Why is Zone V 50% gray?

[Ralph W. Lambrecht]

Why does a digital camera record a Kodak Gray Card (Zone V) as 50% gray?

A Kodak Gray Card represents an 'average' scene of 18% reflectance, but that has nothing to do with a mean or medium on a grayscale. A gray card has an absolute log reflectance density of about 0.75. Medium gray on a 2.2-gamma monitor and a calibrated print thereof is about 0.66. So the image of the card is lighter than the card itself. Why? Who came up with this assumption or standard? Is it a mistake? Did someone assume 'average gray' meant 50%?



Disclaimer:

I'm aware that this is a mainly or exclusively analog forum, but I doubt to get a satisfying answer for my question in a digital forum, since they usually understand little about the Zone System. That's why I post the question here. Besides, I need this to work out some details about making digital negatives, which will be used to make pure-silver contact prints. So, it's an analog question in a way.

[Eric Rose]

There is a company that produces "gray cards" for digital cameras. Evidently digi cams do not read the usual cards we use properly. Check out this link:
http://www.betterlight.com/pdf/whitePaper/wp_gray_cards.pdf

[Ralph W. Lambrecht]

There is a company that produces "gray cards" for digital cameras. Evidently digi cams do not read the usual cards we use properly. Check out this link:
http://www.betterlight.com/pdf/whitePaper/wp_gray_cards.pdf

Many thanks, but that's not the point. The question is, why does it place a regular gray card on 50%, where it ends up at a lower density (Lighter) than the card?

[robc]

A grey card is 2.5 stops less than 100% reflectance. So go do a little test and see how many stops less than 100% reflectance your gray in the print is. i.e. zero your densitometer on paper base white and then see how many stops less the print of the grey card is. Let us know how you get on.



re-read the question and I now see that was your test.

sRGB specifies a screen luminance of 80cd/m2

[Ralph Barker]

It's set for the factory-mis-calibrated monitors of "average" users? ;)

[Ralph W. Lambrecht]

Relative screen luminances are set by the gamma value of the monitor. The gamma value defines the relative luminance of the midpoint. At gamma 2, the 50% patch is 2 stops darker than the 0% patch.

[Ralph W. Lambrecht]

It's set for the factory-mis-calibrated monitors of "average" users? ;)

Is this your assumption, or do you know for sure? Really interested!

[Helen Bach]

Ralph,

When you say that it is represented as '50% grey' what does that mean, and in what colour space?

If you are looking at it in say sRGB or Adobe 1998 then the numerical representation will reflect the non-linearity of the conversion to the colour space. There is then a reverse non-linear conversion from the numbers to the print or display.

So 18% grey can be represented as L=50 in Lab (and somewhere between 110 and 128 in RGB, depending on the colour space) while still being identified correctly as 18% grey, and print or display as 18% grey, or thereabouts depending on this and that, but mainly the other.

Does that make sense?

Best,
Helen

Later:

I've re-interpreted your question. I think that the above was not quite what you wanted, but I read two different questions, I think.

In a print the density of an L=50 patch (ie 50% in the image file) depends on the D-max of the print: it is relative and not absolute.

[Ralph Barker]

Is this your assumption, or do you know for sure? Really interested!
It was an attempt at humor. I'm also curious to hear the real answer.

[Ralph W. Lambrecht]

It was an attempt at humor. I'm also curious to hear the real answer.

I'm afraid, it might be all wrong! We'll see.

[Ralph W. Lambrecht]

When you say that it is represented as '50% grey' what does that mean, and in what colour space?

Gray Gamma 2.2 or 1.8, whatever the monitor is set to and what is tagged to the file.


If you are looking at it in say sRGB or Adobe 1998 then the numerical representation will reflect the non-linearity of the conversion to the colour space. There is then a reverse non-linear conversion from the numbers to the print or display.

Yes, I understand that.


So 18% grey can be represented as L=50 in Lab (and somewhere between 110 and 128 in RGB, depending on the colour space) while still being identified correctly as 18% grey, and print or display as 18% grey, or thereabouts depending on this and that, but mainly the other.

Does that make sense?

Nope, lost you there. 50% always seems to be 50% in my tests, as long as the attached color space matches the file's intention.


