A mirror ball does not know its size

A mirror ball is an equal-area fisheye treats the ball as an instrument: a known sphere, and what it does to the room reflected in it. Every essay in the mirrors field is written that way round, and it is the direction a reader never stands in. The reader has a photograph. The ball in it belongs to somebody else.

So: given the picture, how big is the ball?

The answer has two parts and only the first is the one anybody expects.

The outline gives a ratio and stops #

A ball’s outline in a picture is the boundary of the cone of rays through the eye that graze it. That cone’s half-angle is $\arcsin(r/d)$, and one picture of a ball established what follows: the cone’s axis is the direction of the centre to eight decimal places, its half-angle is $\arcsin(r/d)$ to ten, and neither $r$ nor $d$ separately.

That was said there of an ordinary ball. It is said here of a mirror ball for exactly the same reason, because an outline does not care what the surface is made of. A ball of radius 0.576 m at 2.40 m subtends a half-angle of 13.887°; so does one of 2.016 m at 8.40 m; so does one the size of a planet at the right distance.

The scale ambiguity is not a weakness of the method. It is what one view of anything is: a projection through a point destroys absolute size, and no amount of care with the outline recovers it.

It is worth being precise about how complete the loss is, because “the outline is nearly the same” and “the outline is the same” are different claims and only the second is true. Scale the ball, its distance and the whole room by a factor and the conic the outline traces has identical entries — not similar entries, the same nine numbers. Every derived quantity is therefore identical too: the cone’s axis, its half-angle, the ellipse’s eccentricity, the offset between the ellipse’s centre and the image of the ball’s centre. There is nothing left over for a cleverer reading to find.

What a mirror ball has that a matt one does not #

A matt ball’s picture is its outline and its shading. A mirror ball’s picture is its outline and the room.

That is the whole of the difference, and it is enough. Scale the ball and its distance together and the outline is unchanged — but the room did not scale with it. A ball twice as far away sees the room from twice as far away too, and a room point that filled a quarter of its little world now fills less.

So an identified point of the room, reflected in the ball, is a second observation, and it is one the outline does not already contain.

The forward map is a reflection off a sphere, which is Alhazen’s problem: given the eye, the centre and the target, find the point of the sphere where a ray from one reflects to the other. It is a quartic in general and it is a single root-find here, because the reflection point of a sphere always lies in the plane through the eye, the centre and the target — the configuration is symmetric about that plane, and a reflection point off it would have a mirror twin.

Bracketing between the two grazing angles leaves exactly one root, which is the part that matters. The quartic’s other three roots are on the far side of the ball or behind the eye, and choosing among them is where a solver written from the algebra goes wrong.

The recovery #

Hold the outline’s ratio fixed, slide the ball along its own axis, and watch where the identified room point appears. Push the ball out and the reflection slides toward the ball’s pole, because the room subtends less and less of the ball’s field. That is one-signed over the whole range, so a bracket is safe.

Run the root-find and the distance comes back: 2.400 m against 2.4, and the radius with it at 0.576 m. The residual of the reflection law at the answer is $3\times10^{-14}$ degrees, which is arithmetic rather than geometry.

Between the two balls in the first figure, the reflected point moves 8.70°. That is not a subtle signal. A reader with a protractor and a photograph could see it.

Why a bracket is safer than a formula here #

It is worth dwelling on the monotonicity, because it is the only reason the recovery is a root-find rather than a search.

Fix the ratio $r/d$ and push the ball out. The ball grows to hold its outline, so the angular geometry at the eye is unchanged: the same cone of grazing rays, the same little world. What changes is where the room sits in that world. A room point at 3.8 m was two thirds of a ball-radius away when the ball was small; when the ball is three and a half times bigger it is much closer than a radius, and a point close to a convex mirror images close to the mirror’s rim.

So the reflection slides toward the rim as the ball comes out — one-signed over the whole range, with no turning point to trap a solver. The bracket is checked rather than assumed, which matters: a configuration where the observed reflection lies outside the achievable range is a photograph that is inconsistent with the stated ratio, and returning a nearest answer for it would be reporting a measurement of an impossible ball.

