I asked several photofriends this question: why do we call a 17% grey card "medium grey", why is it not 50%? Untill now I did'nt get a satisfactory answer. Anyone clear the fog?
Medium grey 17%, why not 50?
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I wondered about this for a long time. I read somewhere, it is 18% because it is a geometric center. I'd really like to see an equation for this.... Maybe one of APUG's heavy hitters will join in and explain it to us. Please?
It's 18% gray. Of the zones that are discernable on exposed film and paper (B&W specifically) Zone V is at the center. This zone reflects 18% of the light that falls on it. As it is situated in the middle of exposure/print zones/values, it is chosen as a reference point upon which to base exposures should you choose to do so. That's why a card that reflects the same amount of light as a Zone V exposure is referred to as an 18% gray card.
Nicholas Lindan
Advertiser
The average scene reflects 18%. Metering an 18% card will, on average, produce the same meter reading as metering the actual scene.
You may want to meter a card rather than the scene if the scene is very dark or very light - avoiding the problem that white cats and black cats both come out 18% grey if you depend on the meter reading.
You may want to meter a card rather than the scene if the scene is very dark or very light - avoiding the problem that white cats and black cats both come out 18% grey if you depend on the meter reading.
I tell students that if one took all the shades of grey that are in the average scene and mixed them up like paint, the resulting shade of gray would be "middle gray" and have an 18% reflectance...and that is what our meters assume the scene to be.
Don't know if this is 100% true, but it seems to satisfy them...LOL!
Don't know if this is 100% true, but it seems to satisfy them...LOL!
Well, the average outdoor scene supposedly integrates to 12.5% gray and that is why light meters are calibrated to that number.
18% gray was, IIRC, picked because it is "middle gray" and the progression of grays is a geometric one, with each step being twice or half what the previous one was.
If you carefully read Kodak's instructions on how to use an 18% gray card, you will see that it accounts for the difference between the 12.5% and the 18% numbers.
18% gray was, IIRC, picked because it is "middle gray" and the progression of grays is a geometric one, with each step being twice or half what the previous one was.
If you carefully read Kodak's instructions on how to use an 18% gray card, you will see that it accounts for the difference between the 12.5% and the 18% numbers.
Regardless of what the average scene reflects or what lighmeters are calibrated to, it all has to do with how our eyes compare different brightness levels. The human response to reflection (lightness) is not linear. For example, a surface reflecting 18% of the light that falls onto it, is perceived as being only half as bright (50%) as the illumination itself. The response follows the following equation:
L =116*(R)^(1/3)-16
where L and R are the lightness and the reflection in % respectively.
For example, use R=0.18 (18%) and L will return 50%.
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Of course! All makes sense now ;-) Sorry, just kidding. The explanation was great - the maths just did my head in!
I'm always curious about how we can say that the light is "perceived as being only half as bright". Any idea how these tests are conducted?
If I was asked to look at two stimuli, I dont' think I could say, "ahh yes, that looks exactly twice as bright".
Same thing with decibals, a 3dB increase is perceived as "twice as loud", but how are these types of qualitative measurements reliably made?
Between 90% reflection and 45% of reflection there is 1 stop (1EV of difference). If "middle grey" corresponded to 50% of reflection, that would mean that "middle grey" would be only 1 EV below 100% reflection, i.e. the brightest reflection you can have in nature.
So a card reflecting 18% of the light falling over it is around 2.3 or 2.5 Exposure values below a very bright white reflecting 96% of the light falling over it. The very bright white is the shoulder of your slide, and 2.3 or 2.5 below it is your middle grey.
9% is an EV below middle grey, and 4.5% another one, and there you are, with a very dark object reflecting only around 3% of the light falling on it you reach, more or less, the "foot" of your slide, somewhere between 2.5 and 3 EVs below middle grey.
And any case, if what above is in contrast with what Ralph writes, then disregard it
So a card reflecting 18% of the light falling over it is around 2.3 or 2.5 Exposure values below a very bright white reflecting 96% of the light falling over it. The very bright white is the shoulder of your slide, and 2.3 or 2.5 below it is your middle grey.
