The horizon has a pole

A point and a line are one object establishes that a duality is a choice: pick any invertible symmetric matrix and every incidence theorem acquires a partner. Nothing in that essay singles out a matrix, and nothing could, because the projective plane has no distinguished conic.

A photograph does. Its camera has a focal length and a principal point, and those two fix a conic in the picture that no figure chose — the image of the absolute conic, ω = (K Kᵀ)⁻¹. Under the polarity that conic defines, the horizon of a plane and the vanishing point of that plane’s normal are pole and polar.

This collection has been using that sentence as an aside since its foundation essays. Here it is a number.

What the conic is #

The absolute conic is the set of directions at infinity satisfying x² + y² + z² = 0 — no real points, and the whole of Euclidean structure encoded as a projective object. One conic calibrates the camera is where this collection built it and where the recovery of a focal length from two orthogonal vanishing points was shown to be one statement about it.

Its image in a picture is ω, a symmetric three-by-three depending only on the calibration. It has no real points either, so it draws nothing; what it does is decide, for every point of the picture, a line — and for every line, a point.

Why the horizon and the vertical #

A plane’s vanishing line is the image of its directions at infinity. Its normal’s vanishing point is the image of one direction — the one perpendicular to all of them.

Perpendicularity in the picture is the polarity ω defines: two directions are perpendicular exactly when each vanishing point lies on the other’s polar. Apply that to a whole plane’s worth of directions and the normal’s vanishing point must lie on the polar of every one of them, which means it is the pole of the line they span. That is the horizon.

So the relation is not a coincidence of the ground plane. It holds for any plane in the scene and its own normal, and the ceiling, a wall and a ramp each carry their own instance of it.

The measurement #

The camera is built, the horizon is computed from two of the plane’s own directions, ω is built from the camera’s focal length and principal point, and the pole ω⁻¹l is compared with the vanishing point of the plane’s normal computed independently.

They agree to 2.7 × 10⁻¹² pixels. That is the arithmetic floor on a 690-pixel canvas, and it is what an identity is supposed to look like.

An identity that always held would be worth nothing, so the second half of the measurement builds ω from a focal length deliberately wrong by ten per cent. The pole then lands 1,051 pixels from the vanishing point — a canvas and a half away. The relation is not merely true; it is sensitive, and its sensitivity is what lets it be run backwards.

Where the 1,051 pixels comes from #

The sensitivity is not a fitted number either. The pole of the horizon sits at $f^{2}$ divided by the horizon’s own offset from the principal point, so a relative error $\delta f/f$ in the focal length moves it by

$$2\,\frac{\delta f}{f}\times(\text{the pole's own offset}).$$

Ten per cent in $f$ is twenty-one per cent in $f^{2}$, and twenty-one per cent of a vertical vanishing point sitting some five thousand pixels out is the canvas and a half the measurement reports.

Two readings, and the second is a design rule for the calibration below.

The sensitivity is proportional to the pole’s distance, which is $f\cot\tau$ for a camera tilted by $\tau$. So a nearly level camera has a pole enormously far away and a correspondingly violent response to a wrong focal length — the relation is at its most sensitive exactly where the pole is hardest to mark.

And that cuts both ways when the relation is inverted. Reading $f$ from the two marks gives $\delta f/f = \varepsilon/2D$ for a pole located to $\varepsilon$ pixels at offset $D$ — the same “one over the offset in pixels” form a lamp’s distance obeys — so a distant pole is good for precision and bad for markability, and the two pull against each other.

The best tilt is 45° #

The trade has an optimum, and it is a tidy one. The horizon sits at $f\tan\tau$ from the principal point and the pole at $f\cot\tau$, and the calibration needs both to be measurable: too little tilt and the pole is off the page, too much and the horizon is. Their product is fixed at $f^{2}$, so the arrangement that makes the smaller of the two as large as possible is the one where they are equal —

$$f\tan\tau = f\cot\tau \qquad\Longrightarrow\qquad \tau = 45°,$$

with both marks at distance $f$ from the principal point.

