disc is a cylinder

Vogel's seed head makes the rise fall as one over radius squared, so a disc is not one lattice but a family of them. Feed that into the cylinder's ladder and it predicts where a sunflower's spiral counts change — with nothing fitted, and against a counter that never sees either model.

The plane of stems: divergence across, rise up. Each 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°.

The forks are exact

Where a stem's pattern has to choose between two futures, three spiral families are equally short and the lattice is exactly equilateral. A numerical solver found those points; the numbers it returned turned out to be rational, and chasing that gave a closed form — including the fact that every fork sits at a rational divergence, and the golden angle at none of them.

A cone unrolled: 105 nodes at 137.51°, sector 126°. The counter is shown these coordinates and the sector angle and nothing else. Near the apex it finds 3 and 5; near the base, 8 and 13. The rise falls as one over the distance from the apex, so the pair has to climb.

A cone has a rise that falls

A stem holds one parastichy pair for ever and a seed head changes its pair with radius. A cone does both — it is a cylinder whose rise falls as one over the distance from the apex, and the same blind counter that finds one answer up a stem finds four up a cone.

The rise falls as one over the distance from the apex. A straight line of slope −1 on log axes. The dashed horizontals are the transition rises of the cylinder's ladder, computed with no cone anywhere in them; where they cross, the count changes. Consecutive crossings are 2.62, 2.62, 2.62, 2.62 apart — φ² is 2.618.

Transitions a factor of φ² apart

The ladder's rungs are a factor of 1/φ² apart in rise. A disc's rise falls as one over radius squared and a cone's as one over distance, so the same rungs land a factor of φ apart on a seed head and a factor of φ² apart on a cone — measured, on both, by a counter that has never heard of either.

The exponent sets the spacing, and only the exponent. The curve is φ^(2/p), drawn from the ladder's ratio of 1/φ² and nothing else. The dots are measured: each surface built, a blind counter walked up its axis, the places its answer changed recorded. A cone that elongates and a paraboloid that fills sit on the same point at p = 1 — so the spacing identifies neither the shape nor the way material arrives, only their product.

The shape and the law

A cone's transitions are a factor of φ² apart and a disc's a factor of φ, and the temptation is to read the ratio as the shape. It is not. Five surfaces built and counted show that the ratio measures one exponent, and that the exponent is the shape multiplied by the way material arrives.

What is visible in the outer part of a 4000-element organ. Both surfaces have the same ladder in element number — the rise is 1/(2πi·flare) on a cone and 1/(4πi) on a disc, and c and the internode step both cancel. What differs is where the elements are. Counting outside 50 per cent of the extent, a cone shows 1 change and a disc 2, because half a cone's length holds half its elements and half a disc's radius holds three quarters of them.

Why a cone can be counted once

A pineapple is described as 8 and 13 and the description holds. A sunflower is described as 34 and 55 and the description is a statement about one annulus. Both organs have the same ladder in element number — what differs is where an organ puts its elements.

Three organs, and the exponent each has at each place. Left: the meridian of a cone, an ogive of arc radius 12, and a spherical cap of radius 40, each scaled to its own length. Right: d log r / d log s along it. The cone sits at 1 the whole way; the ogive starts near 2 at its blunt tip and falls to 0.23 by the end of the 100 per cent shown; the cap starts at 1 and falls slowly.

An organ has no single exponent

The earlier work measured that a surface whose circumference grows as a power of arc length puts its transitions a fixed factor apart, and checked it on five surfaces. Every one of them had a single exponent, and no organ does — a fir cone is an ogive, whose exponent runs from 2 at the tip to nearly 0 at the shoulder.

One exponent fitted to an organ that has 4 of them. Each dot is one step between consecutive rings, reporting 2 ln φ / ln(s′/s) — the exponent that step would have if the organ had one. They run from 1.980 to 1.697. The line is what a single fit returns, 1.891, which is their harmonic mean of 1.880 and sits below their plain average of 1.887.

What one exponent reports

Fit a single shape exponent to an organ that has four of them and it returns a real quantity — the harmonic mean of what its individual steps report. Harmonic means sit below arithmetic ones, so the fit understates, systematically, in a known direction, and invisibly.

At 1.0 per cent on each ring, 5 rings show the drift. The gaps between consecutive rings are 1.626, 1.631, 1.657, 1.763. Two rings give one gap and no way to disagree with itself; three give two gaps and a fit with nothing left over. The question is how many gaps it takes for their spread to exceed what the measuring error can explain, and the answer depends on the error as much as on the organ.

How much of a cone to measure

Two rings cannot show a varying exponent — not with difficulty, but in principle, because one gap determines one exponent with nothing left to disagree. Four or five can, if each is found to within a per cent. At three per cent this specimen cannot be told from a power law however many of its rings are recorded.

The memory of a divergence sequence, at 0.75° of scatter. With no noise at all the lag-one correlation is 0.54: the rule corrects itself, so a lattice arrives with a memory in it. Matched at the same recorded scatter, placement noise leaves -0.10, jostle noise leaves 0.66, field noise leaves 0.47. The band is ±0.13, which is what an uncorrelated sequence of this length gives.

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.

What the sequence sees that the scatter cannot. Each point is an ensemble at one amplitude, placed at the scatter it produces. A stem at three quarters of a degree of scatter has a lag-one correlation near zero if its noise arrived after the primordium was placed, and near 0.7 if it arrived before — and no measurement of the scatter can tell those apart. The separation closes above about a degree, because what the other two kinds preserve is the correlation of a lattice.

What one angle says about the next

A tenth of a degree of placement noise moves a stem's divergence scatter from 0.50° to 0.62°, which nobody would report. It takes the correlation between consecutive angles from 0.54 to below zero. The other two kinds of noise, at scatters where no measurement can separate them, leave it at 0.6.

