A scroll round a bend loses its straight-line depth

A scroll through two slits ranges in a straight line drew a pushbroom twice, through a slit leaning 10° forward along the track and one leaning 10° back, and found every point’s two drawings on the same row, separated by 9.169 px for every metre of the point’s depth. Depth proportional to separation, a pixel worth 10.9 cm at every distance, no bias under averaging, and a range limit set only by the length of the roll. It ended by naming the assumption all of that rests on: the eye’s track is straight.

A journey along a river is not straight, and neither is a road through hills. So the measurement it asked for lets the track bend with a stated radius, keeps everything else — 26 px of paper per metre of travel, a 430 px divide across the roll, slits at ±10° — and asks what the bend does to the three things the straight track was measured by.

Where the two slits reach a post from #

A scroll drawn from a bend is built exactly as a scroll is a camera that moves built the straight one. The eye runs along a circle of radius $R$, the paper advances by the distance it travels, and at each position it looks across the track through a vertical slit leaning $\varphi$ from the local perpendicular — from the radial line through the eye, which on a bend is the direction straight across the track. A point is drawn by whichever position of the eye has the point in its slit, and that position is found by searching along the track, not by formula.

On a straight track, a post 60 m out is reached by the forward slit from 10.58 m behind it and by the backward slit from 10.58 m ahead: 21.16 m apart, which the paper records as 550.1 px. Put the same post 60 m outside a bend of radius 100 m, and the two places are 13.16 m apart along the track, and the post is drawn 342.1 px apart. The bend turns each slit toward the post as the eye comes round, so each slit finds it sooner than it would have on the straight, and the two drawings are closer together than the post’s depth says they should be.

The closed form, from the sine rule #

The shortening has an exact expression, and it comes from one triangle: the bend’s centre, the eye, and the point. The point is at distance $\rho = R + D$ from the centre if it lies $D$ outside the bend. The slit makes the angle $\varphi$ with the radial at the eye, so the triangle’s angle at the eye is fixed, and the sine rule fixes the rest. The eye’s bearing, seen from the centre, is off the point’s bearing by

$$\alpha = \varphi - \arcsin\!\left(\frac{R\sin\varphi}{R + D}\right),$$

and the paper records $sR\alpha$ for each slit. The two slits lean equally either way, so the separation between the two drawings is $2sR\alpha$.

The figures are drawn from the search, not from that expression, and the two agree: every plotted point matches $2sR\alpha$ to two parts in a hundred trillion. The two drawings still put every point on the same row, because the two slits still reach it along equal distances. And a bend ten thousand kilometres round gives back the straight track’s $2sD\tan\varphi$ to a part in a thousand, which is what the expression does as $R$ grows: the arcsine’s argument tends to $\sin\varphi\,(1 - D/R)$ and $\alpha$ to $D\tan\varphi/R$.

The rows agree for the same reason as before #

The straight track’s two drawings shared their rows because the two slits reached each point along equal distances, so a matcher could look for a point’s second drawing along a row, the search a point is a line over there describes for any pair of pictures. A bend keeps that. Reversing the slit’s lean mirrors the triangle of centre, eye and point across the point’s own radial line: the forward slit’s eye sits at a bearing $\alpha$ to one side of the point and the backward slit’s at $\alpha$ to the other, the same distance from it. The two reaches are equal, the two rows are equal, and the construction finds them equal to the arithmetic floor at every depth on both sides of the bend. Whatever else the bend does, the search for a point’s second drawing is still a search along a row.

Outside the bend, the separation saturates #

The straight track’s separation grows without limit, in proportion to depth. The bend’s does not.

As a point outside the bend moves further out, $R\sin\varphi/(R + D)$ falls toward zero, the arcsine with it, and $\alpha$ rises toward $\varphi$ and no further. So the separation rises toward $2sR\varphi$ and never passes it. For a bend of 100 m at ±10° that ceiling is 907.6 px. A point 20 m out is drawn 152.4 px apart where the straight track draws it 183.4; a point 256 m out, 653.8 px apart where the straight track gives 2347. However deep the point, the two drawings are never more than 907.6 px apart.

