The outline finds its own centre

Raup’s three numbers place nearly every coiled shell in a space of three: how fast the whorl expands, W; how far the generating curve sits from the axis, D; and how far it travels along the axis, T. The line $W \cdot D = 1$, where successive whorls just touch, is the most useful line in the space, and it is a relation between two of the numbers. Two of them now have a price. The error budget for a nautilus priced W, source by source, and what the axis distance costs priced D: an assumed centre a quarter of the innermost radius off costs D 79.3 per cent at the worst single azimuth, where it costs W 6.88, because a ratio of two distances read from one point is first order in that point’s error and a fitted rate is second.

Both prices were paid on the comfortable case: one number read from a section whose other numbers were known, from a centre simply assumed. The essay ended by asking for the case a specimen presents. A worker with a sawn shell has an outline and nothing else, and has to produce the centre, W and D together from it, with no check on any of them except that together they must redraw the outline. That is a round trip, and it has a failure a per-number budget cannot see: three numbers fitted jointly can trade against each other, so that a fit redraws the section beautifully from numbers that are individually well out.

The section, and the recovery #

The section is a planispiral shell’s median cut: two logarithmic spirals of one growth about one centre, the outer wall at $r = A \, W^{\theta/2\pi}$ and the inner wall at $D$ times it. It is drawn over a stated span of arc, the innermost wall point at radius one, both walls sampled at ninety points a turn, and every drawn point moved by independent noise of a stated share of the rim radius — the unit a person measuring the section holds, as the rim sets the opening argued for a pair of dividers. Two shells are read: the one the shells figures are drawn at, $W = 2.4$ and $D = 0.42$, and a nautilus-like involute one, $W = 3.2$ and $D = 0.3$.

The recovery is told the points and which wall each lies on, in order along it, and nothing else. For an assumed centre the problem is linear: unwrap each wall’s azimuths about the centre and fit $\ln r = \alpha + b\theta + \delta \cdot [\text{inner}]$ by least squares, so that $W = e^{2\pi b}$ and $D = e^{\delta}$. The centre is whatever minimises the residual. The search starts where a worker would put the centre by eye, a quarter of the innermost radius from the truth in one of twelve directions, and is confined to a disc of one innermost radius about that guess.

The confinement is not a convenience, and the reason is instructive. The residual is measured in $\ln r$, and a centre far outside the section sees every drawn point at nearly one distance and one bearing, so the residual falls towards nothing as the centre runs away. An unconfined first version of this search walked the centre off to millions of radii and reported a perfect redraw, with W and D wherever the walk had left them. It is the extreme case of the failure the round trip was set to look for, produced by the search rather than by the shell.

With no noise the recovery is exact. From the guess a quarter of a radius off, it returns the centre, W and D of both shells over half a turn, one and two to a hundred-millionth.

The drawing above is one worked case. Its points carry noise of three thousandths of the rim over one and a half turns, and the recovery finds a centre 0.0076 of the innermost radius from the truth and reads W = 2.4016 and D = 0.4199 — out by 0.07 and 0.02 per cent. Held at the guess, without letting the centre move, the same points read W = 2.550 and D = 0.4117: out by 6.3 and 2.0 per cent. The walls redrawn from the joint reading lie on the drawn points; the fit’s residual is 0.98 of the noise, which is what a model that is right about the shell should leave.

The error is the drawing’s, and scales with it #

A hundred sections a span give each error as a spread rather than a case. Over a turn of the shell the figures are drawn at, noise of a thousandth, three thousandths and a hundredth of the rim gives W to 0.116, 0.351 and 1.22 per cent and D to 0.045, 0.134 and 0.449, with the centre found to 0.0014, 0.0043 and 0.0147 of the innermost radius. Every column scales with the noise, a factor of three for a factor of three and ten for ten, and the fit’s residual is one throughout. That is what a recovery limited by its data and by nothing else looks like: there is no floor set by the method, no trade that grows faster than the noise, and a worker who draws a section twice as carefully reads both numbers twice as well.

