Four marks before anything is said

Before asking what a drawing says, it is worth asking how much it can say, and the answer is a small integer.

Three procedures, one panel established that a correctly executed pavement carries no record of which correct procedure drew it. This is the count underneath that: how many independent statements a pavement of a given size makes at all, and therefore how large a drawing has to be before any reading of it can fail.

Three unknowns, one equation per mark #

A reader holds a row of transversals and wants to know whether they could have been projected. The model has three numbers in it.

Where the horizon is. Not assumed — the reader may not have the orthogonals, and even with them the question of whether the pavement agrees with them is exactly what is being asked. So the horizon is a free number.

And two more that fix the projective map. A projective map of a line to a line has three degrees of freedom; one has already been spent naming the horizon, which is the image of the point at infinity, and the remaining two are a scale and an origin.

Each transversal supplies one equation. So a pavement of n transversals makes n − 3 independent statements, and below four it makes none.

The same number arrives from the invariant side, which is the reason to trust it. Four consecutive transversals give one cross-ratio; n of them give n − 3 consecutive quadruples; and each quadruple is one number that has to come out at 4/3. Two different derivations, one count.

What “unfalsifiable” means here #

It is a strong word and it is meant literally. A pavement of three transversals is not hard to attribute; there is nothing to attribute, because every arrangement of three marks on a line is the image of three evenly spaced ground points under some projective map.

Pick any three marks. There is a horizon, a scale and an origin that carry 0, 1 and 2 exactly onto them, and the construction that finds them is elementary. So a three-tile pavement drawn freehand by somebody who had never heard of a vanishing point is, as a set of marks, indistinguishable from one drawn by Alberti’s own hand.

This is the same structure what one picture of a plane determines counts for a plane rather than a line. There, four points fix a homography and the fifth is the first that can disagree; here, three fix a projective map of a line and the fourth is the first. The pattern is general and it is the reason five marks and the sixth is a title rather than a curiosity.

And this matters because pavements are small #

The count would be a footnote if drawings were large. They are not.

A painted tiled floor typically shows three or four rows of tiles before the furniture, the figures or the edge of the panel cut it off. Four rows is four transversals and one testable statement — a single cross-ratio against a single expected value, with the reader’s own ability to place a mark deciding whether the comparison means anything.

So the honest position on most painted pavements is not that they pass or fail a projective test. It is that they carry between zero and two independent statements, which is not enough to distinguish a construction from a good eye.

Read down that first column and it is zero everywhere, which is the control: a reading that reported a slip where there was none would be reporting itself.

What a longer pavement buys, which is less than it looks #

The obvious expectation is that doubling the number of transversals roughly doubles the reading’s power. It does not, and the reason is worth having.

A pavement’s far transversals are squashed. The gap between the seventh and the eighth is a small fraction of the gap between the first and the second, so a mark misplaced by the same amount on the page produces a much larger error in the ratios there — and a much smaller departure in pixels from where a correct perspective would have put it. The extra statements a longer pavement supplies are statements about the part of the drawing where the least is happening.

Measured across the sweep above, going from four braccia to fourteen at a hand precision of half a pixel moves the detection rate from fifteen per cent to forty-five. Trebling the length does not treble the reading; it moves it by a factor of three at the one precision where the reading is marginal, and by nothing at all at the precisions where it already works or already does not.

Why the returns converge #

The reason a longer pavement buys so little can be written down, and the sum converges.

A slip of a given size in the ground marks displaces the $k$-th transversal by an amount that falls as $1/k^{2}$, since that is how the pavement’s own gaps close. Evidence goes as the square of a displacement over the reading noise, so the evidence carried by the $k$-th mark falls as

$$\frac{1}{k^{4}},$$

and the total over an unbounded pavement is $\sum k^{-4} = \pi^{4}/90 = 1.0823$ times whatever the first mark carries. Ninety-two per cent of everything a pavement of any length can say is in its first transversal, and 99.6 per cent is in its first four. A floor running to the horizon is worth about eight per cent more than a floor of four rows.