The order of the angles carries the count. Three stems, each held at a fixed rise so the pattern sits on one rung of the ladder. At a rise of 0.032 the positions count 3 and 5 spirals and the angles peak at 3; At a rise of 0.013 the positions count 5 and 8 spirals and the angles peak at 5; At a rise of 0.005 the positions count 8 and 13 spirals and the angles peak at 8. Each panel marks the peak and its multiples; the pale strip is what an uncorrelated sequence of this length gives.

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.

The memory belongs to the rise, not to the lattice. The lag-one correlation of a noiseless rising stem, against how fast it climbs the ladder. Below about sixty nodes per rung it is negative; above it, 0.54, 0.74, 0.58, 0.58, 0.55 — flat across a fivefold change in rate. The horizontal line is the same rule with the rise held FIXED, where the correlation is -0.68. So the +0.74 the earlier work called the sequence's own memory is the pattern chasing an equilibrium that is moving under it.

The memory was the rise

The earlier work measured a lag-one correlation of 0.54 in a noiseless divergence sequence and called it the sequence's own memory. Hold the rise fixed and there is no sequence at all — every angle identical — and under a disturbance the correlation is negative. The 0.54 belongs to the pattern chasing an equilibrium that is moving under it.

Two combs, at a rise of 0.005. The autocorrelation of 760 divergence angles from one stem held at a rise of 0.005. The filled teeth are the lags at multiples of 8; the open teeth are the second comb, at the same spacing offset by 5. Reading the spacing off the first and the offset off the second gives the pair 8 and 13, which is what the position counter reports for the same stem — from angles alone, with no coordinate anywhere in the calculation.

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.

Two combs, at a rise of 0.013. The autocorrelation of 760 divergence angles from one stem held at a rise of 0.013. The filled teeth are the lags at multiples of 5; the open teeth are the second comb, at the same spacing offset by 3. Reading the spacing off the first and the offset off the second gives the pair 5 and 8, which is what the position counter reports for the same stem — from angles alone, with no coordinate anywhere in the calculation.

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.

Both statistics, on the same stems, at a rise of 0.005. Five seeded stems at each disturbance, held at a fixed rise. Bars are how many returned the pair the position counter finds; open portions are refusals. The pair comes out from 0.1 to 0.25, and across that whole range the lag-one correlation of the same sequences is -0.33, -0.58, -0.59 — decisive, negative and flat. There is no trade between the two: one stem supplies both. Below the window the sequence has locked onto the sampling grid and is a cycle rather than a sample; above it there is no lattice left, at 116° of scatter.

Two readings from one stem

Three note left with the work in a row have recorded that the two statistics of a divergence sequence want opposite plants — one quiet, one disturbed. Measured on the same stems they do not. The conflict was in the interpretation of a sign, and the window in which both are readable is wide.

Which arrangements carry a comb, and what each one reports. The largest comb mean in five arrangements at a rise of 0.005, all read by the same instrument at the same length, with the sampling band of 0.073 marked. Only the first is a placement rule; the other four are kinematic lattices with no rule in them, differing from one another only in how their azimuth errors are structured. Independent errors and errors with a memory leave nothing to read. A repeating error puts up a comb and names a partner that is not the lattice's. Errors inherited from the contact neighbours reproduce both the comb and the pair.

The comb was never the rule

A control is only as strong as the alternative it builds, and the earlier work built one that varied the rule while holding the disturbance fixed at independence. Five rounds of the angle-sequence thread, with what each claimed and what still stands — and why the next evidence has to come from an intervention rather than from a longer stem.

Every rung has a window it can be read in. For each rung of the ladder, the range of lag windows in which the pair can be read: at least three times the smaller parastichy number, so the main comb has three teeth inside the window, and less than three times the larger, so the larger number cannot itself be a candidate spacing. The band is [15, 24) at 5/8, [24, 39) at 8/13, [39, 63) at 13/21. It is non-empty at every rung, because it is empty only when the larger number is no larger than the smaller. The line at 30 is the window the earlier work used everywhere: it sits inside the 8/13 band and outside the 13/21 one, which is what was mistaken for a ceiling.

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.

The ratio is a U across every rung, and its floor is the number that was reported. The ratio of the second comb to the main comb, on five stems at each of 15 rises spanning two rungs, against the ladder's own coordinate for where each rise sits inside its rung. Both rungs give the same shape: a floor of 0.71 and 0.79 about two thirds of the way up, climbing towards the transition at either end. The dashed line is a transported disturbance with no rule in it at 1.29, which does not vary with the rise at all — a kinematic lattice's angle sequence has no rise in it. Where the rule's curve crosses that line the two accounts are indistinguishable.

The ratio was the floor of a curve

One number was left standing between a placement rule and a transported disturbance, measured at one rise, with the explanation that the geometry there happens to favour the larger parastichy number. Swept across two rungs the number is a U — a floor of about 0.79 two thirds of the way up a rung, climbing past 2.8 as a transition approaches — and the geometry is flat exactly where the curve is steepest.

At 384 azimuths the ratio is 0.62; converged it is 0.82. The comb ratio and the recorded divergence scatter at a rise of 0.005, against how finely the rule samples the circle when it takes its minimum. At 384 azimuths — the grid every flat run in these essays uses, and the grid that earlier work's 0.65 was measured on — the step is 0.94°, which is larger than the 0.25° disturbance the stems carry. The quantisation is white noise, it dilutes both combs, and it does not dilute them equally. The ratio settles at 0.82 from 1152 azimuths up, and the scatter loses 0.19° that belonged to the grid rather than to the stem.

The grid was in the number

The rule places each organ at the least of a profile sampled at a fixed number of azimuths, and every flat run in these essays samples 384 of them — a step of 0.94°, against a disturbance of a quarter of a degree. The quantisation is the larger of the two, it is white, and it moves the discriminator from 0.79 to 0.62. The convergence study this collection had asked for and never done, in the place it turned out to matter.

