The wedge belongs to one length

In a tube the standing members lose turned a ranking round. On a flat base, members standing across a clearance hold eight times what layers lying along it hold. Inside a tube the members are radial fins, they converge as they rise, and the room they need is the room at their tips. Their best height is half the radius, and there they fill exactly half the cross-section and hold exactly half of the ceiling $2\pi R^2/\tau$ that no lining of sheet of thickness $\tau$ can pass. Concentric layers fill all of it.

That essay then tried fins of several lengths, shorter ones filling the gaps between the long ones, and found each extra length worth less than the one before: four lengths at successive halvings held 417 out of a ceiling of 628, and the best of five schedules tried, 443. It called the schedule “a small optimisation problem of its own” that it did not solve, and it closed on a sharper question. Would a fin that tapers close the gap — so that the factor of two turns out to be an artefact of constant thickness — or is the wedge a property of radial arrangement, and the ranking inverted for good?

Neither, as it turns out. Tapering cannot close the gap, and radial arrangement does not cause it. The schedule problem has a closed answer, and the answer puts the factor of two on fins of one length.

The ceiling itself comes from an argument about sheet rather than about arrangement: a sheet of thickness $\tau$ has two faces, so any piece of it filling a volume carries $2/\tau$ of surface per unit of that volume, whatever shape it is bent into.

Why a taper cannot help #

The taper idea is right about where the waste is and wrong about how to stop it.

A fin’s surface is its two faces, so a fin of height $h$ carries $2h$ of surface per unit length of tube whatever its thickness. Thickness buys nothing; count buys everything. And the count is set where the fins are closest together, at their tips: $n$ fins whose tips sit at radius $R - h$ need $n\tau_{\text{tip}} \le 2\pi(R - h)$.

A tapered fin is thinner at its tip than at its root. But the fins here are folded sheet, and a fin cannot be thinner anywhere than the sheet it is made of: the constant-thickness fin already has the smallest tip thickness there is. A fin that tapers the other way, thick at the root and thin at the tip, is a fin thickened where there is room — which adds volume and no faces. It fills more of the wedge and holds exactly what the constant fin held.

So thickness is the wrong currency, and the room near the wall has to be spent on more fins, not fatter ones. That is the intuition behind the taper, and it is the one that leads to the answer: more fins at larger radius means fins of more than one length, the long ones reaching toward the axis and shorter ones standing between them near the wall, where the circumference has room for them.

That is the single-length picture, and it is exactly what a staircase with one step looks like.

A staircase under a line #

Count the fins at each radius and the problem becomes a picture.

At radius $r$ the circumference is $2\pi r$, so at most $2\pi r/\tau$ fins can pass that radius side by side. That limit is a straight line through the origin. The fins actually present at radius $r$ are the ones long enough to reach it; with fins of $m$ lengths whose tips sit at radii $x_1 < x_2 < \dots < x_m$ (as fractions of $R$), the count is a staircase — constant between one ring of tips and the next, jumping up as each new set of shorter fins begins. The staircase has to stay under the line everywhere, and it touches it at each ring of tips, where that set of fins is as crowded as it can be.

The surface is twice the count integrated along the radius, since each fin contributes two faces over the length it spans. The ceiling is twice the area under the line. So the share of the ceiling a set of fins reaches is the area under the staircase divided by the area under the line, and the empty wedges of the cross-section are the triangles between the two.

With one length, the fins’ tips at $x$, the staircase is a single step of height $x$ from $x$ to one, with area $x(1 - x)$ against the triangle’s half. That is largest at $x = \tfrac12$, where it is a quarter against a half — the fins at half the radius, holding half the ceiling, which is the earlier essay’s result read as a picture. The wedge is the triangle under the line to the left of the step, and the triangle above the step to its right.

Equal steps, and what they hold #

With $m$ lengths the area under the staircase is

$$A = \sum_{k=1}^{m} x_k\,(x_{k+1} - x_k), \qquad x_{m+1} = 1,$$

and the share of the ceiling is $2A$. Setting each derivative to zero gives $x_{k+1} - x_k = x_k - x_{k-1}$: the steps are equally spaced, $x_k = k/(m+1)$. The area is then $\sum_k \tfrac{k}{(m+1)^2} = \tfrac{m}{2(m+1)}$, and the share is

$$\frac{m}{m+1}.$$

One length holds a half; two, two thirds; three, three quarters; four, four fifths. The wedges together shrink as $1/(m+1)$: the second length halves what the first left empty to a third, the third cuts it to a quarter, and the staircase with more, equal steps hugs the line ever more closely.