Which leaves the sweep’s threefold rise in detection rate to be explained by something else, and it is not information. A longer pavement supplies more independent statements — $n-3$ of them — and the reading tests the worst of them against a fixed threshold, so more statements means more chances for one to cross it. That is a multiple-comparison effect, and it is why the rise appears only at the precision where the reading is marginal and disappears at the precisions where it already works or already does not.

Stated as advice to a reader: measure the near transversals well rather than measuring many. A fourteen-braccia pavement read carelessly says less than a four-braccia one read under a glass, and no amount of floor compensates for a mark placed to the nearest pixel.

The tolerance is the reader’s, and it is the only choice #

Everything above is stated against a threshold, and it is worth being clear that the threshold is a fact about the reader rather than about the geometry.

The residual is reported as a displacement in pixels on the page: how far the worst transversal is from where a correct perspective would have put it. A fifth of a pixel is a mark read under a glass on a good reproduction. A whole pixel is a mark read across a room. Everything else in this row — the square root of six that separates one kind of hand error from another, the count of three unknowns — comes out of the arithmetic and is not chosen.

That single choice moves the answers a long way, which is why it is printed on every figure that uses it rather than folded into a constant.

The middle of that range is where the pavement’s length does any work at all. Outside it the reader’s eyes decide the answer and the drawing does not.

Why the horizon has to be free #

A reader who is given the horizon has only two unknowns, so the count becomes n − 2 and a three-transversal pavement makes one statement. That looks like a better position and it is a different question.

Handing the reader the horizon means asserting that the pavement was drawn to the orthogonals that meet there. On a correctly constructed panel it was; on a panel where somebody spaced the transversals by a rule and drew the orthogonals separately, it was not, and that disagreement is exactly the evidence the rule that draws another room turns on. Assuming the horizon assumes away the most informative thing the drawing has.

So the count is done twice, and both are worth carrying. Without the horizon a pavement makes n − 3 statements and they are about the pavement alone. With it the pavement makes n − 2, one of which is the agreement between the two halves of the drawing — and that one is worth more than all the others together, because it is the only one that a slipping hand and a systematic rule answer differently.

The count is about the model, not about the noise #

Two things get confused here and separating them is most of what the count is worth.

Identifiability is whether the data can separate the numbers the model asks for at all. It is a property of the arrangement and it does not improve with a better reproduction, a steadier hand or a longer look. Three transversals do not identify anything, and photographing the panel at ten times the resolution leaves them identifying nothing.

Precision is how tightly the numbers are pinned once they are separable. That does improve with the reproduction, and it is what the reader’s ruler is about.

The ladder of assumptions is a ladder of conditioning is this collection’s general statement of the distinction, made about the chain from a projective reading to a metric one. The pavement is the smallest case of it: below four transversals the reading is unidentifiable, at four it is identified and appallingly conditioned, and by ten it is identified and merely poor.

The failure mode the distinction guards against is a familiar one. A reader who measures a three-tile pavement very carefully, fits a horizon and a scale to it, and reports that they agree beautifully has measured nothing: the fit was exact before the measuring began, and its exactness is a fact about the count rather than about the painter.

What the count does not cover #

Three things sit outside it, and two of them are the reason this row has more than one essay.

The orthogonals. A pavement has a second family of lines — the ones running away from the reader — and they carry their own statements: they should be concurrent, and where they meet should be the horizon the transversals imply. The count above is about the transversals alone. Adding the orthogonals adds constraints, and the rule that draws another room is the case where those constraints do all the work.

The tiles’ width. A square pavement’s tiles are square, which is a statement about the two families together and needs the tiles’ diagonals to test. It is the strongest thing a pavement can be asked and it is also the thing that is most often false on purpose: the proportion is the assumption shows how much of a reading depends on a shape nobody stated.