The damage is the sharing; the forgery is the history. Three disturbances of the same size, measured four ways. The two left columns are stems grown by the placement rule and jostled at 0.25° per organ: a disturbance shared between the contact neighbours scatters the lattice by 0.71° against white noise's 0.57°, and one inherited from them — the same sharing, passed on again at every organ — by 0.97°. The two right columns are kinematic lattices with no rule in them at all, where the whole question is what a disturbance can manufacture. The inherited one returns the pair on 8 seeds of 8 with a main comb of 0.205 against a band of 0.073; the shared one, at the same coupling and the same scatter, returns it on 1 and makes a comb of 0.099, which is the band. So sharing an error with the organs you touch does the damage, and only passing it on and on forges the evidence.

What the sharing costs a lattice

A disturbance inherited from the contact neighbours destroys a stem's lattice at half the displacement independent noise needs, and it moves the comb ratio a fifth of the way to a forgery's. Take the inheritance out and keep the sharing, and the damage stays and most of the ratio shift goes — so the two effects have different causes.

The block is one of the two numbers, and not always the smaller. Every lattice a single organ was removed from, one row each: golden, rise 0.020 carrying 3/5; golden, rise 0.013 carrying 5/8; golden, rise 0.008 carrying 5/8; golden, rise 0.005 carrying 8/13; Lucas, rise 0.020 carrying 4/7; Lucas, rise 0.013 carrying 4/7. The two open ticks on each line are that lattice's own parastichy numbers; the filled dots are the blocks the stems that never recovered settled into, one per offset that failed. The claim this table was built to test is that the block is the smaller of the two, which held at the two rises it was first measured at. It does not hold here: 3/5 gives 5, 5/8 gives 5 and 8, 4/7 gives 4 and 7. What survives is weaker and still worth something — every filled dot but 1 sits on one of that row's own ticks, so the orbit carries a count of the lattice it was cut from, and which of the two it carries is decided by the offset rather than by the pattern.

Two accounts of one number

A stem that never recovers from an ablation settles into a repeating block whose length was the smaller of its two spiral counts, at both arrangements it had been measured at. Two different explanations predicted exactly that and could not be told apart. Swept across four rises on the ordinary branch the premise itself fails: at one arrangement the block is the larger number, at another both appear, and the rule that seemed to be there was two measurements.

One rule, one rise, two branches that stay where they were put. The top 70 organs of two stems grown by the same placement rule at the same rise of 0.013, differing only in the stretch of ideal lattice each was started from. The left one was seeded at the golden angle and settles at 136.781° with the pair 5/8; the right one was seeded on the Lucas lattice and settles at 99.785° with 4/7. Neither drifts towards the other: 0.73° and 0.28° from where each was seeded, over four hundred organs. That is what makes an intervention on the right-hand stem a measurement about a different lattice rather than about a different rule — and 4 and 7 are not Fibonacci numbers, which is the property the experiment needs.

A stem on the other branch

Every stem an organ had been cut from carried Fibonacci counts, which is why two rival explanations of the block a wrecked stem settles into had never disagreed. A stem seeded on the Lucas lattice carries four and seven at the same rise, under the same rule. Cut, it settles on seven — the larger number, and not a Fibonacci one.

A bijugate stem answers in pairs, once the half turn is taken out. Removing one organ from a stem grown by a rule that places two at a time, at a rise of 0.0065, where the pattern counts 6 and 10 and has rotational symmetry of order 2. The open circles are the displacement of the next organ as it comes out of the arithmetic, which reaches 180° at offsets where the stem has demonstrably not been disturbed — the two organs of a whorl are interchangeable, so calling the other one "next" is a relabelling and not a movement. The filled circles are the same numbers read modulo 180°, which is the only way a 2-jugate divergence is defined. Read that way the response is a run of equal pairs — 68.0°, 68.0°, 42.9°, 42.9° — ending at 10, the larger parastichy number, with everything past it under 0.5°. Two organs of one whorl give the same answer to the last digit, which is the pattern's symmetry showing up in an experiment.

What a cut costs a whorl

A bijugate pattern is an ordinary lattice seen twice over, so the account that says a wrecked stem's repeating block is the repeat unit of the lattice underneath has a specific prediction here: three and five. It gets six and ten. And the thing a single missing organ does destroy on a whorled stem is the one property its counts cannot see.

The same rule, the same rise, two lattices, two fronts. How many organs back a single removal is still felt, on a stem seeded onto the golden lattice and on one seeded onto the Lucas lattice, at seven rises. Everything but the seed is identical at each rise — the rule, the spacing, the heights, the azimuth grid — and the two fronts differ at every one of them. Which branch has the wider front changes hands four times going down the range, so no function of the rise gives the column. The golden branch carries 5/8 at four of these rises, across a factor of two in the rise, and its front is eight at all four.

Seven rises and two seeds

One organ removed from a stem is felt out to the larger of its two spiral counts. Every test of that has confounded the count with the rise, because on one branch the two move together. Grow a second branch beside the first at the same rise and they come apart — and doing it at seven rises turns a matched pair into a design whose last column changes hands four times.

The newest member of the front is the weakest. For every cell of the design whose rung boundary is inside the range, how far the next organ moves when the organ exactly as many places back as the larger parastichy number is removed — the offset that arrived when the stem entered this rung — against how far below that boundary the stem sits. Each line is one lattice on one branch. The horizontal line is the threshold that decides whether an offset counts as felt, and the three cells below it are the three whose front reads one offset short. Nothing is a different kind of thing: the boundary is a step everywhere, and near the top of a rung its last stair is shallow.

