Field

The real instrument

A lens is a departure from the pinhole, and the departure is the largest systematic error in every measurement made here. It bends straight lines, destroys the invariant, and can be recovered from nothing but the knowledge that some edges were straight — which is the round trip again, on a harder problem.
principal pointk₁ = -0.320, k₂ = 0.110 — barrel distortioncentre line 0e+0 px of sag, outermost 17.8 px

Straight lines that are not

Everybody says the edges of a wide-angle frame bow. Nothing is special about the edge. A radial map moves every point along its own radius, so the only line it leaves straight is one through the principal point, and the bend of every other is decided by how far it passes from that one place.

00.50011.502-0.400-0.20000.200k₁departure of the cross-ratio from the world's value (%)pinhole1.29%the pinhole's own error, on the same four points2e-16 — the control

A lens destroys the invariant

The cross-ratio is the one thing a projection preserves, and nearly everything checkable about a photograph is checked with it. A lens returns it one and a third per cent out where the pinhole is exact to fifteen digits — and the height error that follows tracks a quantity nobody would guess.

fitted k₁ = -0.280000true -0.280000, off by 5e-15

Fitting a lens from straightness alone

No calibration target, no known scene, no camera. Only the knowledge that some edges in the picture were straight — and the coefficient comes back to fifteen digits. Then it comes back with a companion, and the two are correlated at −0.997.

level — the top is cut off0.00° of spreadtilted 13°4.55° of spreadshifted 95 px0.00° of spreada shift moves every point by exactly the shift95.0 px, and no direction at all

The principal point is not the centre

Every textbook that computes a focal length from two vanishing points supplies the middle of the frame as the principal point. On a shifted or cropped picture that is wrong, and it costs one and a half per cent of the answer at a fifth of a frame's shift.

the far field — where the stitch was fitted2.2 m — 3.2 px out24 m — 0.3 px outthe sky registers to 1e-13 pxthe foreground does not — up to 3.2 px

The eye is a place, not a point

Rotate a camera about the wrong point and the sky still stitches perfectly while the foreground slides. The misregistration falls as one over the distance, exactly — which is what says the fault is the pivot and not the lens.

00.50011.5000.5001the lens's radial coefficient, −k₁the worst transversal's distance from a correct perspective, in pxa reader's ruler, 0.2 pxk₁ = −0.40a photographed pavement, against its lens5.7 px of bow at the threshold

The lens a pavement can hide

A photographed pavement reads as a correct drawing up to a radial coefficient of about four tenths — a lens strong enough to bow a straight edge across the page by nearly six pixels and to print as twenty per cent distortion at the frame's corner. The reason is that a pavement sits near the principal point, which is the one part of the frame a radial map barely touches.

the entrance pupilthe stopthe glassthe chief rays, from four object distances3.6e-15 mm apart

The hole a scene actually sees

The stop is not the centre of projection. Model a 50 mm lens with its stop 18 mm behind the glass and the chief rays from every object distance cross the axis at one point 28.1 mm on the other side of the lens — 10.1 mm from the stop and 1.56 times its size — to 3.6 × 10⁻¹⁵ mm. That point is the entrance pupil, and it is where a picture is a projection from.

00.50011.50020406080field angle off the axis (degrees)picture radius, in focal lengthsfolds at 47.49°43.91°50.54°the radial factor reaches zeropinholefolds at 47.49°43.91° and 50.54° share one radius

A barrel model folds at a radius it sets itself

The polynomial every calibration fits to a wide lens stops increasing at a radius fixed by its own first coefficient — 47.49° of field at k₁ = −0.28 — and past it two directions land on one picture radius. The routine that undistorts pictures with it does not refuse there. It hands back wrong directions from 46.75°, by as much as 106.5°, and refuses only at 65.5°: a fifth of the field returned silently wrong.

principal point moved 22.30 px · focal length 0.584 px shorterlines straight to 1e-13 px

A tilted sensor is not a distortion

Tilt a sensor 3° out of square with its lens and every point of the picture moves — up to 7.5 px on the frame drawn here — yet every straight line stays straight to 10⁻¹³ px and the cross-ratio survives to 10⁻¹⁶. The picture is an ordinary pinhole picture whose principal point has moved 22.30 px. A calibration that frees its principal point absorbs it exactly; one that holds the principal point and reaches for tangential distortion terms leaves 1.87 px, and used as a correction it bends straight rows by 4 px.