9% is an EV below middle grey, and 4.5% another one, and there you are, with a very dark object reflecting only around 3% of the light falling on it you reach, more or less, the "foot" of your slide, somewhere between 2.5 and 3 EVs below middle grey.
And any case, if what above is in contrast with what Ralph writes, then disregard it
Diapositivo, it's a very good explanation, but it still makes 18% seem like an arbitrary choice. What am I missing? Ralph?
Vaughn, That's funny I use the same analogy. I say, if you put all the tones in this room in a blender, you'd get middle gray.
18% gray (or 12.5%) are NOT related to "middle" gray of a silver print. (The 18% gray card is an exposure tool, not a printing tool).
We know the log D range of reflected values in a silver print is roughly 2.0 log D (glossy paper approximation).
The half-way point would be 1.0 log D.
The conversion from log D to percent reflectance is:
log D = Log10 (1/percent reflectance)
For an 18% gray card, the log D would be 0.74, which is darker than the middle gray in a silver print on glossy paper.
We know the log D range of reflected values in a silver print is roughly 2.0 log D (glossy paper approximation).
The half-way point would be 1.0 log D.
The conversion from log D to percent reflectance is:
log D = Log10 (1/percent reflectance)
For an 18% gray card, the log D would be 0.74, which is darker than the middle gray in a silver print on glossy paper.
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The 18% gray card reflects 18% of the light that falls on it. This 18% reflectance is the midway point in regards to increments of exposure between white nd black. The 50% refers to a difference in stops, not the actual amount of reflectance. Ten exposure zones between Zone0 and ZoneX. 5 stops from either extreme lands you on Zone V, smack dab in the middle, 50% percent of that range from either end. 18% gray.
That's the best layman's terms I can come up with.
That's the best layman's terms I can come up with.
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Let's see the numbers. What reflectance are we calling zone VIII for instance?
You called me out. 
You caught me away from my library so mind this won't be exact but it will be really close.
Zone 0....0%
Zone I.....1.5%
Zone II....3%
Zone III...6%
Zone IV...10%
Zone V....18%
Zone VI...26%
Zone VII..36%
Zone VIII.50%
Zone IX...68%
Zone X....100%
Mind you a freehand curve on a hand mand eleven point x and y graph but this is real close to the numbers. Never committed the exact numbers to memory but this will put you there. The % numbers regard percentage of light reflected at that zone. And zones 0 and X are probaly a couple of points from the extreme values here so it might be SLIGHTLY skewed, like I said, close not perfect. But very close IIRC.
You caught me away from my library so mind this won't be exact but it will be really close.Zone 0....0%
Zone I.....1.5%
Zone II....3%
Zone III...6%
Zone IV...10%
Zone V....18%
Zone VI...26%
Zone VII..36%
Zone VIII.50%
Zone IX...68%
Zone X....100%
Mind you a freehand curve on a hand mand eleven point x and y graph but this is real close to the numbers. Never committed the exact numbers to memory but this will put you there. The % numbers regard percentage of light reflected at that zone. And zones 0 and X are probaly a couple of points from the extreme values here so it might be SLIGHTLY skewed, like I said, close not perfect. But very close IIRC.
Thanks Chris,
But now I'm having trouble understanding why it goes from 100% to 68%, and then 50%. I guess it's like f/ stops, and how every other number is double/half.
Errr... umm... numbers... as Barbie said, "Math is hard".
But now I'm having trouble understanding why it goes from 100% to 68%, and then 50%. I guess it's like f/ stops, and how every other number is double/half.
Errr... umm... numbers... as Barbie said, "Math is hard".
I'm not understanding this statement.
As I understand it, one of the best reasons for using a reference card is to be able to "place" a subject in relation to a standard.
In the case of a Kodak gray card, if we have a reference shot including the gray card, that "18%" gray shade becomes directly translatable from scene to paper for all the related shots.
If the Kodak card can be considered a "middle grey" subject when it's in the scene, then there is a connection to the print.