In a print the density of an L=50 patch (ie 50% in the image file) depends on the D-max of the print: it is relative and not absolute.

That's a biggy!!! And it might depend on the Dmin too? Oh, I'm getting worried now. It does come down to the absolute vs relative discussion again, doen't it?
But for now, let's stick to the monitor's representation. Why not 0.75 for Zone V? It can do it, and Dmin and Dmax support it comfortably. It this whole thing is really relative, then Zone V should actually be darker than 0.75 on a monitor, because the monitor's Dmax is far beyond the print's Dmax.

[Helen Bach]

OK, my head is beginning to spin with all these numbers.

When you say 50% grey, do you mean K=50%? I guess that because you say that 50% is two stops darker than 0%. Can we determine which value we are using before continuing? How about using the L of Lab?

How are you determining the 'reflectance density' or Dmax of your monitor when you say that it is 0.66 for example?

Thanks,
Helen

[Brian Ellis]

Like everyone else here with the exception of Helen, I don't understand the question very well and I certainly don't have a clue as to the answer.

[robc]

OK, my head is beginning to spin with all these numbers.

When you say 50% grey, do you mean K=50%? I guess that because you say that 50% is two stops darker than 0%. Can we determine which value we are using before continuing? How about using the L of Lab?

How are you determining the 'reflectance density' or Dmax of your monitor when you say that it is 0.66 for example?

Thanks,
Helen

I think he means hex 808080 or R128 G128 B128 which are 50% brightness from black.

I could be wrong though.

[Ralph W. Lambrecht]

Helen

Thanks for your patience. Yes, I meant K=50%. I hope I didn't refer to reflectance density when talking about the monitor anywhere. I measure the relative monitor brightness with a densitometer, set to reading transmission and zeroed at K=0%. When you do that, the 50% patch reads 0.66 'density' for a gamma 2.2 monitor, which is exactly 2.2 stops darker than the 0% patch.

[Ralph W. Lambrecht]

Correct.

[robc]

when you say a digital camera records a grey card as 50% I think we need some more information:

1. did you set the grey card at the correct angles to make the exposure? i.e.

Kodak’s instructions which come with a grey card say:

“Position the grey card in front of and as close to the subject as possible. Aim the surface of the gray card toward a point one third of the compound angle between your camera and the main light. For example, if the main light is located 30 degrees to the side and 45 degrees up from the camera to subject axis, aim the card 10 degrees to the side and 15 degrees up.”

This means you need to buy a sextant to use a grey card as it is meant to be used

2. How were you metering? If you were spot metering from the camera then ignore 1 above because it becomes irrelevant except you need to know the K factor for the camera meter. i.e. how it deviates from a true reading with no adjustment.
If you weren't spot metering from camera then 1 above may not be irrelevant. Thats why we calibrate film speed. You may need to calibrate camera ISO.

3. Are you talking about all digital cameras and have you tested them all?

[Helen Bach]

Everything seems to be in order. With a gamma of 2.2, 50% input brightness is 2.2 stops down in output from 100% brightness, representing an equivalent 22% reflectance. If you wanted 50% input brightness to be 2.47 stops down from 100% (ie representing 18% reflectance) then you would need a gamma of 2.47.

But, on the other hand, you want the display to show shadow detail as well. For a constant gamma (albeit at different values) and a fixed display dynamic range, the lower the gamma the greater the input range you can display. So the greater the dynamic range of the display device, the greater the gamma can be. All to the greater gamma, so to speak.

Of course, gamma need not be constant. In reality you can use a Tonal Response Curve in place of a constant gamma.

How's that?

Best,
Helen

The business about gamma also determines where 18% grey is. If you use a gamma of 2.2 (or if the colour space effectively uses it when converting the file, not just tagging it) then 18% grey should be about 117 in RGB, not 128. So if you take a snap of a grey card with your digicam and it gives you a file with the card at a value of 117, you know that the gamma is effectively 2.2

[Ralph W. Lambrecht]

Kodak’s instructions which come with a grey card say:

“Position the grey card in front of and as close to the subject as possible. Aim the surface of the gray card toward a point one third of the compound angle between your camera and the main light. For example, if the main light is located 30 degrees to the side and 45 degrees up from the camera to subject axis, aim the card 10 degrees to the side and 15 degrees up.”

