Phyllotaxis
the spirals are counted, not admired
Stems and cones

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.
19 min read 9 figures Counted, not admiredOne parameter, one family

Two essays on this site end with a sentence about cones and neither of them computes one.

A stem is a cylinder says that the parastichy pair on a stem is constant because the rise is constant, and adds that a cone is the case in between. A disc is a cylinder derives the rise of a seed head, h(r)=c2/4πr2h(r) = c^2/4\pi r^2, and observes in passing that on a cone the same reasoning gives one power rather than two. Both sentences are correct. Neither was checked, and on a site whose whole premise is that the confident sentences about this subject are the ones nobody checked, that is not a comfortable position to leave them in.

This essay builds the cone.

A cone unrolled: 260 nodes at 137.51°, sector 126°The counter is shown these coordinates and the sector angle and nothing else. Near the apex it finds 5 and 8; near the base, 13 and 21. The rise falls as one over the distance from the apex, so the pair has to climb.5 and 8 near the apex13 and 21 near the baseflare 0.35 · step 1 · 260 nodescounted 5/8 then 13/21
Fig. 1 The cone’s surface, cut along one generator and laid flat — not a projection, because a cone can be unrolled without stretching anything. The counter is shown these coordinates and the sector angle. Near the apex it finds 5 and 8 spirals; near the base, 13 and 21.

What a cone is, to a lattice

A cone of half-angle α\alpha has one number in it, and choosing which number to call the parameter decides how much of the arithmetic is bookkeeping.

The choice used throughout here is the flare, sin⁡α\sin\alpha. A point at slant distance zz from the apex sits at zsin⁡αz\sin\alpha from the axis, so its local circumference is 2πzsin⁡α2\pi z \sin\alpha; and when the cone is unrolled the sector it becomes spans 2πsin⁡α2\pi\sin\alpha radians. So the flare is at once the fraction of a full turn the unrolled surface occupies and the constant relating slant distance to circumference. A flare approaching 1 is a flat disc; a flare approaching 0 is a needle. A stem is not in this family at all, because a cylinder has no apex.

Nodes are placed the way they are placed on a stem, with the two parameters a stem has:

θi=iδ(mod2π),zi=z0+i Δ\theta_i = i\delta \pmod{2\pi}, \qquad z_i = z_0 + i\,\Delta

The azimuth advances by the divergence angle and the slant distance by a constant step. That is a shoot that elongates at a steady rate on a surface that widens at a steady rate — which is what a young conifer cone, a pineapple or a spike of Plantago approximately is.

Unrolled, node ii lands in the plane at polar coordinates (zi, θisin⁡α)(z_i,\ \theta_i\sin\alpha). Distances between nodes are measured there, and that is not an approximation: a cone is developable, so the geometry of its surface is the geometry of that sector. The only care needed is the seam, since the sector’s two edges are the same line on the cone, so angular differences are taken modulo the sector angle and to the nearer side. A parastichy that leaves one edge comes back in at the other, exactly as on the unrolled cylinder.

The rise, which is the whole of it

The cylinder’s second parameter is the rise: the advance along the axis between one node and the next, measured in local circumferences. It is the only dimensionless thing a cylinder offers and it is what the ladder is indexed by.

A cone has one too, and it is not constant. The advance along the meridian is Δ\Delta and the local circumference is 2πzsin⁡α2\pi z\sin\alpha, so

h(z)=Δ2πzsin⁡αh(z) = \frac{\Delta}{2\pi z \sin\alpha}

Three lines and the essay is decided. The rise falls as one over the distance from the apex. Everything the site knows about what happens to a lattice when its rise falls — the ladder, the transitions, the fork tree — now applies to a cone, indexed by position along its axis.

There is a small surprise sitting in that formula, and it is worth taking now rather than later. Substitute zi=z0+iΔz_i = z_0 + i\Delta and let z0z_0 be small compared with the length:

hi≈12πisin⁡αh_i \approx \frac{1}{2\pi i \sin\alpha}

The step has cancelled. The rise at the ii-th node of an elongating cone does not depend on how far apart the internodes are; it depends only on the node’s number and on the flare. Stretch a cone lengthwise and every rise is unchanged, because stretching moves the node outward and widens the circumference underneath it in the same proportion. So the transitions on a cone sit at node numbers, not at lengths — which is a considerably more useful statement for anyone holding one, since counting scales is easy and measuring a cone’s slant distance from a vanished apex is not.

