An upper pane mirrors a sky no curtain can afford

Glass hides a black net and not a white voile put a pane of window glass in front of a sheer and found that it does two things pulling opposite ways. It mirrors the street, which adds to the veil a passer-by sees and hides the room — veil in the sense a sheer’s privacy is the error it multiplies used, light in the view that carries nothing of the room. And it dims the street’s light on the curtain’s threads, which takes veil away and shows the room. For a white voile seen squarely the second wins, and the glass makes the room slightly easier to read; past 57 degrees the first wins. The whole calculation assumed that the pane mirrors a street as bright as a surface reflecting three tenths of the light on the window.

That is right for a ground-floor window, which the passer-by looks at nearly level and whose pane therefore mirrors the houses across the road. It is not right for a window upstairs. A passer-by looks up at it, and a vertical pane sends back the ray that continues upward: an upper pane seen from the pavement mirrors the sky. The essay on glass named this as the input that would change its answers most and left it there.

The answer is not a correction of a few per cent. The sky is too bright for glass to help any curtain, at any angle, and whether an upper pane shows the sky turns out to depend on one number about where the passer-by stands.

The limit on what a pane may mirror #

The break-even condition in the essay on glass compared two veils and found the light ratio cancelling out of it. The pane adds a mirror of brightness $P\,k\,R$, where $P$ is its reflectance at the passer-by’s angle, $R$ the street’s light over the room’s, and $k$ the brightness of whatever it mirrors, measured as a reflectance: $\pi$ times that scene’s luminance over the illuminance on the window. The pane removes $(1-o)\,R_t\,(1-T_d)\,R$ from the curtain’s own veil by dimming the street’s light on its threads, with $o$ the open area, $R_t$ the threads’ reflectance and $T_d$ the pane’s transmittance for diffuse light.

The essay on glass fixed $k$ at 0.3 and solved for the angle. Solved instead for the scene, the condition says how bright a mirrored scene the glass can afford at each angle before it starts hiding the room:

$$k^{*}(\theta) = (1-o)\,R_t\,(1-T_d)\,\frac{1-P(\theta)}{P(\theta)}.$$

For the white voile of the curtain essays, $k^{*}$ is 0.53 straight on, 0.47 at forty degrees and 0.30 at 56.7°, which is where the glass essay’s break-even fell against its street of 0.3. It falls steeply after that, because the pane’s reflectance climbs. A more closed white cloth can afford more: 0.73 at three tenths open, and at the limit of a cloth with no holes at all, 1.04. No sheer, however white and however closed, can afford a mirrored scene brighter than about a white wall lit like the window itself.

That is the ceiling against which the sky has to be measured.

How bright the sky is, in a window’s units #

A sky is not a surface and has no reflectance, but it has a luminance, and the window has an illuminance, and their ratio times $\pi$ puts the sky on the same scale as the street.

The sky chosen is the CIE standard overcast sky: a direction at elevation $e$ is $(1 + 2\sin e)/3$ as bright as the zenith, so the horizon is a third of the zenith and nothing in it depends on the direction round the compass. It is the sky a satin mirrors the top of the sky used for the same reason: it has no sun, and so no accident of the hour or the orientation of the street. No cloth shines under a sky had taken the simpler uniform dome, under which every direction is equal; the overcast one is the smallest step away from it that a real sky takes.

The window’s illuminance comes from what it sees in front of it: the sky above the opposite roofline, the opposite façade below that, and the road. Each direction in front of the window was followed to what it meets in a street twelve metres from façade to façade between terraces ten metres high, with façades reflecting three tenths of their light and a road reflecting a fifth. The façade’s and the road’s brightness depend on the light they receive, which depends in turn on each other, so the three were found together and converged in a few rounds.

The result at a first-floor window, 4.5 metres up, is that the sky’s horizon scores 1.19 and its zenith 3.58. The darkest direction an overcast sky has is more than twice the brightest scene a half-open white voile can afford, and above the limit for a cloth with no holes. A window higher up gets more light from the sky it sees over the roofline, so the same sky scores a little less against it — 1.03 at the horizon for a second-floor window — but the difference is small and the conclusion is unchanged.

No sky is dark enough #

The comparison has one more twist that makes it one-sided. The sky a pane mirrors is not any direction the reader cares to name. It is the direction the passer-by’s line of sight continues in after it is turned round at the glass, and that continues at exactly the elevation the passer-by looked up at. A pane seen at forty degrees from below mirrors sky forty degrees up.