This means you need to buy a sextant to use a grey card as it is meant to be used

Funny, and not all false:), but yes, I did do it that way.


How were you metering?

I delegated the job to Mr. Nikon.


Are you talking about all digital cameras and have you tested them all?

I'm not THAT well off, but I (we) tested 5 different DSLRs from 3 different top-brand manufacturers. Same result from all.

[Helen Bach]

I was in a bit of a rush with the last post, so I didn't write much of an answer to the first part of the question about cameras.

18% grey, in sRGB, should have an RGB value of about 117 to 119. It would show as L=49% or 50% (I will forget about K and I suggest that everyone else does, because CMYK is a whole different can of worms). sRGB uses a gamma of 2.2 in effect (it actually uses 2.4, but with an offset that makes it look like 2.2). An RGB value of 128 (50% of the full input signal) represents an L of 54% in sRGB. It's a mismatch between the perceptual curve that places 18.4% reflectance at L*=50% and the colour space curve that places 21.8% grey equivalent at 50% of the full input signal. Adobe RGB(1998) is similar in that it uses a gamma of 2.2.

Now, exactly where the camera's meter places the grey card (or any card, because if it fills the frame it shouldn't matter what reflectance the card is) depends on the metering constant that the manufacturer programs in to the meter.

That's why it is important to keep track of what number represents what. Easy, isn't it not?

Best,
Helen

[robc]

The obvious has just occurred to me. (I think).

to answer your straight question:


Why does a digital camera record a Kodak Gray Card (Zone V) as 50% gray?

Of course it will render a Kodak 18% grey card as K50%. It would render a 70% grey card as K50%. And it would render a 20% grey card as K50%. Regardless of orientation of the card.

There is a light meter in the camera and, as we all know, camera light meters give enough exposure to render what ever they are looking at as a mid grey. If that always turns out to be K50% then that is correct. The camera does not know what reflectance or colour your grey card is.

And by the way, an 18% reflectance card is only approx 2.5 stops less than 100% reflectance and therefore on a 10 stop scale it is between zone VII and VIII. Only on a 5 stop scale is it the midpoint.

[Helen Bach]

I'll put this as a separate post to keep it simple (well, as simple as I can make it) and expand a little on what 50% means as a measurement of lightness.

The 18% reflectance of a grey card is not 50% grey, but it is 50% lightness. Light is a perceptual property, and measurements of it often attempt to reflect human perception. Sometimes it is done with a spectral response curve (lumens etc) and sometimes it is done with a non-linear response curve, as here with 'lightness' or whatever L you want to call the L of Lab.

18% grey is 18% reflectance. It is a simple physical property, with no perceptual factors.

That's why 18% is also 50%.

Best,
Helen

[robc]

I missed your previous post.

Converting your sRGB(or AdobeRGB) file to Lab mode gives you the grey as L=50. 119RGB.

[Helen Bach]

Converting your sRGB(or AdobeRGB) file to Lab mode gives you the grey as L=50. 119RGB.

I must be explaining myself very badly if you need to mention that.

"18% grey, in sRGB, should have an RGB value of about 117 to 119. It would show as L=49% or 50%"

Oh poo.

[Don Bryant]

Helen

Thanks for your patience. Yes, I meant K=50%. I hope I didn't refer to reflectance density when talking about the monitor anywhere. I measure the relative monitor brightness with a densitometer, set to reading transmission and zeroed at K=0%. When you do that, the 50% patch reads 0.66 'density' for a gamma 2.2 monitor, which is exactly 2.2 stops darker than the 0% patch.

I'm curious to know what kind of instrument you use to measure your monitor values. Notice that I didn't say reflectance since reflectance values of a monitor make no sense.

[robc]

well I created a test file in PS with a gradient from black to white. In sRGB mode the mid point of the gradient(400 pixels along an 800 pixel image) is 128RGB. At that point the RGB brightness is 50% and L=54. Converting the image to Lab mode changes L to 50 and RGB to 119 B(brighness)=47%

I was just playing to see if understand this. I think what I did was what you said.