The same cone, not unrolled — 200 nodes, flare 0.35101 of the 200 nodes face the reader and 99 are behind, drawn open. In the outer half the counter finds 8 and 13; the unrolling changes the visibility and nothing else.8 and 13 in the outer halfapexflare 0.35 · 200 nodes at 137.51°counted 8 and 13
Fig. 2 The same cone drawn as an object rather than as a sector, with the nodes on the far side left open. Half the pattern faces away at any moment, which is why parastichies on a real cone are counted by rolling it along a table.

What the approximation costs

The sentence “a cone is a cylinder whose rise falls” is an approximation, and the honest thing is to measure it rather than to wave at it.

The rise formulation treats a band of the cone at slant zz as a cylindrical patch of circumference 2πzsin⁡α2\pi z\sin\alpha, so that two nodes separated by Δθ\Delta\theta around and Δz\Delta z along are

(C Δθ/2π)2+(Δz)2\sqrt{(C\,\Delta\theta/2\pi)^2 + (\Delta z)^2}

apart. That is exact for a cylinder. On a cone the exact answer is the unrolled straight line, za2+zb2−2zazbcos⁡Δϕ\sqrt{z_a^2 + z_b^2 - 2z_az_b\cos\Delta\phi}, and the two disagree because the local patch has parallel sides and the cone’s does not.

Over every parastichy hop of every offset in a four-hundred-node cone at a flare of 0.35, the worst disagreement between them is half a per cent, and it occurs at the shortest offsets, where the two nodes are furthest apart relative to their distance from the apex.

What the approximation costsThe exact distance across a cone's surface against the distance a local cylindrical patch gives, for every parastichy hop of each offset. The worst disagreement anywhere is 0.50 per cent, at offset 3.offset 30.500%offset 50.354%offset 80.167%offset 130.069%offset 210.027%offset 340.010%worst relative difference over the whole coneflare 0.35 · 400 nodesworst 0.50%
Fig. 3 The exact surface distance against the local cylindrical one, for each parastichy offset in a four-hundred-node cone. The largest disagreement anywhere is half a per cent.

Half a per cent is far below anything that matters here, and it matters that the reason is structural rather than lucky. What the ladder needs is not distances but an ordering — which offset gives the shortest hop — and an ordering survives a good deal of distortion before it changes. That robustness is the same one that makes parastichy counts usable on real material at all, and it is why the counts are hard to pin down near a transition, where the ordering is nearly a tie.

The counter, shown a cone and nothing else

Every count on this site is taken by machinery that has not been told the divergence angle, and the cone gets the same treatment. coneCount() receives an array of unrolled coordinates and the flare of the surface they live on. It does not receive the divergence, the slant step, or the fact that Fibonacci numbers exist.

Its method is the one the cylinder counter uses, with the cone’s metric substituted: for each candidate index offset mm, take the median hop length from node ii to node i+mi+m over the band, and keep the two offsets whose median is shortest. No local-minimum requirement, for the reason the cylinder counter records at length — on a stem or a cone, offset 1 climbs a whole internode and disqualifies itself, so the failure mode the disc counter has to guard against cannot arise here, and imposing a local minimum actively breaks real counts.

Point it at a band near the apex of a cone at a flare of 0.35 and it returns 5 and 8. Point it at a band near the base of the same cone and it returns 13 and 21. This never happens on a stem. Counting up the stem is the essay whose whole result is that three disjoint bands up a cylinder give one answer, and the control that made that a measurement rather than an insensitivity was running the same counter on a disc, where it gives three. A cone is the third case, and it behaves like the disc.

The same counter, on a stem and on a discThe 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.stem, lower third2 and 3stem, middle third2 and 3stem, upper third2 and 3disc, r = 0.18–0.3221 and 34disc, r = 0.45–0.6234 and 55disc, r = 0.82–0.9955 and 89stem at rise 0.050, disc of 1600 points, both at 137.51°counted from coordinates onlyone answer against three
Fig. 4 The contrast the cylinder essays were built on: one answer in every band of a stem, three answers in three bands of a disc, from the same machinery. A cone sits on the disc’s side of this, for the same reason and by a different power.

The test, band by band

The interesting question is not whether a cone’s counts change — the rise falls, so of course they do — but whether they change where the cylinder says they should.

That is a genuine prediction, because the two calculations share nothing but the divergence angle. The ladder comes from comparing ⟨jδ⟩2+(jh)2\sqrt{\langle j\delta\rangle^2 + (jh)^2} across jj on a periodic strip; the cone’s counts come from median distances between unrolled points on a widening sector. There is no wrap in the second and no apex in the first.