The front that reads one short

Eleven cells of a fourteen-cell design put the boundary exactly at the larger spiral count. Three put it one offset earlier, and the tempting move is to lower the threshold until all fourteen agree. Measured instead of tuned, the three turn out to be the three cells nearest below their own rung's boundary — and the last offset of a front is weak because it has only just arrived.

The offset past the front that is felt anyway. The four cells of the design whose response has a hole in it: a run of felt offsets, a stretch of quiet, and then one isolated offset well outside the front at which a removal moves the next organ by tens of degrees. The open circle on each row is the count coming in at the next rung of that branch's ladder, and the filled point is the isolated offset. It sits one inside the incoming count on every row, including on the Lucas branch, where the incoming counts are 7 and 11 rather than the Fibonacci numbers the rule was found on.

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.

$Everywhere a cut of one to five organs can send a 5/8 stem. Every settled divergence reached by any arrangement of up to five organs removed from a stem at the 5/8 rung, on one axis. There are six of them and no more. three are slips of the lattice the stem was cut from: each keeps the lag-5 family intact and sits a whole number of turns of it from the next, which is the ladder marked below the axis with rungs 72.0 degrees apart. The other three keep no lag at all and sit near a fraction of a turn, marked above: 175.0 degrees near 1 of 2 turns, 190.0 degrees near 1 of 2 turns, 235.0 degrees near 2 of 3 turns. A stem that is cut either slides along the ladder it was on or leaves it for a lattice with files in it.$

A file has to close

The three destinations counted with a shared factor sit near a half turn, a half turn and two thirds. Measuring how near is the trap: by distance from the fraction, the golden angle is closer to two fifths than two of them are to anything, and would be reported as having five files it does not have.

Which rises are a lattice, from 0.04 to 0.13. How much the divergence wanders over the last sixty organs, at each rise up the coarse ladder. 13 of the 19 settle, scattering between 0.0000 and 0.3356 degrees. six do not: from 0.09 to 0.115 the divergence sticks on exactly 135.0000 degrees, which is three eighths of a turn, and wobbles about it by 0.79 to 1.60 degrees. A counter shown either kind returns the same pair, so the counts cannot tell them apart. The line is the threshold a rise has to pass before a cut is made on it, and it sits in the gap rather than among the measurements.

A stem coarse enough to cut

Below the 3/5 rung is a 2/3 rung, and it runs from a rise of 0.050 to 0.120. It is not a lattice across all of it: from 0.090 to 0.115 the divergence stops settling and sticks on exactly three eighths of a turn, wobbling by a degree and a half — while a counter goes on reporting 2/3 as though nothing had happened.

What a two-organ cut does at each rise of the 2/3 rung. At every rise the coarse rung is a lattice on, all 36 arrangements of two organs removed, with the ones that never repair counted and split by where they end up. 22 of 324 cuts across the rung reverse the stem's handedness onto the mirror of the divergence they were cut from. 40 fall instead into a cycle whose mean is half a turn, which the lattice they came from has no number for. 4 rises give only the first, 4 give only the second, and at a rise of 0.075 both happen in the same table, which is what says the fate belongs to the cut and not to the rise.

The shallower front turns over

If reversing a stem means rearranging its whole front, then a stem with a shallow front should reverse more often. Measured across three rungs and four hundred and seventy-three cuts: 6.8 per cent at a front of three, 4.7 at five, and none at all at eight — where the nearest approach is two tenths of a degree away and stays there.

Which offsets give short hops, at a rise of 0.013. The two lowest points are at 5 and 8, and those are the parastichy numbers. Offset 1 is high because a hop of one node is at least the rise, which is what makes a stem easier to count than a disc.

Where a handover sits

Inside every rung there is a rise at which the two contact steps change places, so that the shorter hop belongs to the other family below it. Six of the eight rungs on this ladder have one, each has exactly one, and every one of them sits in the coarse half.

Two lines across the 5/8 rung, crossing once. The divergence the rule settles on, against the divergence at which the two contact steps would be exactly the same length. The second is arithmetic on the lattice and no stem is grown for it. Across this rung the balanced line moves 2.281 degrees and the rule's own line moves 1.262, so the shallower line crosses the steeper one, and it does so exactly once at a rise of 0.0154 — 16 per cent of the way down from the coarse end. That crossing is the handover: above it one family has the shorter step and below it the other does. So a rung has one handover, its position is fixed by the arithmetic rather than by any experiment, and a sweep of the rise carries a stem across it at a place nobody chose.

Two lines that cross once

The divergence at which a lattice's two contact steps are exactly equal is a curve across each rung, computable from the geometry with nothing grown. The rule's own settled divergence is a second, shallower curve, and where they cross is where the step ordering changes hands.

$A divergence that does not move across the 5/8 band. Measured at every rise of a band on the golden branch, where a counter returns 5 and 8 spirals throughout. The settled divergence moves by 0.0469 degrees across the whole band, which is a fraction of the azimuth grid step and a fortieth of the slide across the rung it sits in. The ratio of the two contact steps does move: it falls to 1.0013 and the ordering changes hands at a rise of 0.0156, so above that rise the shorter step belongs to the 5 family and below it to the 8 family. Two of the three quantities that vary along a rung are therefore held here and the third is not, which is what makes the ends of this band a matched pair.$

A band that holds the angle still

Around every handover the settled divergence has a shallow floor, so a run of rises either side of it share a divergence to a twentieth of a degree while their two contact steps change places. That is a matched pair with one quantity varying, and it is the design the ablation thread had no way to state.

How many organs a stem needs before it is on a lattice. One row per rise, one mark per starting angle, placed at the organ from which every later divergence stays within a degree and a half of the run's own final value. Where a stem settles at all it does so between 0 and 290 organs in, against the 400 every ablation run here grows before it cuts anything. Not one row needs the length it is given. What changes down the table is the count on the right: how many of the nine starting angles reach a lattice at all, which falls from 7 at the coarse rises to 1 at the finest.