There's no consensus on whether light meters are actually based on an average of 12% or 18%. My Kodak grey card says the card is 18%, and light meters are regulated on an average 18%, and no compensation should be made. Other versions of the notes on the same Kodak grey card actually give different instructions.
I read somewhere that 18% is middle grey for the printing industry and the use of the 18% grey card is in fact to ease printing. If light meters are actually regulated on a 12% average scene, then I suppose the 18% card gives good values because the instructions typically say to orient the card at an angle half-way between the light source and the subject.
It might be that orienting the card this way actually lets less light fall on the light meter than would have fallen if the card was oriented straight at the light meter. So the 18% card actually works well when used the way Kodak say to use it while using a 12% calibrated light meter.
The "middle grey" on film is not really exactly at half of the linear portion of the film curve, if I interpret correctly Minolta thinking. My light meter, a Minolta Spotmeter F, considers the typical slide to begin burning highlights slightly above 2.3 EV above middle grey, and to begin blocking shadows 2.7 EVs below middle grey. So a slide film is supposed - by Minolta at least - to have more "room" below middle grey than above it in the linear portion of the film or, if you prefer, consider middle grey of a slide film to be not "in the middle" of the film response curve. The total contrast of the scene recordable by slide film is this way 5 EVs (linear portion) plus foot and plus shoulder and middle grey is not in the middle of the curve.
That is food for thought: why would film producers design films that have more room "above" the average scene than "below" the average scene? (the answer might be that for film producers "average world reflectance" is actually 12%, but light meters producers know that photographers use 18% as reference and so place this "reference grey" there at 18% and not where the "average world reflectance" as film makers see it put it).
If we see 12% as the exact middle of zone V (the median value of zone V, which is a "zone" not a value) then 24% is middle zone VI, 48% is middle zone VII, the linear part of a slide should arrive unto until 75% or so, a very very light grey, after which the shoulder begins and the film response is not linear any more. The whitest white we can distinguish more or less, that is the white side of the Kodak grey card, is 90%. Frankly I don't think that I would be able to distinguish the difference between 90% and 95%, and none of my films as well.
On the other side, 6% should be the midst of zone IV, 3% should be the midst of zone III, and 1,5% is already in the "foot" of the slide (really the shoulder, as this is a positive). Again, this 1.5% is in the foot so the film response is not any more linear in that region.
If light is even I don't think one can ever see ten zones, nor eight. Maybe 5 or 6. That is, if I scan with my spot light meter a natural scene, with dark and light zones, in even light, I don't think it is easy to see more than 5 or maybe 6 EVs of difference.
But if you have uneven illumination (part sunlight, part shade) and both very dark and very light subjects, then the difference in EV between the dark object in the shade (the black car in the shade rendered as black) and the light object in the sun (the white wall in sunlight rendered as white) can be well above those 5 or 6 "zones" (I would rather just say EVs).
Fabrizio
PS To sum it up, I'm not less confused than most on where this elusive middle grey should be, and I don't use grey cards to determine exposure as I find it an unreliable way to work (a slight inclination of the card gives a different reading). I mainly use a spot light meter and using its instructions works well.
I read somewhere that 18% is middle grey for the printing industry and the use of the 18% grey card is in fact to ease printing. If light meters are actually regulated on a 12% average scene, then I suppose the 18% card gives good values because the instructions typically say to orient the card at an angle half-way between the light source and the subject.
It might be that orienting the card this way actually lets less light fall on the light meter than would have fallen if the card was oriented straight at the light meter. So the 18% card actually works well when used the way Kodak say to use it while using a 12% calibrated light meter.
The "middle grey" on film is not really exactly at half of the linear portion of the film curve, if I interpret correctly Minolta thinking. My light meter, a Minolta Spotmeter F, considers the typical slide to begin burning highlights slightly above 2.3 EV above middle grey, and to begin blocking shadows 2.7 EVs below middle grey. So a slide film is supposed - by Minolta at least - to have more "room" below middle grey than above it in the linear portion of the film or, if you prefer, consider middle grey of a slide film to be not "in the middle" of the film response curve. The total contrast of the scene recordable by slide film is this way 5 EVs (linear portion) plus foot and plus shoulder and middle grey is not in the middle of the curve.
