The centre a scroll does not have

A projection through a centre is exactly the statement that every ray of the picture passes through one point. That is not a description of perspective, it is perspective, and everything this site computes about a picture — the viewing distance, the recovered focal length, the cross-ratio, the height from a photograph — is a consequence of it holding.

So the way to ask whether a handscroll is a projection is to take its rays and see whether they meet.

The fit is the refraction field’s, deliberately #

The solver here is closestPointToRays, imported from lib/refraction.js rather than written again. That is not thrift. It is the site’s existing definition of does this picture have a centre, and asking the question a second way would make a scroll’s answer incomparable with the water’s — which is the one comparison this essay most wants to make.

It solves $\sum (I - d_i d_i^{\mathsf T})\,x = \sum (I - d_i d_i^{\mathsf T})\,p_i$, a 3 × 3 symmetric system, and reports the RMS perpendicular distance from the solution to each ray. For a bundle that really does share a point the answer is zero to the last bit; the refraction field’s control, a stack with no glass in it, comes back at $0 \times 10^{0}$ m exactly.

The closed form, which makes it a measurement #

Every ray of a scroll has a zero x-component: the eye looks sideways, so the whole bundle for one column lies in the plane $x = X$. And the ray for image column $u$ starts at world $x = (u - u_0)/s$.

Put those together and the least-squares problem in $x$ separates from the rest. The fit’s x-coordinate is the mean of the columns’ own $x$ values, and the residual in $x$ is their spread. So for a section sampled evenly across $W$ pixels of paper, the RMS miss is

$$\sigma = \frac{W/s}{\sqrt{12}}$$

the standard deviation of a uniform distribution over the travel the section covers. Nothing about the scene enters, nothing about the focal length enters, and nothing about the depths of the points enters.

Measured against predicted: 7.9738 m against 7.9738 m, agreeing to fifteen digits. That agreement is the result. Without it the number would be a residual — something a better fit might reduce — and with it the number is a fact about the geometry, computed twice by routes that share no code.

Three properties of that line, and each one refuses a way out #

It passes through the origin. A section of no width is one column, which is a pinhole camera: every ray leaves the same point and the fit finds it at $4 \times 10^{-16}$ m. That is the control, and it is what makes the rest of the line mean anything. If a scroll’s rays missed even for a single column, the misses further along would be the solver’s noise rather than the scroll’s geometry.

It is straight. The obvious hope about a picture like this is that the failure is a small-angle effect that saturates — that a scroll is nearly a projection over any reasonable extent and the miss levels off. It does not. Doubling the section doubles the miss, to nine decimal places, for ever.

And it has no threshold. There is no width below which the picture is a projection and above which it is not. Every section of positive width has a positive miss and every section of zero width has zero. A property that holds only in a limit is not a property the object has.

What the number is in the room #

The site’s habit is to turn a computed quantity into something a reader can stand next to, and here the arithmetic is unusually direct because the miss is already in metres of the world.

At the compression these figures use — 26 pixels of paper per metre of travel — a section of paper the width of the figures on this page covers about 26.5 metres of the eye’s journey, and the rays of that section miss any common point by 7.97 metres. Not millimetres. Not a fraction of the scene. Nearly eight metres, in a scene whose pavilions are three metres wide.

Compare the refraction field’s numbers, which are the only others on this site of the same kind: an aquarium misses by 6.4 mm, a swimming pool by 8.3 mm, a flat diving port by 0.5 mm. Those are misses that a reader might reasonably wonder about — small enough that a picture through water looks like a picture, and the field’s finding is that the smallness is beside the point. A scroll’s miss is three orders of magnitude larger and the finding is the same finding. Being nearly a projection is not a weaker version of being a projection; it is a different category, and the size of the number is a fact about the object rather than about the seriousness of the failure.

Why the section a reader has open is not a way out either #

There is a tempting rescue and it deserves a fair hearing, because it is very nearly right.

A handscroll is looked at a shoulder-width at a time. If the section actually in front of the reader is small enough, its miss is small too — the line is straight through the origin, so halving the open section halves the miss. Perhaps the object is a projection locally, and the whole thing is a smooth family of nearly-correct pictures.

The arithmetic says how nearly. A section 60 cm wide at this compression covers 23 metres of travel and misses by 6.6 m. To get the miss down to the width of a pavilion — say 3 m — the open section would have to be 27 cm; to get it below the width of a brushstroke it would have to be a few millimetres of silk.

So the rescue fails quantitatively rather than in principle, and it fails because of the compression. A scroll’s whole purpose is that a great deal of world is laid out on a manageable length of silk, which is to say that $s$ is small, which is to say that a modest window of paper covers an immodest length of track. The property that makes a handscroll useful is the property that makes it centreless, and the two cannot be separated by choosing a better viewing habit.

