A tuck keeps what a gore cuts

What a flat sheet can become sets out the impossibility and the three ways round it. A sheet that cannot stretch cannot become a sphere, because bending preserves Gaussian curvature and a flat sheet has none. A folder can pay for a doubly curved shape in seams, by cutting the surface into gores; in control, by putting the curvature into curved creases; or in strain, by damping the paper and letting its fibres move.

Each of the three removes a mismatch between a flat sheet and a sphere. There is a fourth way to deal with a mismatch of length, and every seamstress, every paper baking case and every gathered sleeve uses it: fold the excess under and keep it. A tuck does not remove the paper a gore would cut away. It hides it, and hidden paper is still paper — it becomes thickness, and the thickness has a computable size.

The length a cap does not have #

A flat disc of radius $s$ has a rim $2\pi s$ long. Lay that disc onto a sphere of radius $R$ so that distances along the radii are kept — the centre at the top, the rim at a geodesic distance $s$ from it — and the rim lands on a circle of the sphere at angular distance $s/R$ from the top, whose circumference is $2\pi R\sin(s/R)$.

The flat rim is longer than that circle by

$$2\pi\bigl(s - R\sin(s/R)\bigr)$$

and that is the paper a cap of angular radius $\alpha = s/R$ has no room for. The strain argument prices the same shortfall as a fraction of the rim, $1 - \sin\alpha/\alpha$, and finds dry paper absorbing about one per cent of it and a hemisphere asking for thirty-six. A gore cuts it out. A wet-folder compresses it. A tuck folds it under.

The shortfall has a shape across the disc as well as a size at the rim. Near the centre it is $\pi s^3/3R^2$, to within a hundredth of a per cent at a twentieth of the radius: a sphere’s circles fall short of a plane’s by an amount that grows as the cube of the distance. So the length a tuck has to hide is almost nothing near the middle of the disc and grows rapidly toward the rim.

A tuck is three sheets where it lies #

A simple tuck takes a length $h$ of the circumference and folds it into a flat flap lying under the surface: the paper goes in, turns back on itself, and comes out again. Where the flap lies there are three thicknesses of paper — the visible surface and the two layers of the flap — over a width of $h/2$.

That fixes the thickness a gathered rim has, and it is a conservation identity rather than a model. The paper is all still there: the rim’s paper length is $2\pi R\alpha$ and the circumference it is gathered onto is $2\pi R\sin\alpha$, so however the tucks are arranged the rim is on average

$$\frac{\alpha}{\sin\alpha}$$

sheets thick. For a hemisphere that is $\pi/2 = 1.571$, exactly. For a cap of 120°, 2.418. For a cap of 60°, 1.209.

The number of tucks does not enter. Eight tucks or twenty-four hide the same total length at the rim and leave the same average thickness; more tucks make each flap narrower and shallower without changing how much paper the rim carries. What the count changes is how the thickness is distributed — eight tall ridges or twenty-four low ones — which matters for how the gathered cap looks and handles and not for how much paper it holds.

Where tucks run into one another #

A simple tuck is three sheets thick, so a rim whose average thickness is three is a rim that is entirely under tucks. Below that average there is flat single-thickness surface between the flaps; at three, the flaps meet edge to edge; past it, they have to overlap, and somewhere the paper is five sheets thick or the flaps are folded into one another.

Setting $\alpha/\sin\alpha = 3$ and solving gives 130.6°. A cap larger than that — more than about two thirds of the way from a hemisphere to a whole sphere — cannot be gathered with simple tucks lying side by side, whatever their number, because there is not enough circumference for the flaps to lie flat against.

That is a genuinely different kind of limit from the strain limit. Strain runs out at about twenty degrees for dry paper, because paper will not compress far. Tucks run out at 130.6°, and they run out because of geometry alone: the flaps collide. Between the two lies nearly the whole of a hemisphere and more, which gathering reaches and stretching does not.

A tuck that follows the sphere is curved #

The shape of a tuck is set by where the hiding has to happen, and this is where the fold becomes interesting rather than merely practical.