What freeing the centre is worth #

Over one turn of section drawn to a thousandth of the rim, the joint fit reads W to 0.12 per cent and D to 0.045. Read at the guessed centre without letting it move, the same drawings give W to 16.8 per cent and D to 0.98. Over half a turn the difference is 0.36 against 56.1 for W; over three turns it narrows to 0.06 against 1.4, because a long span pins W however the centre is placed.

The reason is that the drawing locates its own centre. Over a turn the fitted centre lands within 0.0014 of the innermost radius of the truth, against the guess’s 0.25: about a hundred and eighty times closer than the budget’s gross error assumed a worker would get. The per-number budget priced a centre that nobody refitted. A worker who fits the centre with the numbers pays almost none of that price.

The guessed reading of D is itself already a repair. At a single azimuth the guessed centre costs D 79 per cent, as the earlier essay found; averaged over a whole turn of azimuths, as the fit does, the first-order error cancels and 0.98 per cent is left. Freeing the centre takes that to 0.045.

Each number’s best span #

W improves with every turn shown, from 0.36 per cent over half a turn to 0.06 over one and a half, and then holds: a rate needs arc to be read over, and by one and a half turns there is enough. D behaves differently. It is read best over three quarters of a turn to one and a half — 0.047, 0.045 and 0.047 per cent — and worse beyond, 0.093 over three turns. The noise is fixed against the rim, and the longer the span the smaller the innermost wall is against the rim; D is a ratio of the two walls at every azimuth, and the inner wall’s innermost stretch carries a share of the noise out of proportion to its size.

So the two numbers want different sections. A worker after W should draw as much shell as the specimen shows; a worker after D should draw a turn or a turn and a half and stop, or weight the innermost stretch by its own size. The nautilus-like shell keeps the same shapes with larger errors, W 0.17 and D 0.071 per cent over a turn, since its faster expansion puts the rim further out for the same arc.

How the errors trade #

The trade the round trip was set to watch for is there, and it is between W and the centre, not between W and D. Over half a turn the recovered centre’s misplacement explains 85 per cent of the variance in W’s error; over a turn, 62. The centre is found from the drawing to a thousandth of a radius, but not to nothing, and what is left of its error goes into W at first order. D’s error rides the centre over half a turn — 49 per cent of its variance — and not over a turn, where the centre explains under 3 per cent of it: a whole turn of azimuths cancels D’s first-order term, as it did for the guessed centre.

And W and D do not trade against each other: their errors across the hundred sections are correlated by −0.02. So the round trip has one compensating direction, and it is the one the budget already knew was dangerous, the centre; the joint fit makes that direction short rather than removing it.

An oblique view #

The fit so far has always had the right model. A section photographed a few degrees off its own plane has not: a tilted view squashes the plane across a bearing, and a squashed logarithmic spiral is not a logarithmic spiral. A section seen from the wrong angle priced the tilt for W alone, with the centre known. Here the centre is free to move, and the question is where the model’s error goes.

It goes into W. Over one turn, tilts of five, ten and twenty degrees move W by 0.34, 1.36 and 5.64 per cent and D by 0.01, 0.04 and 0.17: D is a ratio of two lengths on one line through the centre, and a squash takes such lines to lines and scales both lengths alike, so D is nearly immune even jointly. And freeing the centre makes W worse, not better. Read about the true centre, the same tilts move W by 0.16, 0.66 and 2.74 per cent: about half as much. The free centre moves to absorb the squash, finds a place from which the squashed outline looks more like a spiral, and W pays for the move.

That is a compensating failure in exactly the sense the earlier essay feared, and it can be caught or not depending on the tilt. With the drawing noisy to a thousandth of the rim, the fit’s residual is 1.82 times the noise at ten degrees and 6.15 at twenty — the outline visibly fails to redraw, and a worker who looks at the residual knows something is wrong. At five degrees the residual is 1.06 times the noise: the fit redraws the section as well as the noise allows, from a W that is out by 0.34 per cent, three times its own noise error of 0.12. A tilt that small is invisible in the fit and not in the number.