The formula is checked two independent ways before the figure is drawn. The fins are counted one length at a time, exactly as the earlier essay counted its schedules — each length adding as many fins as the room at its tips allows beyond those already there — and the surface agrees with the staircase area at every $m$ to eight. And four thousand random schedules of two, three and four lengths are tried; none holds more than equal spacing does.

The same staircase on a flat base #

The picture also says why the question never arose on a flat base, and that is the connection that makes the tube result more than a correction.

On a plain the room at every height of a comb is the same, because the walls are parallel: the limit on how many walls can stand at a given height is a constant, not a line through the origin. The staircase then sits under a horizontal line, and the best staircase under a horizontal line is no staircase at all — one step, the full height, touching the limit everywhere. Every wall the same length, standing from floor to ceiling, fills the comb completely, which is why standing up beats lying down by eight never had a schedule to choose and why the angle the eight does not know could lean the walls without changing the answer.

The tube tilts the line. Curvature makes the limit on the count grow with the distance from the axis, and a staircase under a sloping line always leaves triangles; the number of lengths a lining needs is set by how much its limit slopes. A gently curved wall — a wide tube lined with short fins — sees a nearly horizontal stretch of the line and loses little with one length; a lining that reaches toward the axis sees the whole slope and needs many.

That is close kin to what two surfaces in one box found when it divided a depth between surfaces: the reward for a share of a budget is a product of a rising and a falling factor, and a product like that is maximised by sharing the budget out evenly. Here the budget is the radius, the shares are the gaps between rings of tips, and even sharing is equal spacing.

Why halving stalled #

The earlier essay’s schedules now read differently. Its first, one length at half the radius, is the $m = 1$ optimum. Its second added a length at a quarter of the radius, putting the tips at a half and three quarters of the way out; the optimum for two lengths puts them at a third and two thirds. Its measured 393 is 62.5 per cent, against 66.7 for two equally spaced lengths.

Successive halving puts the rings of tips at $1 - 2^{-k}$, crowding them toward the wall, where each new ring adds fins over a shorter and shorter stretch of radius. The staircase’s first step is right and every later one is too close to the one before. Summing the areas gives a share that converges to

$$2\sum_{k\ge1}\bigl(1 - 2^{-k}\bigr)2^{-(k+1)} = 1 - \tfrac13 = \tfrac23,$$

so four halving lengths held 66.4 per cent and forty would hold 66.7. The schedule that looked like the natural refinement is the one that never gets past two thirds. The schedule of 0.6, 0.3 and 0.15 of the radius, the best the earlier essay tried, held 70.5 per cent — above halving, because its first step is further in — and three equally spaced lengths hold 75.

The earlier essay’s remark that “more lengths is not better than better lengths” was exactly right and is now quantified. Four halving lengths hold 66.4 per cent; two equally spaced lengths hold 66.7. Two good lengths beat four bad ones, and the difference between the schedules is entirely where the rings of tips are put.

What the factor of two was #

The earlier essay stated its headline carefully: if tapered fins closed the gap, the honest headline would become that constant-thickness fins lose, and if they did not, the wedge would be a property of radial arrangement and the ranking would invert for good. The answer is a third case it did not list. Fins of one length lose by exactly two, fins of any fixed number of lengths lose by less, and radial fins as a family lose nothing in the limit. With equally spaced tips the fins’ count at each radius approaches the circumference’s limit everywhere, the staircase approaches the line, and the lining approaches the ceiling, as concentric layers reach it.

That puts the tube back beside the flat base in an unexpected way. On a flat base the standing members won because their two budgets — footing and clearance — were charged in different directions. In a tube the fins lost because curvature made them compete twice, at the wall and at the tip. Graded lengths separate the two contests again: each set of fins competes for the room at its own ring of tips and nowhere else, since shorter fins begin where the longer ones have left room.

It also gives the half an exact meaning. The earlier essay found that fins at their best height fill half the cross-section and hold half the ceiling, one statement read twice. The staircase is that statement for any number of lengths: the share of the cross-section the fins fill is the share of the ceiling they hold, because surface is $2/\tau$ times the volume of sheet whatever its arrangement, and the empty wedges are exactly the missing share.

