Kodack grey card values

wrote:

If I take a photograph of the 18% Reflectance side of a Kodak grey card and open the image in Photoshop. What values should I expect for RGB in the Info palette?

18% reflectance is 18% L in the LAB mode. If you fill in 18,0,0 in LAB, you’ll get RGB 48,48,48.

—
Johan W. Elzenga johan<<at>>johanfoto.nl Editor / Photographer http://www.johanfoto.nl/

Johan W. Elzenga wrote:

wrote:

If I take a photograph of the 18% Reflectance side of a Kodak grey card and open the image in Photoshop. What values should I expect for RGB in the Info palette?

18% reflectance is 18% L in the LAB mode. If you fill in 18,0,0 in LAB, you’ll get RGB 48,48,48.

Sorry, it’s too early in the morning, so I wasn’t thinking clearly. It’s 50 L in Lab mode, so it’s (depending on the RGB color space) RGB 118,118,118.

—
Johan W. Elzenga johan<<at>>johanfoto.nl Editor / Photographer http://www.johanfoto.nl/

(Johan W. Elzenga) wrote in
news:1g9wa8p.gonffn2uxieuN%:

Johan W. Elzenga wrote:

wrote:

If I take a photograph of the 18% Reflectance side of a Kodak grey card and open the image in Photoshop. What values should I expect for RGB in the Info palette?

18% reflectance is 18% L in the LAB mode. If you fill in 18,0,0 in LAB, you’ll get RGB 48,48,48.

Sorry, it’s too early in the morning, so I wasn’t thinking clearly. It’s 50 L in Lab mode, so it’s (depending on the RGB color space) RGB 118,118,118.

Well, that seems a bit more on the mark 😉

Curious, though. It would seem that hitting the midpoint in RGB would accomplish the same thing, except the midpoint (127,127,127 or thereabouts) appears to be about 1/3 to 1/2 stop brighter (photographic terms). Any idea why this is? Why would the midpoint in LAB not match the midpoint in RGB?

– Al.

—
To reply, insert dash in address to separate G and I in the domain

"Al Denelsbeck" wrote in message

Curious, though. It would seem that hitting the midpoint in RGB would accomplish the same thing, except the midpoint (127,127,127 or

thereabouts)

appears to be about 1/3 to 1/2 stop brighter (photographic terms). Any

idea

why this is? Why would the midpoint in LAB not match the midpoint in RGB?

128,128,128 is closer to the midpoint than that, isn’t it? Punching the numbers
into the CIE Color Calculator:
http://www.brucelindbloom.com/ColorCalculator.html
the luminance changes from 18.4 (L*=50) to 21.6 for sRGB 128,128,128. 21.6/18.4
is 1.17, which represents only 0.17 of a stop, which is not even 1/5th.

Greg.

"Greg" wrote in
news::

"Al Denelsbeck" wrote in message

Curious, though. It would seem that hitting the midpoint in RGB would
accomplish the same thing, except the midpoint (127,127,127 or

thereabouts)

appears to be about 1/3 to 1/2 stop brighter (photographic terms). Any

idea

why this is? Why would the midpoint in LAB not match the midpoint in RGB?

128,128,128 is closer to the midpoint than that, isn’t it? Punching the numbers
into the CIE Color Calculator:
http://www.brucelindbloom.com/ColorCalculator.html
the luminance changes from 18.4 (L*=50) to 21.6 for sRGB 128,128,128. 21.6/18.4
is 1.17, which represents only 0.17 of a stop, which is not even 1/5th.

Well, two things to consider…

1) The amount I stated was only a guesstimate, mostly from staring at photos onscreen too much of the day. It’s also subject to my gamma ;-)…

2) There’s a difference between the luminance change of the actual monitor output, as measured by a light meter, and the onscreen representation of a photograph converted to 24-bit RGB. The monitor has an exceptionally narrow light range, and right at the moment, my pure black screen with no room lights is pegging about 5 seconds at f2.8, ISO 100, which is much higher than full moonlight on a grey card.