A nine-hundred-node cone is cut into geometric bands and each is handed to the blind counter. Beside each band’s answer sits the pair the cylinder’s ladder gives at h(z)h(z) for that band’s mid-slant, with the flare and the step being the cone’s own description and nothing fitted.

Every usable band agrees. The answers climb 5/8 → 8/13 → 13/21 → 21/34 as the bands move outward, and each is the pair the ladder puts at that rise.

five bands up a cone, counted blindEach band's pair is what the counter returns from the coordinates alone; each prediction is the cylinder's dominant pair at that band's rise. 5 of 5 agree, and the answer climbs 5/8 → 8/13 → 13/21 → 21/34.z ≈ 615/8predicted 5/8z ≈ 1128/13predicted 8/13z ≈ 20513/21predicted 13/21z ≈ 37813/21predicted 13/21z ≈ 69621/34predicted 21/34countedfrom the ladderflare 0.35 · 900 nodes · 10 bands5 of 5 bands agree
Fig. 5 Five bands up a cone. The bar is what the counter returns from the coordinates; the note beside it is what the ladder predicts at that band’s rise. Nothing is fitted between the two calculations.

The bands are geometric rather than equal, and that is forced rather than chosen. Equal bands near the base all sit on one rung and crowd every transition into the first band, because the structure is geometric in the slant distance. Geometric bands sample it evenly.

What cannot be measured, and why

Five bands survive out of ten, and the five that do not are the five nearest the apex. That is not a defect in the sampling; it is a real property of cones and it is worth stating plainly.

Transitions on a cone sit at node numbers ik=1/(2πhksin⁡α)i_k = 1/(2\pi h_k \sin\alpha). At a flare of 0.35 the ladder’s rises put them at nodes 3.6, 9.5, 25.0, 65.3, 171.8 and 449.5. The first three transitions of this cone happen among its first twenty-five nodes, where there are not enough nodes in any band to count parastichies in. A cone’s early history is not countable, on the cone or on the model of it, and no amount of care with the bands changes that.

Real cones have the same problem in a worse form. The tip of a conifer cone is a cluster of small undifferentiated scales, and the first several are usually bracts rather than fertile scales at all. So the region where the model says the interesting changes happen is the region a real specimen also refuses to show.

What the cone does offer, and the disc does not, is the other end. Because the transitions sit at node numbers a factor of φ2≈2.618\varphi^2 \approx 2.618 apart, a cone with a few hundred scales has passed only a handful of them — and most of its length sits comfortably inside one rung. That is why a pineapple can be described as “8 and 13” without immediately being wrong, in a way that a sunflower cannot.

Every transition as the rise fallsThe 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 0.3820.0.2500.5000.75011.25-2.50-2-1.50-1-0.500log₁₀ of the rise between nodes (falling to the right is the plant growing)log₁₀ of the larger parastichy number2/33/55/88/1313/21500 rises, shortest vectors recomputed at eachratio 0.3820 against 1/φ² = 0.3820
Fig. 6 The ladder all three geometries move along. A stem sits still on it; a disc travels it at a factor of φ in radius per rung; a cone travels it at a factor of φ² along its axis. The rungs are the same rungs.

Where the flare goes

The flare enters in exactly one place — as a multiplier on the node number at which each transition falls — and that gives it a clean interpretation.

A narrow cone has its transitions late. At a flare of 0.12 they sit at nodes 10.6, 27.8, 72.8, 190.5, 501 and 1311; at a flare of 0.35 they are at 3.6, 9.5, 25.0, 65.3, 171.8 and 449.5. The narrow cone spends nearly three times as many nodes on each rung, because a narrow cone’s circumference grows slowly and its rise therefore falls slowly in node number.

This has a consequence for what a specimen looks like. A slender spike shows a low pair over most of its length; a broad cone climbs faster and shows higher counts sooner. Both are the same lattice family at the same divergence, sampled at different rates.

It also means the flare is recoverable from a count and a node number, which is the sort of thing this site likes: if a cone’s counts change between scale 60 and scale 80, then hh passed through a transition rise there, and sin⁡α≈1/(2π⋅70⋅hk)\sin\alpha \approx 1/(2\pi \cdot 70 \cdot h_k). That is a measurement of the cone’s shape made by counting rather than by measuring, and it does not require finding the apex.