Share the hidden length evenly among $n$ tucks. At flat radius $s$ each tuck hides $2\pi\bigl(s - R\sin(s/R)\bigr)/n$ of circumference, which is an angular width of that divided by $s$ on the flat disc. Near the centre that angle grows as $s^2$, so the edges of each tuck leave its centre line along curves that start tangent to it and open out rapidly toward the rim. A tuck that gathers a disc onto a sphere has curved edges.

A tuck with straight edges — a wedge with its point at the centre — hides a length proportional to s, because a fixed angle cuts off an arc proportional to its radius. That is the wrong law for a sphere and exactly the right law for a cone: a disc with straight wedges folded out of it closes into a cone whose point has all of the curvature.

That is why a pleated paper baking case is a frustum rather than a bowl: its pleats are straight, straight pleats make a cone, and a cone is what comes out. A gathered hemisphere needs every tuck to widen the way a sphere’s circles shrink.

How much each circle has to hide #

The figure below draws the two laws side by side, as the length hidden inside each circle against the circle’s radius, both scaled to what the rim hides.

The sphere’s curve starts flat — almost nothing hidden in the middle — and steepens toward the rim. The straight tuck’s line rises at a constant rate from the centre, hiding a quarter of the rim’s total by a quarter of the way out, where the sphere asks for a sixtieth. So a straight-tucked disc has far too much paper folded away near its centre and a cone’s point where the sphere wants a rounded top.

The broken lines on the same figure are straight tucks that start at more than one radius, each starting where the one inside it has stopped keeping up. They follow the curve piece by piece, and how closely they follow it is a question of how many radii they start from — the subject of a different argument, because each place a straight tuck starts turns out to be a point where the gathered sheet cones.

A tuck is a gore that is folded instead of cut #

The relation between a tuck and a gore is closer than the difference between folding and cutting suggests, and it is worth being exact about.

Between two tucks lies a piece of visible surface shaped like a gore: widest at the rim, narrowing toward the centre, with the tucks’ edges as its sides. A gore construction cuts along those edges and throws away the wedge between gores. A tucked construction folds along the same edges and keeps the wedge underneath. The visible surface is the same set of gore-shaped pieces in both, and those pieces are no more developable folded than cut.

So a tuck does not remove the strain a gore leaves. The gore arithmetic says a gore of an $n$-gore sphere is out by about $\pi^2/6n^2$ at its widest, and a tucked sphere of $n$ tucks has visible pieces of the same width carrying the same strain. What the tuck changes is the seam. A gore’s seam is a cut, which the rest of this subject forbids; a tuck’s seam is a flap, which keeps the sheet one sheet and costs thickness.

That makes the three answers four, and places the fourth precisely. Gores pay in seams, curved creases pay in control, strain pays in fibres, and tucks pay in layers — with the strain of the gore-shaped pieces between them still owed, and dropping as the square of the tuck count exactly as a gore’s does.

The same number, paid two ways #

The fraction of the rim a gathered hemisphere hides is $1 - 2/\pi$, which is 36.3 per cent. It is the same number the strain argument gives for how much a wet-folded hemisphere’s rim would have to compress: $1 - \sin\alpha/\alpha$ at $\alpha = \pi/2$.

That is not a coincidence and it is worth a sentence. The mismatch between a flat disc’s rim and a cap’s circle is one number, fixed by the sphere. A wet-folder asks the fibres to absorb it as compression, and dry paper refuses beyond about one per cent. A tucker asks the sheet to absorb it as layers, and a sheet can pile three layers with no difficulty at all. The same shortfall that is impossible as strain is easy as thickness, which is exactly why gathering is so much older and commoner than wet-folding.

The price of the easy version is the one thickness always charges. A gathered rim is 1.57 sheets thick for a hemisphere, and its flaps are three sheets thick where they lie, so a gathered paper bowl is stiffer and heavier at its rim than at its crown — twice as thick where it is thickest, in the phrase an essay on a different pattern uses, and for the same reason.

What the number of tucks does change #

The rim’s average thickness does not depend on the number of tucks, and it would be easy to conclude that the number does not matter. It matters, and what it controls is the part of the gore arithmetic a tuck does not remove.

