A circle reads a tilted camera's column, and its tilt has to come from the verticals

A circle on the ground names the column and the focal length at once found that the picture of a circle lying on the ground meets the horizon at two complex points, the images of the circular points every circle in a plane passes through, and that for a level camera with square pixels those points are $c \pm i f$: the principal point’s column as the real part, the focal length as the imaginary part. A manhole cover a metre across, eight metres ahead and marked to half a pixel, read the column to 14 px and the focal length to 1.5 per cent, with no second object and no turn to choose.

Every camera in that essay was level. Most handheld photographs are not — tilted a few degrees up or down by the hand — and a photographer standing over a manhole cover tilts the camera a great deal more. That essay closed by counting unknowns: a tilted camera’s principal point is off the horizon, so the circle’s two numbers must now constrain three, the column, the principal point’s height and the focal length, and the building’s right angle — its two vanishing points — would supply the third.

The count was right about the unknowns and wrong about the third number. The building’s right angle supplies nothing the circle does not already hold. Its verticals supply everything.

What the circle reads from a tilted camera #

A camera pitched down by $\theta$, with no roll, still draws the horizon as a horizontal line, now $f\tan\theta$ above its principal point. The circular points of the ground still lie on it, and the circle’s picture still meets it in two complex points. They are

$$x = c \pm i\,\frac{f}{\cos\theta}.$$

The real part is the column, exactly as before. The imaginary part is not the focal length but the distance from the eye to the horizon line — the length of the perpendicular from the eye to that line, which is the focal length’s hypotenuse when the principal point sits $f\tan\theta$ below it. For a level camera the two are the same; tilted, the second is always the longer.

The reason is worth seeing, because it says exactly what the circle cannot know. The circle tells the reader how the eye sees the angles of the ground, and every direction on the ground vanishes somewhere on the horizon line. The angle between two such directions is the angle at the eye between two rays to points on one line. Turn the eye about that line, keeping its distance from it, and every one of those angles is unchanged. So the circle fixes the eye’s distance from the horizon line and the foot of its perpendicular on that line — the column — and leaves the eye free to swing round the line like a door on a hinge.

The figure puts that hinge on the page. The picture is the vertical line on the left, the horizon a row on it; an eye is a point to the right, opposite its principal point at the focal length’s distance. The circle allows every eye on the solid arc of radius $f/\cos\theta$ about the horizon. A camera pitched 20° down, focal length 854 px, puts the arc’s radius at 909. A reader who assumes the camera is level takes the top of the arc, where the principal point is on the horizon, and reports a focal length of 909 px — 6.4 per cent long.

The building’s right angle is already in the circle #

The earlier essay proposed the building’s two horizontal vanishing points as the third condition: they are the images of a right angle on the ground, and the right-angle condition on an eye is one equation. It is one equation, but it is one the circle has already written. Perpendicular is a pairing found that the vanishing points of perpendicular directions on the horizon are paired by a map whose two fixed points are exactly the circle’s complex points; an angle is a cross-ratio puts it as harmonic separation. Any eye that sees the circle’s points where they are sees every right angle on the ground as a right angle, because a right angle on the ground just is a pair of horizon points harmonic to the circle’s two.

So every eye on the arc satisfies the building’s right angle, exactly. The arc is the earlier essay’s arc of eyes the arc every eye stands on again — the semicircle of stations a rectangle allows — but now drawn in the plane through the column, standing up from the horizon instead of lying along it. A level picture’s ambiguity was where along the horizon the eye was; a tilted picture’s is how far round the horizon line it has swung.

The verticals cut the arc #

What the ground cannot say, the building’s uprights can. Their vanishing point lies straight below the principal point for a camera pitched down — in the column the circle has already named — and an eye sees the vertical direction at right angles to every horizontal one, so it sees the horizon and that vanishing point at a right angle. The eyes that do are the circle whose diameter runs from the horizon to the verticals’ meeting point — Thales’s circle, dashed in the figure. It crosses the circle’s arc once, at the camera.

At a 20° pitch the uprights meet 2,657 px below the horizon, and the two circles cross at the true principal point and the true focal length of 854 px. At 45° the uprights meet much nearer, the dashed circle is small and steep, and the crossing is clean. At 5° they meet 9,800 px below the horizon, far out of any frame, and the dashed circle is nearly a straight line through the horizon: the crossing is still exact for exact marks, but it is a shallow crossing, and the verticals have to be read well to place it.