That is food for thought: why would film producers design films that have more room "above" the average scene than "below" the average scene? (the answer might be that for film producers "average world reflectance" is actually 12%, but light meters producers know that photographers use 18% as reference and so place this "reference grey" there at 18% and not where the "average world reflectance" as film makers see it put it).
If we see 12% as the exact middle of zone V (the median value of zone V, which is a "zone" not a value) then 24% is middle zone VI, 48% is middle zone VII, the linear part of a slide should arrive unto until 75% or so, a very very light grey, after which the shoulder begins and the film response is not linear any more. The whitest white we can distinguish more or less, that is the white side of the Kodak grey card, is 90%. Frankly I don't think that I would be able to distinguish the difference between 90% and 95%, and none of my films as well.
On the other side, 6% should be the midst of zone IV, 3% should be the midst of zone III, and 1,5% is already in the "foot" of the slide (really the shoulder, as this is a positive). Again, this 1.5% is in the foot so the film response is not any more linear in that region.
If light is even I don't think one can ever see ten zones, nor eight. Maybe 5 or 6. That is, if I scan with my spot light meter a natural scene, with dark and light zones, in even light, I don't think it is easy to see more than 5 or maybe 6 EVs of difference.
But if you have uneven illumination (part sunlight, part shade) and both very dark and very light subjects, then the difference in EV between the dark object in the shade (the black car in the shade rendered as black) and the light object in the sun (the white wall in sunlight rendered as white) can be well above those 5 or 6 "zones" (I would rather just say EVs).
Fabrizio
PS To sum it up, I'm not less confused than most on where this elusive middle grey should be, and I don't use grey cards to determine exposure as I find it an unreliable way to work (a slight inclination of the card gives a different reading). I mainly use a spot light meter and using its instructions works well.
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Simplest reason it that 18% reflectance just is not the middle value of a glossy silver print.
More detailed explaination is that each type of paper has a different maximum black (remember all the threads on Dmax
). The D-max, then determines what the D-middle (D-Middle = 1/2 D-Max) is going to be. It is not one universal value and is highly dependent on the paper.The "Zone" followers are always posting transmission log D values of their negatiaves and assigning them zones, but they almost never divide up the paper reflection densities into the appropriate zones. I don't know why they leave this step out. But if they did they would find that the middle is about 36% for paper with a D-max of 2.0.
The 18% card would match the middle value of a paper with a D-max of 1.48.
Again, the 18% card is an exposure tool for times when you need an approximate incident reading and have only a reflected meter.
Simplest reason it that 18% reflectance just is not the middle value of a glossy silver print.
More detailed explaination is that each type of paper has a different maximum black (remember all the threads on Dmax
The "Zone" followers are always posting transmission log D values of their negatiaves and assigning them zones, but they almost never divide up the paper reflection densities into the appropriate zones. I don't know why they leave this step out. But if they did they would find that the middle is about 36% for paper with a D-max of 2.0.
The 18% card would match the middle value of a paper with a D-max of 1.48.
Again, the 18% card is an exposure tool for times when you need an approximate incident reading and have only a reflected meter.
I don't think that D-max/min actually matters in my argument.
I see using the gray card in a reference shot as much more than an approximation. In fact it calibrates the print to the scene. It factors out exposure errors and even differences in film or development choices.
Let's say I've done my paper testing and I can program my EM-10 or color analyzer to reproduce middle gray (the Kodak card tone) reliably on whatever given paper is in the enlarger, and I set the enlarger properly. I should get really, really close to a "real world" match every time.
My argument does assume real rather than relative placement and that shadows and highlights are simply allowed to fall off the paper where they may in a straight print.
Mark, I think that you and ic-racer aren't necessarily disagreeing, but that all he is saying is that the grey card won't be the same reflectance as the representation of the gray card in an otherwise satisfactory print. It is still useful for exposure of course.
Do I have it right?
Do I have it right?
Yep, that's true.
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