A level picture shows its rise on its horizon, and hides its slide

The principal point is not the centre measured what the textbook route to a focal length costs on a shifted or cropped photograph. The route takes two vanishing points of perpendicular horizontal directions — the two walls of a building, say — and solves $f^2 = -(v_1 - p)\cdot(v_2 - p)$ for the focal length, with the principal point p supplied from outside. Every textbook supplies the middle of the frame. A shift lens, a view camera’s rise or an off-centre crop moves the principal point elsewhere, and the essay found 1.6 per cent of error at 150 pixels of shift on a 44-degree lens, more on a wider one.

It also found an awkward pairing. The route that does not assume the principal point — three vanishing points and their orthocentre, the construction recovering the camera uses — needs a finite vertical vanishing point, and a shifted photograph is exactly the photograph whose verticals stay parallel, because the shift exists to keep them so. The picture on which the assumption costs most is the picture on which it cannot be avoided by the orthocentre.

It can be avoided another way, for half of what a shift does. The other half cannot be avoided at all, and how much that half costs depends on something the earlier essay did not vary: which way the building is turned.

The principal point is on the horizon #

A shift lens keeps the camera level. Its optical axis is horizontal, so the plane through the eye and parallel to the ground — the plane whose image is the horizon — contains the axis, and the axis meets the picture at the principal point. A level camera’s principal point lies on its horizon, exactly, however far the lens is shifted. The horizon at eye level is the same statement from the other side: the horizon is the image of the eye’s own height.

The horizon is in the picture. Both walls of a rectangular building are horizontal, so their two vanishing points lie on it, and the line through them is it. A reader who has the two vanishing points has the horizon and therefore the principal point’s height — not assumed, measured. What remains unknown is where along the horizon the principal point sits.

That divides every shift into two parts. A rise moves the principal point up or down, across the horizon; the horizon reads it. A slide moves it left or right, along the horizon; nothing in the two points reads it.

The figure draws the unknown part. Hold the two vanishing points fixed and let the principal point be any column on the horizon between them. For each, the right-angle condition gives a focal length, and the focal lengths trace a semicircle with the two vanishing points at the ends of its diameter. Every point on it is a camera that would draw these two vanishing points exactly. The arc every eye stands on found this semicircle for a drawn rectangle, where every station on it reconstructs a genuine rectangle of a different proportion; in a photograph the same arc is a trade between the principal point’s column and the focal length.

The slider slides the true principal point along the horizon by 0 to 200 pixels. The frame’s middle column stays where it is, and the focal length it reads falls further from the true one at each step: 7.7 per cent short at 100 pixels for this turn of the building. A rise would not appear on this drawing at all, because a rise moves the horizon up or down the frame and the semicircle with it, and the middle column, now read on the moved horizon, lands on the semicircle at the same place.

A rise costs nothing once the horizon is used #

The earlier essay’s measurement put the principal point at the middle of the frame in both coordinates, so its error came entirely from height: a pure rise was what it shifted. Read on the horizon instead, the same rise costs nothing.

The dashed curve is the textbook’s cost, and it reproduces the earlier essay’s 1.55 per cent at 150 pixels. It does not depend on how the building is turned, which is worth noticing: a rise moves the assumed point straight off the horizon, the right-angle condition then reads the true focal length shortened by the square root of one minus the rise squared over the focal length squared, and neither vanishing point’s position enters. The solid line is the cost of the alternative, zero to the last bit, at every rise and every turn.

So the earlier essay’s recommendation for the awkward case — read the shift off the vignetting, find a third perpendicular direction, or fit the principal point on several views — is needed for only half of the awkward case. For a camera that was level and risen, which is how a shift lens is almost always used, the horizon is already in the picture and already carries the rise. The textbook’s error was never a cost of shifting; it was a cost of not reading the horizon.

A slide costs in proportion to how far the building is turned #

The other half is the slide, and here the semicircle decides everything. A camera whose principal point is slid along the horizon, read at the middle column, lands on the semicircle somewhere other than where it stood, and the focal length it reads is the semicircle’s height there instead.

