A page is bounded by a divide, not a centre

What perspective gave up listed the page budget among the four things a pinhole destroys: of a depth range from three metres to sixty, sixty-nine per cent lands in the tenth of the page nearest the far end, and a linear depth map would put ten per cent there. It traced that cost, with the other three, to one operation — the divide by depth — and called the trade a single trade with four faces.

The same divide has a face that is not a cost, and it is worth measuring rather than stating, because the obvious way to state it is wrong.

The obvious statement goes: a parallel system cannot draw an unbounded ground on a bounded page, because it draws a stretch of ground at the same size however far away it is; a pinhole can, because everything beyond some distance is packed into a band below the horizon; so a centre of projection is what lets a picture hold a whole world. The first half of that is true and measurable. The second half credits the wrong property.

The ground, and how the page it costs is counted #

Every system is handed the same stretch of world: the wedge of ground a pinhole of the comparison table sees, 46° wide, from two metres in front of the eye out to some distance D. The question is how much page each system spends drawing it — not how much of it each would choose to frame, but the area its drawing of that ground occupies.

The area is counted without a formula. The wedge is cut into a fine mesh, spaced geometrically in depth and evenly across, every cell is projected by the system in question, and the drawn areas of the cells are summed. For a pinhole and for every parallel system the cells’ edges are straight on the page and the sum is exact; for a handscroll the cells’ sides are hyperbolas, as a straight line in a scroll is, and the sum converges as the mesh refines. The closed forms below are then a check on the mesh rather than its source: the level pinhole’s drawn wedge matches its formula to twelve decimal places at 4 m, at 64 m and at 10 km.

Three laws #

Three of the four curves are three different laws, and each is exact.

The pinhole’s page is bounded. Its drawn wedge out to distance D has area $2f^{2}h\tan\alpha\,(1/z_{0} - 1/D)$, which approaches a fixed number as D grows: 434,628 px² for this camera, whatever D is. Everything beyond any given distance is packed into what is left under that ceiling, and the ceiling is never reached.

The handscroll’s page grows as the logarithm of the distance, as $2sfh\tan\alpha\,\ln(D/z_{0})$. It never stops growing, but it grows slowly: from 256 m to a thousand kilometres, a factor of four thousand in distance, its drawn wedge less than triples.

Every parallel system’s page grows as the square of the distance, as $J\tan\alpha\,(D^{2} - z_{0}^{2})$, where J is the page area the system gives one square metre of ground. Out to a thousand kilometres the cavalier drawing of the same wedge would cover about 2 × 10¹⁴ px² — a sheet some two hundred kilometres on a side at an ordinary screen’s density.

The law is easiest to see one doubling at a time — but the first of the three has a picture that makes its arithmetic unnecessary, and it is worth drawing first.

The ceiling is a rectangle on the page #

The wedge of ground is 46° wide and the level pinhole’s field is 46° across, so the wedge’s two edges are exactly the frame’s two edges, drawn as vertical lines. A strip of the wedge at distance z is 2z tan α wide in the world and is divided by z on the page, so every strip is drawn at the same width: 2f tan α, which for this camera is 690 px — the frame’s own width. The strips differ only in how tall they are drawn, and a strip of depth dz at distance z is drawn fh dz/z² tall.

So the whole wedge, out to any distance, is drawn as a column 690 px wide, running from the row of its near edge up toward the horizon and never reaching it. Its height out to infinity is fh/z₀ — 630 px for a camera 1.55 m up with the near edge two metres out — and the ceiling is that rectangle, 690 by 630, which is the 434,628 px² the figure approaches. The near edge at two metres is drawn below the bottom of a 430 px frame, which is why the ceiling is larger than the frame is.

The other two laws have the same derivation with one divide taken away at a time. A handscroll’s strip is drawn at a width that grows with z, since nothing is divided along the roll, so its area falls as 1/z rather than 1/z², and a sum of 1/z is a logarithm. A parallel system’s strip is drawn at a width that grows with z and a height that does not shrink at all, so its area grows as z, and the sum is a square. That is also why three distances in one landscape could give a far band more page than one camera allows: a band drawn from a higher station of its own is drawn with a larger h, and h is one of the two factors of that rectangle’s height.

