What a scene does

A square floor needs more room than a long one

A satin finish on a room's grey faces costs its painted floor a little chroma when the room is narrow and adds some when it is wide, and in a box two and a half tall the turn was found on the floor's shorter side — 0.64 when the longer side is one unit, about 0.51 when it is two. The prediction for a square floor, with no longer side to speak of, was a later turn, near 0.7, because both pairs of walls are then equally close and both carry the lamp down. Solved at six strips a side, the square floor loses for every one of seventy-two paints at 0.5, 0.6 and 0.7 on a side and gains for every one from 0.8, turning near 0.78. A floor 0.7 by 0.7 loses where a floor 0.7 by one gains. The finish brightens the square's darkest floor by more than an oblong's at every shorter side — by a hundred per cent against forty-six at half a unit — and the turn falls steadily as the longer side grows: 0.78, 0.64, 0.51. The longer side matters, and the light-pipe account says why.

Assumes The width moves the turn only when it is the narrower side, A taller box must be deeper before the finish adds colour and The finish stops costing the paint six tenths deep.

A satin finish laid on the grey faces of a room changes the colour of the painted surfaces they surround. The paint itself gains all or nothing found that it nearly always adds a little colour to the paint, and that the one place it takes colour away is a tall narrow space with its floor or ceiling painted. The finish stops costing the paint six tenths deep walked a box one unit wide and two and a half tall from narrow to deep and found its painted floor turning from loss to gain at about 0.64 of the width, and a taller box must be deeper before the finish adds colour found that turn growing faster than the box’s height.

The account that has survived each scan is a light pipe. Glossy walls facing each other carry the lamp’s light down between them, reflection after reflection at nearly the angle it arrived, the way a polished duct carries daylight into a windowless room. The finish multiplies the light reaching the floor, and the paint loses colour when that multiplication favours light that has never touched the paint over light that has been back to it. Close walls make a strong pipe; walls further apart make a weak one.

The width moves the turn only when it is the narrower side solved boxes half a unit and two units wide at six strips a side and found the turn set by the floor’s shorter side — the closer pair of walls — whichever way round the box stood: about 0.64 when the longer side is one unit, about 0.51 when it is two. Two points could not say whether the longer side mattered at all. Its closing section asked about the one floor with no longer side, the square, and predicted a later turn, near 0.7, because both pairs of walls are then equally close and both pipe the lamp down. If the square turned at 0.64, the longer side would not matter once it was at least the shorter.

The square turns at 0.78

Solved at six strips a side in the same box two and a half tall, a square floor loses chroma for every one of seventy-two paints at 0.5, 0.6 and 0.7 on a side — by 0.065, 0.048 and 0.021 per cent on average — and gains for every one at 0.8 and at one, by 0.0065 and 0.047 per cent. It turns near 0.78, fourteen hundredths later than the floor one unit long and twenty-seven later than the floor two long. At the same shorter side the finish brightens the square floor’s darkest reflectance more than an oblong floor’s: by 100 per cent at half a unit against 46 for the floor two long, by 67 at 0.7 against 54 for the floor one long. The tilt across the floor’s reflectance carries the sign of the paint’s change on every square, as on every oblong.

  • The prediction holds in direction and falls short in size. The square turns later than 0.64, by twice the margin predicted.
  • The longer side matters. With the shorter side fixed, a longer floor turns earlier: 0.78 for a square, 0.64 for one unit long, 0.51 for two.
  • The square pipes harder. Both pairs of walls are close and both carry the lamp down, as the prediction’s reason said.
A square floor turns later than an oblong one. Boxes 2.5 tall solved at six strips a side with the floor painted and a satin finish on the grey faces: the mean change over seventy-two paints in the floor's chroma, against the floor's shorter side, for square floors and for floors whose longer side is one unit and two. The square floor loses at every side to 0.7 and gains from 0.8, turning near 0.78; the floor one unit long turns at 0.64 and the floor two long at 0.51.
Fig. 1 Boxes 2.5 tall solved at six strips a side: the mean change over seventy-two paints in the painted floor’s chroma from a satin finish on the grey faces, against the floor’s shorter side, for square floors and floors one and two units long.

