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I heard someone (a photographer) say recently that 18% grey is half way between black and white, not 50%. This seemed a bit illogical to me, and when I asked her why, she said she didn't know. After reading a few online articles, I found that 18% is often referred to as middle grey, and considered to be half way perceptually. Is 18% for some reason half way between black and white, and if so why (maybe these percents work on a non-linear scale for whatever reason... ). If not, why do we think 18% is half way, not 50%. Do we see color non linearly?, do our cameras capture light non-linearly, or is this just a sort of relative brightness illusion.

After reading the question this is supposedly a duplicate of, I still don't see why Ansel Adams choose 18%, was it a visual thing?, or why it was so widely adopted. Is this number arbitrary? just what someone though looked correct... or does it have some valid claim to being the middle grey, due to perception (it appears our eyes see things linearly, do cameras do likewise?) or other technical reasons.

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    \$\begingroup\$ duplicate of What is the 18% gray tone, and how do I make a 18% gray card in Photoshop? \$\endgroup\$
    – MikeW
    Apr 30, 2015 at 23:20
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    \$\begingroup\$ @MikeW : it may be a duplicate question of sorts, but all the answers in the link you've referenced are in need of serious corrections. For example, exposure meters are not calibrated for any reflection, not 18%, not 12%. The simple fact is that 18% is a linear measure (since the there is no "gamma" in Nature), while 50% is perceived (as in L* Lab) brightness. 18% is an average according to what is called "grey world" hypothesis. When comparing incident light to the light reflected from the scene we see that with certain set of calibration in average of 18% is reflected back. \$\endgroup\$
    – Iliah Borg
    Apr 30, 2015 at 23:44
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    \$\begingroup\$ @IliahBorg I'd suggest you answer the duplicate. \$\endgroup\$
    – dpollitt
    May 1, 2015 at 0:05
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    \$\begingroup\$ @MikeW, Although that answer is useful, my question is not a duplicate of the provided question. The answer may apply though... \$\endgroup\$ May 1, 2015 at 0:27
  • \$\begingroup\$ So let's say you've determined the perfect shade of grey for your exposure card. Now the question is how much light do I let fall on it when I register it with my meter? 18%? \$\endgroup\$
    – Octopus
    May 1, 2015 at 18:40

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The story goes that Ansel Adams came up with the "18% gray" figure. Back in the hay day of film photography he was developing the zone system and needed to define a "middle gray". It was a judgment call. Eventually, the idea caught on, but film and camera companies picked their own middle gray. It is a fun fact that your digital camera probably uses something more like 12% gray as middle gray.

Whatever the number, the idea behind middle gray is not that is "reflects 50% of the light". Or even that "it is half way between absorbing all light (pure black) and reflecting all light (pure white)". It has to do with your perception.

Your eyes are logarithmic detectors. That is, if a source gets brighter by a factor of 4, it will only seem brighter by a factor of 2 to you. If it increases by a factor of 32, it will only seem brighter by a factor of 5. If it increases in brightness by a factor of 128, it will only seem 7 times brighter to you.

The above are not the actual numbers. As you can imagine, measuring how bright things seem to people is very tricky, and varies from person to person. The important thing is that it is this weird logarithmic nature of your eyes that keeps middle gray from being 50%.

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    \$\begingroup\$ So 18% grey appears half as bright? \$\endgroup\$ May 1, 2015 at 0:27
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    \$\begingroup\$ Relevant MinutePhysics video describing how digital cameras capture light. Basically, your guess is correct, as @theJollySin says: cameras use linear light, but human eyes use a logarithmic scale. \$\endgroup\$ May 1, 2015 at 2:43
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    \$\begingroup\$ @GiantCowFilms : yes, if you mix 18% of charcoal black and 72% of titanium white (that's how neutral grays were produced, by mixing those 2 components), the result appears to be half the brightness of pure titanium white. Incidentally, it is very close to 2.5 stops (EV) difference. \$\endgroup\$
    – Iliah Borg
    May 1, 2015 at 3:30
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    \$\begingroup\$ @GiantCowFilms - sorry, typed the numbers in the wrong order. must be: 18% titanium white, 72% charcoal black; to get 18% reflection. \$\endgroup\$
    – Iliah Borg
    May 1, 2015 at 4:53
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    \$\begingroup\$ Your eyes are logarithmic detectors +1 That's pretty much the hardest thing about photography. \$\endgroup\$ May 1, 2015 at 14:49
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It's worth looking at a gamma chart for additional perspective as you think about this. Standard display gamma, for example, is 2.2. The curve looks like this:

enter image description here

50% grey, in an 8-bit space, is 127 (horizontal axis). This lines up with ~20% luminance output of the display. Both for display and print the concept of gamma is important as it provides the mapping, or conversion, between the linear (camera/image) data and the logarithmic sensitivity of the human eye.

The human eye can resolve something on the order of 10-14 f-stops of dynamic range at a fixed pupil size. This is up to ~3 stops better than the best DSLRs shooting in 14-bit RAW. Our brain is also capable of using all of that data at once - it's like we have a 16-bit RAW image processor built into our visual cortex[*] and it automatically adjusts the highlight and shadow levels, etc, to get a perfect exposure in real time. ~18% grey is just an empirical value that fits the processing that our eyes will naturally apply to the scene they see.

It is empirical because it works and looks mid-grey in a typical scene. The eye is easily fooled, however, and is extremely context sensitive. The brain will mercilessly photoshop what the eyes see in order to try to make sense of it and greys are routinely imagined to be any shade that makes sense to us. The classic illusion of this is this :

enter image description here

where the A and B squares are identical in brightness. So, yes, the eye is extremely non-linear and, furthermore, is not even uniform in its rendering over our visual field. Darks are brightened, brights are darkened, and the entire scene is heavily compressed into a narrow perceptual range that we can extract detail from.

When shooting high dynamic range scenes this is intuitive, I think, to photographers - we really have to work in post to balance a high dynamic range scene into a form that appears similar to what the eye perceives. When we can control the light, we add LOTS of it - fill, fill, fill. Getting a balanced colour photo that doesn't need a lot of post requires that we add as much light as possible to fill in the dark areas of the scene - reducing the dynamic range as much as possible to produce a scene that is flatter and more uniformly lit (just like our brain tries to do with the scenes we see).


To answer the comment below, this is taken from the image above to make the point :

enter image description here


[*] To be more precise, for those who wish it, some of the initial image processing and compression is done by several layers of specialized cells directly behind the retina before the information is sent to the brain.

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    \$\begingroup\$ I fail to see why you state that A and the B in the photo you posted are of the same color or brightness. B is rgb(82,82,82) or so, while A is rgb(66,66,66) (as are the shaded areas of the squares). \$\endgroup\$ May 3, 2015 at 12:48
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    \$\begingroup\$ @DenisdeBernardy Hmm. I get rgb(78,78,78) for both squares using the colour picker in GIMP and tried painting a line between them. Which parts did you measure exactly? \$\endgroup\$
    – Anko
    May 3, 2015 at 14:26
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    \$\begingroup\$ @DenisdeBernardy I don't mean the letters A and B, I mean the squares upon which they are written. At that, the values you are stating are wrong on both fronts - are you sure you measured correctly? \$\endgroup\$
    – J...
    May 3, 2015 at 14:27
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    \$\begingroup\$ @Anko Indeed, the correct value is x78 (120, decimal). The letter "A" is x51 (81 decimal) and the letter "B" is x42 (66 decimal). \$\endgroup\$
    – J...
    May 3, 2015 at 14:35
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Even beyond perceptual issues, film exposure latitude is another reason to favor 18% gray. If one tried to expose a scene so that the average gray tone on the scene would yield an exposure value of 50%, then anything which was even twice as bright as the average would get totally blown out. If one tried to expose a scene so that the average gray tone on the scene would yield an average value of e.g. 5%, then things that were dimmer than average would hardly get exposed at all. If one uses the guideline that typical 35mm film has five f-stops of latitude, 18% gray will fall almost exactly in the middle of that (2.47 f-stops down from 100%), putting it right in the middle of a five f-stop range.