One doubling at a time #

Cut the ground at two metres, four, eight and so on out to 256, and ask what share of the page each doubling takes.

For the pinhole, each doubling costs exactly half what the one before it cost. The stretch from two metres to four takes 50.4% of the page spent out to 256 m, the next 25.2%, and the last — from 128 m to 256 m, a hundred and twenty-eight metres of ground — takes 0.8%. Continue the doublings forever and the shares sum to a finite page, which is the ceiling above written as a geometric series.

For the handscroll, every doubling costs exactly the same: 14.3% each, a seventh of the page for each of the seven. Doubling the distance adds a fixed amount of page, which is the logarithm written out.

For isometric — and for every other parallel system, since the ratio does not depend on J — each doubling costs four times the one before. The last doubling takes 75.0% of the page and the first a sliver too small to see. A parallel drawing of a ground spends almost all of its paper on its furthest stretch, which is the opposite of a pinhole in the most literal sense.

The three ratios — one half, one, four — are unit-free and they are the three laws in one number each. They are also the three possible answers to one question about each system, and that question turns out not to be whether it has a centre.

What each law is counting #

Each system’s page is a product of two things: how much page one row of ground gets in the direction across the picture, and how much in the direction up it.

A pinhole divides by depth in both. A strip of ground at distance z is drawn narrower as 1/z and shallower as 1/z², so its page area falls as 1/z² and the total converges.

A handscroll divides by depth in only one. Across the roll it is the pinhole’s column, so a strip of ground at distance z is drawn shallower as 1/z². Along the roll it is a scaling, a map along and a picture across, so the strip keeps its width whatever its distance. The area falls as 1/z, and a sum of 1/z grows as the logarithm.

A parallel system divides in neither. A strip of ground is drawn at the same size at every distance, so the area is constant per metre of depth, and a wedge that widens with distance gives the square.

Stated that way, what bounds the page is a divide by depth in both page directions — and a centre of projection is one way of having that, not the only one. Whether it is the only one is a question with a definite answer, and the answer is a camera.

A horizon of finite length #

There is a second way to say the same thing, in the vocabulary of drawing rather than of areas, and it connects the result to one already measured.

As the ground runs out toward infinity, where does its drawing go? In a pinhole, every row of ground at distance z is drawn at a height that approaches the horizon, and every one is drawn the same 690 px wide. So the ground at infinity is drawn as a line segment of finite length — the horizon, across the frame — and the page between the near edge and that segment is finite.

In a handscroll the rows approach a horizon too, because across the roll it is a pinhole. But along the roll nothing divides, so a row at distance z is drawn 2zs tan α wide, and the ground at infinity is drawn along a horizon of infinite length. The page between the near edge and an infinitely long line is infinite, and it grows only as a logarithm because the rows crowd toward that line so fast.

A parallel system has no horizon at all. Nothing is divided by depth in either direction, the ground’s rows are drawn at equal spacing however far out they lie, and a picture with no size–distance signal has no horizon findable from its figures for exactly that reason.

The crossed-slits camera below divides in both directions, so the columns of its far ground converge as well as its rows, and its ground at infinity is a horizon segment of finite length, as a pinhole’s is. Two divides give a finite horizon; one gives an infinite one; none gives none. Where parallel lines meet in a pinhole picture is on that finite segment, and the page it closes off is the page this essay has been measuring.

A camera with two divides and no centre #

A crossed-slits camera draws each point by the ray that passes through two straight slits: a vertical one and a horizontal one, placed at different distances. A point’s column is set by the vertical slit, as $u = u_{0} + fx/(z + a)$, and its row by the horizontal slit, as $v = v_{0} + f(h - y)/(z + b)$. With the two slits at the same distance they cross at a point, and the camera is a pinhole. With them at different distances — seven metres and ten and a half here — every ray passes through both slits and no two of them need meet.

The camera divides by depth in both page directions, with two different offsets, and has no centre. Fitting a common point to its rays over the whole frame leaves them missing that point by 0.46 m on average, which is a length and not a residual.

