The rim sets the opening

The error budget for a nautilus priced seven ways a growth factor read off a sawn shell can be wrong, and the largest was a pair of dividers walked along the spiral, which could read a nautilus 278 per cent too high. The dividers belong to the opening found what decides that entry, extending a measurement in steps. A person sets a pair of dividers to an opening and walks until the curve runs out, and a walk reads the factor exactly at every span when the opening is under $\sqrt{k} - 1$ of the innermost radius — 0.789 for a nautilus growing 3.2 times a turn — and too high at every span when it is over.

That criterion is stated in units of the innermost radius, and the essay ended on the difficulty with it. A shell ten centimetres in radius, read over five whorls, has an innermost radius of a third of a millimetre, and a safe opening of a quarter of one. No hand walks dividers that fine. The first whorls of a real section are often broken, or filled with calcite, and in practice a walk starts at the innermost whorl the section preserves.

This essay carries the criterion to that walk, and asks the practical question: is a pair of dividers set to a width a hand can read on a real section inside the criterion for the whorl it starts on?

The criterion does not change; its unit does #

A logarithmic spiral is the same shape at every size. The part of it outside any radius is the whole spiral scaled up, so a walk started at a preserved radius $r$ is exactly the walk from the true innermost radius, magnified by $r$. Whatever held there holds here in units of $r$: the walk is exact when the opening is under $(\sqrt{k} - 1)\,r$.

What does change is what a person holds fixed. The dividers are set against the section in front of them — a millimetre, two millimetres, a width the eye can check against a rule — and that width is a share $f$ of the section’s outer radius, not of any whorl inside it. The walk runs from the innermost preserved whorl out to the rim, over $T$ turns, and the whorl it starts on has a radius $k^T$ times smaller than the rim’s. In units of that whorl the opening is therefore $f\,k^T$, and the walk is exact while

$$f\,k^T < \sqrt{k} - 1, \quad\text{that is, while}\quad T < T_{\max}(f) = \frac{\log\big((\sqrt{k} - 1)/f\big)}{\log k}.$$

So the criterion becomes a span. For a nautilus, dividers opened to a fiftieth of the outer radius are exact over any walk shorter than 3.16 turns; to a hundredth, 3.76; to a five-hundredth, 5.14. A tenth of the outer radius is exact over walks shorter than 1.78 turns.

The figure’s section preserves its outer three turns. The dividers, at a fiftieth of the outer radius, are 0.66 of the radius of the whorl they start on, under the 0.789 threshold, and the walk of 48 steps reads the factor exactly. Turn the dial to more preserved turns and the walk starts on a smaller whorl: past 3.16 turns its first step is wider than the threshold allows, and it begins to read the factor too high.

The change-over is where the arithmetic puts it #

The closed form rests on nothing but self-similarity, and it is checked by walking.

For six openings from a five-hundredth of the outer radius to a tenth, every walk on the grid of spans from half a turn to six was taken and read. At a five-hundredth the last span read exactly is five turns and the first read too high is six, with the closed form at 5.14; at a two-hundredth, four and five around 4.35; at a hundredth, three and a half and four around 3.76; at a fiftieth, three and three and a half around 3.16; at a twentieth, two and two and a half around 2.37. Every change-over brackets the line.

The tenth of the outer radius is the exception, and an instructive one. Its closed form says 1.78 turns, but under two turns it takes fewer than the nine steps the fit accepts — seven over one turn, eight over one and a half — and at two turns it is already past the line. It has no span it reads exactly at all. A coarse pair of dividers fails in two ways at once: too few steps on a short walk, too wide a step on a long one.

The coarsest pair that can be exact #

The tenth’s failure has a closed form too. A walk needs nine steps, and at an opening of $f$ of the outer radius a walk of $T$ turns takes $(k^T - 1)/(f\,k^T)$ of them, so the shortest span it can read is $T_{\min}(f) = -\log(1 - 9f)/\log k$. That is a twelfth of a turn at a hundredth of the outer radius and half a turn at a twentieth — well below the three quarters of a turn where the budget begins to refuse, and so never the binding limit for a pair a person would choose.

