A nest pays four a level

Two surfaces in one box ends by naming the arrangement that should beat every split. Divide a box’s depth between $m$ surfaces and the total falls by $m$, because a single surface’s ceiling is quadratic in the depth it has. Fold the fine surface inside the coarse one’s panels instead, and the essay argues that the depths no longer add — the coarse level buys a factor, the fine level buys another factor of what the coarse one produced, and the two multiply. A gut lining is folded at the scale of a plica, again at the scale of a villus and again at the scale of a microvillus, and the three factors are what makes it so large.

That argument is right about the factors and wrong about what they are factors of. Worked out on the same box and the same parabola, a nest of packed corrugations does not escape the arithmetic. It pays a fixed price at every level, the price is four, and one level with the whole box beats any nest.

What a nest was supposed to buy #

Start with the one-level result, because everything below is built out of it. How much surface fits in a body folds a sheet of thickness $t$ into a box of depth $D$. With $k$ folds the stack uses $kt$ of the depth on its own thickness, each panel is as tall as what is left, and the surface held is proportional to $k(D - kt)$. That is a downward parabola in the fold count. It peaks at $k = \tfrac{D}{2t}$, and at the peak it holds $\tfrac{D}{4t}$ times the surface of one unfolded panel.

Two things about that peak matter here. It is the same half-share the convergence argument found at every material’s optimum: at the best fold count exactly half the depth has gone to the sheet’s own thickness, and the panels are half as tall as the box. And the factor it reaches — $\tfrac{D}{4t}$ — depends on the depth and on the thickness only through their ratio. A box a thousand sheets deep holds 250 times the flat panel at best, whether the sheet is a micron thick in a millimetre box or a millimetre thick in a metre one.

The thickness in that sentence is the one the sheet has a thickness gave every folded object, and the packing is the packing a leaf’s corrugation achieves in a bud: panels pressed against their neighbours, with no room between them.

A nest takes that sheet, corrugates it within some small depth $d$, and then treats the corrugated sheet as a new material to be corrugated again at a larger scale. If the fine level multiplies the area by one factor and the coarse level multiplies by another, the surface of the whole is their product. That much is simply true, and it is what the intuition about nesting rests on.

The depth a level hands upward #

What the intuition leaves out is how thick the new material is.

A corrugation that has been packed into a depth $d$ is a slab $d$ thick. Its panels stand across that depth and are pressed against one another, so the slab has no room left in it; it is as thick as the depth it was folded into, and a coarser corrugation that wants to fold it has to fold a layer $d$ thick, not a layer $t$ thick.

So the coarse level faces exactly the problem the single level faced, with $d$ where $t$ used to be. Folded into the whole box, it reaches at most $\tfrac{D}{4d}$ times the area of its own flat panel. The fine level, folded into its depth $d$, reached at most $\tfrac{d}{4t}$. The product of the two ceilings is

$$\frac{D}{4d}\cdot\frac{d}{4t} = \frac{D}{16t}$$

and the depth the fine level was given has vanished from it. However that depth is chosen — a tenth of the box, a hundredth, twice the sheet’s thickness — a nest of two packed levels holds at most a sixteenth of the box over the sheet’s thickness. One level with the whole box holds a quarter. The nest loses by exactly four.

The first figure is that sentence drawn. Along the horizontal axis the inner level’s depth runs over two and a half orders of magnitude, and at each depth the fold counts of both levels are searched over whole numbers. The curve never rises above 62.5; it touches 62.5 wherever the two optimal fold counts happen to be whole numbers; and it never comes near the 250 a single level reaches.

Every intermediate depth cancels #

The same arithmetic runs for any number of levels, and the result is short enough to be memorable.

Give a nest of $L$ packed levels the depths $D = d_1 > d_2 > \dots > d_L$, with the innermost level folding the sheet itself, of thickness $t$. Level $i$ folds a composite as thick as $d_{i+1}$ into a depth $d_i$, so its ceiling is $\tfrac{d_i}{4d_{i+1}}$. The product of the ceilings is

$$\frac{d_1}{4d_2}\cdot\frac{d_2}{4d_3}\cdots\frac{d_L}{4t} = \frac{D}{4^L\,t}$$

and every depth between the box and the sheet appears once on top and once underneath. The product telescopes. What survives is the ratio of the box to the sheet, which no arrangement can change, and a power of four, which is set by nothing but the number of levels.