Where the miss goes as the compression changes #

The closed form has $s$ in the denominator, so the miss is inversely proportional to the compression, and that has a consequence worth stating because it runs against the intuition the word compression invites.

A tightly rolled scroll — more paper per metre of world — has a smaller miss. At twice the compression the same width of paper covers half the travel, and the rays of that section miss by half as much. In the limit of a very large $s$ the picture becomes a single perspective view of a very small piece of world, which is a photograph.

A sparsely rolled scroll has a larger one, without limit. A scroll laying out a hundred kilometres of river on eight metres of silk has a miss measured in kilometres, and it is no less a handscroll for that.

So the family runs continuously from a photograph at one end to something with no local coherence at all at the other, and the parameter that moves along it is the one the painter chooses. The gate for this phase measures three points on that family — the site’s own compression, a scroll rolled twice as tightly, and a scroll of a distant landscape at a long standoff — and requires the closed form to hold at each, which is what says the law is a law rather than a coincidence at one setting.

The standoff, incidentally, does not enter. A scroll of a distant landscape and a scroll of a near one at the same compression have the same miss, because the miss is made of the eye’s travel and nothing else. That is a small surprise and it is worth checking against the intuition that a more distant subject should be more forgiving: it is not, and the reason is that moving the subject away moves every ray’s direction without moving any ray’s origin.

What this does to everything the site computes #

A picture with no centre has no viewing distance, and that is not a gap in the analysis — it is the analysis. Four things this site computes for every other picture simply do not exist here:

No station point, so there is no place to stand at which the picture becomes a correct projection of the scene. The viewing field’s entire question has no answer for a scroll, and the honest response is to say so rather than to compute the across-roll distance and present it as the figure’s.

No recovered focal length in the usual sense. The across-roll projection has a focal length and it is recoverable; the along-roll direction has a compression instead, which is not a focal length and has no viewing distance attached. A figure that quoted one number would be quoting half the picture.

No cross-ratio on a drawn line, because four collinear world points do not image to a line at all.

And no height from a single view, by the usual route: the height recovery is a cross-ratio along a vertical against a known reference, and the cross-ratio is not available.

This is why the figures in this field print no single viewpoint — the rays miss by 7.97 m where every other figure on the site prints the distance it is correct from. That strip is under a gate: scripts/viewpointcheck.mjs holds the list of figures allowed to print it, and — since this phase — it demonstrates the miss for a scroll the way it already demonstrated it for water, with the same solver and the same control.

The fit is not the only way to ask, and the others agree #

A least-squares fit is one test, and a reasonable objection to any single test is that it might be answering a question about the estimator rather than about the object. Three independent checks say otherwise, and they are worth listing because none of them runs the solver.

The rays are coplanar in the wrong way. Every ray of a scroll lies in a plane of constant $x$. A bundle through a common point spans directions in all three dimensions from one origin; this one spans a one-parameter family of parallel planes. No point lies on more than one of those planes unless the planes coincide, so the only bundles with a common point are the ones drawn from a single column. That is a statement about the model, needs no arithmetic, and gives the same answer.

The image of a straight line is a curve. A projection through a centre maps lines to lines, because the plane through the centre and the line cuts the picture surface in a line. So a system that curves a straight line either has a curved surface or has no centre, and the surface here is flat. That is a second route to the same conclusion by way of the previous essay.

And the ratio survives in one direction only. A projection through a centre destroys the ratio in which a point divides a segment, in every direction, without exception — that is the exact difference between the perspective and parallel families. A scroll preserves it along the roll and destroys it across. No projection through a centre does that, so the scroll is not one, and the next essay is the measurement.

Three routes, no shared code, one answer. That is the pattern the site uses whenever a result would otherwise rest on a single number coming out of a single solver, and it is the reason the 7.97 m can be quoted as a property rather than as an output.

The one thing that does have a centre #

It would be wrong to leave the impression that nothing about a scroll is a projection, because one thing is, exactly.

Take a single column. Every ray in it leaves the same point, so a single vertical slice of a handscroll is a perfect one-dimensional perspective picture, with a real focal length, a real station point and a real viewing distance. The site’s whole apparatus applies to it without modification.

A scroll is therefore a continuous family of exact projections, indexed by position along the roll, whose centres lie on a line. That is a more precise statement than “a scroll has no viewpoint” and a more useful one, because it says exactly what is available: everything a projection gives, in one dimension, at every point of the picture, with no consistency between neighbouring columns beyond the track being straight.

The next essay measures what that costs and what it buys, and the answer is not symmetric — the roll direction ends up with a stronger guarantee than perspective offers, not a weaker one.

The two other pictures with nothing behind them #