Standing up beats lying down by eight

A nest pays four a level ends on the question its own arithmetic cannot answer. Its levels are packed — plies pressed together, each level’s cost charged against the depth it occupies — and the four it loses at every level comes from that charging. A level whose members stand apart, it says, spends width instead, and whether a nest of those telescopes the same way is a different computation.

It is a different computation, and it does not start where the question expected. Before any nesting, a level whose members stand apart beats a level whose members lie down by a factor of eight — and the eight has neither the clearance nor the thickness in it.

Two ways to fill a clearance #

A level, in this account, takes a base and a clearance above it and returns surface. How much surface fits in a body built the first way and every later result has used it: plies laid parallel to the base, spread through the clearance. Each ply’s own thickness takes from the clearance the others could have used, so $k$ plies of a sheet of thickness $\tau$ in a clearance $c$ hold

$$f_{\text{plies}} = k\!\left(1 - \frac{k\tau}{c}\right), \qquad f^* = \frac{c}{4\tau} \ \text{at}\ k = \frac{c}{2\tau}$$

and the four in that denominator is the price of a quadratic. It is not a constant somebody measured; it is what a parabola’s peak is worth against the line it would follow if the plies cost nothing.

The second way is to turn the members ninety degrees. Walls standing on the base, each the full height of the clearance, each carrying two faces. Nothing a wall does takes height from its neighbours — a wall’s height is the clearance, whatever the count — so what the walls compete for is the base they stand on, and $n$ walls per unit of base hold

$$f_{\text{walls}} = 2cn, \qquad n\tau \le 1 \ \Longrightarrow\ f^* = \frac{2c}{\tau}$$

with no quadratic anywhere. The ratio of the two ceilings is $8$: two for the wall’s second face, four for the clearance the plies have to share.

One has a peak and the other has a wall #

The two optima are different in kind, and the difference is more informative than the factor.

The stack’s optimum is interior. There is a best number of plies — fifty, here — and both fewer and more are worse, because a ply is simultaneously a gain and a charge against the same budget. Fifty-one plies is a slightly worse stack than fifty.

The comb’s optimum is a boundary. Every additional wall is pure gain and no wall costs any other wall anything, right up to the point where there is no base left to stand on. A hundred walls hold two hundred times the base; a hundred and first wall cannot be built. Nothing turns over, because nothing is being shared.

That is the whole mechanism, and it says where to look in any similar question. A quadratic ceiling means the members are competing for the resource that also rewards them. A linear ceiling with a hard stop means they are competing for something else.

Whole members, in case the optimum was an artefact #

Both closed forms allow fractional members, which is a thing a body cannot build, and a factor that only exists between two fictions is not worth having.

At a clearance of 0.25 the stack does best with twelve plies rather than 12.5 and holds 6.24 against a hundred walls holding 50.0 — a ratio of 8.013, the rounding falling the comb’s way by a tenth of a per cent. At every other clearance drawn the rounding costs nothing at all and the ratio is 8.000. Neither ceiling is an artefact of allowing fractional members, and a body building whole villi and whole layers gets the same answer as the algebra.

Eight, at every size the question makes sense at #

The factor’s independence is worth checking rather than reading off two formulas, because two formulas that cancel are exactly the situation in which an error cancels too.

A clearance of 1 out of a sheet 0.01 thick gives 25 and 200. Thinning the sheet to 0.002 gives 125 and 1,000. Shrinking the clearance to 0.4 gives 10 and 80. A clearance of 4 out of a coarse sheet of 0.05 gives 20 and 160. The ceilings run over a factor of a hundred; the ratio is 8.0000 on every row.

So the eight is a fact about sharing rather than about size, and it holds at the scale of a gut lining and at the scale of a heat exchanger with the same force. That is a different kind of statement from most of what this subject produces, where a ratio usually carries the thickness somewhere.

A comb inside a comb gains nothing #

Now the question the nesting essay actually asked. Give the walls combs of their own and see whether the factors compound the way the packed levels’ factors did.

They do not, and the way they fail is more interesting than a loss. A comb of walls at pitch $p$ leaves a gap of $p - \tau$ between neighbours, shared by the inner combs growing from the two faces that bound it, so each inner comb has a clearance of $(p-\tau)/2$ and reaches $(p-\tau)/\tau$. The outer comb reaches $2c/p$. Multiplying,

$$f_{\text{nest}} = \frac{2c}{p}\cdot\frac{p-\tau}{\tau} = \frac{2c}{\tau}\left(1 - \frac{\tau}{p}\right)$$

which is the single comb’s ceiling times a factor strictly below one.

