A guessed veil halves the error

The shadows the unit counts are the ones a room removes put a veiling luminance on a display: the room’s light reflected off the screen, a fixed number of candelas a square metre added to every pixel. On a thousand-candela display seen through one candela of veil, a lightness step at the bottom of the scale lost a fifth of the difference ΔEITP gave it and a step at the top lost nothing measurable. The unit counts shadows most, and shadows are what a room takes.

ΔEITP already takes one argument a relative colour difference does not — the stimulus’s absolute luminance — and that essay proposed a second: a veiled ΔEITP, the same unit with a stated veil added to both stimuli before the perceptual quantiser. At zero veil it is the unit; with the right veil it is exact. It named the awkward possibility too. A unit whose argument is always guessed is not obviously better than one that leaves it out — CAM16-UCS’s adapting luminance is quoted as 100 and almost never measured — and which of those a veiled ΔEITP would be depends on whether the veil is easy to know.

This essay measures what it costs to get the argument wrong, which is the only way to answer that.

A guess beats nothing; only a measurement is exact #

On a 100 cd/m² display in a room putting one candela on the screen, ΔEITP read without a veil is wrong about the darkest grey step by a factor of three. Any declared veil from nearly nothing up to about twice the true one does better than declaring none, on both displays and every veil tried. One middling guess for every room — about one candela — halves the worst error of declaring none. To keep every step within ten per cent, though, a one-candela veil must be known within a factor of 1.18 on a 100 cd/m² display.

• Declaring no veil on a 100 cd/m² display under one candela reads the L 2 step at 10.5 when it is 3.4.* The worst log-error over the scale is 1.14.
• A declared veil beats declaring none from nearly nothing up to between 1.7 and 4 times the true one. More than that, and it is worse.
• In a dark room, declaring a veil that is not there costs: one candela declared where a hundredth is present is several times worse than declaring none.
• Over rooms putting 0.03 to 3 cd/m² on the screen, the best single declaration costs 0.93 against declaring none’s 1.89 on the dimmer display, and 0.30 against 0.61 on the brighter.
• Without a veil, the unit says a shadow step counts 2.5 times a highlight step on a 100 cd/m² display. Through three candelas of veil it counts two thirds of one.

What the veil does to the scale #

The earlier essay reported the veil as a loss at the dark end. It is worth seeing as a whole curve, because the error a declaration makes is the distance between two such curves.

At the top of the scale all four readings agree: a lightness step at L* 90 is hardly touched by a candela of veil on a display whose white is a hundred. At the bottom they fan out. The true step at L* 2 is 3.37; declared without a veil the unit reads it at 10.54, three times too large, because it thinks two dark greys of about a fifth and a third of a candela are being compared as they are, when the eye is really comparing them riding on a candela of reflected room light. Declaring 0.3 candelas reads it with about half the error; declaring 3 reads it too small, by about two thirds as much as declaring none reads it too large.

The shape explains the whole essay. A veil is an offset added to both members of a pair, and ΔEITP’s quantiser is steep near black and flat higher up, so an offset changes the dark steps enormously and the bright ones not at all. Getting the offset wrong in either direction moves the dark steps, and how far depends on how far the declared offset is from the true one — measured, as it turns out, in proportion to the true one.

Which guesses beat none #

This is the error a declaration makes, as a function of the veil declared, for four rooms. Each curve is zero at its own room’s veil and rises on either side.

Every curve has a flat stretch at the left, where the declaration is a small fraction of the truth — declaring a thousandth of a candela is declaring none, and the error there is declaring none’s error: 0.22 in a room putting a tenth of a candela on the screen, 0.54 at 0.3, 1.14 at one candela, 1.89 at three. Every curve crosses that level again at between about twice and four times its true veil. A declaration anywhere between nothing and twice the truth does better than declaring nothing, and for the larger veils the margin runs to four times; beyond the crossing it does worse.

