A start is worth an inverse square

A flat disc gathered into a spherical cap has to lose length round every circle, and straight tucks are the cheapest way to lose it: each one begins at some radius and widens steadily outward, hiding a wedge of paper. The paper reads strain placed the starts of those tucks to keep the strain the paper takes as small as possible, and crossing the curve moves no start found that letting the hiding fall below the sphere’s requirement as well as above it halves every strain without moving any start. That essay ended on the obvious gap in everything before it: a start is not free. Each one is another set of tucks to fold, and every figure so far had counted starts without pricing them.

The price turns out to have a form simple enough to state before any placement is computed. Four starts on a hemisphere leave 1.81 per cent; eight leave 0.43; sixteen leave 0.11. Each doubling of the starts divides the worst strain by just over four.

A chord misses by its width squared #

The model is the one every tuck essay has used, going back to what a flat sheet can become. On a sphere of unit radius, the circle at flat radius $s$ has to hide

$$H(s) = 2\pi\,(s - \sin s)$$

of its length, and straight tucks started at a set of radii hide length along a broken line: straight between consecutive starts, bending upward at each. The strain at radius $s$ is the gap between the broken line and $H$, divided by the circle’s length $2\pi s$.

Between two starts a distance $h$ apart, the broken line is a chord of $H$, and a chord of any smooth curve misses it by at most its width squared times the curve’s bend, over eight: $h^2 |H''| / 8$. That is a standard fact of approximating curves by straight pieces, and here $H'' = 2\pi \sin s$. Dividing by the circle’s length, the worst strain on a piece at radius $s$ is

$$\varepsilon \approx \frac{h^2 \sin s}{8s}.$$

So a piece can be as wide as $h(s) = \sqrt{8 \varepsilon s / \sin s}$ and no wider. A placement for strain makes every piece reach exactly the same worst strain — that is what makes it the best — so the number of pieces is the number of such widths that fit between the centre and the rim:

$$m \approx \int_0^\alpha \frac{ds}{h(s)} = \frac{I}{\sqrt{8\varepsilon}}, \qquad I = \int_0^\alpha \sqrt{\frac{\sin s}{s}}\, ds,$$

for a cap reaching an angle $\alpha$ round the sphere. Turned the other way, $m$ starts leave a worst strain of about $I^2 / 8m^2$, and the only property of the cap that matters is the one number $I$. For a hemisphere it is 1.464, so the law reads $0.268 / m^2$: 1.68 per cent at four starts against the placement’s 1.81, and 0.105 per cent at sixteen against 0.107.

The first figure draws the law against the placements on three caps, and the dial moves the third. On every cap the dots sit just above the line and close on it as the starts multiply. Every doubling divides the worst strain by four, whatever the cap.

Why equal error is the best a placement can do #

The counting step assumed that a placement for strain makes every piece reach the same worst strain, and that assumption is worth a sentence of its own, because the whole law hangs on it.

Suppose some placement left one piece below the worst strain the others reach. That piece has slack: its two ends could move apart until it reaches the common level, which shortens a neighbouring piece and lowers the neighbour’s error. Pass the slack along and every piece’s error can be pulled down a little, so the original placement was not the best. The only placement nothing can improve is one in which every piece sits at exactly the same worst strain, with no slack anywhere — the same equal-error signature crossing the curve moves no start found in the crossing line’s alternating peaks. The earlier essays’ placements were found by search, and the search is built to find exactly that: the smallest strain at which the pieces, each run as far as it can go, number no more than the starts allowed.

Equal error turns the placement problem into a counting problem. If every piece reaches $\varepsilon$, the pieces are as wide as $\varepsilon$ lets them be, and how many fit is fixed by the cap. The best placement and the fewest starts are the same object, read from two sides: fix the starts and ask for the strain, or fix the strain and ask for the starts.

Why the starts spread outward #

The same argument says where the starts go, which the paper reads strain found by placing them and could only describe: they spread toward the rim instead of crowding toward it, against every earlier rule.

The width a piece may have is $\sqrt{8\varepsilon s / \sin s}$, and $s / \sin s$ grows from one at the centre to $\pi/2$ at the rim of a hemisphere, so the pieces widen by a quarter from the centre to the rim. The reason is the one the strain essay gave in words. The hiding curve bends hardest at the rim, which by itself would crowd the starts there, but strain divides the gap by the circle’s length, and the rim’s circle is the longest. The two effects pull in opposite directions and the circle wins, by exactly the factor in the square root.