So the steeper the view, the brighter the part of the sky in the glass, and at the same time the lower the limit it has to stay under, since the pane’s reflectance rises with the angle. The two move apart. At forty-four degrees the first-floor pane mirrors a sky of 2.85 against a limit of 0.44; at seventy-one degrees, a sky of 3.45 against 0.11. From nowhere a passer-by can stand does an overcast sky in the glass come within a factor of two of what a white voile can afford, and a curtain is helped by its glass only where the glass mirrors something that is not sky.

A clear sky does not rescue the argument; it makes it worse. A satin mirrors the top of the sky noted that a clear dome is brightest near the sun and near the horizon, which is nearly the reverse of the overcast gradient, and the horizon is exactly the direction a pane mirrors when it is seen from far off. The CIE’s clear-sky formula makes the horizon several times brighter than the clear zenith, where the overcast horizon is a third of its zenith; that formula is quoted here rather than computed, but the direction is not in doubt, and it is already the overcast sky that fails.

A vertical pane’s reflection climbs across the street #

Whether an upper pane shows sky or building depends on where the ray it sends back goes after it leaves the window, and that ray obeys a short piece of geometry.

The passer-by’s eye is at 1.6 metres, a distance $d$ out from the façade and some distance $s$ along it. The line of sight to a pane at height $H$ rises by $H - 1.6$ over a horizontal run of $\sqrt{d^2 + s^2}$. The pane is vertical, so the mirror turns round only the part of the ray pointing into the façade: the ray leaves the window heading back out over the street, still rising at the same slope, and still running along the street at the same rate relative to its run across it.

Crossing a street of width $W$ the returned ray climbs

$$\Delta z = \frac{W\,(H - 1.6)}{d},$$

and the lateral offset $s$ cancels. Walking along the pavement lengthens the ray’s run across the street and flattens its slope by the same factor, so it reaches the opposite façade at the same height wherever along the street the passer-by stands. The pane mirrors sky if $H + \Delta z$ clears the opposite roofline $H_o$, which is a condition on the distance out from the façade alone:

$$d < \frac{W\,(H - 1.6)}{H_o - H}.$$

For a first-floor window in the twelve-metre street against a ten-metre terrace that distance is 6.33 metres, which is just past the middle of the road. Anywhere on the near pavement, the first-floor pane shows sky.

Closer is more private #

The chart at the head of this essay follows a white voile half open through the day’s standard ratio of thirty to one, and it has the shape the geometry predicts. The ground-floor pane mirrors the opposite façade and a little of the road from every point in the street; their brightness in the window’s units is 0.36, under the voile’s limit, and the glass helps the reading slightly everywhere: a multiplier of 61 against the bare cloth’s 64.

The first-floor pane is different on the two sides of the road. From across the street it mirrors the terrace opposite, which in this street comes out at 0.31 — the very number the essay on glass assumed as its street, now recovered from the geometry rather than put in. The voile’s multiplier there is 60. Step inside 6.3 metres and the pane is suddenly mirroring sky, and the multiplier jumps to 95; closer in it keeps rising, to 116 at three metres, 150 at two and 347 at one, where the pane is seen at 71 degrees and mirrors the sky seventy-one degrees up.

The second-floor pane is too high for the terrace opposite to fill its mirror from anywhere in the street: its edge distance, $W(H - 1.6)/(H_o - H)$, is 28 metres, well past the far kerb, so the pane mirrors sky from every place a passer-by can stand. Its voile reads at 91 from the far side and 315 from two metres out.

So the same upper window is harder to see into from close up than from across the road. From across the road its pane mirrors a building and the curtain behaves as it would at street level. From underneath, its pane mirrors the sky, and the sky dominates the view.

The sky, not the cloth, fills the window #

The bars below split what reaches the passer-by into the three things the essay on glass separated: the room’s image through the holes, the threads lit by the street, and whatever the pane mirrors.

For the white voile seen from three metres the sky is 53 per cent of what reaches the eye, and the room’s share falls from 1.6 per cent to 0.9. From across the street the terrace is a tenth of the view and the room’s share is 1.7 per cent, a shade better than bare. For the black net the change is total: bare, half of what the passer-by sees is room; from across the street a fifth is; from three metres, two and a half per cent, with the sky at 95.

