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).
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.
Simplest reason it that 18% reflectance just is not the middle value of a glossy silver print.
Originally Posted by markbarendt
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.
I don't think that D-max/min actually matters in my argument.
Originally Posted by ic-racer
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?
I aim for 0.75 density for Zone V in a print. See attached:
Originally Posted by ic-racer
Lightmeter calibration or average reflectance has nothing to do with understanding why 18% reflectance is 50% gray.
Handy chart Ralph, printed!
It's not arbitrary. Look at the math I've presented. 18% is perceived as 50%.
Originally Posted by holmburgers
Look at the table in post #26 and use the following equation to calculate the print reflectance:
Originally Posted by holmburgers
R = 1/10^D
where D is the reflection density. For example, Zone VIII has a reflection density of 0.09.
R = 1/10^0.09
R = 1/1.2303
R = 0.81283
R = 81%