And even then, it’s highly subjective. Not only do the monitor specs come into play, but the exposure range of the film that is scanned. Slide films I use are 4-6 stops in latitude, print films about 9. And both of these appear "full gamut" in RGB space. So there’s a teensy bit of play 😉

Bruce’s site is producing some weird results for me, and I’m not sure what he’s working with yet…

– Al.

—
To reply, insert dash in address to separate G and I in the domain

My apologies for the crossposting, thats the first (and last) time I’ll do that.

And, if anyone else needs a reason NOT to crosspost – It’s because you end up with intelligent, knowledgeable people NOT being able to engage in a structured conversation and NOT being able to develop their arguments with each other.

The idea behind the original question was this:

If I expose a roll of film and load the images into Photoshop with one exposure a picture of an 18% reflectance grey card, what should the RGB values be in the Info pallete? The consensus seems to be approx R=119, G=119 and B=119. If I then correct any discrepancies – can I apply the same correction to the rest of the roll of film?

wrote:

My apologies for the crossposting, thats the first (and last) time I’ll do that.

And, if anyone else needs a reason NOT to crosspost – It’s because you end up with intelligent, knowledgeable people NOT being able to engage in a structured conversation and NOT being able to develop their arguments with each other.

The idea behind the original question was this:

If I expose a roll of film and load the images into Photoshop with one exposure a picture of an 18% reflectance grey card, what should the RGB values be in the Info pallete? The consensus seems to be approx R=119, G=119 and B=119. If I then correct any discrepancies – can I apply the same correction to the rest of the roll of film?

In one word: yes. Providing that all the shots were taken under the same circumstances, of course. For example shooting in a studio. And providing that all the images were exposed the same, meaning exposed with manual settings. If the camera is set to automatic exposure, you can NOT assume it will always need exactly the same correction.

—
Johan W. Elzenga johan<<at>>johanfoto.nl Editor / Photographer http://www.johanfoto.nl/

"Al Denelsbeck" wrote in message

2) There’s a difference between the luminance change of the actual monitor output, as measured by a light meter, and the onscreen representation of a photograph converted to 24-bit RGB. The monitor has an exceptionally narrow light range, and right at the moment, my pure black screen with no room lights is pegging about 5 seconds at f2.8, ISO 100, which is much higher than full moonlight on a grey card.

The f-stop change I gave for the monitor would be accurate for a monitor which
outputs 100 Cd/m^2 for it’s pure white. My monitor output’s 75Cd/m^2 when calibrated for 5000K, so the f-stop change for my monitor when calibrated for 5000K would be even *less* than the value I calculated. At 6500K, the output
of my monitor increases to 88Cd/m^2, which is still less than 100, so again, the
f-stop change would still be less. Those of us with LCDs, which are typically
brighter than CRTs, may experience a greater f-stop change than the value I calculated.
(LCDs can typically go over 100Cd/m^2)

Bruce’s site is producing some weird results for me, and I’m not sure what he’s working with yet…

It may help if you ensure that you select "scale XYZ", "scale Y", and "scale
RGB".

Greg.

I wrote:

128,128,128 is closer to the midpoint than that, isn’t it? Punching the numbers
into the CIE Color Calculator:
http://www.brucelindbloom.com/ColorCalculator.html
the luminance changes from 18.4 (L*=50) to 21.6 for sRGB 128,128,128. 21.6/18.4
is 1.17, which represents only 0.17 of a stop, which is not even 1/5th.

I don’t think I did this calculation properly.
I think the proper formula to calculate the relative f-stop change from one luminance to another is:

f-stop(relative) = (ln(Y2) – ln(Y1))/ln(2) =
(ln(Y2) – ln(Y1)/0.693

Where Y is Luminance.

Using this formula the result is 0.23 of a stop.

I’m still not 100% sure of this new formula though. (can anyone confirm?)

Greg.

f-stop(relative) = (ln(Y2) – ln(Y1))/ln(2) =
(ln(Y2) – ln(Y1)/0.693

oops – left out a bracket:

f-stop(relative) = (ln(Y2) – ln(Y1))/0.693

Greg.