The plane is a choice

Every construction in this field so far has been written with the picture plane square across the view. The horizon at eye level, the distance point on the horizon, the measuring point on the horizon: all true, and all true because the plane is vertical and the axis horizontal.

That is a choice, and it is a different choice from where the eye is. A projection has two independent pieces — a centre and a surface — and this essay is about what happens when the second moves while the first stays put.

One eye, two planes, one matrix #

The theorem is short and it is the spine of everything below.

An image point is a ray through the eye, written in the picture’s own coordinates. Changing the picture plane changes the coordinates and does not change the ray. So the map from one picture to the other is

$$H = K_2 R_2 R_1^{\top} K_1^{-1},$$

a product of a rotation and two calibrations — a $3\times3$ matrix acting on homogeneous image coordinates, with nothing about the scene in it.

That last clause is the content. The same matrix moves every point of every scene, so the two pictures are related by one map whatever is in front of the camera: a flat wall, a room, a forest.

Checked on a solid scene rather than on a plane: eighty points spread over three depths, projected through a level camera and through one pitched 18° from the same eye, and the map takes every one of them from the first picture to the second to 2.5e-13 px.

And a moved eye is not #

The counter-case is what makes the theorem a statement rather than a definition, and it is the case people assume is covered.

Move the eye 0.9 m sideways and take the same scene again. Fit a homography to four of the scene’s own points — which always exists, since four correspondences determine one — and apply it to the rest.

The four fitted points come back exactly. Everything else is out by up to 32.0 px.

That gap is parallax: the amount by which a nearer thing shifts more than a farther one when the eye moves. It is not a small residual and it is not reducible by a better fit, because no map of the picture can move two points at the same image position by different amounts, and two points at the same image position at different depths are exactly what a moved eye separates.

So the pair of statements is:

Same eye, different plane: a homography, for any scene.

Different eye: a homography for a plane, and nothing for a solid scene.

The second half is the reason rectifying a photographed façade works and the reason a photograph cannot be re-staged from a different viewpoint by any amount of processing.

What the tilt does to the furniture #

Three things move when the plane leans, and they are three readings of one number.

The horizon leaves the middle of the frame. It drops by $f\tan\theta$ — 213 px at 14°, 416 px at 26°. It is still the image of the ground’s points at infinity; it is no longer at the principal point, because the principal point is no longer at eye level in the picture.

The vertical vanishing point arrives from infinity. At $f/\tan\theta$ from the principal point — 3425 px at 14°, 1751 px at 26°. With the plane vertical it is infinitely far away, which is what “verticals stay parallel” means.

And the verticals converge. 3.59° at 14°, 6.50° at 26°, and exactly 0e+0° with the plane vertical — whatever the field of view, which is the part worth checking because it is the claim a wide-angle lens is usually blamed for.

The identity underneath the three #

The two offsets are not independent, and the relation between them is the pleasant surprise of this essay.

$$(f\tan\theta)\,\times\,(f/\tan\theta) \;=\; f^2.$$

The horizon’s offset from the principal point, times the vertical vanishing point’s offset, is the focal length squared — at every tilt, to 1e-15 of itself across the sweep.

That is the foundations field’s pole and polar arriving in a field with no conics in it. The horizon is the polar of the vertical direction’s vanishing point with respect to the absolute conic, and the product being $f^2$ is what a polar relation looks like when both points are on the principal axis.

It is also useful. Given a photograph with converging verticals, the vanishing point and the horizon are both measurable, and their product gives the focal length — a one-line calibration from two things a straightedge finds, needing no known object in the scene.

The convergence is the third measurement, and it is the one on the page #

The identity is useful and it has an awkward practical edge: the vertical vanishing point is at $f\cot\theta$, which at 14° is 3,425 pixels — some way outside a 690-pixel frame. A reader with a photograph can measure the horizon’s offset with a straightedge and cannot measure the vanishing point at all without extending lines across the desk.

The convergence angle is the quantity that is on the page, and it can stand in. Two verticals at image positions $\pm a$ both run to a vanishing point $f\cot\theta$ away, so the angle between them is

$$\beta \;\approx\; \frac{2a\tan\theta}{f}.$$

Two consequences, and the second is a calibration.