three edges crowded on one side5.69e-3two on one side, one on the other5.66e-4three edges near the centre6.73e-3three edges placed by search9.57e-4the centre free · log scalebest: 5.66e-4

The lines that calibrate a lens

One straight edge through the centre of a picture says nothing about a lens's distortion, and one 180 px from the centre determines k₁ to 9.0 × 10⁻⁴ — the precision rises in proportion to the offset. But distance from the centre is not enough. Crowd three edges on one side and, the moment the distortion centre is also unknown, the coefficient is ten times worse, because a bend on one side looks like a moved centre; put one edge across the centre and it barely changes.

paraxial pupilthe stopa curved front elementcrossings 15.0 · 14.5 · 13.5 · 11.6 mm

The entrance pupil walks with the angle

The place a picture is a projection from is not a point in a wide-angle design. Chief rays traced through a strongly curved front element cross the axis 15.07 mm behind its front vertex when they are nearly on the axis, and 4.23 mm nearer the front at 80° of field. So no pivot makes a wide panorama seam clean: at one metre, pivoting at the paraxial pupil leaves 4.39 arcminutes of misregistration along a seam, and the best pivot still leaves 1.41.

00.50011.5020123where a pinhole would put the point, in focal lengths from the centrewhere the model puts itwhere the polynomial foldsthe division model's horizonboth at k = -0.42fold 42° · horizon 1.54

A model that inverts has a horizon instead of a fold

The polynomial every calibration fits turns around at a finite radius and stops being a map from direction to picture. The division model, chosen because it inverts in closed form, never turns around — it rises for ever toward a horizon at one over the root of its own coefficient, so the whole hemisphere of directions lands inside a finite disc. Fitted to the four fisheye laws over seventy-five degrees it follows every one of them three to five times more closely, and below sixty the polynomial is still the better model.

0102030400.2000.4000.6000.800where along the edge, as a fraction of the visible lengthshare of the edge's response to the coefficient, in per cent93% outside the middle halfone edge, 82 marks, cut into tenthsmiddle two tenths: 1.4%

The response is at the ends and the information is not

A radial map bows a straight edge by an amount that grows as the square of the distance along it, so 93 per cent of an edge's response to the coefficient lies in its outer quarters. Spending the marks there is 16 per cent worse than spreading them evenly, because two clusters say nothing a shifted, tilted line could not say. What identifies the coefficient is a curvature, which needs three places — both ends and the middle, which beats an even spread by 11 per cent.

0.313100510how far away the subject is, in metreshow far the pupil moves forward, in millimetres0.50 mthe whole walk with field anglea 50 mm lens, focused as one pieceequal at 0.50 m

Focusing moves the pivot past its best place

Focusing a fifty-millimetre lens to one metre carries its entrance pupil 2.63 millimetres forward of the camera body, and to half a metre 5.56 — which is more than the whole 5.53 that the pupil walks with field angle, so past a subject at 502 millimetres the focus decides where the pupil is. A panorama head aligned at infinity and used at a metre leaves 7.46 arcminutes along its seam; aligned at four metres it leaves 2.21, better than pivoting at the pupil at all.

0100200300distance from the picture's centre, pxeach edge, by how far it passes from the centreoffset 20 px20–326offset 60 px60–330offset 125 px125–348offset 180 px180–372four edges, 82 marks eachevery one spans 16× in radius

No design separates the two coefficients

Separating a squared term from a fourth-power one was supposed to need marks at radii far apart, which is a statement about where edges are placed. It is not: one straight edge already runs from 20 px to 326. Spreading ninety-six marks over three edges or sixteen changes the answer by 23 per cent, spreading the offsets makes it 19 per cent worse, and the correlation stays at −0.98 whatever is done. What a plumb-line calibration determines is one number, to 1.52 thousandths, and which number depends on the model.

60 px160 px260 px360 pxmarks every 8 px, kept inside the frame47 / 125 / 114 / 122 marks

A known target sharpens the fit and does not separate it

Printed circles of stated size were supposed to break the −0.98 correlation between a lens's two radial coefficients, because a circle puts every mark at one radius and no straight edge can. They do not: every design of circles leaves the pair 0.979 to 0.9997 correlated, and for circles of known size the figure is exactly the cosine between r³ and r⁵. What a known target buys is precision — 3.8 times the straight edges' at the same budget — and only if its size in the picture is known to about a thousandth.

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