Curves and material

How many times can it be halved

The folklore says seven, and the folklore is a statement about one sheet of paper. What actually binds is arithmetic: every halving doubles the layers and the paper spent at the closed end grows as the square of the layer count, so the length needed for twelve folds is nearly a kilometre.

Assumes Four things that are not true and The crease has a radius.

Everybody has been told that a sheet of paper cannot be folded in half more than seven times. It is usually presented as a fact about paper and it is not — it is a fact about a particular sheet, and the interesting version is the one with a formula attached.

How long a strip has to be to fold in half n timesThe length of paper single-direction folding needs, against the number of halvings, for a sheet a tenth of a millimetre thick. The vertical scale is logarithmic and the curve is still steep on it, because the requirement grows as the square of the layer count. The marked lengths are an A4 sheet, a metre, ten metres, a hundred, and the roll that was folded twelve times.24681012-202halvingsmetres of paper (powers of ten)297 mm — 6 folds, 64 layers1 m — 7 folds, 128 layers10 m — 8 folds, 256 layers100 m — 10 folds, 1024 layers1200 m — 12 folds, 4096 layerspaper 0.1 mm thick · L = (πt/6)(2ⁿ + 4)(2ⁿ − 1)the loss is the paper that goes round the closed end, and it doubles twice per fold
Fig. 1 How long a strip has to be to survive n halvings, for paper a tenth of a millimetre thick, on a logarithmic vertical scale. An A4 sheet gets six folds, a metre gets seven, ten metres gets eight, and the roll that was folded twelve times needed nearly nine hundred. The curve is still steep on a log scale because the requirement grows as the square of the layer count.

Where the paper goes

Every halving does two things and only one of them is usually noticed.

It doubles the number of layers, which everybody sees. Sixteen folds would be 65,536 layers, and at a tenth of a millimetre that is a stack six and a half metres deep, which is obviously absurd and is not the binding constraint.

The other thing is what binds. At the closed end of each fold, the paper has to travel round the whole stack. The outermost layer goes round the outside of everything inside it, so it needs extra length — and the extra is proportional to how thick the stack is at that point.

That is the crease radius arriving at a scale where it dominates. One fold of one sheet loses a fraction of a millimetre. Ten folds of a thousand-layer stack lose the same fraction multiplied by the accumulated thickness at every fold, and the accumulation is what makes the answer a surprise.

A crease is not a lineA fold carries the paper round a small radius rather than through a point, and the arc uses more of the sheet than the stack advances by. One crease loses a fraction of a millimetre. A grid with hundreds of them loses that on every line at once, which is why an ambitious tessellation is folded from thin paper and why a large grid comes out short.ρ = 0.12 mmarc 0.377 mm, stack advances 0.240 mmlost per crease (π − 2)ρ = 0.1370 mmone crease, at its real radiuson a 150 mm sheet8 × 81.0 mm — 0.6%16 × 162.1 mm — 1.4%24 × 243.2 mm — 2.1%32 × 324.2 mm — 2.8%48 × 486.4 mm — 4.3%lost along every line of the grid, in both directionswhich is why an ambitious grid is folded from thin paper
Fig. 2 The loss at a single fold, which is where the whole bound comes from. A crease is not a line; it is an arc, and the arc uses more paper than the fold advances by. The quantity is a fraction of a millimetre for one sheet and the sum over a stack is the thing this essay is about.

The bound

For folding in one direction — a long strip halved end over end, which is the case that gets the most folds — the length L needed for n halvings from paper of thickness t is

L = (π t / 6) · (2ⁿ + 4) · (2ⁿ − 1)

The shape of it is what matters. Both bracketed terms are essentially 2ⁿ, so the requirement grows as 4ⁿ — as the square of the layer count. Each fold does not double the paper needed; it quadruples it.

That is the whole result and it explains why the folklore number is so stubbornly small. Going from seven folds to eight is not a matter of finding slightly more paper. It is a factor of four.

For a sheet a tenth of a millimetre thick, the numbers come out as:

halvings layers paper needed
6 64 0.22 m
7 128 0.88 m
8 256 3.5 m
9 512 13.8 m
10 1,024 55 m
11 2,048 220 m
12 4,096 879 m
13 8,192 3,515 m

An A4 sheet is 297 mm long, so it manages six. A metre of paper manages seven, which is where the folklore comes from: a metre is about as much paper as anybody has to hand.

The table also answers the question the folklore never poses, which is what would be needed to do better. Not stronger hands, not better technique: thirteen folds wants three and a half kilometres of tenth-millimetre paper, and no amount of care substitutes for it.