Everything outside the pavement. A drawing usually contains other straight things — a wall, a beam, a window’s head — and those are a second family of marks in the sense the lens a pavement can hide needs. A reading confined to the transversals is a reading that has thrown away most of the page.

So the count is a floor rather than a ceiling. It says how much the transversals alone can carry, which is the right first question because it is the part of the drawing an attribution claim always points at.

Why the fourth mark and not the third #

There is a temptation to say that the cross-ratio needs four points, so of course four is the answer, and to leave it there. That is the right number for the wrong reason, and the right reason is more useful.

The cross-ratio needs four points because a projective map of a line has three degrees of freedom and four points is the smallest set with one more point than the map has freedoms. Change the model and the number changes with it. A reader who knows the horizon has a two-parameter model and needs three marks; a reader who knows the horizon and the scale has a one-parameter model and needs two; a reader who is prepared to assume that the panel was built at a stated viewing distance has no free parameters at all and can test a single transversal.

Each of those is a real reading and each buys its power by asserting something the drawing does not contain. That is the trade the ladder of assumptions prices, and the pavement is where it is cheapest to see: the count of marks a reading needs is exactly the count of numbers it refuses to assume, plus one.

The count against the other invariant readings #

This collection makes several readings of this shape and it is worth putting them side by side, because the counts differ and the differences are informative.

Recovering the camera from the picture it drew needs three mutually perpendicular directions and returns five numbers, so a drawing with two vanishing points is short by one and a drawing with three is exactly determined. Flattening a façade out of the photograph needs four points whose world positions are known up to a rectangle and returns eight, so the fifth point is the first that can disagree. A height, out of one photograph needs a vertical, a horizon and a known length, and returns one number with no redundancy at all — which is why that reading has no internal check and this one does.

The pavement is the cheapest of the four to run and the poorest in what it returns, and both of those follow from the same count. It asks one thing of four marks and it asks it without a camera, a scale, a known length or a second view.

A note on what the marks are #

One practical point, because it decides whether any of the arithmetic above reaches a real drawing.

A transversal is a line, and a reader takes a mark off it by choosing a point on it. Which point does not matter — the reading is along the receding direction, so any point on the transversal gives the same position along that line — but the line has to be found, and on a painted floor it is found from a joint, a change of colour, or the far edge of a row of tiles.

Those three are not the same line. A tile’s far edge, the mortar joint beyond it and the near edge of the next tile are three parallel lines a few millimetres apart in the world, and in the picture they are three transversals whose separation shrinks with depth exactly as the tiles do. Reading the near edges in the foreground and the far edges in the distance mixes two families and puts a systematic error into the residual that looks like a rule.

The fix costs nothing and is worth stating because it is the commonest way a reading of this kind goes wrong: pick one feature of the tile and read it every time. The invariant does not care which one, and it cares a great deal that it is the same one, because the whole test is a comparison between marks and a mark taken from the wrong family is not a smaller error, it is a different measurement.

What a reader should do with a small pavement #

Three steps, in the order that costs least, for somebody standing in front of a painting.

Count the transversals. Below four, stop: nothing about the spacing can be evidence for anything, and any published claim that it is has not done this arithmetic.

Find the orthogonals’ meet. That is one mark, made with a straightedge on a reproduction, and it converts a pavement of n statements into one of n − 2 — including the one statement that is worth having. It also costs nothing and is the step most often skipped.

Then read the four marks. The cross-ratio, against 4/3, at whatever precision the reproduction allows — and quote the precision, because the answer depends on it more than on anything else in the drawing.

The short version #

A pavement of n transversals makes n − 3 independent statements about whether it is a projection, so three make none and four make one. A great many painted floors have four rows of tiles.

Admitting the horizon the panel’s own orthogonals meet at raises the count by one, and the one it adds is worth more than the rest, because it is the only statement a slipping hand and a systematic rule answer differently.