What real linings do #

A body lining a tube with folds would have to choose between these arrangements, and the anatomy of the gut is the standard example of a lining built for surface. Its folding is famously hierarchical — large folds carrying finger-like projections carrying a brush of much smaller ones — which is a hierarchy of scale, a nest of structure on structure, and a nest pays four a level priced exactly that arrangement on a flat base.

The staircase says a different hierarchy is cheaper inside a tube: members at one scale but of several lengths, graded so that their tips sit on equally spaced rings. Nothing here claims that any organ is built that way, or that surface is what any organ maximises; nothing grown has a seam is the reminder that a lining is grown rather than placed, and growth has its own economy. What the calculation offers is a way to ask which economy a lining is in: a cross-section whose members come in a few lengths, with their tips crowded toward the wall, is spending its room the way halving does, and one whose tips are spread evenly is spending it the way the optimum does.

What the staircase cannot show #

It treats the count of fins as a real number. A real lining has whole fins, and at each ring of tips the count is rounded down, which costs a fraction of a fin per ring; with a sheet a hundredth of the radius thick, as drawn, that is a small correction, and with a thick sheet and few fins it is not.

It says nothing about supply. The surface has to be supplied charges a channel against the same room the members use, and the channel grows with what it feeds found the charge rising with the surface served, and the earlier essay hoped that members thinning as they rise would leave channels somewhere to run. Graded lengths do the opposite: equally spaced tips fill the region near the wall as tightly as the region near the axis, which is exactly what leaves no room. A supplied lining has two uses for the wedge, and the staircase has priced only one.

And it keeps every fin radial and flat. A fin that branches — one sheet splitting into two as it approaches the wall — is the continuous version of adding a length, and a folded sheet can make one only at a vertex; whether a branching fin can be folded from one sheet without cuts is a question about crease patterns this essay does not answer.

The lining the calculation assumes #

The tube is a circle of radius $R$ and the sheet has thickness $\tau$; the ceiling $2\pi R^2/\tau$ is the surface a lining of that sheet would carry if it filled the cross-section solid. Every fin is a flat radial slab of thickness $\tau$ standing on the wall, and every fin of one length ends at the same radius.

A length’s fins are placed as densely as their tips allow, less the fins of longer lengths already passing that ring, which is how the earlier essay counted its schedules and how the staircase touches the line at each step.

And surface is the only quantity maximised. A lining that fills its tube solid is useless for anything a tube does; the result is a bound and a ranking, not a recommendation, as the earlier essay said of the concentric layers.

How the figures were checked #

The fins counted length by length are required to agree with the staircase’s area at every number of lengths from one to eight, and equally spaced tips are required to give $m/(m + 1)$ to a part in a billion. Four thousand random schedules each of two, three and four lengths are required to hold less than equal spacing, so a schedule that beat the formula would stop the figure. And halving is required to stay below two thirds while equal spacing passes eighty-five per cent at eight lengths.

Still open: the lining that is also supplied #

The calculation now answers the surface question exactly and leaves the other economy where it was. A lining whose fins are graded for surface and whose wedges are kept for channels is a two-objective problem with a frontier: for each share of the cross-section left open for supply, the most surface the fins can hold. The staircase gives one end of the frontier and the empty tube the other, and the interesting question is whether the frontier is straight — every unit of room given to supply costing a fixed amount of surface — or bowed, with the first few channels nearly free because they sit in the wedges fins of few lengths would have wasted anyway.

The other direction is the branching fin. A fin that splits into two partway to the wall is the natural way to add a length without adding a separate member, and it is the continuous limit of the staircase. Whether it can be folded from uncut sheet, and what the vertex at the split costs, belongs to the same questions how much surface fits in a body opened about corrugations that turn over.

Sideways from here, the staircase is a shape of argument worth recognising. A resource limited by a line through the origin, spent in a few discrete blocks, is best spent in equal blocks, and the share it reaches is $m/(m + 1)$. The same arithmetic prices any quantity that can grow only in steps under a proportional limit, and the halving schedule — which looks like the natural refinement — is the one that stalls.

The habit worth carrying is about optimisation problems left as guesses. When a measurement tries a handful of schedules and reports the best, write the objective down before trying more. Five schedules were drawn and the best reached 70.5 per cent; the objective, once written, was a staircase under a line, and its maximum took one derivative to find.