A separation with a ceiling is a separation that runs out of depth. The geometry is the reason: a point very far outside the bend lies, seen from the bend’s centre, almost along the line of any slit that reaches it, so the two slits reach it from bearings almost exactly $\varphi$ either side of its own — an arc of $2R\varphi$ apart, however much further out the point goes. It is the limit the range a pair cannot see past found in a pinhole pair, arriving by a different route. A pinhole pair’s disparity falls toward zero with depth because its baseline is fixed; a bent scroll’s rises toward a ceiling because its baseline, which on the straight track grew with the depth it measured, stops growing. Either way, depth past a certain point is in the last pixel of a bounded reading.

Reading depth back, and where whole pixels stop helping #

Given the radius, the closed form runs backwards. A separation $\Delta u$ gives $\alpha = \Delta u/2sR$, and outside the bend the point’s distance from the centre is then $R\sin\varphi/\sin(\varphi - \alpha)$. The construction’s own separations for points 20, 60 and 256 m out come back as those depths to nine decimal places. A bend does not destroy depth; it changes the currency it is paid in.

Near the ceiling that currency becomes very poor. Outside the 100 m bend the separation can never reach 907.6 px, and the last pixels below it stand for enormous spans of depth: a separation of 906.6 px is a point 90.2 km out, and one pixel less is 45.05 km; 902.6 px is 18.0 km, and one pixel less 14.95 km; even ten pixels under the ceiling, one pixel spans from 8.9 km to 8.11 km. Read to whole pixels, the outside of a bend cuts depth into shells that grow without limit toward the ceiling — the pattern whole pixels cut space into shells found for a pinhole pair, and the one the straight two-slit scroll was free of. It is the arithmetic of depth is a reciprocal in another form: a reading that approaches a bound spends its last units on the far field.

Inside the bend, it runs ahead and then stops #

Put the scene on the inside of the bend and the same triangle gives the opposite behaviour. The point is at $\rho = R - D$, the eye looks toward the centre, and

$$\alpha = \arcsin\!\left(\frac{R\sin\varphi}{R - D}\right) - \varphi .$$

Now the arcsine’s argument grows as the point comes further in, and $\alpha$ grows faster than depth. Inside a 200 m bend a point 128 m in is drawn 3420 px apart where a straight track would give 1174: nearly three times ahead. And the argument reaches one at $D = R(1 - \sin\varphi)$ — 165.3 m for the 200 m bend — beyond which the arcsine has no value and the construction finds nothing. Every slit leaning $\varphi$ from a radial passes the centre at a distance of $R\sin\varphi$, so the innermost part of the bend, within that distance of its centre, lies on no slit at all and is never drawn.

So a bend cuts the scene into two regimes by which side of the track it is on. Outside, depth is compressed into a bounded separation; inside, it is expanded, without bound, up to a circle the scroll cannot see into.

The bend gives the scroll back a centre line #

There is a way to see both limits as one object, and it answers a question the straight track settled the other way. The centre a scroll does not have fitted a common point to the rays of a straight scroll and found it missing by exactly the spread of the eye’s own track: a scroll drawn from a straight track has no centre, and unrolling more of it only moves its rays further apart.

Bend the track and draw straight across it — slits along the radials — and every ray of the scroll, followed back past the eye, crosses the vertical axis through the bend’s centre. Across the whole bend, every ray’s horizontal line passes that axis at a distance of zero, to the arithmetic floor. The scroll has acquired a centre, though not a point: a line, which the rays of each row meet at their own height. It sits between the straight track’s scroll, whose rays share nothing, and the cylindrical panorama taken from that axis, whose rays all share one point on it.

Lean the slits by $\varphi$ and the rays no longer reach the axis. Every ray’s line passes it at $R\sin\varphi$ — 8.716 m for slits at 5° on a 100 m bend and 17.365 m at 10°, again to the arithmetic floor — so all the rays are tangent to a circle of that radius about the axis. That circle is the one the inside of the bend could not reach into: no slit enters it, so nothing inside it is drawn, and a point just outside it is reached only by slits that barely graze it, which is why the inside separation runs away as the circle’s edge approaches. The outside ceiling is the same circle seen from beyond: a very distant point’s two slits leave it along two nearly parallel tangents, and the eyes they leave from are never further apart than an arc of $2R\varphi$.

