Series

Developability — the series

10 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. cylinderreachablecurved one way onlyconereachablecurved one way, from a pointsphereunreachablecurved two ways — impossiblesaddleunreachablecurved two ways — impossible

    What a flat sheet can become

    A sheet that cannot stretch cannot become a sphere. That much belongs to differential geometry; what belongs to folding is the three ways round it — seams, curved creases, and a few percent of stretch — and what each one costs.

    part 1 · material
  2. curved tucksthe disc is the flat sheet; the dark lines fold each tuck under, the light line folds it in half8 curved tucks, a cap of 90°hidden at the rim: 36.3%rim thickness on average: 1.571 sheetsthe tuck widens as the cube of the radius

    A tuck keeps what a gore cuts

    A flat disc gathered into a spherical cap has more circumference than the cap, and a gore removes the excess while wet-folding stretches it away. A tuck folds it under, which keeps the sheet whole and turns the excess into thickness. At the rim of a gathered cap the paper is α ⁄ sin α sheets thick on average — π⁄2 for a hemisphere — and a simple tuck is three, so single tucks reach a cap of 130.6° before they run into one another. And because a sphere's circles fall short of a plane's as the cube of the radius, a tuck that follows the sphere widens as the cube too: its edges are curves.

    part 2 · material
  3. the bar is the worst gap between the hiding straight tucks do and the hiding a sphere needsa cap of 90°, as a share of what the rim hides — every tuck straight, started at evenly spaced radiifrom 1 radius36.9%straight from the centrefrom 2 radii12.3%2.99 times smaller than 1from 4 radii3.3%3.73 times smaller than 2from 8 radii0.8%3.93 times smaller than 4from 16 radii0.2%3.98 times smaller than 8a broken line through a smooth curve is out by the curvature times the square of the spacing

    A straight tuck is a cone point

    A tuck with straight edges hides length in proportion to how far past its start it has gone, which is a cone's law and not a sphere's. Started at the centre, straight tucks make a cone. Started at several radii, they hide length in a broken line that follows a sphere's cubic, and the worst shortfall falls as the square of the number of starting radii: 36.9 per cent of the rim's hiding from one start, 12.3 from two, 3.3 from four, 0.8 from eight. Every start is three creases at a point, which is a vertex that cannot fold flat — and it is exactly where the gathered sheet's curvature goes.

    part 3 · material
  4. the bar is the worst shortfall in hidden length, as a share of what the rim hidesa cap of 90°, straight tucks started at evenly spaced radii and at the best radii for the same count2 starts, evenly spaced12.3%2 starts, best spaced8.5%31% smaller3 starts, evenly spaced5.8%3 starts, best spaced3.7%36% smaller4 starts, evenly spaced3.3%4 starts, best spaced2.0%38% smaller8 starts, evenly spaced0.8%8 starts, best spaced0.5%40% smaller16 starts, evenly spaced0.2%16 starts, best spaced0.1%41% smallerthe best radii give every stretch between starts the same worst error, which crowds them toward the rim

    Crowd the tucks toward the rim

    Straight tucks started at several radii follow a sphere's hidden length in a broken line, and evenly spaced starts leave a worst shortfall that falls as the square of their number. Evenly spaced is not the best spacing. A sphere's hiding bends hardest near the rim, so the best starts crowd outward — on a hemisphere, four of them at 0.36, 0.60 and 0.80 of the radius — and leave 38 per cent less error than four evenly spaced. As the count grows the saving closes on 42 per cent, a limit set by the square root of how the sphere's hiding bends; on a shallow dish it approaches five ninths. To follow a hemisphere within one per cent takes six rings of tucks instead of eight, and within a tenth of a per cent eighteen instead of twenty-four.

    part 4 · material
  5. accuracy against thicknessthe pile at a start is the gathering's own mean layers there plus the two a tuck addsplacementstarts atdeepest pilemean pileevenly spaced0.20, 0.40, 0.60, 0.803.323.14placed for equal error0.31, 0.51, 0.68, 0.843.373.19a start is three sheets where it sits, over a gathering already 1.57 sheets thick at the rim

    Crowding outward costs almost nothing

    Placing the tuck starts for equal error crowds them toward the rim, where the gathered paper is already at its thickest, and the obvious worry is that the accuracy is bought with depth. Measured, it is not: on a hemisphere the crowded placement's deepest start sits in 3.37 sheets against the even placement's 3.32, because a start is three sheets of its own and the gathering beneath it is only one and a half.

