One picture of a ball

Every quadric has one outline computes the drawn conic from the object: three matrix products, $C = \operatorname{adj}(P \operatorname{adj}(Q) P^{\mathsf T})$, and the ellipse lands on the traced contour generator to nineteen decimal places.

This essay runs it backwards. Given the conic — read off a photograph, with nothing else — what is known about the ball?

The outline is a cone, and the cone is what is known #

Every point of the outline is where a ray from the eye grazes the ball. So the set of grazing rays is a cone with its apex at the eye, and the outline is that cone’s section by the picture plane.

Given the camera’s calibration, the conic in the picture and the cone in space are the same object written two ways: the cone’s matrix is $K^{\mathsf T} C K$, and its axis and half-angle come out of that matrix’s eigenstructure — one eigenvalue of one sign against two of the other, the odd one out being the axis.

So one photograph of a ball gives:

The direction of the ball’s centre, exactly.

The ratio $r/d$, as the sine of the cone’s half-angle, exactly.

And neither $r$ nor $d$ separately, at all.

Where the sign pattern comes from #

The eigenvalue condition is easy to state and worth a moment, because it is the whole of the routine’s domain check.

Write the cone as a quadratic form on directions. In the cone’s own frame it is $x^2 + y^2 - \tan^2\theta\, z^2 = 0$ — two positive entries across the cone and one negative along its axis — so the three eigenvalues are two of one sign and one of the other, and $\tan^2\theta$ is minus the odd one over the average of the pair.

That pattern is what makes a real cone a real cone. Three eigenvalues of the same sign is an imaginary cone, which is what a conic with no real points produces; one zero eigenvalue is a degenerate pair of planes.

So the routine finds the odd eigenvalue out, takes the ratio, and asserts it is positive. If it is not, the input was not the outline of anything and the answer would have been a number computed from a quantity that has no arccosine.

The scale ambiguity, demonstrated rather than asserted #

That third statement is the site’s oldest limit and it is worth showing rather than saying.

Take the ball. Move it a hundred and thirty-seven times further from the eye and make it a hundred and thirty-seven times larger. Compute its outline from scratch, through the same three matrix products.

The two conics agree entry for entry to the arithmetic floor. Not nearly the same ellipse; the same conic, because the two balls are inscribed in the same cone and the cone is all the picture has.

The demonstration is better than an assertion for the usual reason on this site: “one view cannot supply scale” is a sentence that could be true of a method rather than of the geometry, and the identical conic says it is the geometry.

It is also the reason a photograph of a ball is such a poor measurement and such a good calibration target. Poor, because it says nothing about the ball’s size or distance. Good, because everything it does say — a direction and an angular radius — is exact, and neither depends on the ball being at any particular place.

The trap: the drawn ellipse’s centre #

There is a shortcut a reader will reach for and it is wrong by a measurable amount.

The drawn outline is an ellipse. An ellipse has a centre. Surely that centre is the image of the ball’s centre?

No. The foundations field measured the gap: at twenty-three degrees off axis the two are over two pixels apart on a 690-pixel canvas, and the gap grows with the field angle. A centre is defined by midpoints of chords, and a projection through a point does not keep midpoints, so the centre of the image is not the image of the centre.

So the cone route is not an elaborate way of doing something easy. Reading the ellipse’s centre gives a direction that is wrong by a real angle; taking the cone’s axis gives one that is right at the arithmetic floor.

And the trap is well camouflaged. On the axis the two agree exactly, which is where a method gets tested; the error grows toward the edges, which is where the object usually is.

That contrast is worth carrying. The same measurement — read the drawn ellipse’s centre — is exact under a parallel projection and wrong under a camera, and the difference is the last row of the projection matrix.

What the half-angle is good for #

The ratio $r/d$ is an angular radius, which is the thing an astronomer measures and calls apparent size. It is exact from one picture and it is the whole of what the picture says about the ball’s size.

Two things follow.

A ball of known radius gives its distance immediately: $d = r/\sin\theta$. That is one photograph and one length, which is this site’s own recurring shape of measurement — a single view supplies a ratio and a known length turns it into a distance.

And a ball at a known distance gives its radius the same way, which is the arrangement a calibration rig uses.

Neither of those is a new result; what is new is that the ratio comes from an outline rather than from a pair of marks, so it needs no feature on the object and works on a ball with no texture at all.

Two facts about one photograph, and they are not the same fact #

It is worth separating two statements that a reader is likely to run together, because they are about different objects and only one of them is the scale ambiguity.

The first: the picture does not say where on the ball each outline point came from. Two views of a curved surface draw two different curves on it, and one view draws one curve whose identity on the surface the picture never mentions. That is a fact about correspondence and it holds however many pictures there are.

The second: the picture does not say how big the ball is. That is the scale ambiguity, and it is repaired by a second view or by a known length.