The three curves sit one above another. At a shorter side of 0.7, the square floor loses 0.021 per cent while the floor one unit long gains 0.015; at 0.8, the square gains 0.0065 while the floor two long gains 0.082 — twelve times as much. Read along a horizontal line at nought, each curve crosses at its own place: the floor two long first, near 0.51, then the floor one long at 0.64, then the square near 0.78. The square’s curve is also the steepest of the three where it crosses, rising by 0.028 per cent between 0.7 and 0.8, so its turn is the most sharply located of the three. The floor two long has one oddity of its own, a gain at a quarter of a unit that its finer division could not settle, and it is left out of the reading here as it was there.

A floor 0.7 square loses where a floor 0.7 by one gains

The single comparison that settles the question sits at a shorter side of 0.7. The box one unit wide and 0.7 deep was solved at six strips in the depth scan, and the finish adds colour to its floor for every paint. Shorten its long side from one to 0.7, so that the floor is square, and every paint loses instead. Nothing else about the box has changed: the height, the finish, the lamp, the paints and the division are the same. If the turn were a property of the shorter side alone, the two floors would agree in sign, since their shorter sides are equal.

The square floors fill in the plan's diagonal. Every box 2.5 tall solved at six strips a side with its floor painted, placed by its floor's width and depth, the oblong floors also turned a quarter, and the square floors along the diagonal. The squares take colour from the paint at 0.5, 0.6 and 0.7 on a side and add it from 0.8, so the boundary between loss and gain bends out along the diagonal, to about 0.78, past the one-unit box's 0.64 on either side of it.
Fig. 2 Every box 2.5 tall solved at six strips a side, placed by its floor’s width and depth — the oblong floors also turned a quarter — with the square floors along the diagonal, marked by whether the finish adds colour to the painted floor or takes it.

In the plan of floors the squares fill in the diagonal that the width scan left empty. The boundary between loss and gain is not a pair of straight lines at a fixed shorter side, meeting in a corner on the diagonal; it bends outward along the diagonal, from about 0.64 where the longer side is one to about 0.78 where the two sides are equal. A floor 0.7 square lies inside the bend, on the losing side, and its neighbours 0.7 by one lie outside it.

That shape is what the light-pipe account predicts once it is taken literally. A pair of satin walls a short distance apart pipes the lamp down strongly; a pair further apart pipes it weakly. In a box 0.7 by one, one pair is 0.7 apart and the other a full unit apart, so the floor is fed by one strong pipe and one weaker one. In a box 0.7 square both pairs are 0.7 apart, and the floor is fed by two strong pipes. The paint has more never-reflected light to keep pace with, and it needs a wider floor before its own returned light catches up.

Two pairs of close walls pipe harder than one

Two pairs of close walls pipe harder than one. For boxes 2.5 tall at six strips a side, how much brighter the satin finish on the grey faces makes the painted floor at its darkest reflectance, against the floor's shorter side, for square floors and oblong ones. At a shorter side of half a unit the square floor is brightened by 100 per cent and the floor two long by 46; at 0.7, 67 against 54.
Fig. 3 For boxes 2.5 tall at six strips a side, how much brighter the satin finish makes the painted floor at its darkest reflectance, against the floor’s shorter side, for square and oblong floors.

The pipe’s strength can be read directly. At a reflectance near nought almost nothing the floor sends out has been to the paint before, so the finish’s brightening there measures how much more of the lamp the satin walls carry down. At a shorter side of half a unit the square floor is brightened by 100 per cent, the floor two long by 46 and the floors 1.25 and 1.5 long by 59 and 52; at 0.6, 78 against 58; at 0.7, 67 against 54. At every shorter side the square is brightened most, by a margin that shrinks as the floor widens and its walls draw apart.

The comparison at half a unit is the clearest, because it holds three longer sides at once. A floor half a unit across and two long is brightened by 46 per cent, one 1.5 long by 52, one 1.25 long by 59, and one half a unit long — the square — by 100. Shortening the floor’s long side strengthens the pipe at an accelerating rate as the long side approaches the short, which is the second pair of walls closing in until it is as close as the first. The paint’s own returned light does not grow to match: its share is set mostly by the floor’s own size and reflectance, and much less by how the glossy walls are arranged around it.

The square’s brightening is also the most sensitive to its size. It falls from 100 per cent at half a unit to 58 at 0.8 and 41 at one — 42 points over the first three tenths, where the floor two long falls by 14 over the same range, from 46 at half a unit to 32 at 0.8. Every step out from the smallest square widens both pairs of walls at once, so it weakens both pipes together; widening an oblong’s short side weakens only one of its two.

