The arc every eye stands on

A rectangle is drawn on a floor in a picture, and the drawing is known to be a projection of a real rectangle. What does that fix?

The answer is more than nothing and much less than everything, and the shape of the leftover is a circle.

What the four corners give straight away #

Two things, and both are free.

The two pairs of opposite edges meet at two vanishing points, and the line joining them is the plane’s own vanishing line — its horizon. That comes from the drawing alone, needs no camera, and is exact.

What does not come from the drawing alone is anything metric. The relation between two perpendicular directions’ vanishing points is

$$f^{2} = -(\mathbf{v}_1 - \mathbf{p})\cdot(\mathbf{v}_2 - \mathbf{p})$$

which is one equation with three unknowns in it: the focal length and the two coordinates of the principal point. Two vanishing points cannot determine three numbers, and that is the whole of the shortfall.

The arc #

Suppose the camera is level, so the principal point lies on the horizon. Then the relation collapses to a statement about one line, and it says something a straightedge could have said.

The eye, folded flat into the picture about the horizon, sees the two vanishing points at a right angle — because the two directions they belong to are at right angles. And the locus of points seeing a fixed segment at a right angle is the circle having that segment as a diameter.

Which point of the arc is a matter of where the principal point is assumed to be: drop a perpendicular from the assumed centre to the horizon and it meets the arc once. The height of that meeting above the horizon is the focal length, and it is the same square root as before.

That is the classical station-point construction, and it is the algebra drawn rather than a second method. Books give it as a recipe — swing a semicircle on the two vanishing points, drop a perpendicular from the centre of the picture, and where they cross is where the camera stood — and what makes it a recipe rather than a derivation is that the semicircle is presented without the right angle it comes from.

Every station gives a rectangle, and they are not the same rectangle #

Here is the part that is easy to state and worth measuring.

Take any point of the arc. The focal length that goes with it is chosen precisely so that the two edge directions come out perpendicular. So the reconstruction cannot fail to have right angles — it has them by construction, to five parts in ten million million of a degree across the whole walk.

What differs is the proportion.

The rectangle that was actually there is two-thirds as wide as it is deep. Only a fiftieth of the arc puts the reconstruction within five per cent of it.

So a drawn rectangle does not say what shape it is. It says what shape it is given where the centre of the picture is, and the centre of the picture is not something the four corners contain.

What the drawing does say, which is a bound #

The arc is usually read as pure shortfall — four corners fix nothing metric — and that is too strong. Parameterising it turns the shortfall into an inequality, and the inequality is worth having.

Put the two vanishing points at $x_1$ and $x_2$ on the horizon, a distance $D$ apart, and write the assumed principal point as $p = x_1 + tD$. The perpendicularity relation then reads

$$f^{2} \;=\; -(x_1 - p)(x_2 - p) \;=\; t(1-t)\,D^{2}, \qquad f \;=\; D\sqrt{t(1-t)}.$$

Three things fall out of that one line.

The principal point must lie between the two vanishing points. Outside that band $t(1-t)$ is negative and there is no real focal length — which is the band the second figure marks along the horizon, derived rather than drawn.

The focal length is bounded above by half the separation. $t(1-t) \le 1/4$, so

$$f \;\le\; \frac{D}{2},$$

with equality only at the middle. That is a metric statement from the drawing alone, with nothing assumed: the field of view is at least $2\arctan(W/D)$ for a picture $W$ wide. Two vanishing points close together on the page prove a wide lens; two far apart permit a long one and do not require it. So the drawn rectangle does say something about the camera after all — an inequality rather than a value, which is the same shape both vanishing points on the paper reaches from the drawing-board side.

And the assumption is stationary at the top of the arc. Since $f = D\sqrt{t(1-t)}$ has zero derivative at $t = 1/2$, an error in the assumed centre costs a second-order error in the focal length there and a first-order one elsewhere. Ten per cent of the band wrongly assumed costs two per cent of the focal length at the middle and nineteen per cent at $t = 0.2$ — a factor of ten between a rectangle sitting symmetrically in the frame and one off to the side.

