A lumen costs nothing until it does

The wedge belongs to one length settled how much surface radial fins can hold inside a tube. Count the fins at each radius and the count is a staircase under a straight line — the most fins the circumference allows at that radius — and the surface is the area under the staircase. With $m$ lengths of fin whose tips sit on rings at $x_1 < \dots < x_m$ of the radius, the best staircase has equal steps, $x_k = k/(m + 1)$, and holds $m/(m + 1)$ of the ceiling $2\pi R^2/\tau$ that no lining of sheet of thickness $\tau$ can pass.

It said plainly what that left out. A tube has to carry something, and a lining that fills its cross-section with fins leaves nowhere for anything to flow. The surface has to be supplied charged 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. The lining that is also supplied is a problem with two objectives, and the essay closed on its shape: for each share of the cross-section left open for supply, the most surface the fins can hold — is the frontier straight, or bowed, with the first channels nearly free because they sit in the wedges fins of few lengths would have wasted anyway?

The frontier is bowed, and there is a free stretch at its start. But the free room is not in the wedges. It is in the middle.

Where the room is #

Start with the best lining and ask where it leaves the cross-section empty. There are two kinds of empty room, and they are not alike.

Inside the first ring of tips there are no fins at all. Every fin stands on the wall and reaches inward to its tip, and the longest reach to $x_1 = R/(m + 1)$; inside that radius the cross-section is open. That is the triangle at the left of each staircase, and for fins of one length it is a disc of radius half the tube’s — a quarter of the cross-section.

Between the rings, the room is the gaps between neighbouring fins. Just behind a ring of tips the fins are packed as tightly as the circumference allows; further out the circumference grows and the same fins spread apart, until the next ring of shorter fins starts and fills the gaps again. Those gaps are the triangles along the staircase’s diagonal.

The gaps are narrow, and the arithmetic says exactly how narrow. Between ring $k$ and the next, the fins present were packed edge to edge at radius $x_k$, so at radius $r$ each has $2\pi r/n - \tau$ of room beside it, where $n = 2\pi x_k R/\tau$. That is $\tau(r/x_k - 1)$, largest just before the next ring, where it is $\tau(x_{k+1}/x_k - 1)$. With the rings equally spaced that is $\tau/k$: one sheet thickness behind the first ring, half a thickness behind the second, a third behind the third. No gap between fins in the best lining of any number of lengths is ever wider than the sheet the fins are made of. A channel as thin as a sheet is not a channel; nothing can be carried in the wedges.

So the question the earlier essay asked about the wedges has a sharp answer. The empty room is real — it is the whole of the shortfall from the ceiling, $1/(m + 1)$ of the cross-section — but only $1/(m + 1)^2$ of it is the lumen, and the rest is slivers no wider than the sheet. The only room a lining leaves in one piece is the lumen down its middle.

Keeping a lumen clear #

Supply, then, means a lumen: a central disc of radius $\rho R$ that no fin may enter. The fins must have their tips at radius $\rho$ or further out, and the question is the most surface they can hold.

It is the same staircase with its first step pinned. If the lumen is smaller than the unpinned optimum’s own first ring, $R/(m + 1)$, the pin does not bind: the best lining already leaves that much open, and the lumen is free — the surface is still $m/(m + 1)$ of the ceiling. If the lumen is larger, the first ring sits at the lumen’s edge and the remaining $m - 1$ rings are equally spaced from there to the wall, which is what the area under a staircase with its first step fixed is largest at. The surface is then

$$S_m(\rho) = 2\rho(1 - \rho) + (1 - \rho)^2\,\frac{m - 1}{m}.$$

For fins of three lengths and a lumen four tenths of the radius across — sixteen per cent of the cross-section — that is 72 per cent of the ceiling, against 75 with no lumen at all. A random search of fifteen hundred tip schedules outside the lumen finds none better, at every number of lengths and lumen tried.

