18% Neutral Gray Paint?

Discussion in 'Miscellaneous Equipment' started by snikulin, Sep 29, 2015.

  1. OP
    OP
    snikulin

    snikulin Member

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    It's Yashica 12 '1967, about 48 years old.
    I doubt its meter was calibrated seriously during production to begin with and now the needle spring is sure off.
    Also the current air-zinc battery is not a precise substitute for an original mercury one.
    Nevertheless the meter does match my Canon 6D + 50mm/1.4 lens in averaging mode.
     
  2. RobC

    RobC Member

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    meters are not calibrated to a reflectance value. They are calibrated to roughly the midpoint of the log exposure range of the film speed. That means they are are calibrated to a Lux value. You should be able to work that out by knowing that an EV value has a corresponding lux value and then calculating back using the standard meter formula to check if thats what is happening. The midpoint lux value can be read from the ISO film speed diagram in wikipedia. Get your calculator out.
     
  3. DREW WILEY

    DREW WILEY Member

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    If you know anything about paint stores, you're lucky to find anyone sober in one. And no, the kind of color analyzers they use won't be able
    to make a true 18% gray paint, if they even have a clue what you are talking about (they won't). At one time Benjamin Moore published the
    reflectance values of their color chips. But printed samples were made with inks, not paint itself, so are never precisely representative of
    what will be in a can. And you'd want a matte finish (flat - nonreflective, yet washable). But this is getting silly anyway. I'd just put a big
    gray card or Kodak gray scale chart in the scene. The Zone System is all relative anyway. God didn't create the world in eight zones to begin with. The Zone System just an artificial shorthand tool. Take from it what you need, tweak it for your kind of subject matter, and be done with it. It's like playing chord on a piano. Once you actually learn how to play, you forget all that beginner stuff and start making real
    music instead.
     
  4. BrianShaw

    BrianShaw Member

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    How exact do you think it needs to be to achieve the goal the OP stated in post #43?
     
  5. DREW WILEY

    DREW WILEY Member

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    That's the whole point. One could merely go out and buy a sheet of medium gray matboard. Pick out a relatively neutral gray paint chip and
    visually compare it to a small gray card. Whatever. The problem with too much gray around you is that it fatigues the eye. It's a chameleon
    that becomes just the opposite of any strong color around it. "Simultaneous contrast" is the physiological and artistic term for it. That's why color pros use matching booths instead of whole rooms of the damn stuff. What on earth does this have to do with basic black and white technique? Well, once one starts getting a tad proficient at printing (necessary to any real progress in the Zone System), they'll inevitably want to fiddle around toning for permanence, which inherently introduces toning relative to hue. Try to get good at that without some eye
    fatigue. Time to repaint. But at least gray paint is better than gray paint with Damien Hirst polka dots all over the wall.
     
  6. OP
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    snikulin

    snikulin Member

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    Drew, I looked all the stuff you refer too and I think you are right and the whole gray room might be not a healthy idea.
    Apparently the problem is in our wetware.
    Well, it's back to my old set of Kodak cards (I have the "pearl necklace" one).

    Thanks!
     
  7. Ken Nadvornick

    Ken Nadvornick Member

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    Good point, regardless of the direction the OP chooses to go.

    I suspect the color analyzer at the home improvement store would be just fine, if not overkill. I know it worked OK for me. My testing meter read the same. And if the spectral response was not exquisitely identical, it certainly fell within the overall error spread for the entire system I was working with. Meaning, it would not have been an outlier data point, and would also probably have been mitigated elsewhere by a different opposing variable.

    I've posted before concerning the desirability and wisdom of finding the sweet spot in systems. That point being at least one step back from the beginning of diminishing returns.

    There is a reason they don't chalk a baseball batters box using laser beams.

    :wink:

    Ken
     
  8. Bill Burk

    Bill Burk Subscriber

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    snikulin,

    I think since you are planning to learn composition with this, using the painted wall as a neutral background for portrait and still life arrangements... The gray you choose is important if you are in tight quarters... which it sounds like you are.