The essay on glass found that the glass hides a black net and not a white voile, and a net in front gives the figure to the room had found the same asymmetry for a pair of curtains, with the outer layer doing the hiding. With the sky in it, the glass hides every curtain upstairs, and the white voile now pays too: it simply has less to lose, because its own veil was already most of the view.

A black net upstairs is a white voile downstairs #

The black net makes the size of the effect plain. It is barely a curtain at all — eight tenths open, threads returning four hundredths of the light — and bare it multiplies a reading’s error by 2.0, the least private sheer in the curtain essays by a factor of thirty.

Behind a ground-floor pane it reads at 5.9, the essay on glass’s result with the street’s brightness computed rather than assumed. Behind a first-floor pane seen from across the street it reads at 5.3. Seen from two metres below, it reads at 60 — within a few per cent of a bare white voile’s 64 — and from one and a half metres at 92. A second-floor black net seen from the near pavement reads at 80 to 160.

The practical reading is a rule anybody who has walked down a street of terraced houses will recognise without having stated it: an upstairs window is hard to see into from the pavement whatever hangs in it, and the sheerest net upstairs does as well against a passer-by underneath as the whitest voile does at street level. The curtain upstairs is doing less of the hiding than it appears to, and the sky is doing more.

The terrace opposite decides how much of the pavement sees sky #

The edge of the region where an upper pane shows sky depends on three dimensions of the street and on the window’s height, and on no property of any cloth.

A terrace of twenty metres opposite pulls a first-floor window’s edge in to 2.25 metres, so from most of the pavement its pane mirrors a building and the curtain behind it is helped slightly by the glass. A terrace of fifteen metres puts it at 3.3; of ten, at 6.3; of six, at 23 — beyond the width of the street, so the pane mirrors sky from everywhere in it. A pane above the opposite roofline mirrors sky from anywhere, and every window facing a park, a river, a car park or a wide road has its glass mirroring sky from every place a passer-by can stand.

A wider street pushes the edge out in proportion, since $W$ multiplies it. So the privacy of an upper window against the pavement is highest in a narrow street of low buildings opposite, and lowest — in the sense of the glass doing least — where the building opposite is tall and close. A tall building opposite gives the passer-by a dark mirror, and a dark mirror is a clear one.

Walking along the pavement makes it worse #

The lateral offset drops out of whether the pane shows sky, but not out of how much damage the sky does. A passer-by three metres out from the façade walking along it sees the first-floor pane at a steeper and steeper angle, 44 degrees opposite the window and 66 degrees six metres along, while the ray the pane sends back flattens: it mirrors sky 44 degrees up from opposite the window and 23 degrees up from six metres along.

Two things change and one wins. The mirrored sky gets darker as it gets lower — 2.85 falling to 2.14 — but the pane’s reflectance rises faster, and the voile’s multiplier rises from 1.8 times its bare value opposite the window to 2.7 times at six metres along and 4.1 times at ten. The best place on the near pavement from which to see into a first-floor room is directly opposite it.

The same geometry explains why a curtain is gathered so that it is seen edge-on matters little upstairs. Its argument was that a slanting view closes a gathered sheer’s holes; upstairs, a slanting view also turns the pane into a better mirror of a bright sky, and the second effect is larger than the first.

Dusk does not bring the room back #

The essay on glass found that its break-even angle did not depend on the hour, because both the veil the pane removes and the veil it adds scale with the street’s light. The sky in the glass inherits that property, with one qualification.

Under an overcast sky, the sky’s luminance and the window’s illuminance both come from the same cloud, so their ratio — the sky’s brightness in the window’s units — does not change as the cloud dims toward evening. The glass’s multiplier on a first-floor white voile seen from three metres stays between 1.5 and 1.8 times its bare value at every light ratio from thirty to one down to one to one, and the room’s reflectance, which by day is buried under an error five and a half times its size, can be read to within a fifth of itself only at equal light, much as it can bare.

The qualification is the street lamp. After dark the window is lit by lamps and the sky is not lit at all, so the sky’s brightness in the window’s units falls toward nothing, and the upper pane becomes what the ground-floor pane is by day: a dim mirror that helps slightly. The ratio that sets the sky in the glass is not a ratio the hour moves, but the arrival of a light source that is not the sky moves it entirely.

How the numbers were produced #

The pane is Fresnel’s, glass of index 1.52 with both faces counted and no absorption, exactly as in the essay on glass. The sky is the CIE standard overcast sky. The street is twelve metres from façade to façade between ten-metre terraces, infinitely long, with façades of reflectance 0.3 and a road of 0.2, and a passer-by’s eye at 1.6 metres.