On sixteen starts the placement’s pieces follow the formula to within 0.2 per cent everywhere past the first two. The first piece is different, and it is different in a way the formula predicts. The tucks that begin at the centre make a cone, and a cone’s strain is worst at its apex, where it equals the excess $1 - \sin b / b \approx b^2/6$ for a cone reaching radius $b$, not the chord’s $b^2/8$. So the first piece can only be $\sqrt{6\varepsilon}$ wide where the formula would allow $\sqrt{8\varepsilon}$, and on sixteen starts it is exactly that, the dot below the curve at the left of the figure.

The correction belongs to the centre #

The law is an approximation for many starts, and the size of its error matters for anyone counting a handful. Divided by $I^2/8m^2$, the placements’ worst strains are 1.164, 1.078, 1.038, 1.019 and 1.009 at two, four, eight, sixteen and thirty-two starts on a hemisphere: high by about three tenths divided by the number of starts.

The striking thing is that the five caps share that error. At sixteen starts the ratio is 1.0188 on caps of 30, 60 and 90 degrees and 1.0190 and 1.0194 on 120 and 150. The correction does not depend on the cap because it does not come from the cap. It comes from the centre, where every cap looks the same, and most of it is the narrow first piece: losing $1 - \sqrt{3}/2 = 0.134$ of a piece’s width at the centre is losing 0.134 of a start, which multiplies the law by $1/(1 - 0.134/m)^2$ — the dashed curve, which carries most of the gap and leaves a little more to the second piece, whose own curvature is not yet small enough for the chord formula. The practical form is the honest one: $m$ starts leave about $I^2/8(m - 0.134)^2$, close enough to count by.

One start, and where the law gives out #

The law is a statement about many starts, and it is worth seeing what it says about one. A single start at the centre makes one cone of tucks running to the rim, and its worst strain is the excess at the rim, $1 - \sin\alpha / \alpha$ — on a hemisphere $1 - 2/\pi$, 36.3 per cent, the fraction of its rim a gathered hemisphere has to hide. The law says 26.8. The single cone is the case the correction was built for, a first piece that is the whole of the placement, and the gap is the cone’s $b^2/6$ against the chord’s $b^2/8$ stretched across the whole cap.

At two starts the gap is sixteen per cent and at four it is eight; by sixteen it is two. That is the useful range to know about. For a cap folded with three or four rings of tucks, which is what most gathered forms use, the law is a slight underestimate and the corrected form $I^2/8(m - 0.134)^2$ is the one to count with. For a pattern with dozens of rings — a crumpled or finely pleated dome, say — the bare law is as good as the placement.

How many starts a paper needs #

The question a maker actually asks runs the other way. The paper gives some amount and no more; how few starts will do?

The law answers it directly: about $I/\sqrt{8\varepsilon}$ starts, rounded up. Counted exactly, the fewest starts never fall below that figure and never exceed it by more than a start and a half, anywhere from a tenth of a per cent to ten. A tenth of the give costs the square root of ten as many starts, a little over three times as many, and a quarter of the give costs twice as many.

The crossing line fits into the same law with no new work. Allowing the hiding to fall below the curve as well as above it halves the worst strain at the same starts, so a paper giving $\varepsilon$ either way needs the starts a stretch-only paper giving $2\varepsilon$ would need: $I/\sqrt{16\varepsilon}$. Letting the line cross is worth exactly a factor of $\sqrt{2}$ in starts. On a hemisphere, a paper that stretches and compresses two per cent needs four starts with its hiding touching the curve and three with it crossing.

That is the number crossing the curve moves no start said a maker would care about and could not give in general, and three answers, one count asked the companion question of whether the divisions a rim needs fit round it at all. The count here is the other side of that ledger: not whether a given number of starts fits, but how many a given paper demands.

What a start is worth at the margin #

The law also prices a single start, which is the trade the earlier essay named: fewer starts with more strain, or more starts with less, at whatever a start is worth to the person folding it. On a hemisphere, the placements give 7.80, 3.29, 1.81 and 1.14 per cent at two to five starts, and 0.57, 0.43 and 0.34 at seven to nine.

So the third start buys 4.5 points of strain, the fourth 1.5 and the fifth 0.7; the eighth buys 0.14 and the ninth 0.09. The law’s derivative says the same in one line: a start’s worth falls as the cube of the count, $I^2/4m^3$. Whatever a start costs, there is a count past which the next one is not worth folding, and because the worth falls as a cube, the count at which a given price stops paying moves only as the cube root of the price. A start ten times dearer changes the right number of starts by a factor of about two.

