Concept

Surface in a volume — where it appears

The problem an organ with area to provide is solving: how much surface a fold can hold inside a fixed envelope. With a sheet of no thickness the answer is unbounded, and with a thickness the curve turns over and falls, so the interesting quantity is where it turns.

Named by 13 essays across one field — each of them below, with the objects they name alongside it.

0204060801001200510152025foldssurface heldpeak at 64 foldsbox of side 1 · sheet thickness 0.01 · most surface at 64 folds · 128 folds fills the box with sheet alone

Nothing grown has a seam

A gut is a tube and a leaf is a disc, and neither was made by joining anything. The sheets this collection builds by identifying a rectangle's edges are the same objects a body grows, reached by an operation no organism performs — and the difference shows up in where the boundary is and in what has to close.

biology · Surface in a volume
0204060801000510152025foldssurface heldno supply — 50 foldsδ = 0.01 — 25 foldsδ = 0.03 — 12 foldsδ = 0.09 — 5 foldsbox of side 1 · sheet thickness 0.01 · optimum at S ⁄ 2(t + δ), so supply and sheet are charged the same way

The surface has to be supplied

The curve that turns over does so because the sheet's own thickness fills the box it is folding into. A surface in a body has to be reached as well as fitted, and the channel that reaches it takes depth out of the same box on exactly the same terms — so the best fold count and the surface it delivers both fall by the ratio of the sheet's thickness to the sheet and its supply together.

biology · Surface in a volume
0204060801000510152025foldssurface held1 surface — 50 folds2 surfaces — 25 folds3 surfaces — 17 folds4 surfaces — 12 foldsbox of side 1, thickness 0.01 · a ceiling goes as depth², so m sharers of one depth reach one m-th of it between them

Two surfaces in one box

A body folds several surfaces into one volume and each of them does a different job. Dividing the depth between them looks like a fair split costing nothing overall, and it is not: the area a single surface can reach goes as the square of the depth it has, so m surfaces sharing a depth reach a total of exactly one m-th of what one of them would have reached alone.

biology · Surface in a volume
-2.5-2-1.5-1-0.5050100150200250depth given to the inner level, log₁₀ of the boxsurface over the flat sheetone level, the whole boxreaches 250.0two levels, any splitnever above 62.5box depth 1 · sheet 0.001 · one level reaches 250.0, and a nest of two reaches 62.5 however the depth is shared

A nest pays four a level

A corrugation folded inside the panels of another looks like the arrangement that multiplies surface rather than dividing it. It multiplies the factors and divides the depths, and the depths cancel: a packed level can hold at most its depth over four times what it folds, the thing it folds is as thick as the depth the level below was given, and so a nest of L levels reaches at most the box over 4ᴸ sheet thicknesses. One level with the whole box beats any nest of two by exactly four.

biology · Surface in a volume
012340200400600800clearance above the basesurface, as a multiple of the basewalls: 2c ⁄ τplies: c ⁄ 4τeight times lesssheet thickness 0.01 · both lines are straight and their ratio is eight everywhere, so no clearance makes the stack competitive

Standing up beats lying down by eight

A level that fills its clearance with plies lying flat makes every ply share the clearance, and the sharing costs a factor of four. A level that fills the same clearance with walls standing on the base gives every member the whole height and charges them only for footing. The ratio is exactly eight, at every clearance and every thickness — and it is the difference between a cost charged against depth and a cost charged against the space beside it.

biology · Surface in a volume
02468101214010203040506070clearance above the basesurface, as a multiple of the basewalls stop at 40.0sheet 0.01, channel 0.05plies keep risingthey cross at 9.40the comb saturates at twice the reciprocal of its channel's share, and the stack does not saturate at all

The channel grows with what it feeds

A comb of standing walls beats a stack of plies by eight because its members do not share the clearance that pays them. Supply takes that back, and asymmetrically: a wall's channel has to be sized for the surface the wall carries, so it grows with the wall's height and is charged against the pitch, while a ply's channel is a constant charged against the clearance. The comb then saturates at twice the reciprocal of the channel's share, the stack does not saturate at all, and the two cross at a clearance the model gives in closed form.

biology · Surface in a volume
the reach, the member length and the member count, against the anglesurface counted as a multiple of the base, lengths as multiples of the clearance90°60°30°10°1°the angle the members stand atthe reach: 200flat, at every anglehow long each member ishow many of them there areclearance 1, sheet 0.01 · the reach is 200 at every angle; the two factors move by 57

The angle the eight does not know

A comb's members are always drawn standing square to the base, and nothing has asked why. Lean one to an angle and it must be longer to reach the same clearance, which is more surface; it also takes more of the base to stand on, which is fewer members. The two are reciprocal and cancel exactly — the surface a comb holds is the same number from a right angle down to one degree, where each member is fifty-seven times the clearance long and there are two of them where there were a hundred.

biology · Surface in a volume
two linings of one tube, and the ceiling neither passessurface per unit length of tube, from a sheet 0.01 thickradiuslayersfins at bestthe ceilinglayers over fins0.516078.51572.040016353146282.020022526125725132.01004100785027100532.0050the ceiling is the tube's cross-section over the sheet thickness, twice over, and no arrangement of flat sheet passes it