Note that the process of shooting and printing negative film creates non-linear behaviors which are very different from those of digital cameras. Areas of the film which are not exposed to any light should be as transparent as possible, and should cause the resulting print to be solid black. Achieving a good solid black print will require exposing the print for long enough that areas of the film which are sufficiently close to transparent will also print as black. Thus, if one wants prints to have good solid blacks, then things which aren't supposed to be black must have a certain minimum level of exposure to stop them from vanishing into nothingness. On the flip side, it takes a lot of light turn turn film totally black; even part of a scene which is significantly overexposed may retain some detail.

When shooting digital, things are a little different: bright areas are more apt to get saturated (losing all detail), while dark areas are apt to appear "noisy". Generally, dark areas will still contain significant detail even when so badly underexposed that the noise dominates. Because different cameras have different amounts of noise (and the noise level varies depending upon various conditions), the "ideal exposure" mid-point for a digital camera may often be very different from what it would be for film.

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    \$\begingroup\$ I like this artistic approach to why 18% is considered middle grey... seems to explain what Ansel Adams reasoning would have been.... +1 \$\endgroup\$ May 1, 2015 at 18:30
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A little history will help you understand the purpose of the gray card:

In the mid 1930's, Messrs Jones and Condit at the Kodak Laboratory determined that statistically, a typical sunlit scene integrated to a reflectance value of about 18%. About this time, the Western Electric Company brought to market the first light meter. Kodak Labs publish a recommendation; place a Kodak film box in the scene. Seems the box reflected 18% of the ambient light. Now measure the reflected light from the box top and use this reading to set your exposure.

In 1941, Ansel Adams, a prominent landscape photographer, and his friend, Fred Archer, a photo magazine editor, jointly published the Zone System which provided photographers with a method to precisely fine-tune exposure. Their zone system revolves around the use of an 18% placard (battleship gray). This card replaces the Kodak box top. The 18% gray target became the de facto standard. Today film and paper speed as well as the digital chip are calibrated, and film and digital ISO is established using the 18% gray card.

Because of the pitfalls associated with reflected metering, a second measuring method evolved called the incident-light reading method. This method places a transparent sphere placed over the entrance of the light meter. The meter is positioned close to the subject and pointed backwards towards the camera. Thus, the meter measures the light just prior to striking the subject (incident old French word for about to happen).

The incident method yields the same reading as a reflected meter taken from a gray card however, it eliminates most of the pitfalls revolving where to hold and place the meter. In sunlit vistas the photographer can merely turn about and point the meter backwards at an imaginary camera. This method is highly accurate and was adopted by Hollywood camera operators because they are filming a scene and maybe a hundred thousand dollars rides on a correct exposure. .

Technical stuff: When negative film is correctly exposed and processed, an image of the gray card on the film will be rendered to a specific shade of gray. This shade of gay is equivalent to a neutral density filter with a factor or 5.5, it cuts light transmission 2 ½ stops. When written as percentage this value is 18%.

When the image of this gray card on the negative is printed, and if the print paper is exposed and developed to specification, the resulting image of the gray placard on the print paper will have the same 18% reflectivity as the original gray card.

Summation -- The 18% placard is the only tone that: 1. In actuality it has 18% reflectivity. 2. The resulting image of gray card on the negative has a transmission of 18% . 3. On the print the image of the gray card matches the original gray card reflecting 18%.

This 18% value is the key tone or axis of the photographic system - film – digital – and lithography. This is science -- not guess work.

More gobbledygook from Alan Marcus

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I'd like to add that eyes and silver halide emultion naturally have a logarithmic response, unlike CCD/CMOS sensors, for the same underlying reason.

Consider a patch of molecules spread over the focal plane. An incomimg stimulus (two photons hitting the crystal in a certain time window, in the case of film) is recorded by changing the state of that molecule (organic dye molecule or AgX crystal), that unit is now used up. Consider when half the units have already been hit: another stimulus has a 50% chance of hitting one that's already used, so does not add anything. It takes twice as much incoming light to make the same darkening as it does in a pristine area.