In numbers, the crossing has a closed form. With the horizon at row $y_h$, the uprights meeting at row $y_3$ and the circle’s radius $R$, the principal point’s row is $y_h + R^2/(y_3 - y_h)$ and the focal length is $\sqrt{R^2 - (R^2/(y_3 - y_h))^2}$. Read in homogeneous coordinates the formula survives uprights that do not meet at all: on a level camera they are parallel, $y_3$ is at infinity, and the principal point lands on the horizon with the focal length $R$ — the earlier essay’s reading as a special case rather than an exception.

What a tilt costs the level reading #

The plainest question is what happens to a photographer who uses the earlier essay’s reading on a picture they did not hold level.

The arrangement is the earlier essay’s — a 44° field, the principal point slid 100 px, a building 12° off square — pitched down by up to 60°, with a circle a metre across placed where the view meets the ground or eight metres ahead, whichever is nearer. Read as level, the circle’s focal length is out by 1.46 per cent with the camera level, 1.52 at 5°, 1.68 at 10°, 6.2 at 20°, 15 at 30°, 41 at 45° and 99 at 60°. The bias is the factor $1/\cos\theta - 1$, and below about ten degrees it is smaller than the ellipse’s own noise: 0.4 per cent at 5°, under a 1.5 per cent scatter. A handheld picture tilted a few degrees can be read as level without correction. A picture looking down at the thing it measures cannot.

With the verticals, the error stays at the circle’s own noise: 1.5 per cent up to 10° and then better, 0.63 per cent at 20° and about half a per cent beyond, because the circle is now nearer and drawn rounder. The building’s three vanishing points alone — the textbook calibration, with the principal point at the orthocentre of the triangle they make — are out by 74 per cent at no pitch, 11 at 2°, 4.6 at 5° and 2.2 at 10°, and match the circle only from about 20°. At small pitches their third vanishing point is thousands of pixels out of the frame and the triangle is a sliver.

The column, which the tilt never touches #

The circle’s column is untouched by the tilt — it is the real part of the same two points whatever the pitch — and both circle readings share it. It is 13.6 px with the camera level, 14.1 at 5°, 14.6 at 10°, and then it improves sharply: 4.8 at 20°, 2.4 at 30°, 1.7 at 45° and 60°. Looking down, the circle lies a metre or two below the camera rather than eight metres ahead, and it is drawn large and nearly round, its outline far better conditioned than a sliver on the horizon. The earlier essay found that a circle pays for distance as the square; a camera that looks down at a circle has stopped paying.

The three vanishing points give the column to 384 px with the camera level, 48 at 2°, 23 at 5°, 15 at 10°, and then about 11 at every steeper pitch — never as well as the circle once the camera looks down by 20° or more. Their column is the foot of an altitude of a triangle, and moving the vertical vanishing point along its own column barely moves that foot; the circle reads the column directly from its slant.

So the question the earlier essay asked last — whether, for a camera looking steeply down, the circle becomes the better half — has a plain answer. It becomes the better half for the column and the focal length together, by a factor of six on the column at 45°. The verticals remain the only half that knows the tilt.

The tilt is the verticals’ alone #

With the verticals the pitch is read to 0.36° for a level camera, 0.38° at 5°, 0.35° at 20° and 0.14° at 60°. The three vanishing points read it to 0.92° for a level camera — where their vertical point is at infinity and a pitch has to be inferred from edges that do not converge at all — and to 0.37° at every pitch from 2° to 30°. Between the two there is almost nothing to choose once the camera is pitched at all, and that is the instructive result: the pitch comes from how fast the uprights converge, and the circle contributes nothing to it.

Two upright edges of 160 px give the pitch to 0.97° beside the circle and 0.75° from the three vanishing points at a 5° pitch; six edges give 0.38° and 0.37°; sixteen give 0.22° and 0.24°. With few edges the circle’s reading is slightly the worse. It places the eye where the verticals’ circle crosses the arc, and an error in the vertical vanishing point slides that crossing along the arc, which is a direction the circle fixes nothing in; the three vanishing points share the error out over a whole triangle. With ten or more edges the two are the same. The tilt is bought with uprights, and the circle neither helps nor hinders the purchase.

