A section seen from the wrong angle

Every instrument priced so far on a shell section starts from a set of points in the plane of coiling. The fit takes them and an assumed centre; the calipers take three of them on a line; the growth-line count takes the whorl each falls in. All three assume the points are where the shell put them.

A photograph taken square-on gives that plane. A photograph taken from anywhere else gives its projection, which is the same plane compressed along one direction by the cosine of the angle between them — and the picture does not say which it is. A specimen photographed on a bench, a published plate whose camera was not normal to the slab, a section cut a few degrees off: all of them produce a curve that is not the curve, by an amount nobody records.

What the compression is, exactly #

With the tilt axis along $x$ and $c = \cos\alpha$, a point at $(r\cos\theta, r\sin\theta)$ is seen at

$$r' = r\sqrt{\cos^2\theta + c^2\sin^2\theta}, \qquad \theta' = \operatorname{atan2}(c\sin\theta,\ \cos\theta)$$

So $\log r'$ is $\log r$ plus a term of period $\pi$ in the angle, running between $\log c$ and $0$. The fit takes $\log r$ against $\theta$ and finds a slope, so the tilt adds a periodic perturbation to exactly the data the fit is a slope through. Everything below is a consequence of that one sentence.

The tilted picture is still a perfectly good-looking shell. At thirty degrees the compression is 0.866, which is not visible by eye on a spiral whose radius changes by a factor of 3.2 every turn, and the projection preserves the things a person would check: the curve is smooth, it is closed at the apex, its whorls do not cross, and it still looks self-similar.

What it costs each instrument #

Over three whole turns the fit drifts low and by very little: −0.0037 per cent at five degrees, −0.062 at twenty, −0.148 at thirty, −0.815 at sixty. A camera twenty degrees off costs six hundredths of a per cent in the growth factor, which is a hundredth of what a displaced centre costs at a quarter of the innermost radius.

The calipers do not move at all. Over every tilt and every aim read, the worst departure is the last bit a double holds.

Why the calipers cannot see it #

The reason is worth stating as an argument rather than as a result, because it says which measures are immune and which are not.

A projection maps a line through the centre to a line through the centre, and it maps the centre to the centre. The caliper measure takes three points where one such line crosses the curve — at $\theta$, $\theta+\pi$ and $\theta+2\pi$ — and reads $(d_2/d_1)^2$. The scale factor $\sqrt{\cos^2\theta + c^2\sin^2\theta}$ has period $\pi$, so all three points are multiplied by the same number, and a ratio of two lengths taken along that line does not contain it.

That covers the aim as well. A thirty-degree tilt shortens a line drawn across the picture by anything between nothing and 0.866 depending on where it points, and the caliper reading is unchanged at every aim to nine decimal places. What the aim changes is both lengths equally, and the measure is their ratio.

This is the same structural property that made the calipers cheap in three points on a diameter — the centre is needed only to aim the line, not to measure from — arriving from a different direction. A measure built out of a ratio along one line is blind to anything that scales that line uniformly, and an orthographic projection is exactly such a thing.

The growth-line count is immune too, and for a different reason #

The third instrument a section carries is a count, and it survives a tilt for a reason that has nothing to do with ratios.

The angular distortion $\theta' = \operatorname{atan2}(c\sin\theta, \cos\theta)$ is monotone, and it fixes every multiple of a quarter turn exactly. Whorl boundaries are multiples of a whole turn, so they are among the angles it fixes — and a monotone map that fixes the boundaries cannot carry a point across one. Swept over twenty-four views at eight tilts and three growth factors, not one of a hundred thousand points changes the whorl it lies in.

It is not that the distortion is small. Inside a whorl it moves a point by as much as a twentieth of a turn at sixty degrees, which is a visible displacement on any picture. It moves points about within their whorls and never between them, so the quantity what the growth lines carry divides — the number of lines in each whorl — is exactly what it was.

That completes an unusual pattern. Of the three readings a section supports, two are exactly immune to an oblique view and the third, the one that uses the most of the data, is the one it damages. The fit uses every point’s position; the calipers use a ratio along one line; the count uses only which whorl each line is in. Using less of the picture is what makes a measure robust here, which is the reverse of the usual relation between how much data an instrument consumes and how well it does.