The record, and why it is the less interesting half

In 2002 Britney Gallivan, then a high-school student, was set the challenge of folding something in half twelve times. She derived the bound above, worked out that it needed about 1,200 metres of thin paper, obtained a roll of it, and folded it twelve times.

The folding is the part that gets retold. The derivation is the part that mattered, and it is worth being clear about what it settled: the limit is not a property of paper, it is a relationship between length, thickness and fold count, and any of the three can be traded against the others. Halve the thickness and the same length gets one more fold. Quadruple the length and the same thickness gets one more fold.

That reframing turns a piece of folklore into a design rule, and design rules are what a manufacturer needs. The question a production line asks is exactly this one with different words: how many times can this web of material be doubled before the accumulated thickness at the fold exceeds the tooling.

A crease is not a lineA fold carries the paper round a small radius rather than through a point, and the arc uses more of the sheet than the stack advances by. One crease loses a fraction of a millimetre. A grid with hundreds of them loses that on every line at once, which is why an ambitious tessellation is folded from thin paper and why a large grid comes out short.ρ = 0.12 mmarc 0.377 mm, stack advances 0.240 mmlost per crease (π − 2)ρ = 0.1370 mmone crease, at its real radiuson a 150 mm sheet8 × 81.0 mm — 0.6%16 × 162.1 mm — 1.4%24 × 243.2 mm — 2.1%32 × 324.2 mm — 2.8%48 × 486.4 mm — 4.3%lost along every line of the grid, in both directionswhich is why an ambitious grid is folded from thin paper
Fig. 3 The record, and why it is the less interesting half: the paper each fold spends going round its own corner, as the pattern is drawn finer. The bound is not about how strong anybody is — it is about how much paper the turns consume.

Why the square, exactly

The factor of 4ⁿ has a one-line explanation and it is worth having, because it is the difference between remembering a formula and understanding a constraint.

At fold number k the stack is 2ᵏ layers thick, so that fold spends paper proportional to 2ᵏ. Summing over all n folds gives a total proportional to 2ⁿ⁺¹ — which would be linear in the layer count and would give a much friendlier bound.

But the folds are not all at the same place. In single-direction folding, each fold’s closed end sits at a different point along the strip, and the paper spent at each is spent at every layer that passes through it. So the sum is over folds and over the layers each fold has to carry, and a double sum over a doubling quantity is the square.

The loss is quadratic because the thickness and the number of places it has to be paid for are both doubling. That is the sentence the formula encodes, and it is why the bound is so much harsher than the layer count alone suggests.

What the model leaves out

Three things, and the first is the one that changes the answer most.

Folding in alternating directions is worse, not better. Halving a square sheet alternately along its two axes is the way anybody actually tries it, and it does worse than single-direction folding — the bound for alternating folds grows faster, because each fold has to carry the accumulated thickness of a stack that is growing in both dimensions. Gallivan derived both; the single-direction case is the one that reaches twelve.

Paper is not perfectly compressible or perfectly incompressible. The formula treats the stack as n layers of thickness t with no give. Real paper compresses under the force of folding, which helps, and springs back, which does not.

Force is not in the model at all. At eight or nine layers the limit is geometric — there genuinely is not enough paper. At the point where a person gives up on an A4 sheet, the limit is that they cannot press hard enough, which is a different constraint arriving at a similar place. The geometric bound is the one that survives having a machine.

A crease is not a lineA fold carries the paper round a small radius rather than through a point, and the arc uses more of the sheet than the stack advances by. One crease loses a fraction of a millimetre. A grid with hundreds of them loses that on every line at once, which is why an ambitious tessellation is folded from thin paper and why a large grid comes out short.ρ = 0.12 mmarc 0.377 mm, stack advances 0.240 mmlost per crease (π − 2)ρ = 0.1370 mmone crease, at its real radiuson a 150 mm sheet8 × 81.0 mm — 0.6%16 × 162.1 mm — 1.4%24 × 243.2 mm — 2.1%32 × 324.2 mm — 2.8%48 × 486.4 mm — 4.3%lost along every line of the grid, in both directionswhich is why an ambitious grid is folded from thin paper
Fig. 4 What the model leaves out, and what happens when it is put back: a crease is not a line but a turn with a radius. Zero thickness is the idealisation this essay is about, and this is the quantity it was hiding.

The same arithmetic, at design scale

The bound above is about halving a strip, which is a special case. The general phenomenon is what happens to every complex crease pattern, and the arithmetic is the same with different constants.