But $T_{\min}$ rises with the opening and $T_{\max}$ falls, and they meet. Setting them equal gives $f^* = c/(9c + 1)$ with $c = \sqrt{k} - 1$, which for a nautilus is 0.0974 of the outer radius. A pair opened wider than that has no span it can read exactly: every walk short enough to start on a safe whorl is too short to take nine steps. A tenth of the outer radius sits just past the edge, which is why it needed two turns for its nine steps and was exact only under 1.78.

Held against the rim, the error grows with the span #

The earlier essay’s most useful finding was that, over the threshold, a wide opening hurts a short span most: the damage is done in the first step, and a long fit dilutes it. That was with the opening held fixed in units of the innermost radius. Held against the rim, the direction reverses.

At a fiftieth of the outer radius the dividers read the factor 8.0 per cent too high over three and a half turns, 8.8 over four, 26.3 over five and 32.9 over six. At a twentieth, 16 per cent over two and a half turns rising to 127 per cent over six. At a two-hundredth the error appears only over five turns, at 2.8 per cent, and reaches 8.4 over six. Every opening’s error grows with the span once it starts.

The two curves are the same geometry read with different things held fixed. With the opening fixed against the first whorl walked, a longer span adds steps beyond the damaged first one and the error falls — 28.7 per cent over two turns, 0.6 over six. With the opening fixed against the rim, a longer span means the walk starts on a smaller whorl, so the same millimetres are a wider opening where the damage is done, and the error rises.

A broken centre helps the dividers #

Put the other way, the same pair of dividers on the same shell is safer the less of the shell’s centre survives.

With all six turns preserved, dividers at a fiftieth of the outer radius start on a whorl whose radius is a twenty-first of their opening, and their first steps skip across whorls; the fit loses turns and reads the factor 32.9 per cent too high. With only the outer three turns preserved, the same walk is exact. A section whose inner whorls are broken away has been, for this purpose, trimmed to the span its dividers can read.

The practical rule follows directly. A person with a pair of dividers set to $f$ of the outer radius should start the walk no deeper than $T_{\max}(f)$ turns from the rim, and nothing is lost by starting shallower, provided enough turns remain for the rest of the budget.

How many steps a walk takes #

The number of steps turns out to be set by the rim as well. The arc from the start to the rim is proportional to the difference of their radii, and nearly all of it lies in the outermost turn, so a walk at an opening of $f$ of the outer radius takes close to $1/f$ steps whatever its span. At a hundredth it takes 69 steps over one turn, 90 over two, 98 over three and a half and 100 over six.

That is a hundred placements of the dividers, the labour of an afternoon with a sawn shell and a steady hand, and it is the same hundred whether the section preserves one turn or six. The threshold, not the labour, is what decides where to start.

The budget, with dividers a hand can set #

The last question is the one the budget was built to answer: does it refuse the golden spiral, now with the dividers priced as a person would set them?

For dividers opened to a five-hundredth, a two-hundredth, a hundredth or a fiftieth of the outer radius, the budget refuses the golden spiral at every span from three quarters of a turn to six. Past each opening’s exact span the dividers add their error — 17.5 per cent over six turns at a hundredth, 32.9 at a fiftieth — but the gap between 3.2 a turn and the golden spiral’s 6.85 is 114 per cent, and nothing a fiftieth of the rim does over six turns comes near it.

Coarser dividers lose it. At a twentieth of the outer radius the budget stops refusing over walks of five and six turns, where the dividers alone read 92 and 127 per cent too high. At a tenth it stops refusing over four turns and more, and cannot refuse under two turns at all, because those walks take fewer than nine steps. Between two and three and a half turns even a tenth of the rim leaves the budget refusing, with the dividers reading up to 38 per cent too high.

Where to start the walk #

The budget has a best place to be read, and the two curves above put it there. The joint entry for the centre and the tilt falls as the span grows — three entries and one span found it at 188 per cent over half a turn and 7 over two turns — while the dividers’ entry is nothing until the span passes $T_{\max}(f)$ and grows after it. The sum is smallest where the first has fallen as far as it can before the second begins.