The finish across the floor’s reflectance

The finish's brightening across the floor's reflectanceFor the square floor 0.5 on a side in a box 2.5 tall, how much brighter the satin finish on the grey faces makes the floor at each of its reflectances, the other squares drawn faint. It brightens the darkest floor by 99.6 per cent and the brightest by 99.0: a tilt of −0.55 points, and the paint loses 0.065 per cent of its chroma.0.000.250.500.751.00+25%+50%+75%+100%how much brighter the satin finish makes the floorthe painted floor's reflectancesquare 0.5 on a sideside 0.52.5 tall, square floor painted · satin at roughness 0.2 on the grey faces · six strips a side
Fig. 4 For a square floor in a box 2.5 tall, how much brighter the satin finish makes the floor at each of its reflectances, the other squares drawn faint. The handle grows the square from half a unit to one through its turn.

The sign of the paint’s change is read from the slope of this curve. A painted floor’s colour comes from the balance between light that has never touched the paint and light that has been back to it, and the second share grows with the floor’s reflectance. If the finish brightens a dark floor more than a bright one, it has favoured the never-reflected light, and the paint’s colour is diluted; if it brightens a bright floor more, it has favoured the light the paint has already shaped, and the paint gains. For the half-unit square the curve falls by 0.55 points across the floor’s reflectance, and every paint loses; at 0.7 it still falls, by 0.15; at 0.8 it rises, by 0.04, and every paint gains. Moved by the handle, the whole curve sinks as the square grows, and it tips from falling to rising between 0.7 and 0.8.

The tipping is small against the curve’s height: a floor brightened by between 58 and 100 per cent, nearly the same at every reflectance, with a difference between its darkest and brightest of under a point. That is why the paint’s changes are themselves so small — hundredths of a per cent of chroma. The finish brightens the floor greatly and almost evenly; what it does to the paint’s colour is the residue of a tiny unevenness. The square’s curve tips later than an oblong’s because its larger brightening needs a larger floor before the paint’s share of it grows fast enough with reflectance to outweigh the lamp’s.

The tilt still carries the sign

On a square floor the tilt still carries the sign. For the square floors and the oblong floors solved at six strips, the paint's mean change in chroma against the finish's tilt — its brightening of the floor's brightest reflectance less its brightening of the darkest. Every square sits with the oblongs in the two quadrants where the two have the same sign, the smallest square furthest out, with a tilt of −0.55 points and a loss of 0.065 per cent of its chroma.
Fig. 5 For the square floors and the oblong floors solved at six strips, the paint’s mean change in chroma against the finish’s tilt — its brightening of the floor’s brightest reflectance less its brightening of the darkest.

Every square floor sits with the oblongs in the two quadrants where tilt and chroma change share a sign: the three squares that lose have negative tilts, the two that gain positive ones. The smallest square is furthest from the origin of any box solved, with a tilt of −0.55 points and a loss of 0.065 per cent, a little over twice the floor two long at 0.35 deep, which was the deepest loss in the width scan. The relation is not a single line — the squares and the oblongs scatter about different slopes — but its sign has held at every box, every division and every width this thread has solved, and the squares add five more.

That matters because the light-pipe account rests on it. The account says the paint changes colour through the balance between the lamp’s light and the paint’s own; the tilt is a direct reading of which way the finish shifts that balance, computed from the floor’s brightness alone with no paint in it. A box where the tilt and the chroma change disagreed would be a box where something other than the pipe was moving the colour. None has been found.

The longer side moves the turn, and the turn falls as it grows

The longer the floor, the earlier it turns. For boxes 2.5 tall at six strips a side, the shorter side of the painted floor at which a satin finish stops taking its colour, against the floor's longer side: a square, a floor one unit long and a floor two long. The turn falls from 0.78 for the square to 0.64 and 0.51: the longer side matters, and lengthening it brings the turn closer.
Fig. 6 For boxes 2.5 tall at six strips a side, the shorter side of the painted floor at which a satin finish stops taking its colour, against the floor’s longer side: a square, a floor one unit long and a floor two long.

Put together, the three turns make a curve. With a longer side equal to the shorter, the turn is at 0.78; with a longer side of one, at 0.64; with two, at 0.51. The turn falls as the longer side grows, and by less per unit as it grows further: fourteen hundredths over the first fifth of a unit of length, thirteen over the next whole unit. That is the shape a weakening second pipe would give — strong when its walls are as close as the first pair’s, fading as they draw apart, and contributing nothing once they are far — and it points to a limit for a floor infinitely long, a slot, somewhere below 0.5.