The two are independent, and a reader who conflates them concludes that a second picture solves everything. It does not: two pictures fix the ball’s size and position exactly and still say nothing about which points of the surface either outline came from, because those points are different in the two views by construction.

For a ball that costs nothing, since a sphere is determined by a centre and a radius and has no other shape to recover. For anything else it is what the visual hull is about, and it is the term that does not go away.

Why the cone’s axis and not the outline’s #

The eigen-decomposition deserves a sentence, because it is where the exactness comes from.

A cone of rays through the eye is a quadratic form on directions. In its own frame that form is diagonal with two equal entries and one of the opposite sign — two across the cone, one along it — so the odd eigenvector is the axis and the ratio of the eigenvalues gives $\tan^2\theta$.

That is a statement about the cone, and the cone is exactly what the outline determines. Everything about the picture plane’s position and orientation has already been divided out by the calibration, so no choice of picture plane can move the answer.

The ellipse’s centre, by contrast, is a property of the section — of the cone and the plane it was cut by — so it moves when the picture plane moves. Which is the same statement as the picture plane being a choice: a quantity that changes when the glass moves is a quantity about the glass.

What the calibration is doing, and what happens without it #

The step from a conic in the picture to a cone in space is one matrix sandwich, $K^{\mathsf T} C K$, and it is worth being explicit about what that step consumes.

$K$ carries the focal length and the principal point. Without it there is a conic on a sheet of pixels and no way to say what angle any of it subtends — a wide lens and a long one draw ellipses of the same shape at different sizes, and a picture with no calibration cannot tell the difference between a large ball far away through a long lens and a small one near through a short one.

So the angular radius is bought with the calibration, and it is bought entirely with it: every other quantity in the recovery is projective and survives without one. The axis’s direction, in particular, needs $K$ too — the cone’s axis is a direction in camera coordinates, and turning pixels into directions is what $K$ is for.

That puts the essay’s result in the right place among the site’s stratification. What one picture of a plane determines walks the ladder from projective to affine to metric, and this sits on the metric rung: the outline gives a conic projectively, and only a calibrated camera turns it into an angle. An uncalibrated camera looking at a ball has a conic and knows that some quadric produced it, which is almost nothing.

The refusal, and the ball that is not in the picture #

coneFromOutline asserts that the eigenvalues have the sign pattern a proper cone has — one against two — and refuses otherwise.

That refusal has a meaning. A conic in the picture that is not the section of a real cone through the eye is not the outline of anything: a hyperbola, say, or a degenerate pair of lines. Handed one, the routine could return an axis and a half-angle by taking absolute values, and the numbers would be plausible and would refer to nothing.

Which is the same failure mode the shadow’s curvature exhibits one rung back, and the refusal here is what that fit lacks: a check that the input is the kind of thing the model can be about.

Why a ball is the object this works cleanly on #

The recovery above is unusually tidy and the tidiness is a property of the sphere rather than of the method, which is worth saying so that a reader does not over-generalise it.

A sphere has no orientation. Its tangent cone from any point is a right circular cone, so the cone has an axis and one angle and nothing else — three numbers for the direction and one for the size, against a sphere’s own four degrees of freedom, of which the picture fixes three.

An ellipsoid’s tangent cone is not circular. It has an axis, two half-angles and a roll, and none of those is straightforwardly the direction of the centre — the cone’s axis is pulled toward the ellipsoid’s long direction, so reading it as the direction of the centre is a second version of the drawn-ellipse trap, one level up.

So the clean statement — axis is direction, half-angle is angular radius — is a sphere’s own property. For anything else the outline still fixes the cone exactly, and unpacking the cone into facts about the object is where the work is.

That is worth keeping in view because the sphere is the case everybody tests on. A recovery validated on balls and deployed on fairings is a recovery whose cleanest assumption has quietly stopped holding, and nothing in the arithmetic announces it.

What a second picture adds #

One picture gives a ray and a ratio. Two give the ball, and the arithmetic is short: two cone axes are two lines through two eyes, they meet at the centre, and the radius follows from either half-angle.

That is the next rung and it has its own conditioning and its own degeneracy, which are worth an essay because they are not the ordinary two-view ones — there is no correspondence problem here, since a ball’s outline in one picture and its outline in another are guaranteed to be the same ball.

What this does not settle #

It does not treat an uncalibrated camera. The step from the picture conic to the cone needs $K$, and without it the outline gives a conic and nothing angular at all.

It does not treat a general quadric. An ellipsoid’s tangent cone is not circular, so its axis is not the direction of its centre and the eigen-decomposition says something more complicated and less useful.

And it does not treat the outline’s own accuracy. Fitting a conic to a blurred edge is an estimation problem with its own literature, and everything here assumes the conic is given exactly — which is the same assumption every essay in this field makes and states.

A single picture of a smooth object gives the cone of rays that graze it, and that cone is the whole of what is known. Every quantity that is a property of the cone is exact; every quantity that needs the picture plane as well is a quantity about the glass.