Where a handover sits

A rung is a range of rises over which a counter, shown nothing but the point positions, returns one pair of numbers. That constancy is the whole of what the word means, and it is why the ladder is the useful picture of this subject: the counts do not slide, they step.

Inside a rung, though, things move. The settled divergence slides. The front deepens. And there is a third quantity, which this essay is about, that not only moves but reverses: which of the two contact families has the shorter step.

What the two steps are #

Take a stem at a settled divergence and a rise. Every pair of organs k places apart is joined by a step that runs some way around the cylinder and some way up it, and the length of that step is a number the lattice fixes. Rank the lags by that length and the first two are, by definition, the contact families — the two numbers a counter returns.

Which of those two is first is a separate question from which two they are, and it is a question nothing in a count answers. A stem counted at five and eight spirals can have the five-step shorter or the eight-step shorter, and a counter returns five and eight either way.

So somewhere inside the rung the two lengths are equal, and the ordering changes hands there. Call that rise the rung’s handover. It is not a place where anything visible happens: no count changes, no front changes width, nothing about a drawing of the stem looks different. It is a place where a ranking swaps.

Six of eight, and each exactly once #

The ladder here carries eight rungs between rises of 0.0040 and 0.0700 — four on the golden branch and four on the Lucas one. Measured by sweeping the rise at one per cent and reading the hop ranking at each step, six of the eight have a handover inside the range, and each of those six has exactly one.

The two without one are the coarsest rung on each branch — golden 2/3 and Lucas 1/3 — and their absence is a statement about the range swept rather than about them. Both are cut off at 0.0700, which is as coarse as this collection’s stems go before the pattern stops being a lattice worth counting, and the arithmetic says their crossings sit above it.

That each of the six has exactly one is the part worth pausing on, because it is not obvious. Two lengths that cross could cross three times, or five. That they cross once is a consequence of the two curves involved being smooth and one of them being shallower than the other over the whole rung, which is an argument the arithmetic makes and no sweep needed to discover.

How it was measured, and what would have broken it #

The ranking is read from the ideal lattice rather than from a grown stem: at each rise the settled divergence is measured once, the step lengths of the first forty lags are computed from that divergence and that rise, and the list is sorted. That separation matters. A ranking read off a grown stem’s own neighbour graph would fold the rule’s noise into the answer, and near a handover the two lengths differ by parts in a thousand, so any noise at all would decide the ordering instead of the geometry.

Two things would have made the measurement worthless and neither happened.

The first is that a handover could have been an artefact of the azimuth grid. The divergence is quantised — a stem’s angles land on a grid of 1,536 steps — and if the two step lengths differ by a thousandth then a grid step of a quarter of a degree could easily be what decides which is shorter. Re-measured on a grid four times finer, the handovers move by less than one sweep step. They are in the geometry rather than in the discretisation.

The second is that the ranking could have been unstable — the two lengths crossing, recrossing and crossing again over a few rises, so that “the handover” was a region rather than a rise. It is not. On every rung the ratio of the second length to the first falls monotonically to one and rises monotonically away from it, which is the shape a single crossing has and not the shape a fluctuating ranking has.

What remains is a bound rather than a defect: within about one per cent of a handover the two lengths differ by less than the ranking can meaningfully separate, so a rise landing there has no ordering worth quoting. That is not a failure of the measurement; it is what “equal” means when it is approached from both sides. It matters in practice, because one of the census’s own lattices sits within two and a half parts in a thousand of a handover, and the readings scored on that row have been quoting an ordering that is not there.

And every one of them is in the coarse half #

Here is the measurement this essay exists for. Taking the position of a handover as a fraction of the way down its own rung from the coarse end — measured in the logarithm of the rise, because the ladder is geometric — the six sit at:

Six of six in the coarse half, spread from six per cent to forty. Never past the middle, and never at either end.

Six is a small number and it is worth being clear about what it can carry. The claim is not that a handover can never sit past the middle — nothing here proves that — but that on both branches of this ladder, over every rung the range reaches, it does not. Two branches with unrelated seed angles and pairs from 3/5 to 7/11 is a wider base than a single branch would be, and the arithmetic behind it is the same on any branch, which is the reason to expect the pattern to hold where it has not been checked rather than to assume it.

Why the coarse half #

The handover is where the two step lengths are equal, and that condition defines a curve in the plane of rise against divergence: for each rise there is one divergence at which the two families are equidistant. The rule’s own settled divergence traces a second curve across the same rung.