Why the steps are equal from the lumen out #

The pinned optimum follows from the same calculation the earlier essay made, with one variable taken away. The area under a staircase whose steps sit at $x_1 < \dots < x_m$ is $\sum_k x_k(x_{k+1} - x_k)$ with $x_{m+1} = 1$, and moving any one inner ring $x_k$ changes it at the rate $x_{k+1} - 2x_k + x_{k-1}$, so at the optimum every inner ring sits halfway between its neighbours. Without a lumen the first ring is free too, and the whole staircase comes out evenly spaced from the axis. With a lumen, the first ring may not move inward past $\rho$, and when that constraint binds the rest are still evenly spaced — from $\rho$ to the wall, in steps of $(1 - \rho)/m$. Adding up the areas under those steps gives the closed form above.

The constraint binds exactly when the free optimum’s first ring, $1/(m + 1)$, lies inside the lumen, which is where the flat stretch ends. And because the free optimum is a maximum, the frontier leaves the flat stretch level: moving the first ring a little past its best position costs only in proportion to the square of the move. A lumen slightly larger than the free one is very nearly free. For three lengths the free lumen is a quarter of the radius; a lumen of three tenths costs 0.3 per cent of the ceiling, and one of four tenths three per cent. The cost grows as the lumen passes further beyond what the fins would have left anyway.

The frontier, by number of lengths #

Each number of lengths has its own frontier, and each has the same shape. It is flat out to the free lumen, at a lumen share of $1/(m + 1)^2$ of the cross-section: a quarter for one length, a ninth for two, a sixteenth for three, a hundred-and-twenty-first for ten. Then it bends down, meeting zero surface when the lumen is the whole tube.

The dashed line is the limit. As the lengths multiply, $S_m(\rho)$ tends to $2\rho(1 - \rho) + (1 - \rho)^2 = 1 - \rho^2$: exactly the share of the cross-section not given to the lumen. With four hundred lengths every radius is within three thousandths of it. That is the same identity that made the ceiling in the first place — a sheet of thickness $\tau$ carries $2/\tau$ of surface per unit of the volume it fills, whatever shape it is bent into — so fins of unlimited lengths fill everything outside the lumen solid, and a unit of lumen costs a unit of surface, no more and no less.

So the frontier is straight in the limit and bowed for every finite number of lengths, and the bow is not a property of supply at all. It is the price of having few lengths: the earlier essay’s wedges, which a lumen can use up to the free radius and not beyond.

What each unit of lumen costs #

Differentiating the frontier gives the price directly. Per unit of the cross-section’s area, a lumen of radius $\rho$ past the free radius costs

$$\frac{(m + 1)\rho - 1}{m\rho}$$

of the ceiling: nothing at the free radius, rising steadily, and approaching one as the lumen fills the tube. The first channel is free and the last costs full price, and the price in between is set entirely by how many lengths the fins come in. One length’s lumen is free out to half the radius and then gets dear fast; ten lengths’ lumen is free only to an eleventh, and costs nearly full price from there on. The price reaches half the full rate at a lumen of $2R/(m + 2)$ — two thirds of the radius for one length, two fifths for three, a sixth for ten — so the lengths a lining has decide not only where its free lumen ends but how quickly the lumen beyond it becomes expensive.

That inverts the natural expectation about refinement. More lengths hold more surface at every lumen — the curves never cross — but they make the lumen dearer at the margin, because a lining of many lengths has already spent the free room on short fins near the axis. A lining that must keep a lumen of a given size gains less from extra lengths than one that need not: with a lumen of half the radius, one length holds half the ceiling, three lengths two thirds, and unlimited lengths three quarters, against a half, three quarters and all of it with no lumen.

A flat base has no middle #

The free lumen is a property of the tube, and it is worth seeing why a flat base has nothing like it. Standing up beats lying down by eight filled a clearance above a flat base with walls standing on it, every wall the full height, and on a flat base that fills the clearance completely: the limit on how many walls can stand is the same at every height, so the staircase sits under a level line and the best staircase is one step with no room left over. There is no region the walls fail to reach, and so no room a channel can take for free. The surface has to be supplied found exactly that on the flat base: the channel takes its room out of the box on the same terms as the surface, from the first unit.