The error is first order in the slide, with either sign, and its slope near zero is 100·cot 2ψ/f per cent a pixel, where ψ is how far the camera is turned from one wall. That has a plain geometric reading. For a level camera turned ψ, the two vanishing points sit at f·cot ψ to the right of the principal point and f·tan ψ to its left, so the middle of their diameter is f·cot 2ψ to the right. The principal point is that far from the top of the semicircle, on its slope, and a slide moves it up or down the slope in proportion.

The slider shows what that does across ordinary buildings. Turned 30 degrees, a 50-pixel slide costs about 3.4 per cent. Turned 10 degrees — a facade seen nearly square, which is how most buildings are photographed, with the far wall a sliver running towards a distant vanishing point — the same 50 pixels cost 15 to 18 per cent. Turned 45 degrees, the two vanishing points straddle the principal point evenly, it sits at the top of the semicircle, the first-order term vanishes, and 50 pixels cost a sixth of a per cent.

The sign is readable too, and it is the useful part for anyone checking a claim about which lens took a picture. Slide the true principal point towards the distant vanishing point and the frame’s middle is left nearer the close one, down the steep side of the semicircle, so the reading comes out short: a wider lens than was used. Slide it towards the close vanishing point and the middle column sits further up the semicircle, towards its top where the slope flattens, and the reading comes out long, by less. At 30 degrees, 200 pixels towards the distant point cost 17.9 per cent and 200 pixels towards the close one 10.3.

The earlier essay noted that a sideways crop is first order and a rise second. The semicircle says how big the first-order term is and why: it is set by how lopsided the two vanishing points are about the principal point, and a nearly square view of a facade, with one vanishing point inside the frame and the other thousands of pixels away, is as lopsided as an ordinary photograph gets.

Edges read with error leave the horizon as good as the truth #

Every figure so far uses exact vanishing points, and a real reading finds them by extending edges that were marked with some error. The question is whether reading the principal point off a horizon that is itself measured adds anything to the error the edges already cause.

It adds nothing that can be seen. With edges marked to half a pixel, the focal length from the true principal point is out by a median 0.88 per cent, because the vanishing points themselves are uncertain; from the principal point on the read horizon, also 0.88. The horizon’s height is read to about a pixel — a median 1.0 pixel off the true rise at half a pixel of marking — and that error barely touches the focal length. The focal length is set by how far apart the two vanishing points are along the horizon. Moving the assumed principal point a pixel up or down changes the right-angle product by the square of that pixel, a second-order amount, and the first-order part of the product is about positions along the horizon, which the edges measure equally well whichever principal point is used.

The frame’s middle adds its bias on top: 1.56 per cent at half a pixel, which is the 1.31-per-cent rise error and the edges’ own scatter combined. At four pixels of marking the edges dominate everything and all three readings are poor, near 12 per cent. In between, the horizon reading is the one that matches the floor the edges set.

There is a further consequence worth stating. A careful reader who marks edges to half a pixel and reads the principal point off the horizon has a focal length good to about one per cent from two vanishing points, on a shifted photograph the earlier essay judged awkward. The orthocentre construction would have nothing to work with, since it needs a third vanishing point and that is exactly what the shifted photograph lacks.

Which crops are which #

The two halves of the shift sort the ordinary operations on a photograph by whether they hurt.

A tenth of the frame trimmed off one side moves the principal point 69 pixels along the horizon, and what that costs depends entirely on the building. Eleven degrees off square it is 23 per cent of the focal length, enough to call a 35-millimetre lens a 27. At 31 degrees it is under five per cent; at 43.5, under one; past 45 it changes sign and grows again towards the other wall seen square. The rise of the same 69 pixels, read at the frame’s middle as the textbook does, costs a third of a per cent at every turn — and read on the horizon, nothing.

So the two operations the earlier essay called equally innocent in an image editor are not merely different in order. One is fully recoverable from the picture and the other is not recoverable at all, and the one that cannot be recovered is the one whose cost varies by a factor of thirty with the view.