Where the round trip breaks #

The recovery has a floor of its own, set by the drawing rather than the model. On the nautilus-like shell over three turns, noise of a hundredth of the rim is larger than the innermost wall’s own radius — the inner wall starts at radius one and the rim is at 109 — and the innermost stretch of that wall is no longer drawn at all, only scattered. The fit then reads D 79 per cent out. Its residual gives it away, in the opposite direction from the tilt: 0.59 of the noise, a fit better than the noise allows, which means the fit is absorbing the noise into the centre rather than reading through it. At a thousandth of the rim the same section reads D to 0.25 per cent with a residual of one.

So the residual is worth reading in both directions. Well above one says the model is wrong; well below one says the drawing is not resolving what the model needs, and a worker should drop the stretch of wall the noise has swallowed rather than fit it.

Which side of the line #

The contact line is where the two numbers matter together, and the shell the figures are drawn at sits just outside it: $W \cdot D = 1.008$, its whorls clear of each other by eight parts in a thousand. A reading places it on the right side of the line or the wrong one, and that is a decision rather than an error bar.

Read jointly over a turn, a hundred sections drawn to a thousandth of the rim put it on the wrong side none of the time; to three thousandths, four times; to a hundredth, 37 times, since the product’s spread, 1.2 per cent, is now larger than its distance from the line. Over a turn and a half, none, none and nine. Read at the guessed centre, the same drawings put it on the wrong side about half the time at every span and every noise — 48 in a hundred over a turn at a thousandth of the rim. A worker who assumes a centre cannot say which side of the line a shell this close to it lies on; a worker who fits the centre can, from a drawing no better than a thousandth of the rim.

What the round trip changes about the budget #

Three points on a diameter found a different escape from the centre for W: calipers read two diameters on one line and never use the centre’s coordinates, so their error is second order in the aim. The joint fit is the other escape, and it reaches both numbers: it uses the centre, but it finds it.

The two budgets were built for a worker who assumes a centre and reads one number from it, and for that worker they are right: 79 per cent for D at one azimuth, 6.9 for W. A worker who fits the centre along with the numbers is in a different position altogether. The gross centre error that dominated both budgets largely disappears, and what is left is the noise of the drawing, the span chosen, and the model’s own fit to the specimen. That moves where the band nobody can be placed in lies: a band drawn from the budgets’ worst cases is far wider than the one a joint reading of a well-drawn section supports.

It does not remove the budget’s other entries — the counting clock, the septa, the change of law through growth — and it replaces the centre entry with one the budget never had: the tilt, which the free centre enlarges and can hide. A section should be photographed square to its own plane, and a worker reading W should look at the residual before believing the number.

What this reading assumes #

That the section shows both walls over its whole span. On an involute shell like the nautilus the inner wall lies against the previous whorl, and parts of it are not drawn; a real recovery would fit the stretches the drawing shows, with fewer points and a correspondingly larger error. That the walls are logarithmic spirals to the precision of the noise — a shell that changed its law during growth is not, and its W is a mean over the span read. And that the section is planispiral: T, the third of Raup’s numbers, is invisible on a median cut of a planispiral shell, so this is a round trip in two of the three.

Readings that would undo it #

A noiseless section from which the recovery, started a quarter of a radius off, does not return W and D to rounding. Errors in W and D that are correlated across sections. A span over which D is read better than at one turn by a factor of two. An oblique view that moves D by as much as W. Each would mean the joint reading here is wrong rather than incomplete.

Still open: the third number off a second cut #

A median section of a helicoid shell shows the coiling in the plane and not the translation along the axis. T needs a second cut, along the axis, where each whorl’s cross-section appears at a height, or a view of the shell’s profile. The next measurement is the round trip in all three: a shell built at stated W, D and T, sectioned along its median plane and along its axis, the three numbers and the axis’s position recovered together from both drawings — asking whether the axis, which both cuts share, is located as well as the centre was here, and whether T trades against W the way the centre does, since a misplaced axis would tilt every whorl’s height into its radius.