How long a stem takes to settle

Every result here is grown on a stem that has settled onto a lattice, and settling has always been tested for and never timed. Timed, it takes between nothing and two hundred and ninety organs — against the four hundred every ablation run grows before it cuts anything, and the nine hundred the noise runs carry.

A wall and not a budget

Below a rise of about 0.005 this collection's stems stop settling onto a lattice, and the limit has been written up four times without anybody asking which kind of limit it is. Grown three times as long, the table is identical row for row: not one stem that failed to settle succeeds. The floor is a wall.

The settled divergence down the golden branch. Every rise from 0.07 down to 0.00482, plotted against the divergence the rule settles on, with each rung drawn in its own stroke and the branch's limit angle marked. The curve does not slide: it turns three times in four rungs, climbing across one and falling across the next, so a value it takes on one rung it takes again on another. That is what makes a matched pair possible — two rises, different counted pairs, one angle — and it is the whole reason the design exists on this branch. The widest excursions from the limit angle, coarse rung first, are 3.195°, 3.352°, 0.961°, 0.422°.

The angle the ladder returns to

Down the golden branch the settled divergence climbs across one rung and falls across the next, turning three times in four rungs. That is why the same angle is reached at two different rises — and why the Lucas branch, which turns once, almost never offers the same thing.

Two rungs, one angle

Five pairs of rises settle on the same divergence while a counter returns different pairs at them, and four of the five agree to 0.0000° — the same value of a quantity read on a grid of 1,536 azimuths. The rises differ by factors of 1.48 to 5.09.

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.

The last of three quantities

The pair, the divergence and the step ordering move together when the rise is swept, and for a long time no result could be attributed to any of them. Two designs later, two are ruled out as sufficient and the third has never been held still — because holding it is what a rung already does.

Where the two contact steps change places, on every rung. One row per rung of the two branches, drawn from its coarse end to its fine one on a logarithmic axis. The mark on each row is the rise at which the two contact steps change places — the rung's handover — and the shaded stretch is the band around it over which the settled divergence holds still. six of the eight rungs have a handover and every one of those sits between 6 and 40 per cent of the way down its rung, never past the middle. The two rungs without one are the coarsest on each branch, whose crossing is above the range this ladder reaches.

Four crossings nobody visited

Six rungs of the ladder carry a handover and two of them had a band built on them. The other four are here: 70, 126, 16 and 124 rises wide, found by sweeping at a ratio rather than at a fixed step in the rise, which is why the fine ones had been stepped over.

What each band holds, and what it moves. One row per band. The counted pair is held by construction and the settled divergence is held to a twentieth of a degree; the quantity a band exists to move is which of the two contact steps is the shorter. five of the six do move it — the ordering changes hands exactly once inside, and at both ends the two steps differ by enough for an ordering to mean anything. The remaining one changes hands three times and its two steps are never more than 0.0 per cent apart, so it holds all three quantities and is a control rather than an experiment.

A band that moves nothing

One of the six bands holds the counted pair, holds the divergence, and does not move the ordering: its two contact steps stay within four parts in a thousand of each other across the whole of it, so the ordering changes hands three times and neither end has one worth the name.

Two ways of predicting how wide a band is. A band ends where the settled divergence has moved 0.05° from its value at the handover, so the width should follow from how fast the divergence changes there. Reading that rate as the rung's average slope predicts widths that are wrong by factors of 0.20 to 5.92 — wrong in both directions, so no constant rescues it. Reading it as a curvature about a stationary point gives 0.41 to 1.08, with five of the six inside a third. The difference between the two is the difference between a curve and its average, and a band is exactly where the two are least alike.

How wide a band should be

A band ends where the divergence has slid a twentieth of a degree, so its width should follow from how fast the divergence slides. Predicted from the rung's slope that is wrong by factors of 0.20 to 5.92; predicted from a stationary point it is 0.41 to 1.08, and the outlier is the rung that has no stationary point.

The families left standing, on both sides of every handover. One row per band, with every offset cut at rises spread across it and always at both ends and at the handover itself. The last column is every family left standing anywhere on that band. On three of the four bands that wreck at all it is a single family, unchanged across a rise at which the two contact steps swap places — so the step ordering is not what decides which family survives, and the result now rests on four counted pairs rather than on two. The shortest-hop reading scores 44 of 114 across the whole set, which is what a reading looks like when the quantity it is stated over is not in the mechanism.

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.

Every rise of the 8/13 band, cut at every offset. One column per rise of the band, coarse on the left and fine on the right, and one row per offset that wrecks anywhere on it. A filled cell is the family the cut stem keeps; a pale cell is an offset that recovers at that rise and has no survivor to report. The band holds 126 rises and 1890 cut stems. three of the six offsets change their answer somewhere inside, three never do, and the vertical rule is the handover — the rise where the two contact steps change places. Not one of the 19 changes is at it.

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.

Offset 7 across the 8/13 band, rise by rise. The family this one offset keeps at each of the band's 126 rises, coarse on the left. It wrecks at 126 of them and keeps the 4-family and the 8-family at different rises. The ticks below mark one islands — runs of 2 rises where the coarse family comes back inside the fine one. The handover is the taller rule and the change of answer is nowhere near it.

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.

Offset 4 across the 8/13 band, rise by rise. The family this one offset keeps at each of the band's 126 rises, coarse on the left. It wrecks at 98 of them and keeps the 8-family throughout. The ticks below mark no island: the answer changes once and stays changed. The handover is the taller rule and the change of answer is nowhere near 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.

Which offsets wreck across the Lucas 7/11 band. One row per offset and one column per rise, with a mark where a single removal at that offset wrecks the stem. The set is not the same at every rise: on this band one offset wrecks at only 27 of its 124 rises, in several separate stretches, while others wreck at all of them. So a census taken at one rise of a band and a census taken at another are censuses of different sizes, and every claim of the form "at every offset that wrecks" is quantified over a set the rise decides.