Across the 5/8 rung the balanced curve moves 2.281° and the rule’s own moves 1.262°, so the rule’s line is the shallower one. It starts above the balanced line at the coarse end, crosses it, and finishes below. Where it crosses depends on how the two slopes compare, and on this ladder the balanced line is consistently about twice as steep — which puts the crossing nearer the coarse end than the fine one, on every rung.

So the answer to “why the coarse half” is arithmetic rather than biological or dynamical. It is a statement about the relative slopes of two curves, one of which is fixed by the lattice and one of which is where the placement rule settles.

What it is for #

Two things, and the second is why it is worth an essay rather than a footnote.

The first is that a handover is a place to put a controlled experiment. Straddle it and the counted pair is held, the settled divergence is nearly held, and the step ordering flips — which is a matched comparison with one quantity varying, and it is exactly the design the ablation thread needed and had no way to state before the rung’s interior was mapped.

The second is that a handover explains a puzzle in the tables. Every census here grows one stem per rung, and the position that stem sits at was never recorded. Positioned now, eight of the ten lattices that produce a wrecked stem sit past their rung’s handover — which is not a coincidence but a consequence of this measurement. If a handover is always in the coarse half, then a rise picked anywhere near the middle of its rung is past it, and picking rises near the middle is what a person does when they want a rise that reliably produces a pair.

That turns an accident of sampling into something predictable: any census built the sensible way, by anybody, will sit past the handover on nearly every row.

The name, and why it needed one #

This collection has got along without a word for it until now, which is itself worth a paragraph. The quantity has been visible in the tables since the hop ranking was first computed — every ranking ever printed here has a first entry and a second — and the reversal was noticed once, in passing, as a thing that happens along a rung. It was not named, and unnamed quantities do not get measured across a ladder.

The consequence was concrete. The reading that a wrecked stem keeps its shortest hop was stated, scored and refuted without anybody establishing where the census sat relative to the reversal, because there was no noun to hang the question on. With one, the question asks itself: which side was each row on? That is the whole content of the next two essays, and neither needed a new experiment.

What a handover is not #

It is not a transition. A transition is where the counted pair changes and the ladder steps; a handover is inside a rung and changes nothing a counter can report. Two rises either side of a transition are different lattices; two rises either side of a handover are the same lattice with its two nearest-neighbour families ranked the other way round.

It is not a fork. A fork on the van Iterson diagram is where three families are equidistant, and those have a closed form and sit at exactly rational divergences. A handover has two families equidistant, which happens along a whole curve rather than at isolated points, and the rule crosses that curve once per rung.

And it is not, on the evidence so far, a place where the mechanism does anything. Cutting an organ out of a stem either side of a handover leaves the same family standing, at every offset that wrecks, on both bands measured. That is a negative result about the ablation rather than about the handover, but it does bear on how much weight the ordering should carry: a ranking that reverses without changing an outcome is a ranking the outcome does not depend on.

What a reader can check #

The whole of this is arithmetic on two numbers, which makes it unusually easy to verify without any of the machinery here. Pick a rise, take the settled divergence at that rise, and compute for each lag k the horizontal offset — k times the divergence, folded to within half a turn and divided by a full one — and the vertical offset, k times the rise. The step length is the hypotenuse. Sort. The first two entries are the pair a counter will return, and which of them comes first is the ordering this essay is about.

Doing that at 0.018 and at 0.008 on the golden branch gives five first and then eight first, with the same pair reported both times. Doing it at a hundred rises in between finds exactly one place the order changes. That is the measurement, and it takes a spreadsheet.

The part that is not arithmetic is the settled divergence, which has to come from somewhere. Here it comes from growing a stem under the placement rule and averaging its last sixty divergences, which is the same measurement the rest of this collection uses and which carries the rule’s own choices with it. Substituting the ideal Fibonacci limit angle instead — 137.5077° at every rise — moves the handovers by several per cent of a rung and does not move any of them out of the coarse half. So the finding survives the substitution, and the substitution is worth naming because it is the one a reader is most likely to make.

What is left #

The obvious extension is the rungs this range cannot reach. Both branches’ coarsest rungs have their crossings above 0.0700, and both branches stop below about 0.005 for a reason that has now been measured: a stem below that does not settle onto a lattice at any run length. So the six rungs here are not six of many available; they are six of eight, and the other two are outside the window rather than unexamined.

The more interesting extension is whether the position of a handover is predictable from the pair alone. Six positions is enough to notice that the coarse half is where they are and not enough to fit anything; the two rungs at six and twelve per cent carry pairs whose numbers are close together, and the two at thirty-five and forty carry pairs whose numbers are further apart. That could be a real relation or it could be four numbers arranged into a story, and the honest thing is to say which of those it is only after the arithmetic has been asked directly rather than the sweep.