And now the number that spoils it #

The recovery is exact in arithmetic and that is not the same as useful. The question a reader should ask of any recovery is what one pixel of measurement error costs, and here the answer is unpleasant.

At the configuration above — a 58 cm ball at 2.4 m, a room point at 3.8 m and 38° off the ball’s axis, a 900-pixel focal length — the reflected point moves 0.148 pixels per centimetre of the ball’s distance. Turn that round: one pixel of error is about seven centimetres of ball.

That is workable. What is not workable is what happens when the room gets further away.

The size comes from the near room #

Sweep the room point’s distance from 30 m to 3 km and fit the slope: −1.04. The sensitivity falls as $1/L$.

The mechanism is worth naming rather than fitting. The recovery works by parallax across the ball: moving the ball changes where a fixed room point lands in it. A room point at infinity is not fixed in any useful sense — its direction from every part of the ball is the same direction, so the image of it sits at the same place in the ball’s little world whatever the ball’s distance. The parallax is gone, and with it the size.

At 100 m the figure is $9.8\times10^{-4}$ pixels per centimetre: one pixel of measurement is ten metres of ball. That is not a weak measurement. It is no measurement.

The practical statement is blunt. A mirror ball photographed outdoors has an outline, a reflection, and no recoverable size. Everything in the picture is far compared to the ball, so nothing in it carries the parallax the recovery needs. A mirror ball photographed in a room, with a hand or a chair or a doorframe near it, has a size — and the closer the near thing is, the better.

The law is derivable without the quartic #

The exponent is fitted because Alhazen’s quartic is in the way, and the leading-order sensitivity can be had without going near it — which turns $-1.04$ from a measurement into a prediction and supplies the coefficient as well.

Move the ball by $\Delta d$ along the eye’s axis. A room point at distance $L$ from the ball, at angle $\psi$ from that axis as seen from the ball, changes its direction from the ball by $\Delta d\,\sin\psi/L$ — the ordinary parallax of a displaced observer. The ball then renders that direction at picture radius $\rho = R_{\text{img}}\sin(\theta/2)$, so the drawn point moves by $(R_{\text{img}}/2)\cos(\theta/2)$ per radian of direction. Multiplying,

$$\frac{\text{pixels}}{\Delta d} = \frac{R_{\text{img}}}{2}\,\cos\frac{\theta}{2}\, \frac{\sin\psi}{L}.$$

One over $L$, with no quartic anywhere in it. The quartic decides where on the ball the reflection happens; it does not enter the rate at which that place moves, because the rate is a composition of a parallax and the ball’s own equal-area rule.

Put the essay’s own configuration in — a ball drawn 222 px across the radius, a room point 1.4 m beyond the ball at 38° off the axis — and it gives 0.159 pixels per centimetre against the measured 0.148. Seven per cent, from an expression with three factors in it, which is what a leading-order derivation should manage.

The coefficient is more useful than the exponent, because it is a design rule with four terms in it.

Use a ball that is large in the frame. The sensitivity is proportional to its drawn radius, so filling more of the picture with the ball is worth exactly as much as it sounds.

Identify a near room point. One over $L$, which is the essay’s own conclusion, now with the constant attached.

And put it about 110° off the ball’s axis. The two angular factors pull against each other — $\cos(\theta/2)$ wants the reflection near the picture’s centre and $\sin\psi$ wants the room point well off the axis — and their product $\sin(\psi/2)\sin\psi$ peaks at $\psi = 109.5°$. So the most informative feature is one behind and to the side of the ball, not the one squarely reflected in its middle, which is where an eye would naturally pick.

That last is the sort of thing a fitted exponent can never say, and it is the argument for pushing a derivation one step past the power law even when the exact expression is out of reach.

An exponent, and what it is worth knowing #

A fitted exponent is a weaker statement than a derivation and a stronger one than a plot, and it is the right instrument here for a reason worth naming.