A crease is not a lineA fold carries the paper round a small radius rather than through a point, and the arc uses more of the sheet than the stack advances by. One crease loses a fraction of a millimetre. A grid with hundreds of them loses that on every line at once, which is why an ambitious tessellation is folded from thin paper and why a large grid comes out short.ρ = 0.12 mmarc 0.377 mm, stack advances 0.240 mmlost per crease (π − 2)ρ = 0.1370 mmone crease, at its real radiuson a 150 mm sheet8 × 81.0 mm — 0.6%16 × 162.1 mm — 1.4%24 × 243.2 mm — 2.1%32 × 324.2 mm — 2.8%48 × 486.4 mm — 4.3%lost along every line of the grid, in both directionswhich is why an ambitious grid is folded from thin paper
Fig. 5 The loss for a grid rather than a stack: how much paper a crease radius costs across a sheet divided into n parts each way. Every crease spends (π − 2)ρ, and a grid of 48 spends it hundreds of times. The number at the foot of each row is what an ambitious design comes out short by.

A tessellation on a 48-grid has a great many creases, and a complex base has more. Each one spends the same fraction of a millimetre, and the sum is millimetres across the sheet — which is why the finished model is smaller than the geometry said and why the discrepancy grows with ambition.

The halving bound is the extreme case of the same accounting, made vivid because the layers double. In a design the layers do not double, they accumulate in patches, and the loss is spread rather than concentrated. Both are the same charge for the same thing: paper that goes round a corner instead of straight through it.

The reason the halving version is famous and the design version is not is that the halving version has a clean answer. Ask how many times a strip can be halved and there is a number; ask how much a complex base comes out short and the answer is a distribution over the sheet that depends on every crease’s local layer count.

Why the record was set on a roll and not a sheet

The section above says alternating folds do worse and leaves it there. The companion bound Gallivan derived for that case puts a number on how much worse, and the number explains the roll.

For a square sheet halved alternately along its two axes, the side needed for nn folds goes as πt23(n1)/2\pi t \cdot 2^{3(n-1)/2}. The side therefore multiplies by about 2.83 with each extra fold, so the area multiplies by eight. A strip halved end over end multiplies its length by four, and its width does not change at all — so its area multiplies by four.

Each extra fold costs twice as much paper the alternating way, and the factor compounds. At six folds the two are within a small multiple of each other; by twelve they are not close.

Put the twelve-fold case in units anybody can picture. Alternating, at a tenth of a millimetre, wants a square about twenty-nine metres on a side — some eight hundred square metres of paper. Single-direction wants a strip eight hundred and seventy-nine metres long and as narrow as one likes: at half a metre wide that is four hundred and forty square metres, at a centimetre wide it is under nine.

That is the whole reason the record is a roll. It is not that a long strip is easier to handle, and it is not a trick that evades the bound — the bound is obeyed exactly either way. It is that the strip’s width never enters the arithmetic, so a folder can buy the length the formula demands without buying the area a square would demand alongside it.

And why the folklore number is about squares

The comparison also explains why the received figure is seven rather than something else, and why it feels like a fact about paper.

Everybody who has tried this has tried it on a square or a rectangle, alternating directions because that is what a sheet invites. On the alternating bound a sheet a metre square manages about nine folds at a tenth of a millimetre, and an A4 sheet manages six or seven — which is exactly the folklore, arrived at by the harder of the two routes.

So the number in circulation is the alternating bound evaluated at the paper people have, and the single-direction bound was never in the story because nobody folds a sheet end over end without a reason to. The reframing this essay is about therefore has two halves rather than one: the limit is a relationship between length, thickness and fold count, and it is a different relationship depending on how the folding is done. Fixing the paper and changing only the direction of the folds moves the answer by several folds, at no cost in material at all.

The one limit no idealisation removes

Most of the constraints on this site are geometric, and geometric constraints have the property that a cleverer construction can sometimes route around them. This one cannot be routed around, and that is what makes it worth a rung of its own.

The paper spent at a fold is not a modelling artefact or a manufacturing tolerance. It is the arc length of a curve that has to exist because the material has thickness and does not stretch — the same non-stretching that every theorem here is built on. A thinner sheet pushes the number up by one or two; nothing pushes it up by ten.

So this is the place where the idealisation is not a simplification but a change of subject. A zero-thickness sheet folds in half indefinitely, and the statement is not an approximation to the real behaviour — it is a completely different behaviour, in which one of the subject’s most famous facts does not exist at all.