Between two tucks lies a visible piece of gore shape, and its width at the rim is set by how many tucks share the rim. The piece is developable, the sphere under it is not, and the mismatch across the piece — the gore strain, about $\pi^2/6n^2$ at its worst — falls as the square of the number of pieces. Eight tucks leave pieces strained by about 2.6 per cent at their widest; sixteen tucks leave 0.64 per cent. Twelve, the globe maker’s number, leave about 1.1 per cent, which is roughly what dry paper absorbs without complaint.

So the tuck count trades in exactly the currency the gores did — strain in the visible pieces against the number of seams — with the seams folded rather than cut. More tucks make a rounder cap from the same thickness of rim, and what they cost is more creasing and shallower flaps that are harder to fold accurately.

The flaps’ size follows from the same arithmetic. On a hemisphere of unit radius the flat sheet’s rim is $\pi^2$ long and the sphere’s rim $2\pi$, so the tucks hide $\pi^2 - 2\pi$, about 3.59, between them. Eight tucks hide 0.45 each, and a flap folded in half lies two layers deep over half of what it hides — 0.22 of the rim apiece, or 28.5 per cent of the rim altogether. Sixteen tucks make flaps 0.11 wide that cover the same 28.5 per cent. The share of the rim under a flap is fixed by the cap; the number of tucks only decides how finely it is divided. Halving a flap’s width is harmless on a large cap and a real difficulty on a small one, where a flap a few millimetres wide has to be folded in half in paper a tenth of a millimetre thick.

What the tuck drawing cannot show #

The patterns draw tucks on the flat sheet and cannot show the gathered cap, because the gathered cap is not a surface anything here computes.

The visible gore-shaped pieces are developable and a sphere is not, so a tucked disc does not lie on a sphere; it approximates one, with each visible piece bending as a cylinder or cone and the tucks forming ridges between them. How the paper actually settles — which way each piece bends, how the flaps lie, whether the rim ripples — is a question about a real sheet under its own weight and stiffness, and the drawing shows only where the creases go.

Nor can it show whether a particular tuck can be folded. A curved tuck is three curved creases meeting at the centre, and where curved creases meet is a subject with its own conditions. The drawings put the creases where the length has to be hidden and do not check that a sheet will follow them there.

The gathering the arithmetic assumes #

Distances along the radii are kept. The disc is laid onto the cap with every radius running along a meridian at its full length, so all the mismatch is circumferential and all of it is hidden in the tucks. Other ways of laying a disc on a cap distribute the mismatch differently.

A tuck is a simple flap of three layers, with no thickness allowance at its folds and no paper lost to its crease radius. A real flap takes a little more length than it hides.

The hidden length is shared evenly among the tucks. Uneven tucks change the pattern and not the rim’s average thickness, which the conservation identity fixes.

And the sphere is the target. The arithmetic is specific to constant positive curvature; a different doubly curved shape has a different shortfall law and different tuck edges.

How the shortfall was checked #

The hemisphere’s two exact numbers are checked as exact: the hidden fraction $1 - 2/\pi$ and the average rim thickness $\pi/2$, each to a part in a trillion.

The cubic law near the centre is checked against the series it comes from — $\pi s^3/3R^2$ — at a twentieth of the radius, where they agree to a hundredth of a per cent. That is the statement that makes a sphere’s tuck curved, checked before any tuck is drawn.

The collision angle is found by bisection on the average thickness reaching three, and lands at 130.6°.

Still open: where a straight tuck puts the curvature #

The drawings show that straight tucks from the centre make a cone, and that straight tucks started at several radii follow a sphere in a broken line. What they do not say is where the broken line’s curvature goes.

A straight tuck that starts partway out has a point: three creases meeting at the place it begins, with the flap folded out of the paper around it. Three creases at a point is a vertex that cannot fold flat — nothing meets at three — and the gathered sheet around it is not flat either. Each start is a point where the curvature a sphere spreads evenly has been gathered into a cone, and how well a pattern of such points approximates a cap, as the number of starts grows, is the natural next computation.

The other direction is the curved crease. A curved tuck is three curved creases, and a tucked cap is therefore a curved-crease construction that also hides paper — a combination of two of the answers, which the first account of this subject noted nobody appears to have studied as a combination.

The habit worth carrying is a question to ask of any mismatch a material cannot absorb. Can it be hidden instead of removed? A length that is impossible to lose may be easy to fold away, and the price of folding it is then a thickness that can be computed exactly.