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.

The hops the correction is fitted over, with the new one at the near end. Each cluster of rows placed by the lag it kept and the angle of the hop that lag keeps. The four the correction was fitted over run from 19.5 to 39.1 degrees; the new one sits at 12.78 degrees, a third smaller than any of them. A fifth point beyond the near end of a fitted range is a test of the fit, where a fifth point between two old ones would mostly have been a restatement.

A fifth cluster

A correction to the exchange's size was fitted over four hop clusters and its own file said so. A search turned up a fifth, at a hop smaller than any of the four, and the rule is right on it — which is what a prediction being confirmed looks like when the confirmation is worth having.

What is the same at 0.00998 and at 0.00997. Six quantities read on the two rises the transition sits between. Five of them are identical: the two walls the slot has, the divergence the intact stem settles to, the block the cut opens, which of the three removals wreck the stem, and the lags the doubled cut leaves rigid. The sixth is how far the first organ placed after the doubled cut moves, and it goes from 163.59 to -9.14 degrees. The lattice is the same on both sides; where one organ goes is not.

A transition and not a slope

The question was whether a fourth cell's cost declines smoothly to nothing or falls in one step. It falls in one step, and the answer decides whether a word in the collection names something or is a threshold on a continuum.

Twenty rises at the fine end of both branches, cut at every offset. One row per rise searched, coarse at the top of each block. The bar names the counted pair the stem shows and the numbers on the right are the lags its wrecking cuts leave standing. A pale row is a rise whose settled divergence has left the branch it was started from by more than 20 degrees, which is what happens below the ladder's finest rung — the pairs there are 2 and 4, 8 and 16, 11 and 22, which are not two consecutive terms of any additive sequence. One rise on the Lucas branch keeps a lag of 11 while still on it.

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.

Where the doubled cut's run finishes, at every rise of the sweep. The divergence the wrecked run ends at, rise by rise. It takes only a handful of values and jumps between them at rises one part in a thousand apart, 4 times inside the 9 rises of the finest sweep alone — where the walls, the block, the intact stem's divergence and the rigid lags are all held. So the end of a wrecked run is not a stable quantity on this rung, and no statement about where the stem finishes is available on either side of the transition. The first organ's displacement is the reproducible half of the same measurement.

The end of a wrecked run

The obvious follow-up to a transition in where one organ goes is whether the stem also finishes somewhere different. On this rung the question has no answer: a wrecked run's final divergence takes four values and changes between rises a thousandth apart, three times inside a nine-rise sweep.

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 two widest bands on the ladder, each cut at every rise. One block per band, one row per offset that wrecks anywhere on it, one column per rise, coarse on the left. A filled cell is a cut that wrecks, and its tone is the family left standing; a pale cell is a cut that recovers. The golden 8/13 band above changes its answer at three of its six offsets, 19 times in all. The Lucas 7/11 band below changes it nowhere: every cut that wrecks on it keeps the 7 family, at every offset and every one of its 124 rises.

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 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.

What nine rises find on each band, against what all of them find. Two bars per band: the changes of surviving family a nine-rise design finds, and the changes the full sweep finds. On the Lucas band the two agree exactly, at none and none. On the golden band they agree that something changes and disagree about how much — 5 against 19 — because one step of that design is 16 rises and the band carries features one to three rises wide. A sample was never wrong about whether; it was wrong about how many.

When nine rises are enough

A coarse design was shown to be misleading on one band and it has been criticised on that ground ever since. On the second band it is exactly right, and the difference between the two cases is a property of the band rather than of the design — which is the awkward part.

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.

Where a slot loses a wall

Two rungs were reported to go free at 84 and 85 per cent of themselves — the second removal stops costing anything over the larger one alone. Three samples a rung cannot say whether that is a transition or a slope, and twenty-nine more say it is a transition one grid step wide.

All three bands cut at every rise, offset by offset. One row per wrecking offset on each band cut whole, one cell per rise, coarse on the left. A pale cell is a rise at which that offset's cut recovers and has no survivor; a dark cell is a cut that wrecks and keeps one of the band's own counted pair; a warm cell is a cut that keeps a family off the pair. The vertical rule on each row is that band's handover, where its two contact steps change places. The golden 8/13 band changes the family it keeps 19 times, the golden 5/8 twice and the Lucas 7/11 not at all.

The third band, cut whole

Two bands cut at every rise disagreed about whether the family a cut keeps ever changes, and three explanations were available for a difference between two things. The cheapest third band settles which of them survives, and it settles it against the account nobody was betting on.

The 17 rows of the exchange table, gathered by the lag they kept. One bar per surviving lag, its length the number of rows the table holds at that lag, with the hop that lag's stems keep written beside it. The hop is nearly constant inside a lag, so a correction fitted over 17 rows is fitted over four hops — which is the denominator that matters and is much smaller than the row count suggests. Adding a lag to the table is worth more than adding rows at a lag already in it.

A family that is a multiple

When a wrecked cut keeps a family that is not one of the lattice's counted pair, the first case looked like a rule: it was half of one of them. The second case is four times the other, which makes the rule a coincidence and leaves a weaker statement that is probably the true one.

Four accounts of which bands speckle, scored on the three cut whole. Each candidate explanation of why one band's cuts change the family they keep and another's do not, against what the three bands cut whole actually do. A tick is an account that puts that band on the side the sweep does. The branch the band sits on is right on all three; the size of the counted pair, the number of wrecking offsets and how much of its rung the band spans are each wrong on two. Three bands can eliminate and cannot confirm, and this eliminates three of the four.