The width essay’s two points had left open whether its shallower turn for the floor two long was real or an artefact of its long faces being cut into coarse strips. The square removes that doubt from the other direction: its faces are the squarest any box in the scan has had, its division the most faithful — division being what the tall room gains once its walls are divided showed can reverse a reading — and its turn the latest. The ordering of the three turns follows the longer side, not the quality of the division.

What this means for a real room

A stair core, a lift shaft and a light well are often close to square in plan and much taller than they are wide. The census says that a glossy grey surround greys a coloured floor in a square shaft whose side is under about three tenths of its height — 0.78 in a box two and a half tall — where an oblong shaft of the same shorter side but four times as long would colour it from about a fifth. A square lift lobby 2.4 metres across and nine tall is on the greying side; a corridor 2.4 metres wide and as tall, but twenty long, is past its turn. The changes stay small, hundredths of a per cent of chroma, and the room’s own reading moves by tens of per cent under the same finish, as a grey finish mirrors the lamp onto the paint found. What the scan maps is where in the space of shapes a satin surround stops helping the paint’s own colour, and the square is the shape that resists it longest.

How the boxes were solved

Each box is 2.5 tall, its floor a square 0.5, 0.6, 0.7, 0.8 or 1 on a side, its whole ceiling emitting, its floor painted and its other faces grey at a reflectance of a half, with a satin finish — a microfacet lobe of roughness 0.2 on a dielectric of index 1.5 — on the grey faces. Every face is divided into six strips along each side, the division the width scan found necessary — more than the slot itself needed, whose sign the slot’s loss survives every finer division found steady from three strips to eight — and the unknowns are the radiances leaving each patch towards each patch it sees, the lobe averaged over the cones the two patches subtend, solved by Gauss–Seidel to a residual of 10⁻¹¹ with and without the finish at twenty reflectances on Chebyshev nodes. Each of seventy-two paints is read in CIELAB against the lamp on the painted floor.

The oblong floors are the width and depth scans’ six-strip boxes. The square’s turn is the straight-line crossing of nought between 0.7 and 0.8; the floor two long’s between 0.5 and 0.65; the floor one long’s between 0.6 and 0.7.

What the scan does not reach

Two points around each turn. Every turn is a straight-line crossing between two solved boxes a tenth or more apart; the square’s could lie anywhere from about 0.75 to 0.8.

One height. The squares are two and a half tall; the height scan’s square law was found on floors one unit long, and whether a square’s turn grows with height at the same rate is not solved.

The slot. The limit the turn approaches as the longer side grows is suggested by three points and not computed; a box with its longer side several times its shorter would locate it.

The floor only, one roughness, a closed box. The ceiling, a glossier finish and an open wall all moved the earlier turns — an open room, as an open room hands over sooner found, shifts where a finish’s effect changes sign — and none is solved here.

Still open: where a very long floor turns

Three longer sides give three turns that fall and flatten, and the flattening says the second pair of walls stops mattering somewhere beyond two units. That limit is the turn of a slot: two satin walls facing each other across the floor’s shorter side, the other pair so far off that they pipe nothing.

The calculation is the 2.5-tall box with its floor’s longer side at three and four units, the shorter side scanned from 0.35 to 0.65 at six strips a side, the turn read off as here. The prediction is that the turn settles near 0.45 — below the floor two long’s 0.51 by about as much as the step from one unit to two lowered it, halved — and that it is reached by four units, because a wall pair four units apart pipes the lamp to the floor about as weakly as an open end. If the turn keeps falling past four units, the far walls still matter at a distance sixteen times the near pair’s, and the light pipe is carrying light along the room as well as down it. A long floor never loses its colour to the finish solves it.

A second instance tests a rule more than a finer first one

The habit here is about how a rule drawn from a few cases gets tested.

The width scan had a rule — the shorter side sets the turn — supported by two oblong boxes and refined by a finer division that had already overturned one earlier rule. The natural next step was more oblongs. The square was a better test, because it is the one case where the rule’s two readings make different predictions: a rule in the shorter side alone puts the square’s turn at 0.64, a rule in both sides puts it later. A case chosen where readings disagree tests more than many chosen where they agree.

The failure mode is to test a rule with more examples of the kind that produced it. The test is to find the case where the rule and its nearest rival part company and solve that one. Here it was a square floor, and it showed the shorter side setting the turn only as a first approximation, with the longer side moving it by more than a quarter of a unit between a square and a floor two long.

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