Now the two-photon complicates things, but the overall shape of the distribution is the same kind of curve. I recall reading about a "zombie door" in a math column. Imagine a red carpet leading up to a wall with a door in one narrow spot. Zombies are regularly spaced in line walking down the carpet, and each is randomly (uniform distribution) positioned side-to-side.

The distribution of zombies coming through the door is called an inverse log. Now imagine there is a row of doors all across the wall, like a subway turnstyle bank. Each door can only be used once. Later, after the exposure, you note how many turnstyles have been used vs remain unused.

Without modern electronics, it's difficult to look at a patch and say that's xx% optical density, but coarse grained film and a microscope will let you count how many black dots (exposed crystals) are in a sample square. I don't know how he determined a linear coverage based on the test exposures: emperically, doing one-stop increments you can find the capabilities of the media, and point to the one in the middle. But how do you know that's 18% on a linear scale, without a densitometer? Maybe mixing pigments in ratios, so it came from a tradition of painting.

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  • \$\begingroup\$ That's really neat! \$\endgroup\$
    – SilverWolf
    Jun 29, 2018 at 18:01
  • \$\begingroup\$ nice story, too bad it's not true :P film response curves differ, but the main, most important section is mostly linear. after all, the typical goal is to reproduce the original view rather than create special effects. yes, the "saturation" section becomes log-like like you described, but it is not what we normally use. either way, all this is completely unrelated to the "18% grey". because "18% grey" is perceived as "near half brightness" even without any image capturing device, e.g. in camera obscura or just a live view. \$\endgroup\$
    – szulat
    Apr 25, 2019 at 23:32
  • \$\begingroup\$ @szulat how is it not true? That is the physical response for film. You can verify by making a test strip where each zone doubles the exposure time. That’s why we measure exposure in stops: a geometric series. The last para desribes how that gives you 18%. This can be seen directly on an old CRT monitor, where a signal strength of about 18% looks the same as a checkboard pattern of full on/off seen at a distance so it blends together. The phosphor molecules are exactly like silver halide crystals. \$\endgroup\$
    – JDługosz
    Apr 26, 2019 at 6:01
  • \$\begingroup\$ that's not only factually incorrect, it's illogical. we can't say that all nonlinear phenomena magically match each other just because it sounds great. the behavior of film photosensitive molecules in the saturation region decreases the sensitivity resulting in a gradual overexposure (log-like response). fine. but if you want to explain the CRT nonlinearity by saying that phosphor is "exactly like silver halide", the response would diminish with increasing signal, which is exactly the opposite of what happens in reality. CRT response (and sRGB nowadays) is exponential and not logarithmic. \$\endgroup\$
    – szulat
    Apr 26, 2019 at 9:05
  • \$\begingroup\$ so, moving on to your example, in CRT the input signal of 0,5 ("half brightness") gives the archetypical ;-) 18% light intensity (technically it's more like 21%), but we perceive it as "half" because that's how our vision system works. remove all photography, silver halide, cmos, crt and stuff and our eyes still see 18% as about 0,5 intensity. this is about biology, not technology. \$\endgroup\$
    – szulat
    Apr 26, 2019 at 9:16
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My understanding is that 18% grey is considered the average reflectance of light from the world around us - not snow (about 90%) or a black cat in a coal mine at the other end, but an average on an average day. Grass, for example, reflects about 18% grey, so if you are reading your meter you can take a reading off the grass and then calculate from there. Caucasian skin is considered around 36% grey, so you can meter off your hand and then compensate back to 18% from there by opening or closing a stop - this is for film, either negative or transparency.

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  • \$\begingroup\$ Ah, so you're saying its convenient... good point \$\endgroup\$ May 2, 2015 at 13:39
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Ok...a card with 18% reflectance appears to the human eye to be middle grey. More formally stated: an object with a relative luminance of 18%( relative to a reference white ) will have a lightness of 50%. This is not some random number. It is a consequence of our non linear perception of brightness.

https://en.m.wikipedia.org/wiki/Lightness

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The other answers are not wrong. "18%" is also related to the video world: if the gamma of a monitor is 2.4 and you give it a 50% signal then the light output is 0.5^2.4 = 19%.