The branch is what is left

Four accounts of why one band's cuts change what they keep and another's do not were written down before a third band was cut. Three of them are now wrong on a band each, and the survivor is the one with no mechanism behind it.

What nine rises find on each band, against what the whole band holds. The coarse design cuts nine rises of a band, evenly spaced in the logarithm of the rise, and asks whether the family a cut keeps changes anywhere. On the golden 8/13 it finds five of nineteen changes and on the Lucas 7/11 it finds none of none, so it had never been wrong about whether anything changes. On the golden 5/8 there are two changes and it finds neither, both of them at single rises with the offset recovering on either side. Its record on that question is now 2 of 3.

Nine rises were not enough

A coarse sample of a band had never been wrong about whether anything changes inside it, and that record was the argument for trusting a negative from it. The third band cut whole makes it two of three, and the missed feature is one rise wide.

A change with nowhere to be

The claim this whole thread rests on is that a survivor does not change where the two contact steps change places. Nineteen located changes never put one there. The twentieth is flagged at a handover, and it is flagged because the offset stops wrecking for thirty-four rises.

One step of the grid, again

A search of the fine end stepped from one rise to another a hundred grid steps away, and the family a cut keeps changed somewhere between. Cutting the rises in between puts the change inside one step, with the lattice identical on both sides.

The two shortest hops across the Lucas 7/11 rung, and where they cross. The lengths of the two shortest hops at every one of the 378 rises of the rung, fine on the left, computed from the settled divergence rather than measured off a cut. They cross exactly once, between 0.00804 and 0.00803, which is this rung's handover. The rise at which a cut first keeps the longer family is 164 steps of the grid further down and is marked separately; nothing happens to either length there.

The hops cross once

Walking a whole rung at the grid its rises are named on costs a few hundred stems and no cuts at all, and it answers a question nobody had asked: whether a rung has one handover or several. It has one, and the ladder had recorded it in the wrong place.

Two rises on the Lucas 7/11 rung, each located to one step of the grid. The whole of the Lucas 7/11 rung, coarse on the left, with the two rises this round located. The two contact steps change places between 0.00804 and 0.00803, found by walking all 378 rises of the rung and evaluating hop lengths, with no cut stems at all. The family a cut leaves standing changes between 0.0064 and 0.00639, found by cutting 89 stems. They are 164 steps of the grid apart, which is 45 per cent of the rung's whole span, so the geometry crossing is not what moves the survivor.

Two rises far apart

The whole handover thread rests on the claim that where the two contact steps change places is not where the survivor does. On one rung both rises are now located to a single step of the grid, and they sit forty-five per cent of the rung apart.

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.

Where every wrecked run finishes, rung by rung. One row per rung of the ladder, one mark per cut cell of the slot design, placed at the divergence that run ended on. The open mark on each row is the divergence the intact stem of that rung settles to. On the Lucas 1/3 and golden 2/3 rungs every wrecked run finishes at the same value; on the golden 5/8 they finish at 13 values spanning 215 degrees. 27 cells recover, and each of those finishes at its own settled divergence to within 0.03 degrees.

The column nobody read

Every cell of the slot design carries where its run finished as well as how far its first organ moved. One rung's worth had been plotted and called unusable. Reading all eight says the endpoint is exact on two rungs, wanders on six, and is worst on the one it was read on.

A wrecked run goes somewhere

Where a wrecked stem finishes was called unstable. Half of them finish within a degree of a destination measured from intact stems started at arbitrary angles — two tables that share no run, no design and no question.

The twelve wrecked runs that finish at a half turn. Wrecked runs whose final divergence is a half turn, which is two files of organs rather than a spiral. Every one is on a rung at the coarse end of its branch, where the front is short enough that removing one organ reaches past it. The value is within 0.35 degrees of 180 on eight of the twelve.

Two files, and a way back

Twelve wrecked runs finish at exactly a half turn, which is a pattern with no spiral in it, and every one is at the coarse end of the ladder. Three finish at the divergence they would have had anyway, after being thrown a hundred degrees off it.

The Lucas 4/7 band cut at every one of its 86 rises. One row per wrecking offset on each band, one cell per rise, coarse on the left, with each band's width in proportion to the rises it holds. A pale cell is a rise at which that offset's cut recovers and has no survivor to report; a dark cell keeps one of the band's own counted pair and a warm cell keeps a family off it, so a change of answer is where the tone changes. The Lucas 4/7 is the test the branch account most needed: 86 rises, 774 stems, 144 wrecks across two offsets and not one change of the surviving family. Across the one bands drawn, 144 cuts wreck of 774 grown. The dashed rule on each row is that band's own handover, where its two contact steps change places.

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.

The two bands that answer nothing, 586 stems and no wreck anywhere. The g35 and L34 bands cut at every rise, with one row per offset tried and one cell per rise, coarse on the left. A pale cell is a cut whose stem recovers and a mid cell is one that neither recovers nor leaves an orbit to count; the warm tone means a cut that wrecks, and it appears on the reference row below, which is offset 4 of the Lucas 4/7 wrecking at 83 of its 86 rises. g35 grew 490 stems across 70 rises at seven offsets and L34 grew 96 stems across 16 rises at six offsets, and wrecked none of them, so this is a zero that was measured rather than a band nobody cut. The dashed rule on each block is that band's own handover, where its two contact steps change places.

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.

Four accounts scored on the four bands of six that can score them. One column per band of the ladder and one row per candidate account of why a band's wrecking cuts change the family they leave standing. Each cell is what that account predicts of that band, in plain type where the band agrees with it and pale where it does not, and the header of each column is what the band actually does. The two bands with no wrecked cut are shown silent, because a band with no surviving family has no answer to change; over the four that can answer, the branch the band sits on is right 4 times of 4 and the other three are wrong twice each. A warm block marks each cell where the account and the band disagree.