The practical form of that is a single guess made for every room, because a display tolerance is usually written once and applied wherever the display is used. Over rooms from a dim edit suite, 0.03 candelas on the screen, to a bright office, 3 candelas, the best single declaration on a 100 cd/m² display is about 0.8 candelas, and its worst error is 0.93 — against 1.89 for declaring none. On a 1,000 cd/m² display the best is about 1.3 candelas and its worst 0.30 against 0.61. In both cases a guess in the middle of the range halves the worst case.

That settles the awkward possibility as the earlier essay posed it. A veiled ΔEITP with a guessed veil is better than ΔEITP without one, in the worst room as well as on average, provided the guess is not wildly above the rooms it will meet. The one way to make it worse is to guess large for a dark room.

The cost of a phantom veil #

A declaration of one candela in a room that puts a hundredth on the screen — a grading suite with the lights down — reads every dark step too small. Its worst error is several times declaring none’s in that room, because declaring none was nearly right there and a candela of imagined veil erases the shadow differences the dark room actually shows.

So the direction of a wrong guess matters. Guessing low costs at most what declaring none costs, since declaring none is the lowest guess there is. Guessing high has no such floor. A tolerance written with a veil for offices and applied in a grading suite would pass shadow errors the suite’s colourist can see. The guess that halves the worst case across rooms is safe precisely because it sits in the middle of the rooms it is meant for; it is not a default that transfers to a room it was not chosen for.

How well the veil has to be known #

Halving the worst error is not the same as removing it. An error of 0.93 in the logarithm is still a factor of two and a half on some step, and a tolerance quoted to two significant figures needs more than that.

To keep every step within ten per cent, a one-candela veil must be known within a factor of 1.18 on a 100 cd/m² display — to within about a sixth, above or below. Within 25 per cent, a factor of 1.48. On a 1,000 cd/m² display the same veil can be off by ×1.48 for ten per cent, because a brighter display’s darkest greys are brighter and a candela matters less to them. The larger the veil, the more closely it must be known; small veils on bright displays hardly matter at all, which is why those curves leave the top of the axis.

A factor of 1.18 is not a guess. It is a measurement, and a cheap one: a luminance meter aimed at the screen with the display showing black reads the veil and the display’s own black together, and with the display switched off reads the veil alone. A calibrated luminance meter reads a candela to far better than a sixth. The argument is one that can be measured with the instrument already on a colourist’s desk, in the room where the tolerance will be applied, which is exactly what CAM16-UCS’s adapting luminance is not.

What the unit without a veil says about shadows #

There is one thing a veil changes that no single number captures: the balance between shadow and highlight differences, which is what a tolerance written in ΔEITP implicitly sets.

On a 100 cd/m² display, ΔEITP without a veil says a lightness step in the deep shadows counts 2.53 times a step in the highlights. Through a tenth of a candela of veil it is 2.27; through one candela, 1.23; through three, 0.67 — the shadow step counts less than the highlight one. On a 1,000 cd/m² display the same veils take the ratio from 3.57 to 2.56. The unit without a veil reports the leftmost value in every room, so a tolerance written in it treats shadow errors as two and a half times as serious as highlight errors on a display where, in an ordinary lit room, they are two thirds as serious.

That is a larger consequence than any single step’s error, because a tolerance is a rule for trading errors against each other. A lit room brings the units’ medians together found that a room changes what CAM16-UCS says through the viewer’s adaptation; here the same room changes what ΔEITP should say through the stimulus. The units part by hue, not by light level compared the two units as though neither had a room; both have one, entering in different places.

What a tolerance should carry #

A display tolerance in ΔEITP should state the veil it assumes, as it states the display’s peak luminance. How bright is white followed the move to encoding that names candelas; a tolerance that names candelas for the display and nothing for the room has named half of what reaches the eye. The veil is a number in candelas a square metre, measurable in a minute, and the error from leaving it out is larger at the dark end than the tolerances themselves.