The derivation would have to go through Alhazen’s quartic and would produce an expression nobody can check. The plot shows a curve that flattens, and a curve that flattens on linear axes is indistinguishable by eye from a curve that flattens to something. The exponent settles which: −1.04 over two decades is $1/L$ with the fit’s own noise on it, and $1/L$ tends to zero.

That distinction — a term that falls away against a term that flattens onto a floor — is the one an error with two terms makes into a measurement, and this row’s sensitivity is one of the laws it reads. There is no floor here. The size is not merely hard to recover from a distant room; it is not there.

The reflection that says nothing at all #

There is a second degeneracy and it is the one every photographer walks into. A room point on the ball’s own axis reflects at the ball’s pole for every distance the ball could be at. It is a one-parameter family, not an answer.

The solver refuses that configuration rather than returning the pole. That is the right behaviour and it is worth a check of its own: a routine that returns a number where there is a family of them is how a degeneracy becomes a row in a table.

And the photograph everybody takes of a mirror ball is the one with the photographer centred in it.

What this shares with the rest of the row #

Read next to the dome port and the pane of glass, the shape of the answer is the same one three times: the instrument returns a ratio, and the size comes from the room.

A dome port’s picture depends on its decentring over its radius and on nothing else. A pane’s displacement is thickness times $(1 - 1/n)$, a product of the two numbers a reader wants separately. A mirror ball’s outline is $r/d$. In every case the missing scale has to be imported from something the instrument is not.

The exception in this row is the caustic, and the reason it is an exception is the design rule the rest of the row is measured against: a caustic is a length on the table. It is not a ratio of anything. That is what makes it the only one of the five that answers without borrowing.

The trap next door #

There is a second reading of a ball’s picture that looks easier and is wrong, and it is worth restating here because a mirror ball invites it more than a matt one does.

The drawn outline of a ball is an ellipse, and the ellipse has a centre. That centre is not the image of the ball’s centre — it is out by pixels, and the gap grows with how far off-axis the ball is. One picture of a ball measures it at 2.25 pixels for a ball a metre and a half off the optical axis.

So the cone route is not an elaboration of an easy method. The easy method is wrong, and it is wrong by an amount that is invisible unless somebody computes the honest answer to compare against.

Two pictures, and what they do not fix #

Two pictures of a ball settles the matt case: two cones from two eyes intersect at the centre, the radius follows, and a third picture adds nothing because a sphere is four numbers and two cones supply five constraints.

That route works here too and it needs a second photograph. The recovery in this essay needs one photograph and one identified point of the room, which is a different kind of expense — and it is the cheaper one whenever the room is known and moving the camera is not.

What neither route escapes is the near-room requirement, and it arrives in the two-picture case as the baseline: two eyes and the ball in a line have cone axes that are one line, and a point on a line is not determined by that line. The degeneracy has a different name and the same content. A recovery of size needs two places, and how far apart they are is the whole of its conditioning.

A note on what “the room” has to be #

The recovery above assumes an identified point at a known position. That is a strong assumption and it is worth being honest about how strong.

It does not need the whole room. It needs one point whose distance and direction from the camera are known — a corner of a sheet of paper on the table, a mark on a ruler, the near edge of the table itself. Anything that supplies a length.

Which is the same requirement, stated in the same words, as the one every other essay in this row arrives at. The instrument returns a ratio. A length has to come from somewhere, and the somewhere is never the instrument.

What the reader is left holding #

Three numbers, and it is worth separating them because they are usually run together.

The ratio $r/d$ is free. It comes off the outline, it is exact, and it survives any amount of scaling of the whole arrangement.

The size costs one identified room point, and the price is set by how near that point is: about seven centimetres of ball per pixel at arm’s length, ten metres of ball per pixel at a hundred. That is a factor of a hundred and fifty in the answer for a factor of thirty in the room, which is the $1/L$ law appearing as a bill.

And the shape is assumed throughout. Everything above takes the ball to be a sphere. A fitted radius is wrong before it is uncertain is what happens when that assumption is examined, and its finding transfers here directly: a fit inside a shape family returns a confident answer for objects outside it, and the residual does not say so.