Its page, doubling by doubling, is not a geometric series from the start. The first doubling costs almost as much as the one before, 0.96 of it, because at two metres the two slits’ offsets are a large fraction of the distance and the camera is far from a pinhole. By the last doubling the ratio is 0.52. Far enough away, the difference between dividing by z + a and by z + b stops mattering, and the camera’s page behaves exactly like a pinhole’s.

Out to a thousand kilometres the crossed-slits camera’s page is 44,237 px², against 44,213 px² at ten kilometres. It is bounded. A camera whose rays share no point spends a finite page on an infinite ground, exactly as a pinhole does, and for exactly the reason the laws above give: it divides in both directions.

Where the pushbroom sits among them #

The crossed-slits camera places the handscroll precisely, which is worth a paragraph because the handscroll was the system that made the centre a scroll does not have a measurement.

A handscroll is a crossed-slits camera with one slit sent to infinity. Its horizontal slit is the eye’s height, at the eye’s own standoff; its vertical slit is infinitely far away, so the column coordinate does not divide by anything and becomes a scaling. Bring that slit in from infinity and the column coordinate starts dividing by depth, the growth law changes from logarithmic to bounded, and the pushbroom becomes a crossed-slits camera. Bring it all the way to the other slit’s distance and the camera becomes a pinhole.

So the three laws are one family with a single knob — how far away the second slit is — and the knob’s three positions are the three systems. At infinity, a logarithm. At any finite distance, a bound. At the other slit’s distance, a bound and a centre. The centre arrives at one setting of the knob and the bound arrives at all of them but one.

What the bound actually bought #

This changes the reading of what a pinhole’s trade was.

What perspective gave up priced four losses and named the gain as the station point: a picture that is a projection from one place, which is what makes a camera recoverable and a photograph evidence. That stands. What does not stand is an easy addition to the gain — and it fits the world on a page — because the fit belongs to the divide rather than to the centre, and a system can have the fit without the centre.

The distinction matters for how the earlier costs are read. The page budget that essay priced — sixty-nine per cent of a depth range in the last tenth of the page — is the other side of this essay’s bound. A picture that packs an infinite ground into a finite page must pack the far ground into very little page, so the depth range crowds toward the horizon. Every system with two divides pays that price and gets that bound; the pinhole pays for its centre separately, in the currencies a yes in the table is a price measured.

It also locates the elevation, which is the degenerate member of the parallel family. An elevation draws the ground edge-on, as a line, and spends no page on it at all — its curve cannot be drawn on the figure’s logarithmic axes, and its doublings are all zero. That is bounded in the most trivial possible way, by drawing nothing, and it is bounded because the ground’s normal is perpendicular to the picture rather than because of anything the projection does with depth.

What this does not say #

It does not say a crossed-slits picture is a good picture. Its rays share no point, so no place exists from which it is a correct view, straight lines are drawn as curves, and none of the recoveries a photograph permits apply. The camera is here as a separating example, the smallest object that divides in two directions without a centre, and not as a proposal.

It does not say anything about the sky, or about verticals. The wedge is ground. A vertical wall receding from the eye is drawn by a pinhole with a bounded page too, but a building rising out of the ground is drawn at a height that divides by depth once, and a sky above the horizon is drawn by a pinhole over an unbounded page unless it is cropped. The bound is a statement about planes that recede, and a picture holds more than those.

And “bounded” is not “small”. The pinhole’s ceiling of 434,628 px² is larger than the 296,700 px² of a 690 × 430 frame, because the near edge of the wedge at two metres already spills past the frame’s sides. Every system’s page depends on where the wedge starts, and the near ground is where a pinhole spends most of its page — half of it on the first doubling. What the bound says is that the far ground costs a pinhole almost nothing, not that the near ground costs it little.

Still open: what the tenth row of the table holds #

The crossed-slits camera has now been measured on one question, and it is not on the table. It belongs there, and putting it there is the open task: price it in length, area and angle on the same four hundred boxes, run the same midpoint test along each of its two page directions — it divides in both, so neither should keep a midpoint — and ask the question the exclusion is about in the sharpest form available. The exclusion says a centre and a measure do not share a row. A camera with two divides and no centre has neither; the measurement to make is whether a row with no centre can have any measure the pinhole lacks, or whether giving up the point while keeping the divides buys nothing but the loss of the point.