In a tube the standing members lose

Members standing across a clearance beat layers lying along it by eight, and every drawing of that argument has a flat base under it. Curve the base into a tube and the ranking inverts: radial fins converge, so the room they need is the room at their tips, and their best arrangement fills exactly half the cross-section. Concentric layers fill all of it. The eight becomes a half, and the half is exact.

biology · Surface in a volume
123456780.40.50.60.70.80.91how many lengths of finshare of the ceilingtips equally spacedheights halvingtwo thirdsradius 1, sheet thickness 0.01 · the share of the ceiling 2πR²/τ that fins of m lengths hold

The wedge belongs to one length

Radial fins inside a tube reach at best half of what any lining of sheet could hold, because converging fins leave empty wedges behind their tips. Tapering the fins cannot help: the tip already sets the count, and a fin cannot be thinner there than the sheet it is made of. Fins of several lengths can. Counted along the radius they are a staircase under a straight line, and the staircase with m steps is best with its steps equally spaced, where it holds exactly m ⁄ (m + 1) of the ceiling. The factor of two belonged to fins of one length, not to fins.

biology · Surface in a volume
fins of several lengths with a lumen kept clear down the middlethe diagonal is the most fins the circumference allows at each radius; the staircase is what the fins achieveradius, axis to wallfree lumenlumen radius 0.40R: 16% of the cross-sectionfins of 3 lengths hold 72.0% of the ceilinga lumen out to 0.25R would have cost nothingthe most any lining could hold now: 84.0%

A lumen costs nothing until it does

A tube lined with radial fins of m lengths holds at best m ⁄ (m + 1) of the most surface its sheet could carry, with the fins' tips on equally spaced rings. A tube also has to carry something down its middle. Keep a central lumen of radius ρR clear and the best lining puts its first ring of tips at the lumen and spaces the rest evenly to the wall, holding 2ρ(1 − ρ) + (1 − ρ)²(m − 1) ⁄ m of the ceiling — until the lumen is smaller than R ⁄ (m + 1), where it costs nothing at all, because the fins' own optimum already leaves that much empty. Past it, each further unit of lumen costs surface at a rate rising from nothing towards one. With unlimited lengths the frontier is the straight line 1 − ρ²: the bend is the price of having few lengths, and the free lumen is the wedge they wasted.

biology · Surface in a volume
the most surface a lining can hold for each lumen, and the lumen that surface needsthe straight line is the demand; each dot is a lining that holds what its own lumen can supply00.2500.5000.750100.2000.4000.6000.8001share of the cross-section given to the lumenshare of the ceiling the fins hold1 length2 lengths3 lengths10 lengthsunlimitedleft of its dot a lining would hold more than its lumen can supply; right of it, the lumen is wider than its surface needs

Supply decides whether lengths pay

A tube lined with fins has a frontier of surface against the lumen left down its middle, and a lining that must supply its own surface needs a lumen that grows with the surface. The two meet at one point per number of fin lengths. When a channel is narrow for what it serves, the meeting is on the flat stretch and supply costs nothing; when it is wide, every lining meets near the full lumen, and the lengths that were worth half again are worth under one per cent.

biology · Surface in a volume
the surface a supplied lining holds when its supply has to reacheach unit of surface at radius r needs a channel 1 + λ(r − ρ) times as wide as at the lumen's edge00.2500.5000.7501-2-101log₁₀ of the channel's width per unit of surface, in sheet thicknessesshare of the ceiling the fins holdone lengthunlimited lengthsthe same, reach freethe upright line is the channel of a fifth of a sheet per unit of surface; its dot is the unlimited lining there

Reach taxes the extra lengths

A tube's fin lining that must supply its own surface was priced with a channel of the same width for every unit of surface, wherever it stood. Charge instead for how far the supply has to carry, and the extra lengths of fin — which multiply toward the wall — pay more for each unit of surface than a single length does. Where supply was cheap they are worth half again rather than two thirds; the passage to lengths buying nothing moves to channels half as wide; and where supply was dear, nothing changes at all, because a thin ring has no reach inside it.

biology · Surface in a volume
one supply bill, written per unit of surface and as a branching tree3 lengths of fin at a channel of 1 per unit of surface, reach costing λ = 200.2500.5000.75010.8000.9001radius, from the lumen's edge to the wallextra channel at each radiuseach unit for its own reacharea 0.1170a tree sized to its flowarea 0.1170ticks on the axis are the rings of fin tips; the tree's channel steps down at each, the per-unit charge steps up

A shared supply is the same bill

A tube's fin lining charged for how far its supply has to carry was expected to do better under a supply that branches, one vessel running out along each fin and feeding everything it passes. It does not, and it cannot: with channels sized to the flow they carry, the per-unit charge and every branching tree are the same integral. Sharing pays only when a channel's width grows more slowly than its flow, as Murray's and Poiseuille's vessels do, and then branching buys back exactly what those economies would otherwise have charged, to within a fifth of a point.

biology · Surface in a volume

Named alongside it

The objects these essays reach for when they reach for this one.

Trade-offScalingSurface areaThicknessConstraintOptimisationPacking ratioConservationEfficiencyIdealisationBoundaryCorrugation

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