    Here's advice from William Mortensen in Pictorial Lighting, Chapter 3, Equipment, The Background...

    "The background should be no smaller than ten feet wide and eight feet high. (For general use. For portrait heads only it may be somewhat smaller.)"

    "It should be either matt-white or matt-gray, depending on the size of the shooting space. If there is room to pull the camera back thirty feet from the background, make your background white."

    "Eighteen by ten, with the background across the end, is the smallest feasible space."

    "The side walls and ceiling, at least immediately adjacent to the background, should be light in color; tan or gray rather than white. If possible, the floor also should be white or light grey."
     
  9. Bill Burk

    Bill Burk Subscriber

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    A couple things I didn't mean to sound like I was saying (when I said the gray you choose is important)... I don't think the shade of gray must be a particular percent. I think 18% is fine. I don't think you need it to be neutral across the entire visible spectrum (Unless you can. The paint DREW WILEY made would be wonderful but is the formula available?)

    You're just learning, so a little non-neutrality and inaccuracy of shade can be tolerated. I learned a lot with my two-tone painted Zone Board and it's neither neutral nor accurate (at least not now 8 years later - it started out 16% and 36% but I doubt it's still those shades)

    I meant to say it's important in a straightforward sense, whether the wall is white or gray... Since you might not have thirty feet in your small room to back up (for those occasions when you want a dark background)... Because it will be easier to work with a middle gray because you will only have to drop its value by a couple f/stops to make it black, and you will only have to raise its value by a couple f/stops to make it white.

    For a meter target, I'd take advantage of the recent Internet lore that meters see 12% and that's a better target, closer to what the meter reads than 18%. But don't paint the background 12%... Make a small portable target or gray card that's 12% and use that at the subject position to take a meter reading (12% is better to help you learn to verify your understanding of metering, 18% will just give you readings that don't agree with any other meter readings you might take).
     
  10. Bill Burk

    Bill Burk Subscriber

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    I don't mean that the meter is calibrated to 12% reflectance.

    I mean that 12% reflectance is a better test target than 18% when you want to take meter readings off a test target. I verified this with a light meter that has both spot and incident modes and a test target with several target patches 1/6 stop apart surrounding 18% gray. The 12.7% target patch reading in spot mode closely matches the reading in incident mode.

    My test doesn't prove that 12.7% is the theoretical correct percent, it just gives me a patch that's pretty close to the right percent to cause the meter to agree from one mode to the other. There's a 14.3% patch that's "too light" for the two modes to agree.

    I'm using this as a sanity check on any math that I'm playing with, because any result far off of 12% needs to be explained.
     
  11. Bill Burk

    Bill Burk Subscriber

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    I followed the discussions behind wikipedia and found Jeff Conrad was behind some of the edits.

    He wrote an article on the homepage of LFPhoto.info that covers the idea of 18% gray and K and C.

    Jeff Conrad's article about exposure metering

    While yesterday I claimed that 12% is close... Now I found a formula that shows the implied relationship between incident light and reflected light.

    Reflectance = ( K x pi ) / C

    It's interesting that it works out near 18% when you use C for the flat disc on an incident meter and close to 12% when you use C for the hemisphere disc.

    So maybe it boils down to 18% being correct for flat copy flatly illuminated. And 12% is correct for 3-D subjects.

    (Sekonic L-758DR Lumisphere C = 340, Flat diffuser C = 250 and K = 12.5)

    (12.5 x 3.14) / 340 = 11.5% and (12.5 x 3.14) / 250 = 15.7%

    Very interesting...
     
  12. RobC

    RobC Member

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  13. Bill Burk

    Bill Burk Subscriber

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  15. Bill Burk

    Bill Burk Subscriber

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    I figured out what's going on here... they are numeric factors without units, but... K is 1.16 when the units of luminance in the equation are footcandles, and K is 12.5 when the units are candles per square meter.
     
