A scroll can be asked its own radius

A scroll round a bend loses its straight-line depth ended with a reader holding a scroll and not holding its radius. Every depth in it was read with the radius given, and the radius is not something a painting comes with.

The bend leaves two marks, and they depend on the radius differently. A point’s own two drawings, through slits leaning forward and back along the track, are separated by $2sR\alpha$, with $\alpha$ set by the radius and the depth together. Two points a metre apart at that same depth are drawn a distance apart along the roll that works out at $sR/(R \pm D)$ px per metre — the same two numbers in a different combination.

Two measurements and two unknowns. The question is whether they separate, how many points it takes, and what a reader’s inevitable errors do to each answer.

Neither mark says anything on its own. The separation confounds the radius with the depth, and so does the scale; a reader given one of them and asked for either number has two unknowns and one equation.

They separate, and the algebra says which mark does the work #

Write $\sigma$ for the along-roll scale measured beside a point: how many pixels of paper a metre of ground occupies there. Then $\sigma = sR/(R+D)$ outside the bend, so

$$R + D = \frac{sR}{\sigma}$$

and the ratio inside the disparity’s arcsine — $R\sin\varphi/(R+D)$ — becomes $\sigma\sin\varphi/s$. That depends on the measured scale alone. So

$$\alpha = \varphi - \arcsin\!\left(\frac{\sigma \sin\varphi}{s}\right), \qquad R = \frac{u}{2s\alpha}, \qquad D = R\,\frac{s - \sigma}{\sigma}$$

with $u$ the measured separation. No search and no iteration: one point, one neighbour at its depth, and both answers fall out.

Six points from five metres to a hundred and twenty return a radius of 200.000 m and their own depths to the ninth digit, and the same holds inside the bend, where the signs turn over and nothing else changes.

The structure of that answer matters more than the answer. The measured scale sets the bracket $\alpha$ by itself, and the measured separation then only sets how large $R$ is. Everything the radius is known from arrives through $\sigma$; the separation contributes a multiplier and no information about the bend’s tightness at all. That is not obvious from the two formulas, and it is what governs everything below.

How many points it takes, and what more of them buy #

One point and a neighbour. That is a surprising answer for a recovery problem and it is worth being clear about what it does and does not mean.

It means the reading is pointwise: each point’s two marks are read alone, and nothing is fitted across the picture. So more points are not needed to solve. What they buy is a test.

Every point beside a circular track returns the same radius, so a spread in the radii several points return is a statement that they were not all drawn from one curvature — which is a necessary condition for the varying-curvature track a scroll round a bend could not settle, and not a measurement of it.

The distinction between agreeing and being fitted to agree is worth holding onto, because it is what makes the test worth running. Nothing in the reading pushes the six points toward a common answer. Each is read alone, from two numbers measured beside it, and they land on 200.000 m together because they were drawn from one circle. A fitted radius would agree by construction and would report a residual that a reader would then have to interpret; this agrees or does not, and the disagreement is in metres of radius rather than in pixels of anything.

The caution there is real and worth stating rather than implying. On a track whose curvature genuinely changes, the eye travels through a range of curvatures while a point is in view, and the marks are an integral along the track rather than a value at a place. Nothing here models that. What can be said is that points disagreeing about the radius were not drawn from one circle; what cannot be said is what curve they were drawn from.

The two answers are not equally held #

A reader measures $\sigma$ by finding two things a known distance apart at a common depth and counting pixels between them. That measurement carries error, and the error reaches the two answers very differently.

The depth is held. A scale read one per cent too large under-reads the depth by one per cent, at every distance from the track and on both sides of the bend. The amplification is −1.01 at forty metres, −1.015 at ten, −1.008 at fifty: one, to a per cent or two, everywhere.

That asymmetry is unusual enough to be worth a sentence of context. Most recoveries of this kind trade one quantity against another — a single view fixes shape and not size, and every quantity in the answer inherits the same conditioning. Here the two answers come out of the same two measurements and one of them is robust while the other is not, which means a reader can quote the depth from a picture whose radius they would not quote at all.

