One turn per survivor

A stem that never repairs after an organ is removed keeps exactly one family of the lattice it was cut from. Its hop — the angle from an organ to the one p places above it — is where it always was, to hundredths of a degree, while the divergence itself swings through eighty. That is the finding the previous rung of this ladder makes, and it is a description of the wrecked stem rather than a prediction about it.

This essay is the prediction. If a family of period p is still standing, then the angle the wrecked stem settles at cannot be arbitrary, because p steps of the new divergence have to land where p steps of the old one landed. Write the old divergence d and the new one d′, and the constraint is that p times their difference must be a whole number of turns. Which is to say the change in the divergence is 360° divided by p, or twice that, or none. There are no other options. A wrecked stem’s angle is picked from a list with one small whole number in it.

Why the constraint is forced rather than fitted #

It is worth being slow about the step from one family survived to the angle is one of a list, because it is the whole content of this essay and it takes ten seconds to derive and rather longer to believe.

The p-hop of a lattice is p times its divergence, reduced to a turn. That is not a modelling assumption; it is what a lattice is. Organ zero sits at some azimuth, organ one sits a divergence further round, organ p sits p divergences further round, and the only thing that makes the p-family a family is that this is the same for every starting organ.

So if the p-hop after the disturbance equals the p-hop before it, then p times the new divergence and p times the old one differ by a whole number of turns, since two angles that are equal on a circle differ by whole turns and by nothing else. Divide by p and the difference in the divergences is a whole number of turns divided by p. There is no room in that for a small correction, a fitted parameter or a residual — either the family is preserved or it is not, and if it is, the divergence has moved by 360° ÷ p times a whole number.

That makes the measurement in this essay a genuine test rather than a consistency check, because the two halves are measured differently. The rigidity of the hop is measured on the positions, organ by organ, as a spread. The slip is measured on the divergences, as a mean over a stretch. Neither computation sees the other’s answer. They agree at nineteen offsets on six lattices.

What the numbers come out as #

The prediction is arithmetic, so it is worth reading the measurements one at a time rather than as a scatter.

At the 5/8 lattice the surviving family has period five and the divergence moves by 71.95° and 72.00° at the two wrecked offsets, against 360 ÷ 5 = 72.00° exactly. At the 8/13 lattice, where the survivor is the eight-family, the moves are 44.97° and 45.03° against 360 ÷ 8 = 45°. On the Lucas branch, with a surviving seven-family, they are 51.38° against 360 ÷ 7 = 51.43°.

Every row closes to within 2.97° of a whole number of turns, and every row but one closes to within 0.71°. The loose one is the Lucas branch at a rise of 0.020, where the surviving family has period four, so a residual of 2.97° in the closure is a residual of 0.74° per organ — and that stem’s settled divergence sits 0.74° from the nearest point of the grid its azimuths are computed on. The departure is the grid, and it is the largest anywhere in the census.

A wreck is a dislocation #

The way to read one turn per period is not as a coincidence of small numbers. It says the wrecked stem is the stem it was, with one extra turn threaded through every p organs.

Think of the surviving family as a helix running up the stem, unchanged. In the undisturbed pattern, the organs between one member of that family and the next are laid down in a fixed arrangement. In the wrecked pattern they are laid down in an arrangement that has been rotated by one full turn relative to the family’s own step — which is invisible in the family and impossible to hide anywhere else.

That is a dislocation with a stated size: a defect specified by two whole numbers, the period of the family it leaves alone and the number of turns it inserts. It is not damage in any sense that would need a description with parameters in it.

The reason it is worth insisting on is that “the pattern is disturbed” and “the pattern has acquired one extra turn per five organs” are different claims about what a meristem is doing, and only one of them can be checked on a plant that has been photographed once.

Two rows that are not one turn #

The claim as measured is seventeen of nineteen, and the two that are not are worth more attention than the seventeen that are.

One is zero turns. At the 5/8 lattice at a rise of 0.008, with the organ six places back removed, the surviving family is the eight-family and the mean divergence moves by 0.02° — nothing at all. The stem has not gone anywhere. It is nevertheless wrecked: its divergences run an exactly repeating orbit of period eight, swinging across sixty degrees, about a mean that is where it always was.

That case is a warning about instruments rather than about plants. A survey that recorded a stem’s average divergence and compared it with an undisturbed control would call this one repaired. Everything that distinguishes it lives in the sequence.

The other is two turns. On the Lucas branch at a rise of 0.013, cut five places back, the divergence moves by 102.78° where one turn per seven organs would be 51.43°. Two turns per seven is 102.86°, and that is what it is.

So the second whole number is genuinely free, and the observed distribution of it is seventeen ones, one zero and one two. That is not a claim about which values are possible; it is a count over a small census, and a census of a different range of rises would be entitled to a different tally.

The destinations follow #

Putting the two rungs together gives something that was not available from either alone: a list of the places a wrecked stem can settle. If the outcome is determined by p and j, then the divergence must be d + 360° · j ÷ p, and the p values available are the handful of lags the rule can hold rigid.

That predicts a small, closed set of destinations — and a set that does not grow when the damage does, because damaging a stem harder does not create new lags.

The main destination at the 5/8 lattice is 208.78°, which is 136.78° + 72.00°: the five-family held, one turn inserted. The next is 280.43°, which is 136.78° + 143.65°, or two turns per five, the same p with the other j. A third at 136.99° is one turn per — nothing: a stem whose mean did not move, the zero-turn case again. The list is short because the arithmetic is short.