The convergence is not a property of the tilt alone. It carries $a$, the verticals’ own separation on the page, so two edges near the middle of the frame converge less than two at its sides at the same tilt — and a photographer judging “how much is this leaning” is reading a product of the tilt and the framing. The measured 3.59° and 6.50° at 14° and 26° correspond to verticals about 107 pixels either side of centre; the same tilts on verticals at the frame’s edge would give twice as much.

And it completes the identity without leaving the paper. The horizon’s offset is $h = f\tan\theta$ and the convergence gives $\tan\theta = \beta f/2a$, so eliminating the tilt:

$$f^{2} \;=\; \frac{2\,a\,h}{\beta}.$$

Every quantity on the right is a measurement inside the frame — the half-separation of two drawn verticals, the horizon’s drop below the principal point, and the angle between the verticals. Against this sweep: $2 \times 107.3 \times 213 / 0.0627 = 729{,}500$ against $f^{2} = 729{,}154$, which is four parts in ten thousand and is the arctangent this expression linearises.

So a photograph of a building with converging verticals states its own focal length three different ways: from the two vanishing points, whose product is $f^{2}$ but one of which is off the page; from the horizon and the convergence, which are both on it; and from three perpendicular directions, which is the general recovery. The middle one is the reader’s route, and it is the one no treatment writes down, because the convergence angle is usually treated as a defect to be removed rather than as a measurement.

It also says which photographs refuse. $\beta$ appears in the denominator, so a nearly-level camera gives a nearly-zero convergence and the recovery becomes ill-conditioned — the same pole the ramp’s own vanishing line meets at zero gradient, and the same answer: the construction is exact everywhere and useful over a range, and the range excludes exactly the pictures that did not need correcting.

There is a matching refusal at the other end, and it is less obvious. As the tilt grows the horizon leaves the frame — $f\tan\theta$ passes half the frame width at $\theta$ equal to the half-field of view — so a steeply tilted photograph has a large convergence and no measurable $h$. Between the two refusals lies a working band, and its width is the camera’s own half-field: the recovery works when the tilt is smaller than the half-angle of the lens and larger than about a degree. A wide lens therefore has a wide band and a long one a narrow band, which is the reverse of how the two are usually ranked for architectural work and is a fair summary of what this field keeps finding — the lens that makes the picture harder to look at makes it easier to measure.

Why the scene does not appear in the matrix #

The scene-independence is the property everything else depends on, and it is worth one section on why it holds, because the neighbouring case where it fails is only one word away.

A point’s image records the direction from the eye to it and nothing else. Two points on the same ray at different distances have the same image, exactly — which is the whole of what one view supplies — so a picture is a picture of the ray bundle rather than of the scene.

Changing the picture plane re-coordinatises that bundle. Nothing about which rays exist changes, so the map between the two coordinate systems is a map of the bundle, and the bundle is the same for every scene.

Move the eye and the bundle itself changes: a different set of rays, with different points on them. There is no map of the old picture that produces the new one, because the new one contains information — the depth-dependent shift — that the old one does not have.

So the theorem’s scene-independence and the counter-case’s parallax are the same fact told twice. A picture knows the rays and not the points, and an operation that only re-describes the rays is available while an operation that needs the points is not.

Four operations inside one group #

Once the plane is a free choice, several separately-named techniques turn out to be the same operation.

Tilting the camera rotates the plane about a horizontal axis. Keystone, three-point perspective, converging verticals.

Shifting the lens moves the principal point without rotating the plane. Verticals stay parallel, and the whole image translates, because $c_x$ enters the projection additively.

Cropping takes a rectangle out of the picture the plane already carries. Which is the same operation as a shift, exactly — and the figure below measures it: the shifted frame’s points sit at one constant offset from a wider picture’s, spread 1e-13 px over the whole scene.

And rotating the camera about the eye — panning to stitch a panorama — is a plane change with the axis vertical, which is why a rotation homography stitches and a step sideways does not.

All four are elements of the group of homographies of the picture, and the group is the answer to “what can be done to a photograph without moving the photographer”.

The frame is not the plane #

A distinction the group makes obvious and ordinary language hides: the picture plane is unbounded, and the frame is a rectangle drawn on it.