That last one is the essay’s closing observation with a number on it, and it is the practically useful part. A reader with a photograph and one rectangle in it should ask where the rectangle’s two vanishing points sit relative to the frame’s middle, and trust the reconstruction in proportion to how nearly the middle bisects them. It is also why the same assumption behaves so differently on different pictures, and why the principal point being taken as the centre is sometimes harmless and sometimes not — the cost is not a property of the assumption, it is a property of where on the arc it lands.

The proportion inherits all of it, since the reconstructed shape is a function of $f$ and the two vanishing points. Its 0.071-to-1.120 range across the arc is the image of $f$ running from zero up to $D/2$ and back to zero, and the fiftieth of the arc that lands within five per cent of the truth is narrow for the same reason the focal length’s own dependence is steep away from the middle.

None of which repairs the shortfall. Three unknowns and one equation is still three unknowns and one equation, and a box supplies the missing two where a rectangle cannot. What the parameterisation adds is that the one equation is not empty: it excludes a band, it caps the focal length, and it says where along the remaining freedom an assumption is cheap.

Where along the arc the picture actually was #

There is a natural question and it has a clean answer: which point of the arc corresponds to the assumption everybody makes?

Assuming the principal point is the middle of the frame puts the station at the point of the arc directly over the middle of the frame. For a rectangle that sits symmetrically in the picture that is near the top of the arc, where the focal length is largest and the proportion is least sensitive to being wrong. For a rectangle off to one side it is well round the arc, and both the focal length and the proportion are worse behaved there.

So the quality of the assumption is a property of the picture’s composition rather than of the camera. A rectangle photographed head-on and centred is well conditioned; one photographed obliquely at the edge of the frame is not, and nothing about the two pictures announces the difference.

Two rectangles close it #

One rectangle is one pair of perpendicular directions. Two rectangles on the same floor, at different angles, are two pairs — and perpendicular is a pairing says two pairs determine the involution, which determines both numbers.

Drawn as arcs it is the same statement and it is prettier. Each rectangle gives a circle whose centre is on the horizon. Two circles whose centres are both on one line meet at two points, one each side of it, and the one above the horizon is where the eye stood.

That is the arithmetic solved by drawing, and the two routes agree because they are the same constraint written twice: the involution’s two parameters and the two circles’ one crossing.

What a second measurement could have been instead #

The second rectangle is one way to close the family. It is not the only one, and the alternatives are worth listing because each of them corresponds to a piece of information a real photograph might contain.

A known proportion. If the rectangle is known to be a square — a floor tile, a window, a sheet of paper — the proportion is given and the arc collapses to the one station that reproduces it. That is one number supplied, and one number was what was missing.

A vertical. One edge in the picture known to be vertical, and therefore perpendicular to both of the rectangle’s directions, adds a third direction and turns the problem into the three-point recovery. That is more than enough, and it is why a picture with a doorway in it is a much better picture to measure than a picture of a floor.

Or a circle. The image of a circle lying in the same plane supplies the plane’s imaged circular points directly, which is two constraints — the same two the second rectangle gives, arriving from a curve rather than from four corners.

All four routes — second rectangle, known proportion, vertical, circle — close the same one-parameter gap, and each of them is one number or two. Nothing closes it for free.

What the arc is, exactly #

Worth being precise about, because “the eye is free along an arc” is the kind of sentence that is right and vague.

The arc is a rabatment — the eye folded into the picture plane about the horizon, so its position on the paper is a picture of a position in the room rather than a place a mark could go. Un-fold it and the arc becomes a circle in the horizontal plane through the eye, standing over the two vanishing points, and the eye is somewhere on that circle.

The free parameter, then, is one number: where along the horizon the principal point is. That single number carries both the focal length and the proportion, and the two move together in a way worth watching — the focal length is largest at the middle of the arc and falls to nothing at either end, while the proportion runs monotonically from one extreme to the other.