The radius is not. The amplification has a closed form — $\tan(\varphi - \alpha)/\alpha$ — and it is large wherever the point is close to the track. For a point ten metres from a five-hundred-metre bend it is 49.99: a scale read one per cent too large reports a radius half again too big. For a point forty metres from a two-hundred-metre bend it is 4.99, and for one a hundred and twenty metres out it falls to 1.7.

The memorable form of that number is $R/D$, and it is an approximation rather than the thing: 4.989 against 5.000, 49.985 against 50.000. The difference is under half a per cent and the ratio is the form to carry, but the two are not the same quantity and the essay that needs three digits needs the tangent.

That the two amplifications differ so sharply follows from the algebra above. The depth comes out of the difference between the measured scale and the paper’s own, which is a quantity of the same size as its own error. The radius comes out of the arcsine’s departure from $\varphi$, and near a straight track that departure is nearly zero — so a small absolute error in $\sigma$ is a large relative error in $\alpha$, and $R$ is $u/2s\alpha$.

Inside the bend, the sign turns and the magnitude does not #

Everything above is stated for ground outside the bend. Inside it the formulas differ only in two signs — the scale is $sR/(R-D)$ and the angle is $\arcsin(\ldots) - \varphi$ — and the round trip closes just as exactly. What changes is worth reporting, because one thing reverses and one thing does not.

The magnitude of the amplification is the same. A point 40 m inside a 200 m bend amplifies a scale error by 5.02 against the outside’s 4.99; one 20 m inside, by 10.02. The $R/D$ approximation holds on both sides and to the same accuracy.

The sign reverses. Outside the bend, reading the along-roll scale too large reports a radius too large. Inside it, reading it too large reports a radius too small. The reason is that the two sides put the ground on opposite sides of the track’s own arc: outside, a metre of ground occupies less paper than a metre of track, and inside it occupies more — 130.0 px against 26 for ground 80 m inside a 100 m bend.

That matters for a reader with a scroll of a river, where both banks are in the picture and one of them is inside every bend. The two banks amplify the same error in opposite directions, so a radius recovered from the outer bank and one recovered from the inner bracket the truth rather than agreeing wrongly — and the size of the bracket is a reading of how badly the scale was measured. It is the nearest thing to a self-check the arrangement offers.

And the inside has an edge the outside does not. A slit leaning $\varphi$ from the perpendicular stops reaching ground closer to the bend’s centre than $R(1 - \sin\varphi)$ — 165.3 m for a 200 m bend at ±10°. Past that there are no marks to read, so the recovery is not badly conditioned there; it is unavailable. Outside the bend there is no such edge, and the reading degrades smoothly instead.

What a painter’s habit does to it #

One more thing was asked when the bend was first drawn: whether a painter’s straight-track habits would show up as a bend that was never there.

The habit to model is keeping the scale along the roll constant by eye. A painter who has drawn straight-track scrolls knows that a metre of ground is a fixed number of pixels of paper, and on a bend it is not — but the departure is the sort of thing an eye evens out. So take a true bend, leave the separations exact, and soften only the scale’s departure from the paper’s own.

The answer is a clean no, and it is better than a no.

The reported radius is inflated by very nearly the reciprocal of what the painter kept. Keep nine tenths of the scale’s departure and a 200 m bend reads as 222 m; keep half and it reads as 400; keep a twentieth and it reads as 3,991; keep none and the recovery refuses to report a radius at all, because a scroll whose along-roll scale is exactly the paper’s is a scroll of a straight track.

And the depth survives. Across that whole range — a radius wrong by a factor of twenty — the reported depth falls only from 40.0 m to 33.5 m.

So the habit erases a bend rather than inventing one, and it cannot do otherwise. Inventing a bend would require the along-roll scale to depart from the paper’s by more than it truly does, and a habit of keeping the scale constant is the opposite operation. A reader who suspects a painter of evening out the scale should read the recovered radius as a lower bound on how straight the track was and the recovered depths as very nearly right.

That is the useful form of the result. The quantity a painter’s eye is likely to corrupt is exactly the quantity the radius depends on entirely, and exactly the quantity the depth barely depends on — so the answer a reader most wants from a scroll is the answer the painting is least likely to have spoiled.