The plane of stems: divergence across, rise upEach 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°.-2.50-2-1.50-1-0.500100120140160180divergence angle (°)log₁₀ of the rise between nodes1,2,32,3,548 × 150 lattices, each solved548 runs drawn
Fig. 7 The plane a cone travels through. A stem is a point in it; a cone is a vertical path, moving downward as the distance from the apex grows, and the regions it crosses are the rungs its counter reports.

What the model is not being asked

It is worth being explicit about what the divergence angle is doing here, because a reader arriving from the popular literature will expect it to be the answer to something.

It is an input. The cone is built at 137.508° because that is the angle the site’s other lattices are built at, and every result above would hold at 99.5° or at 151.1° with a different ladder underneath it. Nothing on this page bears on why plants are at one angle rather than another; that question belongs to the dynamical model and to the branch tree, both of which produce the angle rather than assume it.

What the cone contributes to that argument is negative and useful. If the parastichy pair on an organ is set by the rise, and the rise is set by the organ’s shape and the rate it adds material, then the counts on a plant are not a direct readout of the divergence angle at all. Two plants at the same divergence can be counted differently because one is a spike and the other a head; two plants at different divergences can be counted the same because their rises put them on the same rung. Every inference from “this plant has 8 and 13” to “this plant is at the golden angle” runs through the ladder, and the ladder needs the shape.

That is the sense in which this essay is about counting rather than about cones. The instrument is the same one the site has used since the foundation phase; what changes is the surface it is pointed at, and what is learned is how much the surface contributes to the reading.

What a real cone does that this one does not

The model here is a right circular cone with a constant slant step, and real organs depart from it in three ways that are worth naming.

A cone is not a cone. A conifer cone is closer to an ogive — pointed at the tip, nearly cylindrical through the middle, tapering again at the base. The local circumference therefore does not grow linearly with slant distance, and where it stops growing the rise stops falling and the counts stop climbing. That is not a failure of the argument; it is the argument applied to a different C(z)C(z), and the essay after next does exactly that for a general surface of revolution.

Scales are not points. They have size and they compete for space, so a real cone is closer to a packing than to a lattice of centres. The effect is small where the pattern is regular and is not small at the tip and the base, where it is not.

The step is not constant. Internodes lengthen and shorten through a cone’s development, so the slant advance is a function of time as much as of position. That objection is the largest of the three and it is the subject of the rising-phyllotaxis essays, which put a rate into the model and ask whether the pattern keeps up with it.

None of the three moves the counts, and the reason is the one already given: what is being predicted is an ordering rather than a distance.

The spiral counts, band by band, in one headThe same flower gives 13/21, 21/34, 34/55, 55/89 at different radii. A caption saying "34 and 55 spirals" is a statement about one annulus.0255075406080fraction of the head's radiusparastichy numbers found in a band there1000 primordia at the golden angle4 different pairs
Fig. 8 The disc’s version of the same behaviour, from the foundation phase. A cone and a head both change their counts along their axis; what separates them is one power in the rise law, which is the next essay.
The exponent sets the spacing, and only the exponentThe 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.1231234rise exponent p, where the rise falls as z⁻ᵖratio between consecutive transitions along the axis5 surfaces, 16 measured transitionsworst disagreement 0.8%
Fig. 9 Where a cone sits in a wider family. Its rise exponent is one; a filling disc’s is two; and the spacing of the transitions along any of their axes follows from that one number.

What this adds

Three things, in increasing order of how much they change.

The smallest is that the sentence two essays ended with is now computed rather than asserted, and it came out right. That is the least interesting possible outcome and it is still worth having on a site whose subject is claims that were never tested.

The middle one is that a cone is a genuine third case rather than a rhetorical one. A stem does not move along the ladder, a disc moves along it at one rate, a cone at another — and the same instrument, pointed at all three, gives one answer, three answers and four answers respectively. Any account of phyllotaxis that treats “the parastichy pair” as a property of a plant rather than of a place on a plant has to explain all three.

The largest is the one the next essay is about. The rate at which a geometry moves along the ladder is a number, it differs between the disc and the cone by exactly one factor, and that factor is φ\varphi.

Where this goes next

2 and 3rise 0.090 · divergence 137.51°counted 2 and 3, opposed 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. 501001501020304050radius in the disclarger parastichy number, measured on the disc21/3434/5555/8989/144prediction from the cylinder, counts from the disc15 of 16 bands agree 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. 0.2500.5000.75011.25-2.50-2-1.50-1-0.500log₁₀ of the rise between nodes (falling to the right is the plant growing)log₁₀ of the larger parastichy number2/33/55/88/1313/21500 rises, shortest vectors recomputed at eachratio 0.3820 against 1/φ² = 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.