Choosing the rectangle is a crop, or a shift, or a change of film format. Nothing about the projection changes; a different part of the same picture is kept.

Choosing the plane is a tilt or a rotation. The projection changes — different coordinates on the same bundle of rays — and every point in the frame moves by an amount depending on where it is.

Both are homographies of the picture, so both are in the group; they are different subgroups of it, and the difference is visible in the measurements. A shift’s offsets are constant over the scene — 1e-13 px of spread, which is the definition of a translation. A tilt’s offsets spread by 25.8 px over the same scene, which is what a projectivity does.

The practical form of the distinction is the one every architectural photographer knows and states as a rule of thumb: shifting costs field and tilting costs shape. Shifting keeps the geometry and throws away the parts of the image circle the frame no longer covers. Tilting keeps the frame and converges the verticals. The rule of thumb is exactly the subgroup statement, and the numbers above are its price list.

It also explains an operation that looks like a third option and is not. Cropping the top out of a tilted picture does not straighten anything — the plane is still tilted, so the verticals still converge, and the crop has merely chosen a smaller rectangle on a leaning plane. To straighten, the plane has to change, which is the next essay.

What a plane change cannot do #

The boundary is the reason the theorem is worth stating carefully, and it has two halves.

It cannot move the eye. Every picture in the group is a projection of the scene from the same point. The reader’s viewing position for a corrected picture is computed from its own focal length and displayed width and will generally differ from the original’s — but the scene’s relationship to the observer is fixed, and no processing supplies the view from somewhere else.

And it cannot un-mix the depths. Because the map is scene-independent, it carries no information about depth; because it is a bijection of the picture, it cannot add any. Everything a photograph left out stays left out.

Which is exactly what makes the group useful. A transformation that changes nothing about the geometry’s content is a transformation that can be applied without asking what is in the picture, and that is why keystone correction is a slider in every photo editor and re-staging from a different position is not a feature anywhere.

What a reader can check in a photograph #

Three things follow that need no computation, and they are the reason this is worth knowing rather than merely true.

Converging verticals mean a tilted plane, not a wide lens. The convergence is 0e+0° with the plane vertical at any field of view, so a wide-angle picture with parallel verticals is perfectly possible and a long-lens picture with converging ones is too. Blaming the lens is blaming the wrong parameter.

The horizon’s position measures the tilt. It sits at the principal point when the plane is vertical and drops by $f\tan\theta$ as the camera points up, so a horizon low in the frame is a camera pointing up and nothing else. Combined with the vertical vanishing point it gives the focal length.

And a picture with no convergence carries no evidence about the tilt. That is the refusal the machinery makes rather than a caution: with the plane vertical the imaged verticals are parallel and their meeting point does not exist, so there is nothing to compute from. A method that returned a tilt anyway would be reporting the intersection of two nearly-parallel lines, which is the failure mode the projector’s recovery refuses by name.

The other kind of map, for contrast #

There is a second family of maps of a picture on this site, and the census that separated them is worth invoking because the difference is exactly the one this essay turns on.

A central collineation — a homology or an elation — has a line of fixed points, and the census found three of them: a shadow, a floor anamorph, a mirror. Each changes where a plane is seen from: the plane stays put and the projection centre moves.

A general projectivity has no such line. A rectification is one: both ends of the map are pictures of a plane rather than the plane and a picture of it.

A plane change is in the second family. Both ends are pictures, taken from the same point on two different surfaces, and there is no line of the world sitting in both to be fixed. Which is a compact way of saying what has and has not changed: the eye has not moved, and neither picture is the world.

Where the plane’s freedom is used #

Three places on this site turn out to be this essay’s theorem applied, and none of them announces it.

Keystone correction of a projector. A projector is a camera run backwards, and correcting its keystone re-casts its picture onto the plane the wall actually occupies — at a cost of 18.7% of the panel’s pixels at 15° off square, because correction cannot add light outside the thrown quadrilateral.

Stitching a panorama. Frames from one point differ by plane changes, so they compose exactly; frames from a moving camera do not, and the error is the parallax measured above.

And the standard construction’s own conditions. “The horizon is at eye level” is a statement about a vertical picture plane, and the figure that measures it says so in its title. This essay is the rung where the condition is removed rather than assumed.