Cutting a 3:2 frame to 16:9 by keeping the top band, which is what happens when a horizon is being placed, moves the principal point 36 pixels from the new middle, all of it vertical. The textbook pays 0.09 per cent for that, the horizon nothing. A shift lens near full rise moves it 138 pixels, vertically: 1.31 per cent and nothing. Cutting the same frame to a square keeping the left side moves it 115 pixels along the horizon: 9.10 per cent, and the horizon cannot help because the error is entirely in the direction the horizon does not measure. A tenth trimmed off the left costs 2.23 per cent either way.

That gives a short rule for a reader handed a photograph whose history is unknown. If the edges of the frame might have been trimmed top or bottom, read the principal point’s height off the horizon and the trimming does not matter. If the sides might have been trimmed, the two-point focal length carries an unknown error whose size is set by the building’s turn: small for a corner seen near 45 degrees, large for a facade seen nearly square. Quote it with that condition, or find something else that fixes the column.

What two vanishing points can and cannot say #

Put together, a level photograph of a rectangular building fixes three things and leaves one free. It fixes the horizon, and on the horizon the principal point’s height, and with them a semicircle of possible cameras on the two vanishing points as diameter. It leaves free where on the semicircle the camera stood: the principal point’s column and the focal length, traded against each other one for one. An angle is a cross-ratio and one conic calibrates the camera say the same thing algebraically — two perpendicular vanishing points impose one condition on the camera’s calibration, and a level camera with unknown principal point and focal length has three unknowns, of which the horizon supplies one.

That is the honest summary of what a shift costs. The rise, which is the shift photographers actually use, is fully recoverable from the picture, and the textbook’s error on it was a failure to read what the picture holds. The slide is not recoverable from two vanishing points by any method, and the sensible response is not a better formula but a statement of the condition: this focal length is right if the frame was not trimmed sideways, and wrong by about cot 2ψ per cent for every focal length’s hundredth of trimming if it was.

A height from one photograph is unaffected by any of this, which is a small mercy worth repeating from the earlier essay: heights come from the horizon and the vertical vanishing point, neither of which involves the principal point. It is lengths that need the focal length, and with it the column.

The level camera the argument assumes #

The camera is level. Everything here rests on the principal point lying on the horizon, which is true only when the optical axis is horizontal. A camera tilted up and then shifted down to compensate, or a photograph straightened in software after being taken tilted, does not have that property: its principal point is off the horizon by f·tan of the tilt. Such a picture has converging verticals, though, so the orthocentre construction is available to it, and the awkward pairing the earlier essay found does not arise.

The two walls are perpendicular and horizontal. A building whose corner is not a right angle gives two vanishing points on the horizon that the right-angle condition misreads; the semicircle is then the wrong curve, and the error is the angle’s own, not the principal point’s. Every reading from two vanishing points carries this assumption, and none of the measurements above relax it.

The lens draws straight lines straight. A lens with barrel distortion bends the edges a reader extends, and its vanishing points move with it. Straight lines that are not is about that; distortion is radial about the principal point, so a slid principal point also moves where the distortion is centred, and a correction applied about the frame’s middle leaves exactly the decentring residual the earlier essay described.

The edges are marked independently. The noise figure draws each edge’s ends with independent errors. A reader who marks edges systematically too steep or too shallow shifts every vanishing point together, and a shared bias in the vanishing points is a bias in the focal length that no number of edges removes.

Still open: what else fixes the column #

Two vanishing points leave the principal point’s column free, and the cost of guessing it is largest exactly where photographs most often sit, with a facade seen nearly square. Something else in the picture has to fix it, and the candidates are not equally good.

A third direction that is perpendicular to both walls and not vertical does not exist for a building. But many scenes contain a second rectangular object turned at a different angle to the first — a kerb, a parked car, a paving pattern, a second building — and each supplies its own two vanishing points on the same horizon and its own semicircle. The true camera lies on both semicircles, and two semicircles on one line meet in at most one point above it. The measurement that settles what that is worth takes a level, slid camera looking at two rectangular objects turned by a stated angle relative to each other, reads all four vanishing points from edges marked with half a pixel of error, intersects the two semicircles, and asks how well the column and the focal length come back as the two objects’ turns are brought together — since two objects turned alike draw two nearly identical semicircles whose intersection is nearly undefined — and whether a difference in turn of twenty degrees already does as well as knowing the principal point outright.