Six bands, one table

Every rung of this ladder that carries a handover now has a band grown on it and cut at every rise it holds, and four accounts of which bands change their answer are scored on all six at once. The survivor is right on every band that can test it, and the same table read one cell differently kills it.

What nine rises find on six bands, against what those bands hold. The coarse design cuts nine rises of a band, evenly spaced in the logarithm of the rise, plus both ends and the handover, and asks whether the family a cut keeps changes anywhere on it. Scored on six bands it is right about whether on five and wrong on the 5/8, where it finds none of 2. On the one band where it finds anything it finds 5 changes of 19, so its record is a record about whether and never about how many. The figure under each band's name is that design's step on it, in rises.

The coarse design scored

Every claim this thread has made about an uncut band rests on a sample of nine rises, and its record was the argument for trusting it. Six whole bands close that record, and two of its six correct verdicts are correct only because there was nothing on those bands to find.

How broken the wrecking is on four bands, from 2.43 rises a stretch to 126. Every offset that wrecks on every band that wrecks, placed on a logarithmic axis of rises wrecked per separate stretch — the smaller the number, the more broken the wrecking. A filled mark is an offset that wrecks at every rise of its band and an open one comes and goes. The range runs from 2.43 on g58 offset 3 to 126 on g813, a factor of 52, and two of the four bands have no offset that wrecks everywhere at all. A band with no offset that wrecks at every rise is said so at the right.

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.

Six rungs walked at the grid, 10 crossings between them. Each row is one rung, drawn from its coarse end on the left to its fine end on the right and scaled to its own width so positions inside different rungs can be compared. The shaded stretch is the band the ladder grows around the rise it recorded as that rung's handover. five of the six rungs carry exactly one crossing, and on each of those the whole stretch a second one could sit in has been walked at the grid and closed at both ends. The 3/4 rung of the Lucas branch carries five, spaced 22, 12, 15 and 20 grid steps apart, and its band holds three of them. The rule on each row is a located crossing.

Five rungs walked

Six rungs of the ladder carry a handover and only one of them had ever been walked at the resolution its rises are named on. Walking the other five costs 1,224 grown stems and no cuts at all, and it returns a crossing count per rung — five ones and a five.

A reading that steps against an ordering that slides, on six rungs. Every hop length here is arithmetic on the settled divergence, and the settled divergence is a mean of angles read off a lattice placed on a fixed set of azimuths — so it is read in steps rather than continuously. Between two steps the ordering slides with the rise; at a step it jumps. The bar is how much a jump is worth in slides on each rung, measured inside the stretch that was walked at the grid so that the six are read over comparable regions. The account is a threshold at one and it is right on all six: the 3/4 rung sits at 12.71 and carries five crossings, and on three rungs the reading does not step inside the window at all.

One crossing or two

One rung of the ladder changes hands five times where the other five change hands once, and the difference is not in the geometry. It is a divergence read in steps against an ordering that slides, and the account is a threshold at one that is right on all six rungs.

Six handovers relocated, in steps of the sweep that recorded them. The ladder finds a handover by stepping at a ratio of one per cent, which never lands on the grid the rises are named on, so a recorded handover is the nearest rise the sweep visited to a crossing nobody had located. This is the difference, in units of the sweep's own step at that rise: 0.082 to 0.835, mean 0.374. All six are positive and all six are inside a single step of the sweep, and both of those are predictions rather than summaries: the ladder reports the first rise it visits at which the ordering has already changed, so the rise it records must sit on the fine side of a crossing and within one of its own steps. The dashed rule is one step of the sweep, which is the bound the sampling predicts.

The handovers corrected

Six recorded handovers, relocated to the grid against where a one-per-cent sweep put them: all six sit on the fine side of a crossing and all six inside a single sweep step. Nothing about the rung explains the size of the discrepancy, which is what a sampling artefact is supposed to look like.

19 changes and 2 at the coarse step, 21 and 2 at half of it. The changes of surviving family each band makes, at the step the sweep has always used and at half of it. The golden 8/13 band goes 19 to 21 and the golden 5/8 band 2 to 2. Halving the step cannot lose a change, since the fine grid holds every coarse rise with the same answer; what it can do is find one, and it does so on one band and not on the other. So whether a tally is a floor or a total is a property of the band rather than of the sweep — a floor where there are islands narrower than the step, and exact where there are none.

A count or a floor

Nineteen changes of surviving family on the widest band have been quoted as a number since the band was cut, with nothing to say whether a finer grid would find more of them. Halving the step finds twenty-one there and nothing at all on the next band along.

The rise at 0.005845 the coarse grid steps over, under offset 6 at both steps. Offset 6's cut drawn at every rise between 0.00596 and 0.00573, coarse above and fine below, with a cell per rise coloured by the family the cut leaves standing and pale where the cut recovers. It changes its answer 3 times at the coarse step and 5 times at half the step, and the gain is the single rise 0.005845, where the 4 family is kept with 8 on both sides. That island is 0.86 parts per thousand of the rise wide, against 1.47 for the smallest step the coarse grid takes anywhere on the band, so no coarse rise could have landed on it.

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.

The 125 steps of the golden 8/13 band, in units of the grid they are rounded to. The sweep's rises are held to five decimal places, so every step it takes is a whole number of units of that fifth place. Each bar is how many of this band's 125 steps are that many units: 99 of 1, 26 of 2. One unit is 1.653 parts per thousand of the rise at the handover of 0.00605, against a nominal step of 2.00, so a sweep asked for at one part in a thousand would land on the same rise twice and the finer grid has to be laid at 7 places instead.

The last unmoved setting

Halving a band sweep's step is only a halving if the sweep steps where it says it does, and this one does not: its rises are rounded to five decimal places, so at one handover the grain is 1.65 parts per thousand against a nominal step of two. Moving the last setting nobody had moved found the setting was never what it was called.

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