So the relation calibrates best from a camera tilted at forty-five degrees, and degrades in the same way toward both ends: a level camera has an unmarkable pole, a steeply tilted one an unmarkable horizon, and neither is a different failure from the other. That is an unusually clean piece of advice for a photographic measurement, and it is decided entirely by the invariance of the product.

It also says what the relation costs on the photographs people actually take. Architectural photographs are tilted by a few degrees, so their poles sit at ten or twenty focal lengths — hundreds of thousands of pixels — and the calibration is unusable however sensitive it is. The relation is at its best on a picture nobody frames deliberately: a camera pointed halfway down, which is a snapshot of a floor rather than a photograph of a building.

Which is worth knowing before reaching for it. The identity holds everywhere and is measurable in a narrow band, and the band is centred on a tilt that no convention recommends.

The band’s width follows from the same product. Requiring both offsets to be inside a frame of half-width $W/2$ needs $f\tan\tau < W/2$ and $f\cot\tau < W/2$, which together demand $f < W/2$ — a field of view of at least 90°. So the relation is measurable on a wide-angle picture near 45° of tilt and on nothing else, which is a narrower door than the identity’s generality suggests and is worth stating beside it.

Run backwards, it is a calibration #

Two marks on a photograph — the horizon of a plane, and the point where that plane’s normals converge — and one equation in one unknown.

With the principal point at the origin, ω⁻¹l works out to x − pₓ = f²a/σ and y − p_y = f²b/σ, where (a, b, c) is the horizon and σ = a·pₓ + b·p_y + c is the horizon’s own value at the principal point. Either equation gives f².

Across seven cameras from a wide one to a long one, the recovered focal length matches the true one to the last bit. Nothing about the scene enters: no known length, no right angle in the world, no square, no repeated interval. Two straightedge readings and a principal point.

Which of the two equations carries the reading #

The two equations are not equally informative, and the honest form says which one is doing the work.

A camera with no roll has a horizontal horizon: a is zero and b is one. Then the x equation is 0/0 — the vertical’s vanishing point sits directly below the principal point, so x − pₓ is zero as well — and the whole reading comes from the y equation. Averaging the two would be reporting a division nobody did.

So what is quoted beside the recovered focal length is the share, a²/(a² + b²), and on a level camera it is 1.0 × 10⁻³⁴. The information is entirely in one equation, and that is a fact about the picture rather than about the arithmetic: a level horizon says nothing about the horizontal placement of the vertical’s vanishing point, because there is nothing to say.

What it costs to be wrong about the principal point #

The relation involves the principal point twice — once in σ and once in the shift — so it is not free of the assumption every recovery here has to make somewhere.

That is the same dependence the principal point is not the centre measures for the lens field’s own recoveries, and it is worth naming rather than glossing: this is a two-mark calibration given a principal point, not a three-parameter one. A photograph with an unknown shift needs a third piece of evidence, which is what the three-vanishing-point recovery supplies and what makes it the more general instrument despite needing a box.

Every plane, not just the ground #

The relation is about a plane and its normal, so a picture with several planes in it carries several instances, and they are independent readings of the same focal length.

A ramp has its own horizon — the ramp has its own horizon is where that was established — and the pole of the ramp’s horizon is the vanishing point of the ramp’s normal, which is a direction nothing in the picture is drawn along. So the relation predicts a mark that no edge in the drawing points at, which is a much stronger test than predicting a mark the uprights already converge on.

What the relation says about a picture with no ground #

A photograph of the open sea, or of a wall filling the frame, has no plane whose horizon can be found — and the relation then says nothing, which is the right answer and worth stating as one.

That is not a limitation of this particular recovery. It is the same limitation every single-view calibration on this collection has, expressed in the vocabulary of the polarity: ω has six coefficients up to scale, five free parameters, and a picture supplies constraints on it only where the scene contains structure the eye can recognise as perpendicular or parallel. No structure, no constraints, and the conic stays unknown however sharp the photograph is.

What the polarity adds is a clean account of which structure buys what. A pair of perpendicular directions buys one constraint. A plane and its normal buys one constraint, and is easier to find. A circle whose image is a conic buys two, which is why the conic a circle becomes is the strongest single piece of evidence a photograph can contain. The ladder is the same one the ladder of assumptions is a ladder of conditioning walks, seen from the dual side.