A nest of two combs never reaches what one comb reaches, and it approaches that value only by spreading the outer walls until they have stopped multiplying anything. Every bit of room the inner combs gain is room taken from the outer comb’s wall count, at exactly the rate that cancels. The two levels are not compounding; they are trading.

Set beside the packed case, that is a clean contrast. Packed levels lose a factor of four each, so a nest of two is four times worse than one level with the whole box. Unpacked levels lose nothing per level and gain nothing either: the product is bounded by the single level and creeps up to it. Nesting is a mistake in one architecture and a wash in the other, and in neither is it the multiplication the idea of levels suggests.

Where the four went #

It is worth saying plainly what happened to the factor that the whole nesting argument was built on, because it did not go anywhere — it was never a property of levels.

The four is the peak of a parabola, and the parabola exists because a ply both delivers surface and consumes the clearance that lets other plies deliver it. Turn the members and that coupling disappears: a wall consumes base, and base is not what pays it. The telescoping in the packed nest — depths cancelling from level to level — is a consequence of each level handing the next a composite whose thickness is its own depth, which is again the same budget being spent twice. A comb hands its inner combs a gap, and a gap is not the clearance the outer comb was paid out of.

Two surfaces in one box found the sharpest version of the same idea: a ceiling quadratic in a resource means dividing the resource $m$ ways divides the total by $m$ rather than leaving it alone. That essay divided the resource between surfaces. This divides the members between two ways of spending one resource, and finds the arithmetic hinges entirely on whether the spending is against the paying budget.

Which of the two an organ is #

The distinction is not a modelling convenience. It separates real structures, and it separates them along a line that is easy to see in a drawing and easy to miss in a description.

A lining thrown into folds — a stomach’s rugae, a pressed sheet, a bellows — is a stack. Its members lie along the surface they are multiplying and each fold’s material occupies depth the next fold would otherwise have had. A lining carrying fingers — a gut’s villi, a gill’s lamellae, a root’s hairs — is a comb. Its members stand off the surface, each has the whole clearance, and what limits their number is how densely they can be rooted.

Both are called folding, and the word is doing a lot of work. They are not two patterns within one economy; they are two economies, and the ratio between their ceilings is fixed at eight before any pattern is chosen. The same corrugation in four places makes the case that convergence on a geometry can be checked rather than admired, and this is the check one level up: an organ whose whole job is area should be a comb, and the argument says by how much.

What the model cannot do is say that any particular organ is one, because nothing here is measured from a specimen. What it can do is say what a measurement would have to find. A structure whose surface multiple rises linearly with its own depth is behaving as a comb; one whose multiple peaks and falls as it is made finer is behaving as a stack; and the two are distinguished by a series of sections rather than by a single photograph.

There is a third possibility the arithmetic quietly allows and no organ appears to use: a stack built at fewer than its best fold count. The parabola is symmetric about its peak, so twelve plies and eighty-eight plies in the same clearance hold the same surface, and a lining could reach a given area either by folding coarsely or by folding finely. Which side of the peak a structure sits on is not visible in its area and is very visible in its cost to build, and nothing in this account prefers one.

The stack, for comparison at its own best #

Nothing above is an argument that the stack is a bad object. It is what a pressed lining is, and its turnover is the result the whole field started from.

The curve rises, turns over and reaches zero, and the zero is the honest end of the idealisation: at enough folds the box holds sheet and nothing else. That behaviour is what makes the packed model’s four inevitable, and it is also why the comb has no equivalent picture — a comb never fills its clearance with material, because its material stands in the base.

Both figures are the packed architecture doing what it does. Read beside the comb’s straight line and its hard stop, they make the point the comparison is really about: these are two different economies, not two settings of one.

The winner is not a folded object #

There is something uncomfortable about the result, and stating it is more useful than working around it.

A comb has no creases. Its walls are not a folded sheet; they are surfaces standing on a base, joined to it and to nothing else. Every quantity this field has developed for a folded structure — a hinge with a radius, a fold count with an optimum, a crease that remembers, a packing ratio — applies to the stack and not to the comb. Nothing in a body folds on a line gives a crease its width, and a comb has no crease to give it to. Four materials, four optima finds each material’s best fold count from that width, and a comb has no fold count to find.