So a nest’s ceiling does not depend on how it is built. It depends on how many levels it has, and it falls by four with each one. For a box a thousand sheets deep the ceilings are 250 for one level, 62.5 for two, 15.6 for three and 3.9 for four — and at four levels the nest holds less than four times the area of the flat sheet it started from.

The search is what makes the table evidence rather than algebra. At two levels every one of 36 intermediate depths is tried with every whole fold count at both levels; at three, every pair of depths from the same list. None of the nests found rises above its ceiling and the best of them meet it, so the telescoping is something the search observes rather than something assumed on the way in.

Why the price is four and not some other number #

A factor of four per level looks arbitrary until it is split into its two halves, and each half is a result already established.

The first half is the thickness. At a level’s own optimum, half the depth it is given goes to the thickness of what it folds. That is the half-share — a benefit proportional to a count against a cost proportional to the same count taken from a fixed budget, which optimises where half the budget has gone.

The second half is the height. With half the depth spent on thickness, each panel is half as tall as the depth, so the surface each fold contributes is half what it would have been in a box with nothing in it.

A level’s own factor is therefore its fold count times a half, and its fold count is its depth over twice the thickness it folds. The two halves make the four. A nest pays both halves at every level, because every level is packed and every level’s optimum spends half its depth on the layer below.

The same pair of halves is what a corrugation costs when its shrinkage is priced in paper rather than in depth, and it is why the pile rather than the panel is what a thick fold is up against: a packed level’s composite is a pile, and the next level folds the pile.

That also explains why no clever choice of depths helps. Giving the inner level more depth makes its factor larger and makes the composite it hands upward thicker by the same proportion, which costs the coarse level exactly what the fine level gained. The exchange is one for one, which is why the intermediate depth cancels, and why the only thing left to pay is the pair of halves at each level.

Three levels, written out #

The table in the figure is worth reading as a ledger, because it makes the cancellation concrete in a way the formula does not.

The outer level has the whole box and folds a composite a tenth of the box thick: a ratio of ten. The middle level has a tenth of the box and folds a composite a hundredth thick: a ratio of ten. The inner level has a hundredth and folds the sheet, a thousandth: a ratio of ten. Multiply the three ratios and the answer is a thousand, which is the box over the sheet — and it is a thousand for every choice of depths, because the depths cancel in the product whatever they are.

Each level, folded as well as whole numbers allow, reaches half of five: two and a half. Three levels at two and a half give 15.6. A single level with the whole box and the bare sheet, folded 500 times, reaches 250. The nest has used three levels of folding, 5 + 5 + 5 folds at their three scales, to hold a sixteenth of what one level of 500 folds holds.

Shift the split and the whole-number rounding moves but the ceiling does not. With depths of one, a fifth and a fiftieth, the three ratios are five, ten and twenty; the levels reach 1.2, 2.5 and 5, whose product is 15.0; and the ceiling is still the thousand divided by sixty-four.

Splitting costs a sharer, nesting costs a level #

Two surfaces in one box priced the other way of putting several surfaces in one volume, and the comparison is sharper now that both prices are known.

Splitting divides the depth between $m$ sharers side by side. Each sharer’s ceiling goes as the square of its depth, so the total across all $m$ falls to one $m$-th of a single surface’s ceiling.

Nesting puts the surfaces inside one another. Each level’s ceiling goes as its depth over what it folds, the depths cancel, and the total falls to one $4^{L-1}$-th of a single surface’s ceiling.

For two surfaces the split costs a factor of two and the nest costs a factor of four. For three, three against sixteen. For four, four against sixty-four. Of the two arrangements this model can represent, nesting is the more expensive one, and increasingly so — which is the opposite of what the picture of a villus inside a plica suggests, and the opposite of what the previous argument concluded.

That does not make the previous argument’s instinct wrong about bodies. It makes it wrong about packed corrugations, and it says precisely what an arrangement has to do differently to reward nesting: stop charging each level the full thickness of the one inside it.

A rocket that cannot drop its stages #

The shape of this result has a well-known cousin in a different subject, and the comparison says exactly where the difference lies.

A rocket is also a nest of stages, and staging a rocket is what makes large velocities possible. Each stage multiplies the velocity it adds by something, and the multiplication is worth having because a spent stage is thrown away: the next stage does not have to carry the empty tanks of the one before it. The dead mass leaves the problem at each level.

A nested corrugation is a rocket that cannot drop its stages. The fine level’s packing — the half of its depth spent on thickness, the half-height panels — is carried into the coarse level as part of the material being folded. The coarse level pays for the whole of the fine level’s slab, dead half included, and then spends half of its depth on that slab’s thickness. The dead fraction compounds instead of being discarded.