Where the room is unknown, declare a middling veil rather than none. About one candela on a 100 cd/m² display halves the worst case over rooms from a dim suite to a bright office, and it can only do worse than declaring none in rooms darker than it assumes — which, for content delivered to lit rooms, is the rare case.

And count the room in the budget. A delivery tolerance is three tolerances priced a colour’s journey to a reader as a profile, a gamut mapping and a room, and a tolerance has no light level showed why the room’s share cannot be read off a unit that has no argument for it. For a display judged in ΔEITP, the veil is the room’s stage written as a number. A budget that carries it can say how much of a shadow error the room will hide and how much it will reveal; a budget that does not has quietly assumed the dark room every time.

And where shadows are what is being judged, measure it. A black that is not black is the standing reminder that a display’s own black is never zero either; the veil and the display’s black add, and a meter reads their sum. The unit is exact when given it and useless for shadows when not.

How the steps were read #

Each step is a pair of greys one unit of CIELAB lightness apart at L* 2, 5, 10, 20, 35, 50, 70 and 90, relative to a D65 white, scaled to the display’s white in candelas, with the veil — D65 light of the stated luminance — added to both members, and read in ΔEITP through the perceptual quantiser. The true reading adds the true veil and the declared reading the declared one. The error of a declaration is the largest, over the eight steps, of the size of the natural logarithm of the declared reading over the true one.

Declared veils run on a logarithmic grid from a thousandth of a candela to ten, with none first. The tolerance factor is found by bisection as the largest factor f for which declaring the true veil times f and divided by f both keep every step within the stated error. The single best declaration is the one whose largest error over true veils of 0.03, 0.1, 0.3, 1 and 3 candelas is smallest.

What this leaves out #

A veil is not uniform. Room light reflected off a screen depends on where the lamps and windows are, and a glossy screen reflects them specularly, so the veil varies across the screen by more than the factors here. A measurement at the centre is a measurement at the centre.

The steps are greys. A coloured step’s shadow behaves like a grey’s in how the veil lifts its luminance, and the veil also desaturates it, which a lightness step cannot show. The shadows the unit counts are the ones a room removes measured lightness only, as does this.

And the error is ΔEITP’s error, not a reader’s. A veiled ΔEITP is exact about what ΔEITP says through a veil, and whether ΔEITP through a veil matches what a reader sees through one is the same open question it always was.

Still open: whether the display’s own black belongs in the same argument #

A display’s black is not zero: an LCD leaks a fraction of its white, an OLED’s black is dark until the room’s light lands on it. The display’s own black and the room’s veil both add a floor to every pixel, but they differ in how they scale — the display’s black is a fixed share of its white and the room’s veil a fixed number of candelas.

The prediction is that the two can be folded into one declared floor without loss, because ΔEITP sees only the sum at each pixel; and that the tolerance factors here apply to the sum. The awkward case is a display whose black level changes with content — a local-dimming panel, whose black under a dark scene is far lower than under a bright one — where the floor is not one number at all. The computation is the veil census with a content-dependent black added, the question being how large the floor’s variation across a scene can be before a single declared floor does worse than none.

An argument is worth what it costs to know #

The habit is about judging a model’s extra argument by how well it can be known, not by whether it can be.

A unit with an extra argument is exact when the argument is right and wrong in a characteristic way when it is not, and the fair comparison is not with perfection but with the unit that leaves the argument out. Measured that way, the veil earned its place twice: a guess in the middle of the rooms halved the worst error, and a cheap measurement removed it. CAM16-UCS’s adapting luminance, the argument the worry was modelled on, fails the second test — nobody reaches for a meter to find what a viewer is adapted to — which is why a guessed default is all it ever gets.

The failure mode is to dismiss an argument because it is usually guessed, without asking how well it has to be known and what the guessing costs. The two numbers here — twice the truth to beat nothing, a sixth of it to be trusted — are what that question returns, and they are properties of the unit that should travel with it.