By a magical coincidence, analog video signals and digital image files (sRGB) or signals (BT.1886) are coded almost perceptually uniform. A 50% signal gives an 18-19% Luminance, but that is perceived as approx. 50% Lightness. A quantity with "-ness" in the name is always about human perception.

Later the video industry has tried to quantify Lightness perception and perceptual uniformity. Peter Barten (Philips) has laid much of the ground work, resulting in a PhD thesis and a summary paper (SPIE 2004). This work has been used by the Dolby company for standardising the "Perceptual Quantizer", this is the OECF for HDR TV as written in SMPTE standard 2084. Later Poynton, Nijland and myself have published a new formula for the same OECF curve and named it the "Barten Lightness" function (SMPTE MIJ 2015). It assumes perfect adaptation of the eyes to the average Luminance level, wherever that is.

This formula shows that human Lightness perception follows a gamma curve (1/2.07) in low light (< 0.1 nit) and a log curve in bright light (> 1 nit). With this formula the "50% Lightness = 18% Luminance" relationship is only exact for 18% vs. 100% of 0.57 nit. For example, 10.9% of 10 nit or 5.6% of 100 nit or 2.4% of 1000 nit or 0.9% of 10000 nit Luminance are also perceived as 50% Lightness. Again, this is after perfect eye adaptation to sequential stimuli, not when shown side by side.

If you want to know more then I suggest you look up Charles Poynton's PhD thesis, our Lightness formula is on page 93.

Please note that Dr. Barten has only investigated black-and-white perception, so anything derived thereform is only valid for the greyscale. Application of perceptual uniformity to color imaging is a different matter, and we have also taken a shot at that. This was all done in the context of high dynamic range and wide color gamut television.

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  • \$\begingroup\$ @ StessenJ -- A tip of the hat from Alan Marcus \$\endgroup\$ Apr 26, 2019 at 3:46
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Your eyes aren't logarithmic detectors. If only life were such a party; it'd sure be convenient to be able to walk into and out of my dark room without having to wait for my eyes to adapt. Instead, it's a mess of iris adaptation, Michaelis–Menten kinetics, power factors, context effects and more.

Jones and Condit were concerned with the brightness characteristics of scenes, not with reflectivity. Think about it, most of the scenes they studied were taken outside, often with visible sky in it. If they came up with the 18% figure, show me a citation.

I think it's a convention / tradition, but I don't know the ultimate source. Possibly 18% is standardised from some ad hoc object used previously.

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  • \$\begingroup\$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. \$\endgroup\$
    – Community Bot
    May 30, 2022 at 18:59
  • \$\begingroup\$ Iris adaptation is common knowledge and doesn't need a reference. MM kinetics is its own reference, just look it up, there are numerous articles on the subject. Power factors, context effects and such are common knowledge in the field and every colour appearance model deals with them to some extent, see e.g. CIE 248:2022. J&C again is its own reference. It's fine to ask for references if people are just making stuff up, but when everything is either commonly known or so easily confirmed it's just being rude for the sake of being rude. \$\endgroup\$ Jul 2, 2022 at 15:39
  • \$\begingroup\$ Also keep in mind that this not an answer, but a comment, principally directed at all the logarithmic nonsense. It's just posted here because the system wouldn't let me post it as a comment for no good reason. I stated as much, but apparently someone edited my comment to delete that message. \$\endgroup\$ Jul 2, 2022 at 15:42
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It baffled me for years too. Quite simply, the average amount of light reflected off objects around us is 18%. Some things are darker, some things are brighter. But 18% is average. Our eyes will perceive this average reflectance as the midtone to the highlights and shadows around us. Some other people have used maths and graphs to explain the difference between linear and log style data. But I'm just happy to know 18% of the direct light on my subject is being reflected at me, and that will give me my midtone to which my highlights and shadows can dance.

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  • \$\begingroup\$ Any source to cite? what surfaces is 18% grey derived from? What do you mean by average? who took this average? \$\endgroup\$ Feb 3, 2016 at 3:54

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