What it costs to see, on a plant #

The list of destinations is the part of this that an experiment could use, and it is worth pricing.

To place a wrecked stem on the list, a measurement needs two things: the divergence the plant carried before the intervention, and the divergence it carries afterwards, to better than half of 360° ÷ p. At a five-family that is 36°, which is a coarse measurement by any standard — this collection has priced divergence measurements at tenths of a degree elsewhere and found them expensive. At a thirteen-family it is 14°, still coarse.

The catch is the other half. The list is d + 360° · j ÷ p, and reading a measured stem against it requires knowing p — which means knowing which family survived, which means the measurement of the previous rung. So the two halves are not independent instruments for the same fact; they are two measurements that have to be made on the same plant and then checked against each other.

That is the shape a good prediction has, and it is also the reason this is not yet an experimental protocol. Both halves are cheap on a computed stem and only one of them is obviously cheap on a real one.

The one that is a mirror #

There is a case in this collection that looks like an exception to all of the above and is not. At the coarsest lattice, 3/5, a cut of two organs can leave a stem settled at 220.31° against a grown 139.69° — which is 360° minus the original, to the last digit. The counted pair is unchanged and the order of the shortest hops is unchanged. It is the same lattice wound the other way.

A reversal is not a slip. The change in the divergence is 80.62°, and the five-family’s hop is not preserved — it is reflected, which is a different operation and one that carries every family with it at once. So the mirror sits outside the description here, and the honest statement is that this essay covers the stems that stay on their own handedness and says nothing about the ones that do not.

The blind reading agrees, and says something extra #

The whole argument so far is in the divergences. There is an independent reading in the positions, and it is worth taking because it is the one a photograph supports.

Hand a counter the wrecked stem’s positions and nothing else. At the 5/8 lattice cut four places back it returns 5 and 12; the same lattice cut five places back gives 5 and 13. Both keep the five. Their conjugate numbers are 12 and 13 rather than the 8 the undisturbed stem showed — and that is what a slip of one turn per five organs has to do, because inserting a turn between consecutive members of the five-family changes the pitch of every other family in the pattern while leaving that one alone.

So the counted pair is a second route to the same statement, and it carries a detail the divergences do not: which of the two families was replaced. A wrecked stem is half a lattice preserved exactly and half a lattice rebuilt at a different pitch, and the count tells apart the two halves without any knowledge of what was done.

The prediction that has not been made yet #

One consequence of the arithmetic has no measurement behind it and is worth writing down as a claim rather than leaving implicit.

If the destination is d + 360° · j ÷ p, then a lattice whose surviving family has a large period has a fine grid of available destinations, and a lattice whose survivor is small has a coarse one. At a four-family the options are 90° apart; at a thirteen-family they would be 27.7° apart. So the finer the lattice a stem carries, the more nearly a wreck can leave it where it was — and the harder it becomes to tell a wreck from a recovery by measuring the mean divergence.

That is a statement about the instrument as much as about the stems, and it points the same way as the zero-turn case above: the coarser the pattern, the more a wreck announces itself in the mean, and the finer the pattern, the more it hides in the sequence. Nothing in this census reaches a survivor of period thirteen, so the claim stands unmeasured.

What this does not say #

It does not say every whole number of turns is available. The census contains zero, one and two, and one is nearly all of it. Whether three ever happens, and whether the value is decided by the offset or by the lattice, are not answered by nineteen rows.

It does not say the slip is the whole of the wreck. The wrecked stem is an orbit, not a lattice: its individual divergences swing across tens of degrees and only their mean sits at d + 360° ÷ p. What sets the shape of the motif about that mean is not touched here.

It does not say the surviving family is chosen by anything the arithmetic knows. The constraint is conditional — given a survivor of period p, the slip is fixed — and which p the stem ends up with is still unexplained.

And it does not say a plant has a preferred defect. The claim is about what a placement rule does to a cylindrical lattice when an organ is taken out of it, and it becomes a claim about a meristem only if that rule is what a meristem does.

The check that would refuse it #

Two assertions carry this, and they are deliberately separate.

The first is the closure: for every wrecked stem in the census, the period of the surviving family multiplied by the change in the divergence must be within three degrees of a whole number of turns. Three degrees is set by the loosest row, which is a lattice whose settled angle sits three quarters of a degree off the grid it was computed on, and every other row clears the bound by a factor of four. A stem that had drifted to some angle unrelated to its survivor’s period would fail this immediately, and that is exactly what a rule with no conservation in it would produce.

The second is that the whole number is one at all but a few offsets. It is stated as at least sixteen of nineteen rather than all, because two rows are not one and pretending otherwise would be a claim the measurement does not support — and because an assertion written to be satisfied by any tally is not an assertion.

There is a third guard, and it is arithmetic rather than physics. The mean divergence of a wrecked stem is taken over a whole number of the orbit’s own periods, not over a fixed window. A mean over a window that ends part-way through a repeating motif weights some of its angles twice, and at a block of seven over a hundred and twenty organs that bias is 3.5° — larger than the quantity being compared against zero. The first version of this measurement used a fixed window and reported two of the Lucas rows as failing to close, which is a result about arithmetic that would have read as a result about stems.