A reading in three steps #

Put together, the reading of a picture with a circle on the ground and a building beside it has three steps, and each takes one thing from one object. The building’s two horizontal directions give the horizon, a line, and nothing else is asked of them. The circle, cut by that line, gives the column and the radius $R$ — the eye’s distance from the horizon line — and it gives both whatever the camera’s tilt. The building’s uprights give the one number left, where on the arc of radius $R$ the eye is, and they give it through a single vanishing point that may lie far outside the frame or at infinity without breaking the arithmetic.

A reader who has no uprights, or does not trust them, still has the column exactly and has the focal length bracketed: it is $R$ for a level camera and less for any tilted one, never more. Below ten degrees of tilt, $R$ is within one and a half per cent of the focal length — inside the ellipse’s own scatter for a metre-wide circle eight metres off — so the earlier essay’s reading stands for a hand-held picture as it was. Beyond that, the uprights are not an optional refinement. A photographer looking down at a manhole cover from a standing height, which is how most such pictures are taken, has a focal length forty per cent too long without them.

What the reading leans on #

The camera does not roll. A camera turned about its own axis draws the horizon as a slanting line, and the circle’s two points then lie on a slanting line too — the reading is the same with the horizon’s own direction in place of the rows. What changes is that the verticals no longer meet in the principal point’s column; they meet on the perpendicular from the principal point to the horizon, which is the same statement drawn at an angle.

The circle is on the ground, and the uprights are upright. The earlier essay’s warning about a disc lifted off the ground stands: its circular points are on its own plane’s vanishing line, not the horizon. The new warning is the uprights’. A building’s corners are usually plumb to a few millimetres in a storey; a lamp post, a door frame in an old house or a tree is not, and an upright that leans by a degree moves its vanishing point and every number read from it. The ramp has its own horizon is the same mistake made by the ground.

The horizon is the building’s. It is the line through the two horizontal vanishing points, and on a camera pitched down it is above the frame — 100 px above its top edge at 20°, 640 at 45°. The building’s edges still converge on it; what matters is that they are long enough and spread enough to place it, which at these pitches they are.

The lens draws straight lines straight. Straight lines that are not is about the lenses for which that fails, and a camera looking down at a circle near the bottom of a wide frame is where barrel distortion bends an ellipse most.

Where the tilt was all along #

One conic calibrates the camera treats a camera’s calibration as one conic in the picture, the image of the absolute conic, with five numbers in general and three for a camera with square pixels and no skew. A circle on the ground gives two of them: the conic’s two points on the horizon. A right angle on the ground gives none that the circle has not given, because it is a statement about the same two points. The third number is not on the ground at all. It is how the ground’s horizon sits against the vertical, and only something vertical can say it.

That is why the earlier essay’s level camera needed nothing but the circle and a horizon: being level is the third number, assumed. A level picture shows its rise on its horizon made the same assumption pay for a shift lens’s rise; the principal point is not the centre is the reminder that an assumption about where the principal point lies is a measurement someone did not make. A tilted camera makes the assumption visible by breaking it, and the building’s uprights are the measurement.

Still open: whether two circles at different distances read the tilt without a building #

Everything here took the tilt from a building’s uprights, and many pictures of the ground have none: a car park, a square with its kerbs and covers, a field marked with circles. Two circles on the ground share their circular points, so they give the horizon between them — the line through the two pairs of complex points — and each gives the same column and the same radius $f/\cos\theta$. They agree with each other about everything the arc allows, and that suggests they cannot cut it.

But a camera looking down sees a near circle and a far circle with different shapes, and the difference is a statement about the eye’s height above the ground against its distance from the horizon line — the swing round the hinge. The measurement that settles what that is worth places two circles at stated distances, reads the horizon from their four complex points instead of from a building, and asks whether the two ellipses’ sizes and drops below the horizon, given the circles’ real diameters, fix the pitch; how far apart in depth the two must lie before they do; and whether one circle of known diameter at a known distance below the eye does the same, since a known distance is a length the ground supplies and the circle alone does not.