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Photo Forum / Digital Photography / Digital Photo / November 2007How to understand 18% gray?
why 18% gray is in the mid of the full scale of brightness from darkest to brightest? what's the math behind it? >why 18% gray is in the mid of the full scale of brightness from >darkest to brightest? what's the math behind it? It really means that 18% of incident light is reflected. Human perception of brightness is logarithmic, though, and not linear. Basically, logarithmic means we are multiplying instead of adding. Compound interest is logarithmic. So is the frequency of the notes on a piano. So is the decibel scale for measuring sound pressure. Likewise with the light plotted on a typical histogram. You can make (or imagine) a set of brightness patches by starting with 100% reflection and halving each time. The next is 50%, then 25%, 12.5%, 6%, 3%, and 1.5%. That's 7 patches. It is a logarithmic sequence; the factor is 1/2. The first patch is white, the last very dark gray, virtually black. In the middle it will be middle gray, 12.5% actually, a bit darker than 18% gray. You can fiddle with sequences like this to make as many patches as you like, with different gradations of brightness between them. Using a factor of 0.75, we get 13 steps, with the middle being 18%: 100 75 56 42 32 24 18 13 10 7.5 5.6 4.2 3.2. Ed In article <9278fad4-e7ff-4737-a7bd-bf06adbc7831@i12g2000prf.googlegroups.com>, > why 18% gray is in the mid of the full scale of brightness from > darkest to brightest? what's the math behind it? It can be mid-way for exponential gamma. The exponential gamma has its origins in old tube TV circuits but it remains in use because it models visual perception reasonably well. It shifts more digital levels into areas where the eye is most sensitive. Few things in the natural world are linear. > why 18% gray is in the mid of the full scale of brightness from > darkest to brightest? what's the math behind it? See: doug.kerr.home.att.net/pumpkin/Scene_Reflectance.pdf and doug.kerr.home.att.net/pumpkin/Exposure_metering_18.pdf Cheers  Signaturecmyk Despite the science behind it, in reality it is the art that counts. For most situations 18% works well. However for some situations and for some expectations of a result some other gray will work better. In other words, don't worry so much about the science of photography that it gets in the way of the art. :-) > why 18% gray is in the mid of the full scale of brightness from > darkest to brightest? what's the math behind it? SignatureJoseph Meehan Dia 's Muire duit > why 18% gray is in the mid of the full scale of brightness from > darkest to brightest? what's the math behind it? While the "logarithmic" answer people have given is correct, it's easier to see if you remember that a one stop difference means a twice the brightness. So if you have mid-gray at 18%, you get five (that's all, folks) useful tones on a print: 4.5%, 9%, 18%, 36%, and 72%. If you toss in 0% and 100%, that makes 7 (although you probably won't be able to see the difference between 4.5% and 0). David J. Littleboy Tokyo, Japan >> why 18% gray is in the mid of the full scale of brightness from >> darkest to brightest? what's the math behind it? [quoted text clipped - 10 lines] > David J. Littleboy > Tokyo, Japan Perhaps a usefulness not yet explained or stated is for matching colour. I use a white, black and grey card (home made) in the first shot of a scene. I use it with the "levels" function of Photoshop and so far (maybe 1000 frames) it works better than nearly any other post shoot white balance. I don't believe digital photography has the same use for a grey card as film shooters have. Douglas >>> why 18% gray is in the mid of the full scale of brightness from >>> darkest to brightest? what's the math behind it? [quoted text clipped - 7 lines] >> 100%, that makes 7 (although you probably won't be able to see the >> difference between 4.5% and 0). Note that Dave M. has provided a much better discussion of this issue than the above. > Perhaps a usefulness not yet explained or stated is for matching colour. I > use a white, black and grey card (home made) in the first shot of a scene. > I use it with the "levels" function of Photoshop and so far (maybe 1000 > frames) it works better than nearly any other post shoot white balance. I don't recommend this. I have two Kodak gray cards here that have radically different color compositions: one is somewhat blue, the other quite warm. David J. Littleboy Tokyo, Japan > >> why 18% gray is in the mid of the full scale of brightness from > >> darkest to brightest? what's the math behind it? [quoted text clipped - 20 lines] > > Douglas I'm too lazy to look it up now, but weren't the definitions for ASA and ISO speeds based on an exposure just above the toe of the characteristic curve? I always assumed the 18% value was related to that. On the other hand, didn't the ISO for digital speed try to stick as close to that for film speed as possible? >> >> why 18% gray is in the mid of the full scale of brightness from >> >> darkest to brightest? what's the math behind it? [quoted text clipped - 10 lines] >> > David J. Littleboy >> > Tokyo, Japan stuff sniped One other reason I've not seen mentioned yet is that it's the level all our exposure meters, even the ones in our digital cameras are set to "see". That is if you expose a white or black object with you built in meter, it tries to give you an expose that will make it 18% gray. So including a gray card will give you something to set your meter by. John Passaneau On Nov 27, 11:16 pm, Don Stauffer in Minnesota <stauf...