The fit’s real cost is a band, not a number #

Three whole turns is the best case and nobody measuring a fossil is guaranteed it. Over a partial span the perturbation does not average away, and where along the shell the span begins decides what the fit returns.

At twenty degrees the band of answers over twenty-four starting points is 23.7 per cent wide at half a turn, 5.9 at one turn, 1.5 at two and 0.66 at three. Half a turn of shell photographed twenty degrees off gives a range of growth factors wider than everything else priced so far put together — the centre at a quarter radius at 4.56 per cent, the dividers, the sampling failure in a spiral with no clock at four per cent — combined.

The band’s mean sits within a few thousandths of a per cent of the right answer at every span. So a tilt is a spread and not a bias, and that is exactly what makes it invisible: a person who fits one shell once gets a number drawn from a wide distribution centred on the truth, with nothing to indicate the width. Averaging over many shells would recover the growth factor and still report a scatter that is not the animals’ variation.

There are spans a tilt cannot touch #

The fit’s slope is a weighted sum over the span, with weight $(\theta - \bar\theta)$. Putting that weight against a term of period $\pi$ leaves

$$\int_0^L (\theta - L/2)\,e^{2i\theta}\,d\theta \;\propto\; \sin L - L\cos L$$

which vanishes exactly when $\tan L = L$. So there is a discrete set of spans at which an oblique view costs the fit nothing whatever the tilt and whatever the starting point, and the first are at 0.7151, 1.2295, 1.7354 and 2.2387 turns.

At those spans the band collapses: 0.00129 per cent at 0.7151 turns against 11.4 per cent an eighth of a turn below it, 0.00055 against 3.31, 0.00033 against 1.56. The collapse is three or four orders of magnitude and it happens at a root of a transcendental equation, which is not a place a trend would put it.

None of the roots is a whole turn or a half turn. That is the part with a practical edge to it: every span anybody would choose — one turn, two turns, half a turn — is a local worst rather than a local best, because the perturbation’s period is half a turn and a span of whole half-turns weights its two halves most unevenly. Choosing a round number is choosing the one span where the error is largest.

The residual is the measurement, not the verdict #

A fit to a square-on section leaves a residual of 2 × 10⁻¹⁶ — the machine’s last bit, and nothing else. A tilted one leaves 0.00199 at five degrees, 0.03291 at twenty, 0.42152 at sixty.

Those are not arbitrary numbers. The perturbation runs between $\log c$ and $0$, so the largest departure from a line through its middle is $|\log\cos\alpha|/2$: 0.00191, 0.03110, 0.34657 for the same three tilts. The measured residual sits within a quarter of that over the whole range and within five per cent of it below twenty degrees, where the periodic term is nearest a cosine.

The residual is not the test established that this quantity cannot tell a logarithmic spiral from an Archimedean one — it fits a circle exactly and refuses a golden spiral that is right to three decimal places. That finding stands. What this adds is that the residual is not useless; it is answering a different question from the one it was being asked. It does not say whether the curve is the right kind. It says how far the points depart from a straight line in $\log r$ against $\theta$, and an oblique view is a departure with a shape.

The tilt comes back out #

The map from tilt to residual is monotone, so it inverts. One measured residual goes in and one angle comes out, with nothing minimised and nothing fitted — five degrees returns 5.000, twenty returns 20.000, sixty returns 60.000, the worst of eight readings being three thousandths of a degree out.

That makes an oblique view the rare error a section reports about itself. A displaced centre does not: how far a centre must move found centres that turn a shell grown at 3.2 into a golden spiral, and the only thing that refuses them is a separate span condition. A resolution limit in a growth-line count does not report itself either. A tilt does, in a number the fit already prints.

What a person should therefore do #

Three steps, and only the last is new. Read the residual first, before the growth factor. If it is at the machine’s floor the view is square-on and the fit can be believed. If it is not, invert it for the tilt, and then either correct the picture or stop using the fit.

And if the residual is not at the floor and the tilt is not the reason — if the section is worn, or noisy, or the curve is genuinely not logarithmic — then the fit’s answer carries a band whose width the span decides, and the caliper reading does not. That is the practical division of labour between the two instruments, and it is sharper than the one three points on a diameter arrived at: there the choice turned on which error the section carried, and here one instrument is exactly immune to an error the other is badly exposed to.