Why the answer had to be a conic #

The arc is a circle, and that is not a coincidence of the level camera. It is the shape a right angle always makes.

Every constraint in this subject that says “these two directions are perpendicular” is the same constraint: two vanishing points are conjugate with respect to one conic in the picture, the image of the absolute conic. One conic calibrates the camera states it in that form and shows the orthocentre construction is the same relation with a square root in it.

Read the constraint as an equation in the camera’s unknowns instead of the scene’s, and it is one linear equation in the entries of that conic. The set of cameras satisfying it is therefore a plane section of the three-dimensional space of possible conics, and its intersection with the level-camera assumption is a curve — the arc.

Two rectangles give two such equations and the intersection is a point. Three would over-determine it, which is what the spare rectangle in perpendicular is a pairing is doing.

That is why the answer had to be a conic and could not have been an interval. The constraint is quadratic in the picture’s coordinates, so the locus it defines is a conic, and the only question was which one.

What this does not say #

It says nothing about a tilted camera. The right-angle-on-a-circle argument needs the principal point to be on the horizon, which is the level condition, and a tilted camera has the same shortfall stated differently: the eye is free on a curve in space rather than on a circle in the picture plane, and the drawing does not say which curve without the tilt.

It says nothing about the aspect ratio of the frame. Everything here is about the principal point, which is a property of the camera; the frame is a crop and moving it moves the principal point relative to the picture without moving anything in the world.

And it does not make the assumed centre unreasonable. Assuming the centre of the frame is the centre of the picture is usually nearly right, and the next rung is about how nearly.

The count, in one line #

It is worth setting the arithmetic out plainly, because the whole essay is a consequence of it and the counting is not hard.

A square-pixel camera has three unknowns: the focal length and the two coordinates of the principal point. Each pair of perpendicular directions in the scene supplies one linear constraint on them. So:

• one rectangle — one constraint, two unknowns left, a surface of possible cameras;
• one rectangle on a level camera — one constraint plus the levelness, one unknown left, the arc;
• two rectangles on one plane, level camera — two constraints plus levelness, none left;
• a box — three constraints, none left, and one to spare.

That table is the answer to every question of the form “is this enough”. It is the same counting a line is a space of its own recommends doing before any test is run, applied to the camera rather than to a quadruple of points, and it is what says in advance that the arc had to be one-dimensional.

The arc’s own ends #

The two ends of the arc are the two vanishing points, and it is worth saying what a station there would mean.

At an end, the folded eye has come down onto the horizon, which means the focal length has gone to zero — an infinitely wide camera, with the whole world crushed into a point. That is not a picture; it is the degenerate limit the construction runs to, and the sweep stops just short of both ends for that reason.

Between the ends, the focal length rises to a maximum at the top of the arc and falls again — so two different assumed centres give the same focal length, one each side of the middle, with different proportions. A focal length alone therefore does not locate the station on the arc, which is a small trap worth naming: two cameras of identical focal length, both consistent with the same four drawn corners, disagreeing about what the rectangle is.

That is a genuine ambiguity within the ambiguity, and it says the free parameter is properly the position along the arc rather than the focal length. Naming the parameter correctly is what stops a reader from thinking that a known focal length would close the family.

The transferable form #

Two rungs of this site have now ended with a curve rather than a number, and the second one arrives by a different route from the first.

A picture that leaves something free leaves it free along an object, and finding the object is more useful than bounding the number — because the object says how a second measurement will close it.

The ball stands at a focus leaves the lamp on a focal hyperbola, and the branch says at once what a second observation has to supply. Here the eye is left on a circle, and the circle says at once that a second rectangle will do — because two circles with collinear centres cross once above the line, and no more measurements are needed than that.

Had the answer been “the focal length is somewhere between five hundred and eleven hundred pixels”, both facts would have been invisible.