The window’s illuminance was found by following every direction in its forward half-space, in steps of half a degree in both angles, to the sky, the opposite façade or the road, weighting each by the cosine of its angle to the window’s normal. The façade’s brightness was set from its own mean illuminance over its height and the road’s from the illuminance at the street’s centre, and the three were iterated to agreement. The mirrored scene’s brightness is $\pi$ times its luminance over the window’s illuminance, and it was put into the pane-and-curtain reading as the scene the pane mirrors, with nothing else in that reading changed.

What the calculation is required to do. An unobstructed window under the overcast sky with a black road must get $\pi/6 + 4/9$ of the zenith’s luminance and open ground $7\pi/9$, which are the integrals done by hand. The height at which the mirrored ray meets the opposite façade must agree with a ray stepped across the street independently, at several positions along the pavement as well as opposite the window. The brightest affordable scene at the essay on glass’s break-even angle must come out as that essay’s street of 0.3, to six figures. And two things that could have failed: from the near pavement a first-floor pane must mirror sky and make a white voile harder to read, while the same voile at street level must be helped.

What the account assumes #

The street is symmetric and uniform, and its façades reflect three tenths of their light, the value what a sheer hides is decided by the furniture showed matters as much as the cloth when it sits behind the curtain rather than across the road. The opposite façade’s illuminance is taken as this façade’s average, both terraces are the same height, and nothing breaks the roofline. A real street has gaps, dormers, trees and parked vans, and each changes which pavement positions see sky; the edge distance is the answer for the uniform terrace and a guide to the rest.

The sky is overcast and uniform in azimuth. A clear sky makes every conclusion here stronger for the upper floors, since its low band is brighter and the ground-floor pane never mirrors it; a sunlit façade opposite could raise the ground-floor pane’s mirror past the limit too, which the overcast sky cannot do.

The veil and the image are what the curtain essays made them: a room reflecting three tenths of its light, the threads’ reflectance and forward scatter from a cloth is more opaque than it is closed, and the curtain lit through the glass by a diffuse street at the day’s thirty-to-one. The window’s own illuminance and the curtain’s are taken as the same.

What the figures cannot show #

The sky in the glass is featureless, and that matters to an eye in a way the arithmetic cannot follow. A sheer hides whichever side is darker and a figured sheer is a negative from one side both found that what the eye reads in a curtain depends on what lies behind it; the reflected street in the essay on glass was an image with edges in it that pull the eye away from the room. A reflected overcast sky is a smooth pale field with no edges at all. The eye may be better at looking through it than through a reflected street of the same brightness, and the multiplier counts both alike.

The figures also cannot show a passer-by moving. A reflected sky does not move with the viewer the way two sheers make a moiré that walks with the viewer found a fringe does; it simply brightens and dims as the viewing angle changes, and the room’s faint image stays fixed behind it. Whether that relative motion helps the eye separate the two is a question about vision, not about the light.

Who found which part #

The overcast sky’s luminance distribution is Moon and Spencer’s, from 1942, adopted as the CIE standard overcast sky in 1955 and in lighting design ever since. Fresnel’s reflection coefficients are from 1823. Architects’ daylight calculations follow directions from a window to the sky and to obstructions in just this way, and the street canyon is their standard case.

What is added here is the limit on the scene a pane can mirror and still help a curtain, $k^{*}$, which is the essay on glass’s break-even solved the other way round; the observation that no overcast sky comes within a factor of two of it; the climb of a vertical pane’s reflection across a street and the cancelling of the passer-by’s position along it; and the resulting rule that an upper window’s glass hides its curtain from the near pavement and not from across the road, with the edge between the two set by the terrace opposite.

Still open: a pane that faces the sky #

Every window here is vertical. A roof light, a dormer’s side window or a sloping conservatory pane is not, and a pane tilted back from the vertical by an angle $\beta$ sends the passer-by’s ray upward at their elevation plus $2\beta$. A pane tilted back twenty degrees and seen from a street at ground level would mirror sky forty degrees up even when looked at level.

A tilted pane mirrors sky from everywhere, and at a brighter part of the dome than a vertical one does, so every curtain behind one would be hidden by its glass from every place in the street. How that interacts with the pane’s own angle to the line of sight, and whether a pane tilted forward, as on some shopfronts, can be made to mirror the road and help the curtain instead, is the same geometry with one more angle in it, and it has not been drawn.