That is worth carrying back to the rest of the site. Every crease pattern here is verified against theorems that assume zero thickness, and those theorems are correct and useful. What they do not contain is any of this, and a complex tessellation with two hundred creases is spending the loss above at every one of them, which is why an ambitious design comes out short.

What this does to the rest of the subject

Two consequences follow for essays elsewhere on this site, and both are the kind that change a number rather than a conclusion.

Complex bases need thin paper, and the requirement is quantitative. A base folded on a 32-grid stacks a dozen or more layers in places, and the thickness at those places is what decides whether the model closes. Designers reach for tissue-foil and thin kami not out of preference but because the layer count is fixed by the design and the only free variable left is t.

Manufacturing meets this before anybody else does. A production line folding an airbag or a solar array is halving a web of material, and the material is much thicker than paper — so the fold count available is far smaller and the bound bites immediately. Thickness accommodation exists because the alternative is to accept the bound, and accepting it means a mechanism with three folds where the geometry wanted eight.

A crease is not a lineA fold carries the paper round a small radius rather than through a point, and the arc uses more of the sheet than the stack advances by. One crease loses a fraction of a millimetre. A grid with hundreds of them loses that on every line at once, which is why an ambitious tessellation is folded from thin paper and why a large grid comes out short.ρ = 0.3 mmarc 0.942 mm, stack advances 0.600 mmlost per crease (π − 2)ρ = 0.3425 mmone crease, at its real radiuson a 200 mm sheet8 × 82.4 mm — 1.2%16 × 165.1 mm — 2.6%24 × 247.9 mm — 3.9%32 × 3210.6 mm — 5.3%lost along every line of the grid, in both directionswhich is why an ambitious grid is folded from thin paper
Fig. 6 What this does to the rest of the subject: the same arithmetic at a stiffer paper, where the arc a crease turns through is nearly a third of the panel. Every technique for building a fold in something thick is a way of paying this bill somewhere else.

Neither of those is news to anybody in either field. What the halving bound adds is the arithmetic underneath the received wisdom, and the arithmetic is what makes it possible to say how much thinner rather than merely thinner.

What no figure here can show

The figure plots a bound and not an experiment. Nothing on this page was folded, and the curve is a formula evaluated rather than a set of measurements — which is a real difference, because the formula’s assumptions about compressibility are exactly where a measurement would disagree with it.

It also plots one thickness. Paper thickness spans a wide range — tissue at 0.02 mm, copier paper at 0.1, card at 0.3 — and the bound scales inversely with it, so the whole curve slides by more than an order of magnitude across ordinary materials. The figure states its thickness on its face for that reason.

And it says nothing about what the folded object is. A strip folded twelve times is a stack of 4,096 layers a few centimetres across, and every one of the essays on this site about layer ordering has an opinion about which layer goes where. This one does not: the folding is a simple accordion or roll and the ordering is whatever the sequence produced, which is a question the machine ladder answers and this bound does not ask.

The idealisation, named

Even this essay’s model is idealised, and it is worth naming what it assumes so the number is not read as more solid than it is.

The stack is uniform. In reality the layers at the folded end bunch and the ones at the open end splay, so the “thickness” the fold has to carry is not 2ᵏt exactly.

The fold is a semicircle. The π/6 in the formula comes from treating each fold’s loss as a half-circle of the accumulated thickness. A real fold is tighter at the inner layers and looser at the outer ones, which is the same taper that thickness accommodation exists to manage.

The paper does not tear. At twelve folds the outer layers are under real tension and the inner ones are in compression, and a material that failed under either would stop earlier for a reason the geometry never mentions.

None of those changes the shape of the answer. All three change the constant, and the constant is not what the result is about.

Who worked this out, and when

Britney Gallivan derived the single-direction and alternating-direction bounds in 2002 as a high-school student in Pomona, California, and published them the following year. Before that the “seven folds” claim circulated with no derivation attached and was treated as an empirical curiosity.

What she contributed was to notice that it is not a limit at all in the form it was stated — it is an equation with three variables, and fixing two of them gives the third. That is a good example of a whole class of folklore: a number everybody repeats, which turns out to be one point on a curve nobody had drawn.

Where this goes next

This closes a phase whose subject was cost — what a question costs to answer, what a construction costs to run, what a machine can and cannot reach. This rung is the same accounting applied to the material rather than to the mathematics, and it lands on the one bound in the subject that is neither a search nor a theorem: a length, a thickness, and a number of folds, related by an equation anybody can evaluate and nobody can escape.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

What links here

Every essay whose body links to this one.

The objects this essay names

Each one links to every other essay that touches it.

Combinatorial explosionConservationCrease radiusError propagationIdealisationThickness