What a pixel is worth, on each side #

The depth one pixel of separation is worth is the reciprocal of how fast the separation grows, and it turns the two regimes into instrument terms. On the straight track it is 0.109 m at every depth — the flat line the two-slit scroll was valued for. Outside a gentle 500 m bend it is already 0.132 m at 50 m and 0.248 m at 250 m. Outside a 100 m bend it is 0.248 m at 50 m and 1.355 m at 250 m, rising without limit as the separation approaches its ceiling, which is a pinhole pair’s behaviour in a scroll.

Inside a 200 m bend it goes the other way: 0.061 m at 50 m, better than the straight track, and 0.0050 m at 150 m, twenty times better — falling toward nothing as the slit’s reach runs out at 165.3 m. The inside of a bend is, in this one respect, a better range-finder than the straight track, right up to the edge of what it cannot see.

The scale along the roll depends on depth #

The straight track’s other result was about the roll itself. A map along, and a picture across found a scroll to be a scale drawing along its length: a metre of travel was 26 px of paper, and so was a metre of the ground beside the track at any depth, which is why a segment’s midpoint along the roll landed on its image’s midpoint.

On a bend that fails, and it fails by a law. The paper advances by the eye’s travel, $R$ times the angle it turns through, but ground $D$ outside the bend turns through the same angle over $R + D$ times it. So a metre of that ground occupies $sR/(R + D)$ px of paper: 23.64 px at 10 m and 14.44 px at 80 m outside a 100 m bend, against 26 on the straight. Inside the bend the ground’s arc is shorter than the track’s, and a metre of it occupies more paper: 28.89 px at 10 m and 130.0 px at 80 m. The measurement, from two points a metre apart along their own arc, matches $sR/(R \pm D)$ to the last digit.

A scroll of a winding river is therefore a scale drawing only of its own track. The banks outside each bend are drawn shrunk and the banks inside drawn stretched, by an amount that depends on how far each thing stands from the line the painter travelled. The gentler the bend, the less it matters: a bend of 500 m keeps ground 20 m out within 3.8 % of the track’s scale. A pond in a scroll is not an ellipse measured a pond’s width along the roll as a direct measurement of its size; on a bend, the pond’s width is that measurement divided by $R/(R \pm D)$, and the reading needs the bend’s radius to mean anything.

A smaller slit angle moves both limits #

The slit angle was the straight track’s one design choice, and on a bend it moves both of the new limits.

At ±5° the straight-track separation halves to 4.549 px per metre, as before. Outside the 100 m bend the ceiling halves too, to 453.8 px, because it is $2sR\varphi$; a point 20 m out is drawn 75.8 px apart against the straight track’s 91.0. Inside the 200 m bend the unreachable circle shrinks, because its radius is $R\sin\varphi$, and the slit now reaches to 182.6 m instead of 165.3; a point 128 m in is drawn 1636 px apart against 582.3.

So a shallower slit trades the same way it did on the straight track — precision for reach — with one change. On the straight track, the two-slit scroll found that the slit angle only redistributed a fixed budget of distinguishable depths set by the roll’s length. On a bend the outside’s ceiling is proportional to the slit angle, so a shallower slit lowers the ceiling as well as the rate; what it buys back is only on the inside, where the unreachable circle shrinks and the slit reaches 17 m further in.

What this does not settle #

A circle. The bend has one radius all the way round. A river’s bend tightens and relaxes, and a track whose curvature changes along its length would give each column its own $R$; that was not drawn.

A level track. The eye stays at one height. A road that climbs as it bends adds a slope to the triangle, and the rows of the two drawings would no longer agree.

The radius is known. Every depth here is read with the radius given. A reader holding only the scroll does not have it.

Still open: whether the scroll reveals its own bend #

The bend leaves two marks on a two-slit scroll, and they depend on the radius differently. The separation of a point’s two drawings is $2sR\alpha$, with $\alpha$ set by the radius and the depth together. The scale along the roll is $sR/(R \pm D)$, set by the same two numbers in a different combination.

The question that leaves is whether the two together let a reader recover the bend: whether a handful of points, each with a measured separation and a measured along-roll spacing to a neighbour at the same depth, fix both the radius and their own depths, how many points it takes, and whether a painter’s straight-track habits — scale kept constant along the roll by eye — would show up as a bend that was never there.