Where this sits among the other dualities #

The census in a point and a line are one object has eight rows, and this is the last of them and the only one whose duality is not a choice.

The distinction is worth stating precisely, because “the horizon is dual to the vertical” is the kind of sentence that sounds like the others. Under the standard duality — the identity matrix — the horizon’s dual point is somewhere with no meaning at all. It is only under ω that it lands on the vertical, and ω is the camera’s. Change the lens and the same drawing’s horizon has a different pole.

That is what makes the relation evidence: a duality fixed by the instrument is a duality that can be measured to find the instrument.

What a straightedge can and cannot do here #

The pole of a line with respect to a drawn conic can be constructed with a straightedge alone, which is the whole content of the polar with a straightedge: two secants, a complete quadrangle, and the line through its remaining diagonal points.

ω is not drawn. It has no real points, so no secant meets it and the quadrangle cannot be built. The pole of the horizon must therefore be computed from the calibration rather than constructed on the page, which is the exact sense in which this relation is the calibration: the construction and the knowledge of the camera are the same thing.

Turned round, that is why the recovery works. Marking the horizon and the vertical’s vanishing point on a photograph and solving for f is constructing the conic the straightedge could not reach.

The same relation on a picture that is not level #

A camera with roll turns the horizon and makes both equations informative, and the share becomes a number worth reading rather than a machine zero. The two readings of f² then agree to the precision of the marks, and their disagreement is a check nothing else supplies: the relation is one equation in one unknown twice over, and two readings of one unknown is a residual.

That is the shape this collection asks of every recovery — recovering the camera fits three vanishing points and reports how far the third is from where the first two put it, and what one picture determines is the census of what such readings can reach at all. Here the redundancy is smaller and cheaper: two marks, two equations, one number and one residual.

What it cannot do is find the principal point. Both equations were written with the shift assumed, and a shift moves the pole in exactly the direction a change of focal length does not, so the two are separable only with a third mark. That separation is what the box in the three-point recovery is buying, and naming the price is the difference between a calibration and a claim.

Why this is not the same as the two-vanishing-point calibration #

One conic calibrates the camera recovers a focal length from two vanishing points known to be perpendicular. This recovers one from a line and a point known to be perpendicular. They are the same equation with different evidence, and the difference in evidence is the whole practical point.

Two perpendicular horizontal directions require two families of parallel edges in the scene at a known right angle — a building corner, a floor tile, a box. A horizon and a vertical require a ground plane and something standing on it, which is nearly every photograph ever taken. The relation does not become more powerful; the evidence becomes easier to find.

And the failure modes differ. Two nearly parallel horizontal families give a badly conditioned right angle and a focal length that swings; a horizon fitted from a few short uprights gives a badly conditioned line, and the pole moves along the direction the line is least determined in. Both are conditioning rather than ambiguity, in the sense an ambiguity is not an uncertainty makes precise: more evidence shrinks them.

What the relation needs from the drawing #

Two marks, and it is worth being precise about how they are obtained, because neither is free.

The horizon comes from two vanishing points of directions in the plane, or from a fit to several — horizon from uprights is the practical version and it comes with a scatter. The vertical’s vanishing point comes from two or more uprights, and on a nearly level camera it is far below the frame and badly determined along the direction the uprights are nearly parallel in.

So the recovery inherits two fits, not two exact marks, and its precision is theirs. What it does not inherit is any assumption about the scene’s contents beyond “there is a plane and things standing on it”, which is the whole of its advantage.

The short version #

A picture fixes a conic its author did not choose, and under the polarity that conic defines, a plane’s horizon and its normal’s vanishing point are pole and polar. Checked against the camera that drew the picture, the two agree to 2.7 × 10⁻¹² pixels; built from a focal length ten per cent wrong, they are 1,051 pixels apart.

Because it is sensitive, it inverts. Two marks a reader can make with a straightedge on a photograph — the horizon of any plane, and where that plane’s normals converge — give the focal length exactly, with nothing known about the scene. And on a level camera the whole of that reading comes out of one of the two equations, which the figure says rather than hides.