How far off a camera actually is #

The tilts drawn here run to sixty degrees, which is further than anyone photographing a specimen would be. The useful part of the range is the first ten degrees, and it is worth reading off directly.

At five degrees the compression is 0.9962 and the fit over three whole turns is out by 0.0037 per cent — nothing. At ten it is 0.9848 and 0.015 per cent. Those are the numbers that make an oblique view sound like a non-problem, and they are the numbers for the best case: a long span, and a whole number of turns of it.

Over half a turn at ten degrees the band is 5.8 per cent wide. That is larger than the centre error what the centre costs measured at a quarter of the innermost radius, from a camera angle a person would not notice and would not think to record. The lesson is not that tilts are large; it is that a short span converts a small distortion into a large uncertainty, which is the same relation how far a centre must move found for the centre and for the same reason — a short arc has little leverage on a slope, so anything periodic in the angle is poorly averaged over it.

Where this is weakest #

A real oblique photograph is not an orthographic projection. A camera at a finite distance adds perspective, which is a larger distortion and a different one: it does not scale a line through the centre uniformly, so the caliper immunity does not survive it. What does survive is the direction of the argument — the calipers are immune to anything that scales a line through the centre uniformly, and the question for a real photograph is how close it is to being such a thing.

The honest statement is that this prices one specific and very common distortion exactly, and names the property that makes a measure immune to it. A photograph taken at ten times the specimen’s width is orthographic to within a per cent of the compression; one taken at twice is not.

What this changes about a published figure #

A growth factor quoted from a photographed section carries an uncertainty nobody has been recording, and its size depends on two numbers the caption almost never gives: the span of arc fitted, and whether the camera was normal to the slab. A quoted 3.2 from three whole turns is worth what it says. The same 3.2 from a partial whorl, photographed on a bench, is a draw from a band several per cent wide.

The remedy costs nothing. A residual printed beside the growth factor settles the second question outright, and the span is already known to whoever did the fitting. Both belong in the caption, and neither is a new measurement.

What is claimed, in one line #

An oblique view of a shell section compresses the coiling plane by the cosine of the tilt, which adds a term of period $\pi$ to the data the growth-factor fit is a slope through — opening a band of answers 23.7 per cent wide on half a turn at twenty degrees, centred on the truth, and vanishing exactly at the spans where $\tan L = L$; the caliper measure is unmoved to the last bit at every tilt and aim, because a projection scales all three of its points by the same factor; and the fit’s residual is $|\log\cos\alpha|/2$, so the tilt inverts out of it to three decimal places.

What none of this establishes #

That any published section was photographed off the normal — nothing here measures a specimen, and every reading is from a curve built to a stated tilt and then read as a section would be. That an orthographic projection is the right model for a real camera. That the tilt recovered from a residual would survive a section which is also noisy, worn or genuinely non-logarithmic: the residual is one number and this asks it to carry one cause, which is exactly the overloading the essay above warns about in the opposite direction.

What would withdraw it #

A caliper reading moved by a tilt at any tilt or aim. A fit unmoved by one. A band whose mean is further from the true growth factor than a fifth of its own width, which would make the tilt a bias rather than a spread. A root of $\tan L = L$ at which the band does not collapse, or a collapse anywhere else. A residual departing from $|\log\cos\alpha|/2$ by more than a quarter, or an untilted view leaving a residual above the machine’s floor. Each is checked every time the measurement runs.

Still open: whether a tilt can be corrected rather than only detected #

The residual names the tilt but not its axis, and a correction needs both. The axis is in principle recoverable from the same perturbation — its phase angle rather than its amplitude — and a section corrected for both would return the fit to the answer the calipers already give, which is a round trip worth having for its own sake. What is not obvious is whether the phase angle is stable enough to read on a span short enough to need correcting: the amplitude is a peak-to-peak and survives a short span, while a phase angle is a position and may not. The next measurement builds tilted views at known axes, recovers the axis from the residual’s phase angle, corrects the picture and asks whether the corrected fit meets the caliper reading.