[neil poulsen]

It comes down to how you metered. You indicated that you delegated the job to Mr. Nikon. Were you using spot metering? Did the gray card fill the frame?

[Ralph W. Lambrecht]

I'm curious to know what kind of instrument you use to measure your monitor values. Notice that I didn't say reflectance since reflectance values of a monitor make no sense.


I use a regular densitometer set to transmission mode. Mine is from Heiland, but I did it with an Agfa densitometer before and see no difficulty is using any other brand.

[Ralph W. Lambrecht]

It comes down to how you metered. You indicated that you delegated the job to Mr. Nikon. Were you using spot metering? Did the gray card fill the frame?


Neil, I think we've been there. The metering is not the issue. Use a white wall and you get the same result. The point is the camera puts an average reading at 50% gray and prints it lighter than the density of a gray card. Therefore, a gray card is not accurately represented by a digital system, and the question is WHY?

[robc]

I need some clarification here.

First, is a grey card a true middle grey? If so then where's the proof. Ralph has already said:



A Kodak Gray Card represents an 'average' scene of 18% reflectance, but that has nothing to do with a mean or medium on a grayscale.


If you don't think its a "mean or medium on a grayscale" then why do you expect it to be reproduced the same as its colour when you know that a light meter will change it to a middle grey in the sRGB colour model?

[Ralph W. Lambrecht]

Not bad, not bad at all.

[Ralph W. Lambrecht]

Guys and Gals

I like to thank everyone who has participated in trying to solve this. Your posts have all been valuable to me. I have collected enough info to go back to the drawing board, have designed a few tests to shed some light on this issue and will share the results as soon as the data is evaluated.

Many thanks again.

[Helen Bach]

Neil, I think we've been there. The metering is not the issue. Use a white wall and you get the same result. The point is the camera puts an average reading at 50% gray and prints it lighter than the density of a gray card. Therefore, a gray card is not accurately represented by a digital system, and the question is WHY?

It's important to be absolutely clear about what you mean by 50% grey.

An 18% grey is considered to be perceptually 50% (linear vs logarithmic), as I've already explained, and that 50% is what you appear to be referring to because the common colour spaces used in still photography use perceptual values and so are non-linear, not linear. It is perfectly correct.

What the camera meters to depends on the meter calibration constant. For example, for some reflected-light meters and 35 mm film cameras the constant may be equivalent to metering off a 12.5% grey (many complications to that simple statement).

I've only explained (or at least I hope that I've explained) why a monitor displays it 'lighter' than the original: we haven't really got on to prints yet.

Is there a problem with my explanation of why 18% is also correctly represented as 50%, or why a monitor set to a gamma of 2.2 displays '18% grey' lighter than you think it should be?

Best,
Helen

[robc]

If there is a problem, then it is because I always thought that a Kodak 18% grey card is not a mid grey (50% lightness grey). This stems from understanding that it is derived from tests to determine the average luminance of a typical scene and not the the mid point of a perceptual grayscale. Your statement "An 18% grey is considered to be perceptually 50% (linear vs logarithmic)" is at odds with that.

Do you have references stating that an 18% grey card is a 50% lightness grey?

[Ralph W. Lambrecht]

Helen

Your explanation is absolutely fine. It helped me to understand some of the relationships. Many thanks for it. I'm going to try some transfer function now. I will also try an ICC profile, which Richard is offering to make on his RHDesigns website.

This issue is more complex than anything I have encountered in photography ever.

Thanks for the help.

[Helen Bach]

I don't have a lot of confidence in my ability to write in an easily-understood manner.

"Do you have references stating that an 18% grey card is a 50% lightness grey?"

Well, if you convert a reflectance of 0.184 (18.4%) to CIELAB L* it comes out as 50.

If you go to Bruce Lindbloom's site, http://brucelindbloom.com/ then > Calc > Companding Calculator and enter 0.1842 into the 'Y' input, you should get L* = 50.000 out - in fact I think that they are the default values when the calculator window opens. Fish around on Bruce's site and you will find plenty of other info.

Munsell N5 - which is middle grey on the Munsell scale - has 19.8% reflectance, so that isn't quite the same, but it is close. It's based on a slightly different method than CIELAB L*a*b*, having an L* of 51.6 insteads of 50.