The tube is different because its limit is a sloping line. In a tube the standing members lose found the fins crowding together as they converge, so that the room they need is the room at their tips, and fins that reach towards the axis cost more than they give. The best lining therefore stops short of the axis, and the space it stops short of is the free lumen. What made the tube a worse place for standing members is what makes it a better place for supply: the convergence that shortens the fins leaves the middle open, and the middle is where a tube carries what it carries.

What a real tube does #

The cross-sections make the free lumen visible. Fins of one length at their best height leave the middle half of the radius open, and that open disc is the thing a tube exists for; fins of three lengths push their longest members three quarters of the way in and leave a lumen a quarter of the radius. A lining built for surface alone chooses its lumen by accident of its number of lengths, and a lining that needs a particular lumen should choose the number of lengths whose free radius matches it — one length for a lumen of half the radius, two for a third, three for a quarter — since that is the lining at which the lumen costs nothing and the fins hold the most they can.

The gut is the standard example of a tube lined for surface, and nothing here claims that its folds are fins of a few lengths or that surface is what it maximises; nothing grown has a seam is the reminder that a lining is grown rather than placed. What the frontier offers is a way to read a cross-section. A lining whose longest members stop at about $R/(m + 1)$, for the number of lengths it has, is spending its lumen for free; one whose members stop well short of that is paying surface for a lumen it could have had cheaper with fewer lengths.

What the staircase assumes #

Fins are flat radial slabs of the sheet’s thickness, standing on the wall, as in the earlier essay, and every fin of one length ends at the same radius. The count at each radius is treated as a real number; with a sheet a hundredth of the radius thick a ring holds hundreds of fins and the rounding is small, and with a thick sheet it is not.

The lumen is a disc on the axis. A tube carrying something might carry it in several channels, or off-centre, and the lining would then have to leave those regions clear instead; a lumen that is not a disc at the centre would pin different fins’ tips at different radii, and the staircase would no longer describe it. The central disc is the case the argument can settle, and the one a tube’s own symmetry suggests.

And supply is only room. The channel grows with what it feeds found that a channel must widen with the surface it serves, so the lumen a lining needs is itself a function of how much surface it holds. The frontier here is the menu; the demand curve that picks a point on it is that essay’s, and putting the two together is a small calculation this essay has not made.

How the frontier was checked #

For each number of lengths and lumen radius, the pinned staircase’s area is required to equal the closed form, and fifteen hundred random schedules of tips outside the lumen are required to hold less, at four numbers of lengths and four lumens each. At four hundred lengths the frontier is required to lie within three thousandths of $1 - \rho^2$ everywhere and never above it. On the unpinned optimum the widest gap between neighbouring fins is required to be exactly one sheet thickness for every number of lengths from one to eight.

Still open: where supply and surface meet #

Setting the channel’s demand against this frontier would close the problem. A lining serving $S$ of surface needs a lumen whose area grows with $S$; the frontier gives the most $S$ for each lumen; the lining that satisfies both is where the demand curve crosses the frontier. Whether that crossing lies on a flat stretch — so that the right number of lengths makes supply free — or on the bend, where each unit of lumen costs surface, depends on how steeply demand rises, and the earlier essay’s measured charge is the number to put in.

A branching fin is the other open direction. A fin that splits into two partway to the wall is the continuous limit of adding lengths, and the staircase says what it would buy: every extra length closes a share of the shortfall and shrinks the free lumen. Whether a folded sheet can branch without a cut, and what the vertex at the split costs, belongs with the corrugations that turn over in how much surface fits in a body.

Sideways from here, a nest pays four a level priced a hierarchy of scale on a flat base; a tube’s hierarchy of lengths is a different one, and the free lumen is a property only the tube has, since a flat comb has no middle to leave open.

The habit worth carrying is about waste that looks usable. Before counting empty room as capacity, measure its narrowest dimension. The slivers between fins of one length are a quarter of the cross-section and looked like room for supply; every sliver of them is thinner than the sheet, and the only usable room was the disc the fins never reached.