Reading it the other way round #

There is a second use of the same algebra that costs nothing and is worth naming, because a reader of a real scroll is more likely to be in this position than in the first.

A reader who knows the radius — a scroll of a named river, a road whose course survives — does not need the along-roll scale at all. The disparity alone gives the depth, by the closed form the bent track was measured with. And a reader who knows one depth, from a building of known size or a stated distance, can invert the first equation for $\sigma$ and check it against what the paper actually shows. The gap between the two is a direct measurement of how much the painter evened out, on that stretch, in the one currency the habit acts in.

That makes the flattening measurable rather than merely suspected, provided a single independent distance is available anywhere in the picture. It is the same move a reference length makes in any single-view recovery: one known distance converts a family of consistent answers into one answer, and here it converts a suspicion about the painter into a number. A pond in a scroll is one such thing if it is round: its width along the roll is a direct measurement of its size only when the track is straight, and on a bend that width is the same measurement divided by $R/(R \pm D)$ — so a pond of otherwise-known size reads the along-roll scale off, at its own depth, without any pair of points needing to be identified.

What this settles and what it does not #

It settles that a two-slit scroll carries its own radius. The two marks separate in closed form, from one point and a neighbour, exactly.

It settles which mark carries it. The along-roll scale fixes the angle by itself; the separation only sets the scale of the answer. Everything the radius is known from arrives through the scale.

It settles the conditioning. The depth is read with the error it was measured with, and the radius with that error amplified by $\tan(\varphi - \alpha)/\alpha$, which $R/D$ approximates within half a per cent.

It settles the painter’s habit, in the direction that helps. Evening out the scale reports a gentler bend and never a tighter one, and leaves the depths nearly intact.

It does not read a real painting. Every scroll here is constructed, and nothing about any surviving handscroll is claimed. Whether any painter drew a two-slit pair at all is a question no measurement here has ever claimed to answer; the two-slit scroll is an arrangement that makes depth recoverable, introduced as one.

It does not survive a track that climbs. The eye stays at one height throughout. A road that rises as it bends adds a slope to the triangle, the two drawings stop sharing their rows, and the whole reading rests on their sharing them.

And it does not extend to a track whose curvature varies, for the reason given above: the marks would be integrals along the track and the recovery is pointwise.

Where the handscroll’s geometry runs out #

The handscroll has now been taken as far as the pushbroom goes.

It began by saying what a scroll is — an eye on a track, recording one vertical line at a time, orthographic along the roll and perspective across it. It found that a straight world line is drawn as a hyperbola and that the sag is a depth signal. It fitted a centre to the rays and found the miss to be the standard deviation of the eye’s own track. It found the roll direction an exact scale drawing and the across direction a picture. It drew a round pond and found a quartic that no conic fits, widest exactly on its centre’s row. It drew the scroll twice through leaning slits and found depth proportional to separation rather than reciprocal to it. It bent the track and found what that costs. And it has now asked the bent scroll for its own radius and been given it.

The next question would be one of those with the track climbing or its curvature changing, and both of those are new geometry rather than a further reading of this one — the first breaks the shared rows that everything here depends on, and the second breaks the pointwise recovery. There is nothing further to ask of a level circular track. What remains belongs to a reader with a real scroll, and that is not a question about projections.

Still open: what a scroll of a climbing track would lose first #

The one continuation that is genuinely a next question rather than a new subject is the level track, and it is worth stating precisely because the obvious guess about it is probably wrong.

Every reading since the two-slit pair was introduced has leaned on one fact: the two drawings of a point share a row exactly. They share it because the eye is at one height, so the vertical angle to a point is the same from both slit positions. On a track that climbs, the two positions are at different heights, the vertical angles differ, and the rows come apart by an amount that depends on the slope and on the point’s depth and height together.

The obvious guess is that this simply adds noise. It may instead add a third measurement: a row difference that depends on the slope is a slope signal, and a scroll of a climbing road might report its own gradient the way this one reports its own radius. The measurement that settles it lets the track rise at a stated gradient, records the row difference against the point’s depth and height, and asks whether it separates from the other two marks — or whether it is confounded with the bend, so that a climbing straight track and a level bend leave the same three marks and no reader can tell them apart.