If you want proper printed-on-paper references I'll fish some out.

As far as 'average scene reflectance' goes, just because 18%, or thereabouts, is perceptual middle grey doesn't also stop it from being the 'average scene reflectance'. Or it might not be. But that is such a can of worms.

Best,
Helen

[robc]

Thanks,

I will read up on it.

[Ralph W. Lambrecht]

I don't have a lot of confidence in my ability to write in an easily-understood manner.

I think you're doing just fine.


If you convert a reflectance of 0.184 (18.4%) to CIELAB L* it comes out as 50.

If you go to Bruce Lindbloom's site, http://brucelindbloom.com/ then > Calc > Companding Calculator and enter 0.1842 into the 'Y' input, you should get L* = 50.000 out - in fact I think that they are the default values when the calculator window opens. Fish around on Bruce's site and you will find plenty of other info.

Munsell N5 - which is middle grey on the Munsell scale - has 19.8% reflectance, so that isn't quite the same, but it is close. It's based on a slightly different method than CIELAB L*a*b*, having an L* of 51.6 insteads of 50.

Very valuable info. Thanks for that.


If you want proper printed-on-paper references I'll fish some out.

Yes, please, please, please.


As far as 'average scene reflectance' goes, just because 18%, or thereabouts, is perceptual middle grey doesn't also stop it from being the 'average scene reflectance'.

I think you hit the nail on the head here. There might be no argument after all. Two approaches coming up with the same result. Perfect.

[Paul Coppin]

Everything seems to be in order. With a gamma of 2.2, 50% input brightness is 2.2 stops down in output from 100% brightness, representing an equivalent 22% reflectance. If you wanted 50% input brightness to be 2.47 stops down from 100% (ie representing 18% reflectance) then you would need a gamma of 2.47.

Crap. I thought I understood computers and even tonal relationships quite well, but apparently, I'm a dolt.
Did I just understand that if I wanted to represent, on the monitor, the actual tonal value of an 18% gray card (or its equivalent reflectance) I should have the monitor set to a gamma of 2.47...? (this may explain some issues I have with print density, relative to monitor, or, yes, truly, I am a dolt...:) )

[Ralph W. Lambrecht]

Yes, or accept that the monitor shows a gray card lighter than in real life. BTW, print your monitor image, that might get the gray card back where you want it. So much for prints that match the monitor. They call it a fully 'calibrated system'. But that's hype. One can't match the two precisely, but one can get close, if sacrifices in the shadows are made. Theoretical gamma goes to infinity, actual monitor gamma still outperforms any output on paper. Analog has the same dilemma when trying to matcha print to a slide.

[Helen Bach]

You asked for some references. Color Science by Wyszecki and Stiles is a good comprehensive, quantitative source. Section 6.3 deals with brightness and lightness scales.

Here are some of the formulae quoted there.

Priest, Gibson and McNicholas, 1933, the original Munsell system:
V = 10*Y^0.5

This gives Y = 25% for V = 5

Munsell, Sloan and Godlove, 1933, a modification of the original Munsell system:
V = (1.474*Y-0.00474*Y^2)^0.5

This gives Y = 17% for V = 0.5

Newhall, Nickerson and Judd, 1943, the current Munsell system:

Y=97.5*( 1.2219*V-0.23111*V^2+0.23951*V^3-0.021009*V^4+0.0008404*V^5)/100

This gives Y = 19.3% for V = 5. There's a slight catch with the Munsell system that I wasn't aware of until I did the calc for myself - the usual system is relative to 97.5% reflectance for V = 10, not 100% - that's the way that the Munsell calculator at munsell.com works - and most texts give Y = 19.7% for V = 5. The above formula is equivalent to 100% reflectance being V = 10.

CIELAB 1976
L* = 25*Y^0.3333 - 16

This gives Y = 18.4 for L* = 50. Bruce Lindbloom goes into this in more depth on his website.

Best,
Helen

[Ralph W. Lambrecht]

Helen

Thanks for that. Now, I have some reading to do.

[Helen Bach]

Just spotted a typo. The date of the first Munsell formula by Priest, Gibson and McNicholas should have been 1920, not 1933. Must try harder.

Best,
Helen