@usfamily.net> wrote: > > >> why18% gray is in the mid of the full scale of brightness from > > >> darkest to brightest? what's the math behind it? [quoted text clipped - 25 lines] > characteristic curve? I always assumed the18% value was related to > that. what i can remember is the 0.1 plus filmbase + fog density on which the ISO speed was based. > On the other hand, didn't the ISO for digital speed try to stick as > close to that for film speed as possible? >why 18% gray is in the mid of the full scale of brightness from >darkest to brightest? what's the math behind it? It isn't exactly in the middle, really. Any given scene has a darkest tone you want to record and a brightest tone you want to record, and the middle tone for that particular scene is halfway in between. Now, it's not halfway between in linear-light space, because the eye responds logarithmically. Suppose you have a scene whose brightness ranges from 1 to 100 on some arbitrary (but linear) light meter scale. To the eye, a brightness of 50 is "one step" darker than 100, and 25 is an equal-sized step darker than 50, 12 is another step darker again, and so on. So the midtone for this particular scene is actually a brightness of 10, in the sense that it is just as much brighter than the minimum (10X) as the maximum is brighter than the midtone (10X). One general way to calculate this is: midtone = sqrt(min_bright * max_bright) This value is called the "geometric mean" of min and max brightness. You can also average logarithms: midtone = exp( (log(min_bright) + log(max_bright)) / 2) The magic value of 18% comes from somewhere else. Someone did a study and found that the average reflectance of a large number of photographic scenes was 18%. So if they set their incident light meter to assume that light from the scene was 18% of the light falling on the meter, they would (on average) get a good exposure. Similarly, if they had an 18% reflectance grey card, and measured the light on it with a reflected light meter, that would also be a good exposure. It happens that, in many scenes, the average scene brightness is actually pretty close to the middle tone (halfway between darkest and lightest areas of interest), so it's also useful to expose as if it was the middle tone, giving equal range above and below it. But that's not necessarily true, and spotmeters and the zone system are tools for more accurately figuring out what the tonal range of your scene is, and how to place that relative to what your film (or sensor) can capture. Dave > why 18% gray is in the mid of the full scale of brightness from > darkest to brightest? what's the math behind it? Go back 100 years or so and ask. SignatureNeil reverse ra and delete l Linux user 335851 > why 18% gray is in the mid of the full scale of brightness from > darkest to brightest? what's the math behind it? The gray card is a development of the "artificial highlight" method of exposure metering. Before the gray card was invented, Kodak recommended using a 90% white card and setting your light meter at 1/5 of the normal film speed rating. The gray card simplifies this because you don't need to change your meter setting. 90/5 = 18 Peter. Signaturepirwin@ktb.net thanks for all your explains and that actually made me in the way of understanding. now, another question poped up in my mind: how to map a 18% point ( or a mid-tone point ) in a 1 to 255 scale in digital photo file? there are four possible answers i can guess, 1, 255/2 = 128 2, 255*0.18 = 45 3, 0.18^(1/2.2)*255 = 117 4, 0.5^(1/2.2)*255 = 186 which one is right? > thanks for all your explains and that actually made me in the way of > understanding. now, another question poped up in my mind: how to [quoted text clipped - 7 lines] > > which one is right? any clue? >> thanks for all your explains and that actually made me in the way of >> understanding. now, another question poped up in my mind: how to [quoted text clipped - 9 lines] > >any clue? I don't understand the question. I know "poped" means to make holier, but what does "in digital photo file" mean? And how is 255 supposed to relate to a percent? Ed >> thanks for all your explains and that actually made me in the way of >> understanding. now, another question poped up in my mind: how to [quoted text clipped - 9 lines] > >any clue? Here's a table that might interest you. The two columns marked "Levels" list the