  16. RobC

    RobC Member

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    No, K is whatever the meter manufacturer says it is and you can't change that. It is a "CONSTANT"

    I think you and maths are mutually exclusive.

    Which meter are you using and when was it made?
     
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  17. DREW WILEY

    DREW WILEY Member

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    That's why I've stuck with either Pentax or Minolta spotmeters. They're apparently calibrated to exactly the same standard.
     
  18. michael_r

    michael_r Subscriber

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    This is how it is laid out in Richard Henry's book. I can't say whether or not he was correct, except that he based the meter calibration section of his book on correspondence with H. Kondo, who was THE guy in Japan at the time on their meter standards and testing committee etc.

    The Minolta and Pentax meters at the time seem to have used K=14. Apparently as of the mid 1980s, new Japanese meter models were supposed to switch to 1.16 (12.5) to come in at the most recent recommended ANSI standard. I don't know when/if that change occurred. I bought my Minolta Spotmeter F in the early 1990s, and the manual seemed to have had its last revision in 1986, but K was still 14 for that meter.
     
  19. RobC

    RobC Member

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    Without knowing which formula the meter is using for its calculations its impossible to say what the K factor for it is unless the K factor for the meter was published in its specs.

    I saw one older formula which said K= 3.66 ( If memory serves me correctly) but it was different formula from todays formula for spot meters.
     
  20. michael_r

    michael_r Subscriber

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    The complete ANSI formula reduces to 2EV=[(Luminance * Speed)/K]

    Therefore luminance and K must change proportionately as the units for luminance are adjusted. Equivalent K values should be the following (for example):

    Luminance in cd/m2: 12.5 (actually 12.4 but rounded to 12.5 in the Japanese standard back then)
    Luminance in footlamberts: 3.65
    Luminance in cd/ft2: 1.16

    The roundings make things confusing, but this is just illustrative anyway since I'm getting this all from Henry which uses older versions of the standards.

    For my Minolta Spotmeter F, while K is not explicitly given, you can work backwards with the standardization formula to solve for K, using a data table in the manual showing how to use the meter to measure luminance. The table shows the luminance values in cd/m2 for EV values. Plugging them in gives me a K of 14.
     
  21. Stephen Benskin

    Stephen Benskin Member

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    Bill, this might help.

    Defining K, part 4.jpg
     
  22. Stephen Benskin

    Stephen Benskin Member

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    K Equation - 1.jpg

    And here from ANSI PH3.49-1971

    Units and Values of Exposure Parameters.jpg

    K = 3.33 footlamberts (fL)

    K three three three.jpg

    You can also find 3.33 in the Units and Values of Exposure Parameters table above.

    It is a factor, but it can be thought of as a light loss factor. The meter needs to take into consideration the optical system of the camera. The K equation (see above) can be broken down into three parts, the optical system (light loss through the optics), the exposure constant, and variables related to the meter. The light loss portion is the constant "q" in the exposure equation. The relationship between the three constants K, P (exposure constant) an q can be expressed as P / q = K.
     
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  23. Tommy Isaacson

    Tommy Isaacson Member

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  24. wiltw

    wiltw Subscriber

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    I once made a checkerboard pattern of alternating black squares and white spaces, with an even number of each on a page. Printed it on my printer on photo paper and I took a reflected light reading of that and it DID NOT MATCH a reflected light reading from an 18% tonality card in the same light.
     
  25. RalphLambrecht

    RalphLambrecht Subscriber

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    I used some flat white and black paint, and painted small patches with a 50/50 mix then added black in 10% increments and measured the samples with a densitometer. after a bit trial and error, I had the perfect ratio of black-and white paint to make an 18% gray.took a bit of effort and patience but worked in the end.of course you have to do that again using different paints.
     
  26. chriscrawfordphoto

    chriscrawfordphoto Subscriber

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    That's because a checkerboard is not 18% reflectance; its 50%. For 18% reflectance, you'd need 100 squares, with only 18 of them white!