Stated that way, the escape from the factor of four is also clear. A level that could shed its packing before the next level folded it — whose panels stood apart rather than pressed together, so that the composite it handed upward was not a solid slab — would not pay the full price at the next level. That is not an arrangement this box model contains, and it is where the next question lies.

What the curve leaves out #

The first figure shows a ceiling and not a structure, and it is worth being plain about the gap.

It says nothing about whether any organ is packed. A villus stands in a lumen that is mostly empty, a microvillus stands in a brush border with fluid between its members, and neither is a panel pressed flat against its neighbour. The whole of this result concerns an arrangement in which every level is packed to its own optimum, and the more a real level departs from that, the less the result applies to it.

Nor can it show what nesting is for. A nested surface may be worth having because each level serves a different job, or because a fine level can be renewed without disturbing a coarse one, or because the finest level has to be small for reasons of chemistry rather than geometry. None of those is area in a box, and a model that prices area in a box is silent about all of them — which is the division of labour the organism is not the model asks of every figure in this field, and the reason a lining that was never cut out of anything is not obliged to be optimal by anybody’s arithmetic.

What the picture establishes is narrow and firm: if the levels are packed, the nest’s ceiling is set by the number of levels alone, and it is lower than a single level’s by a power of four.

The box the levels are packed in #

Three assumptions are carrying the result, and each can be named rather than implied.

Every level is packed. The panels of a level touch, so the level’s composite is as thick as the depth it occupies. That is what makes the fine level’s depth reappear as the coarse level’s thickness, and it is the assumption the whole telescoping rests on.

Every level folds as a corrugation in one direction, into a depth. The model is the box a single corrugation is packed into: $k$ folds, $k$ thicknesses stacked in the depth, panels as tall as what is left. A level that folded in two directions, or into a volume rather than a depth, would have a different ceiling and possibly a different price.

The composite folds like a sheet. A corrugated slab is treated as a material with a thickness and nothing else, so that folding it has the same arithmetic as folding paper. Real composites bend less easily across their corrugations than along them, and a sheet folded twice carries creases the first fold put there; both would change the numbers and neither is modelled.

How the search earns its numbers #

One level is checked against its closed form. The best whole fold count for a box a thousand sheets deep reaches 250.00 against $\tfrac{D}{4t} = 250$, which is the parabola the rest is built from, confirmed before it is used.

Two levels are searched, not substituted. At 48 depths for the inner level, spread evenly on a logarithmic scale from twice the sheet’s thickness to half the box, both levels’ fold counts are searched over whole numbers. No depth produces a nest above 62.5, and the value 62.5 is met, so the ceiling is both respected and reached.

Three levels are searched over every pair of depths from a list of 36, and none exceeds $\tfrac{D}{64t}$. The fourth row uses equal ratios, where the continuous optimum sits, and is labelled as such in the figure.

The telescoping is checked as an identity. In the written-out nest the ratios of depth to folded thickness multiply to the box over the sheet to a part in a billion, and each level’s factor is required to stay under its own ceiling. A nest in which some level beat its ceiling would mean the one-level arithmetic had been misapplied, and the figure would not draw.

Still open: a level with room between its folds #

The factor of four comes from packing, and the obvious question is what a level costs when it is not packed.

A level whose folds stand apart — a comb of fins rather than a pressed zigzag — does not spend half its depth on the thickness of what it folds, because there is space between its members for that thickness to sit in. It spends width instead: the fins cannot be closer together than their own thickness allows. So an unpacked level is charged in the other direction of the box, and a nest of unpacked levels is a different product with a different price, which may or may not telescope.

That is a computation rather than a survey, and it is the one that would say whether the three levels of a gut lining are an arrangement geometry rewards or one it merely permits. The answer turns on whether a level’s cost is charged against the depth it occupies or against the space beside it, and the packed model here is the case where it is all charged against depth.

Sideways from here, the rocket comparison points at the supply channels a surface needs, which are exactly the kind of dead fraction a level carries upward. A nest whose every level needs its own supply pays that charge in the coefficient at every scale, on top of the four.

The habit worth carrying is a check on any argument that says factors multiply. Ask what each factor is a factor of, and whether the thing it acts on already carries the previous factor’s cost. When it does, the product telescopes, the intermediate quantities cancel, and what is left is a fixed price per stage.