One integral per cap #

Because the whole cap enters through $I$, caps of different depth can be compared at a glance. A shallow cap of 15 degrees has $I = 0.261$ and needs one start at two per cent; a hemisphere, 1.464 and four; a cap of 150 degrees, 2.13 and six. The integral grows almost in proportion to the cap’s angle while the cap is shallow, because near the centre $\sin s / s$ is nearly one, and then it flattens: from 150 degrees to a whole sphere it gains only seven per cent, reaching 2.278 at the far pole.

The flattening has a plain reading in the paper. Near the far pole the hiding curve is nearly straight — almost all of each circle has to be hidden, and the fraction changes slowly — so a chord fits it with almost no error, and the last part of a deep cap needs almost no starts at all. The paper’s difficulty in becoming a sphere is concentrated round the equator, where the circles are longest and the curve bends hardest at once.

The rule underneath, and where else it holds #

The rule that the pieces should be as wide as $\sqrt{8\varepsilon / |{\rm weight}|}$, with the weight what the error is divided by, is the equidistribution rule of approximation by splines: place the knots of a piecewise-straight approximation so that every piece makes the same error, and their density goes as the square root of the curvature. It is what an adaptive mesh does, and what a polygon approximating a circle to a tolerance does — the number of sides it needs grows as one over the square root of the tolerance, for exactly the reason the starts here do.

The connection worth drawing within this subject is to a curve has no panels, which approximated a curved crease by straight segments and found their kinks summing to a constant of the curve. That is the same broken line approximating a different curve, and the same law would say how many segments a curved crease needs to stay within a given departure from its smooth shape. Tucks and curved creases are both cases of making a curve out of straight pieces, and in both the price of accuracy is an inverse square.

What the law does not price #

What a start costs. The law prices a start in strain bought; it cannot say what a start costs in folding time, in thickness at the tuck, or in the error that comes from folding one more wedge by hand. The trade is stated so that whatever that cost is, it can be set against the curve.

How the strain is shared. The strain here is the gap round a circle divided by its length, spread evenly round it. Real paper takes it unevenly — more at the tucks, less between them — and puckers rather than compresses past some point, which crossing the curve moves no start named as the open half of its own result.

Whether the tucks fit. A count of starts is a count of rings, and each ring holds some number of tucks that must fit round its circle; that is the question three answers, one count asked, and a placement this law approves can still fail it.

A perfect cap and straight tucks #

The sheet gathers into an exact spherical cap, every circle of the flat disc landing on a circle of the sphere, and the only freedom is how much length each circle hides. Tucks are straight, widening at a constant rate from their start, which is what makes the hiding a broken line; curved tucks could follow the curve exactly and are a different construction. The paper has a single give, the same everywhere and in every direction, and is measured as strain round circles; grain, as the fifth thing that is not true put it, would make it depend on direction.

The placement counts, the law predicts #

The count of starts is computed by the placement itself, not by the law: the fewest starts whose pieces all stay within the strain, found by running each piece as far as it can go. The solver used here is checked against the placement the earlier essays drew, at two, four and eight starts, to a part in a million, before any figure is drawn.

Every placement must lie above the law and close on it, its ratio to $I^2/8m^2$ falling at every doubling and below 1.02 by thirty-two starts. The caps must share the correction, their ratios at sixteen starts within a hundredth of one another. The pieces past the first two must follow $\sqrt{8\varepsilon s/\sin s}$ to within six per cent, and the first must be $\sqrt{6\varepsilon}$ wide to within eight; the measured agreement is 0.2 per cent and better. The counted starts must never fall below $I/\sqrt{8\varepsilon}$, nor exceed it by more than a start and a half, at thirty-three values of the give.

Still open: a price for a start #

The law turns the trade the earlier essay named into arithmetic, and what it still needs is the other side: a measured cost of a start, in minutes, in thickness or in error, so that the count at which the next start stops paying can be named for a real paper and a real hand. The cube-root dependence says that cost need not be known well; a factor of ten in it moves the answer by two.

The second lead is the tucks themselves. Every essay on gathered caps has made them straight, and the law says what straightness costs: the broken line’s chord error. A tuck that bends to follow the curve would remove that error entirely and replace it with whatever bending a tuck costs the paper, and whether a few curved tucks beat many straight ones is exactly the comparison an inverse square makes easy, because it says how many straight ones the curved ones would have to beat.

Sideways from here, the same law prices where a ring of divisions belongs, whose ring rule placed divisions by excess rather than by strain; its rings, counted against give, should follow a different power, and the difference would say how much of that rule’s over-division is the price of ignoring the circle’s length.

The habit worth carrying is about counting before placing. When a placement makes every piece reach the same error, the number of pieces is an integral, and the integral can be read before anything is placed. Every earlier essay here found the starts by search; the search was right each time, and the number it was finding had a formula.