number of levels that fstop range is divided into; hence, in the brightest zone an image from 12-bit linear data can have 2048 distinctly different values for brightness, while in an 8-bit gamma corrected image there are 69 possible distinct levels in that zone. Fstop | Linear Linear Gam.Cor. Gam.Cor. Gam.Cor. | 12 bit Analog Analog 8 bit 8 bit Range | Levels Value Value Value Levels ------|----------------------------------------------- 1 | 2048 1.0 1.0 255 69 2 | 1024 0.5 0.72974 186 50 3 | 512 0.25 0.53252 136 37 4 | 256 0.125 0.38860 99 27 5 | 128 0.0625 0.28358 72 20 6 | 64 0.03125 0.20694 53 14 7 | 32 0.015625 0.15101 38 10 8 | 16 0.007812 0.11020 28 8 9 | 8 0.003906 0.08042 21 6 10 | 4 0.001953 0.05868 15 4 11 | 2 0.0009765 0.04282 11 3 12 | 1 0.0004883 0.03125 8 2 The useful dynamic range basically ends when one full fstop is divided into fewer than 8 values. Hence a 12-bit linear data set has about a 9 stop useful dynamic range, while an 8-bit gamma corrected data set has about an 8 stop range. (With fewer than 8 levels the quantization distortion becomes too apparent, and the results are referred to as "posterized".) To answer your question above requires knowing where you are going to set the white and the black points of the "display" (paper or monitor). For example, when printing on paper with a 5 fstop dynamic range, the printer would normally be adjusted to make 255 be "white" and 52 black. The middle zone, 3, would have values ranging from 100 to 136. Does "3, 0.18^(1/2.2)*255 = 117" look grey to you? SignatureFloyd L. Davidson <http://www.apaflo.com/floyd_davidson> Ukpeagvik (Barrow, Alaska) floyd@apaflo.com >thanks for all your explains and that actually made me in the way of >understanding. now, another question poped up in my mind: how to >map a 18% point ( or a mid-tone point ) in a 1 to 255 scale in digital >photo file? there are four possible answers i can guess, >1, 255/2 = 128 >2, 255*0.18 = 45 >3, 0.18^(1/2.2)*255 = 117 >4, 0.5^(1/2.2)*255 = 186 #1 would be correct if the file used 8-bit linear coding, but nobody does that because 8 bits simply doesn't give enough dynamic range or intensity resolution in the shadow areas. On the other hand, 16-bit linear coding is sometimes used. #2 makes no sense to me. #3 is close to correct assuming a particular exposure (see below). The nonlinear function y = x^(1/2.2) is how intensity is encoded in video, in JPEG files, and lots of other places. This is generally not adjustable, nor not adjustable very much. #4 tells you what you'd expect to find in the image file for a scene brightness of 50% of peak (e.g. one stop down), not 18%. Now more about exposure: You or the camera has a choice of what brightness in the image gets considered "1.0" and mapped to 255 in the file. Since images have areas of widely varying reflectance illuminated by different amounts of light, 255 doesn't necessarily represent "100% reflectance"; it's just the brightest tone that the camera has decided to handle. So the "0.18" in this expression is best regarded as "0.18 times peak white in this photo", not any absolute reflectance. It's actually more complex, too, because cameras may depart from this power function in the highlight and/or shadow areas in order to capture more brightness range at lower contrast in those areas. The "shoulder" and "toe" in film response does the same thing in chemical photography. Since the response tends to be more predictable in the midtones, one good way to expose images in the first place is to adjust the exposure so that something you have decided is of average brightness (e.g. an 18% grey card) ends up in the middle of the tonal scale where you've decided it should be (theoretically 117, but 115 or 120 or even 128 work too). Exposing so a grey card ends up in the middle of the code value range, or the middle of the film density range as measured by a densitometer, is very common. In the electronic world, it's sometimes more convenient to expose so that the brightest thing in the scene that you want to preserve information from ends up near, but not beyond, 255. You can do this approximately using any digital camera that has a histogram display. Some cameras also flash or put zebra stripes in areas that were overexposed in the preview image, which gives better visualization of what will be recorded correctly and what will be clipped. Dave if you look at a photographic grayscale of 11 steps, the centre mid-gray (the sixth step) is indicated as 0.7. this is sensitometric density value, a log value of opacity. opacity = 100% of illumination upon X% reflected or transmitted light. if X=18%, then opacity = 100/18=5.5 therefore sensitometric density = log5.5 = 0.74...-round off to 0.7. hence 18% reflection gray is 0.7 sensitometric density value and is the mid-gray between the white and black of a grayscale. > why 18% gray is in the mid of the full scale of brightness from > darkest to brightest? what's the math behind it? |
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