    part 5 · material
  6. the same arithmetic three waysm divisions leave a residual of f ⁄ m inside each piece, whatever the divisions are made ofthe material givesdivisions neededas goresas tucksas curved creases2.0%1919 cuts, 59.7 of seam19 tucks, 3 sheets deep19 creases, no cut and no pile5.0%88 cuts, 25.1 of seam8 tucks, 3 sheets deep8 creases, no cut and no pile10.0%44 cuts, 12.6 of seam4 tucks, 3 sheets deep4 creases, no cut and no pile20.0%22 cuts, 6.28 of seam2 tucks, 3 sheets deep2 creases, no cut and no pilecap of 90°, rim excess 36.3% · the count is ⌈f ⁄ ε⌉ in every column; only the cost of a division changes

    Three answers, one count

    Seams, curved creases and a few per cent of stretch are the three ways round the sphere, and a tuck is a fourth. All four dispose of one quantity — the excess circumference a flat disc has over the sphere's circle — and all four dispose of it by dividing the circle. So the number of divisions needed is the same whichever answer is chosen: nineteen for a hemisphere in a material that gives two per cent, eight at five, four at ten. What differs is what a division costs, and one of the four runs out.

    part 6 · material
  7. 00.20.40.60.811.21.41.600.10.20.30.4arc from the pole (radians)excess, as a share of the circle7 rings5.0% stretchfirst at 0.35last at 0.98of the way to the rima ring goes in wherever the residual excess would otherwise pass what the material takes

    Where a ring of divisions belongs

    A pattern that divides the circle everywhere as finely as its rim requires is over-divided for most of its radius, because the excess grows from nothing. Putting a ring of new divisions in wherever the residual would otherwise pass what the material takes gives seven rings on a hemisphere at five per cent of stretch, at 0.35, 0.50, 0.62, 0.72, 0.81, 0.90 and 0.98 of the way out — and the first of those sits where a completely different criterion put its first tuck start.

    part 7 · material
  8. 00.20.40.60.810%2%4%6%radius on the flat sheet, as a share of the rim'sstrain left in the paperplaced for length: 5.34%evenly spaced: 2.55%placed for strain: 1.81%a cap of 90°, 4 starting radii · strain is the shortfall at a circle over the circle's length

    The paper reads strain

    Two placements of a gathered cap's tuck starts agreed on their first radius and nobody knew why. Both were proxies. What the paper takes is the shortfall at a circle divided by the circle's length, and judged by that the first start must sit exactly where the excess reaches the material's give — and past it the starts spread outward, not in. Four radii placed for length leave a hemisphere 5.34 per cent strain; the same four placed for strain leave 1.81.

    part 8 · material
  9. 00.20.40.60.81-2.0%2.0%radius on the flat sheet, as a share of the rim'ssigned straintouching, all stretchallowed to crossa cap of 90°, 4 starts where the strain placement puts them · above the zero line the paper stretches, below it compresses

    Crossing the curve moves no start

    The broken line of hiding that straight tucks make on a gathered cap has always been drawn touching the sphere's curve at every start, so the paper between starts can only stretch. Let it cross, and the best line turns out to be the old one lowered by a cone: the starts stay exactly where they were at every count, the worst strain halves exactly — 1.81 per cent to 0.90 on a hemisphere at four starts — and a paper that stretches one amount and compresses another places its tucks exactly as a stretch-only paper giving their sum would.

    part 9 · material
  10. the worst strain straight tucks leave, against the number of starts placed for straindots: the placement's own worst strain; dashed: I² ⁄ 8m², with I = ∫ √(sin s ⁄ s) ds over the cap-5-4-3-2-1starts, on a doubling scalelog₁₀ of the worst strain2481632the dialled cap: 90°I = 1.464small dots: caps of 30° and 150°every doubling of the starts divides the worst strain by four, on every cap

    A start is worth an inverse square

    Straight tucks gathering a flat disc into a cap leave a worst strain that falls as the starts multiply, and every essay on them counted starts without saying what one buys. It buys an inverse square. A chord of the hiding curve misses it by its width squared times the curve's bend over eight, so starts placed for strain leave a worst strain of I² ⁄ 8m², with I one integral of the cap's shape — 1.464 for a hemisphere. A paper that gives ε therefore needs about I ⁄ √(8ε) starts, a tenth of the give costs three times the starts, and letting the hiding cross the curve saves exactly a factor of √2. The same argument says why the starts spread outward, and by how much.

    part 10 · material

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