Measured on the grid of spans, that is exactly what happens. For dividers opened to a five-hundredth of the outer radius the budget is smallest over five turns, at 23.2 per cent; at a two-hundredth, over four turns, 23.8; at a hundredth, over three and a half, 24.2; at a fiftieth, over three, 25.1; at a twentieth, over two, 29.3. Each is the longest span on the grid still read exactly — the last one under the closed form. So the rule for where to start a walk is the same rule as for where it is safe to: as deep as $T_{\max}(f)$ turns from the rim and no deeper.

At that best start the budget is a fifth to a quarter of the gap between the nautilus and the golden spiral. A section read that way does not merely refuse the golden spiral; it refuses it with room for errors this budget has not priced.

What a hand can do #

The earlier essay’s example was a shell ten centimetres in radius. On it a fiftieth of the outer radius is two millimetres and a hundredth is one, both openings a person can set and check against a rule. At one millimetre the walk is exact over the outer 3.76 turns, and the budget refuses the golden spiral from three quarters of a turn outward whatever the section preserves. So the answer to the question the earlier essay left is yes: dividers set to a width a hand can read are inside the criterion for any whorl a person would sensibly start on, provided the walk starts within about three and a half turns of the rim — and a section that preserves more than that can simply be walked from further out.

Worked through on that shell: dividers at one millimetre start safely on any whorl at least 1.26 millimetres in radius, which is 3.76 turns in from the rim. The best start on the grid, three and a half turns in, is the whorl of radius 1.71 millimetres; from there the walk takes 98 steps and reads the factor exactly, and the whole budget comes to 24 per cent. A section whose centre is broken away to a radius of two or three millimetres loses nothing at all by it: the walk would have started about there anyway.

That is also why the historical method was not the problem, as the essay before this one put it. The 278 per cent in the original budget was nine steps along five turns, dividers opened to 37 innermost radii, which is to say about a ninth of the outer radius — nearly as coarse as the tenth priced here, and walked over five turns, where a tenth reads the factor twice too high. With dividers set as a careful person would set them, the method reads a nautilus exactly over the part of the shell a careful person would read.

What the pricing assumes #

The centre and the tilt are priced, as in three entries and one span, in units of the innermost radius of the arc read. A section missing its inner whorls is therefore taken to have its centre found to a quarter of its innermost preserved radius, which is an assumption about how a centre is located on a broken section, not a measurement of it; how far a centre must move is what that error costs when it is larger. If a broken section’s centre is found worse than that, the joint entry grows at the short spans, where it already decides the three-quarter-turn limit.

The spiral is exact, with a fixed factor of 3.2, and every reading is of the curve rather than of the septa that divide it. The nautilus question and the budget’s ontogeny entry allow the factor to change with age, and a walk whose preserved turns cover a change in growth rate is reading two spirals, which no opening fixes.

And nothing here measures how many turns a real section preserves, or how wide a real pair of dividers is set. Those are the two numbers that place any actual measurement on the figures above, and both belong in the record of it.

Findings that would overturn it #

An opening at which a span shorter than $\log((\sqrt{k} - 1)/f)/\log k$ is read too high, or a longer span with nine steps or more is read exactly. An opening of a fiftieth of the outer radius or finer at which the whole budget fails to refuse the golden spiral between three quarters of a turn and six. Either would mean the self-similarity argument, or the walk that tests it, is wrong.

Still open: a walk that is not along the curve #

Every walk here follows the spiral’s own curve, which is what dividers along a sawn edge do. The other common way to read a section is radial: a rule laid from the centre outward, and the radius read where it crosses each whorl. It has no opening to set and no first step to damage, but it has a centre to find, which dividers do not need, and its readings come one per turn rather than a hundred per walk. The measurement is the radial reading priced the same way, over the same preserved spans, against the same budget — and whether a person with a section and a choice of instrument should walk the curve or read across it, and from which whorl.