So the architecture that wins this comparison by a factor of eight is the one this subject has the least to say about, and the factor is partly an artefact of that: the stack is charged for its folding and the comb is charged for nothing at all, because nothing in the model corresponds to what joining a wall to a base costs. A wall has a root, the root is a junction, and a junction is exactly the sort of thing the pile, not the panel shows to be where the real cost of a pattern hides.

That is a limit on the claim rather than a repair of it. The eight is correct for the two objects as modelled, and the honest reading is narrower than it first appears: charging a level against the space beside it rather than against the depth it occupies is worth a factor of eight, before anything is charged for the joins the second arrangement needs and the first does not. A body choosing between them is choosing between a cost this account computes and a cost it does not.

What the lines cannot show #

The figures draw two idealisations and neither is an organ.

They cannot show what holds a wall up. A wall standing in a clearance is a cantilever, and a cantilever of height $c$ and thickness $\tau$ buckles or flops at some aspect ratio the model knows nothing about. The comb’s advantage grows linearly with the clearance and its mechanical viability does not, so somewhere the model’s answer stops being available for reasons outside it.

Nor can they show whether the base is free. A comb’s walls are rooted in the base, and the roots take area out of the very surface the base was providing — which the model charges at $n\tau \le 1$ and then forgets. A structure whose members are thick relative to their spacing loses a noticeable share of its base to its own footings, and the loss is charged to the same budget twice.

Nor can they show the third dimension honestly. Both architectures are described per unit of base and per unit of the direction along the walls, which treats a wall as infinitely long and a ply as infinitely wide. A real villus is a finger rather than a wall and pays for its ends; a real lining’s plies are bounded and pay for their rims. Nothing grown has a seam is where boundaries in this subject stop being a detail, and neither figure has one.

And nothing here says which architecture is reachable. A body arrives at a structure by growing it, and growing a comb and growing a stack are different developmental problems with different costs that no counting of surface can see.

What the model assumes #

A member’s only charge is its own thickness. No joints, no roots, no substructure holding the walls apart or the plies at their spacing. Every one of those would be charged against the same budget as the member and would reduce whichever architecture carries more of them.

Every wall is the full height of the clearance and every ply spans the full base. That is the best case for both and is what makes the two ceilings comparable.

The base is flat and the clearance is uniform. A curved lining or a tapering clearance changes both ceilings, though not obviously their ratio.

The clearance is given rather than paid for. Two surfaces in one box prices what it costs to divide a depth between several surfaces, and nothing here asks where this clearance came from or what else wanted it.

And the sheet has one thickness. Where a nest appears, the inner members are made of the same sheet as the outer ones, which is the assumption that makes the gap arithmetic close.

How the numbers were checked #

The ratio is computed at four sizes rather than once, spanning a factor of ten in clearance and twenty-five in thickness, and required to be eight on every one to within a part in a billion. An algebraic slip in either ceiling would show as a ratio that moved.

The stack’s peak is checked against its closed form, and the comb’s ceiling is checked to be a boundary rather than a peak: one more wall than the footing allows must deliver nothing rather than slightly less.

The nest’s reach is checked against $(2c/\tau)(1 - \tau/p)$ at two hundred pitches, and separately required to stay below the single comb at every one of them. A nest that beat one comb anywhere would break the second check first.

And the whole-member comparison recomputes both architectures with integer counts, requiring the loss to rounding to stay under one per cent for the comb and two for the stack, so the factor survives being built.

Still open: a comb that has to be fed #

The comb wins this comparison because its members do not share the thing that pays them. That is exactly the assumption a body cannot keep.

A wall of height $c$ carries $2c$ of surface, and whatever supplies that surface — a vessel, a duct, a nerve — has to be sized for what it serves. So a taller wall needs a fatter channel, and the channel runs beside the wall, which is to say it is charged against the pitch: the one budget the comb was winning on. A ply, by contrast, serves a fixed area however deep the stack is, so its supply is a constant charged against the clearance, exactly as the surface has to be supplied computes it.

That asymmetry is the next computation and it is the one that decides whether the eight survives contact with an organ. A charge that grows with height, divided into a benefit that grows with height, is a ratio that saturates — so a supplied comb should have a ceiling that no clearance can pass, while a supplied stack keeps growing. If both are true there is a crossover, and the architecture a body should use depends on how deep it is going.

The habit worth carrying is a question to ask of any ceiling with a small integer in it. Ask which budget the members are competing for, and whether it is the same budget that pays them. When it is, the ceiling is quadratic and the integer is the peak of a parabola; when it is not, the ceiling is linear with a hard stop, and no amount of arranging will make the two behave alike.