Feedback Please: The Kodak 18% Grey Card and Metering, a new look.

[Bill Burk]

I used this graph to arrive at the densities that might result given a Zone V exposure which would hit 0.8 density when the film is developed to 0.57 contrast which is close to that which would give us the 7 1/3 stops of subject range fitting Grade 2 paper. I used my assertion to propose the three densities, I didn't base it on any standards.

You might propose that the recommended meter reading taken from one of those cards might fall elsewhere... Where would you think the exposures would fall?

 

[RobC]

well there you go. I consistently get my zone V on 0.6 above fb+fog or thereabouts and I'm not using a grey card. But I develop for a 10 stop range to fit into 1.3log density range above fb+fog on neg. So my midtone value hits the middle of the used portion of my neg film curve.
But if I metered a grey card whatever its percentage reflection and placed on zone V, i.e. used actual meter reading, I would expect it to be 0,6 density or close above fb+fog. Given that I or you could use any %age reflection card, meter it and use given reading, it shows that just metering a grey card using metered value has nothing to do with colour reproduction at same tone as card. You must know how close to middle of your curve the %age reflection of your card really is. I still think it should be 8% so that it matches what your meter is using as its %age reduction based on its K factor. And if I were to meter an 18% grey card and open up 1 stop it would produce at a very close tone to the card in the print at G2. After all 18% reflectance is only 2 1/2 stops less than 100% which would be white if you are capturing a full range of tones in direct light.
And that makes a complete nonsense of using a grey card at all.

Looks to me that your chart is developing to manufacturer recommended (or close) for ISO speed.
If I were doing that then my mid point would be a bit higher density coming close to yours.
But your zone X is a whole 0.4log higher neg density than mine so you would stuggle to get a full range of tones on your neg from black to white to print without burning in or going to softer thean G2 paper. Each to his own as they say.

What you are saying is that your negs are developed to higher contrast and hence density range than mine but thats your choice and I know people don't seem to interpret the zone system in the same way as I do.

And that only goes to show that its all a pretty pointless discussion becasue we're all doing it a different way anyway.

 

[RalphLambrecht]

As Jack Dunn states, Lg “is not, as is sometimes mistakenly assumed, located precisely at the luminance level corresponding to the arithmetic average of a hypothetical scene containing evenly distributed tonal areas.

From Jack Holm's paper, Exposure Speed Relations and Tone Reproduction:

"Two significant assumptions which are often neglected in exposure determination concern the scene range and mean 'reflectance. They are as follows:

That the luminance range of a statistically average scene is 160:1 (log range 2.2), and the resulting exposure range on the image capture medium is 80:1 (log range 1.9), corresponding to a camera flare factor of 2.

That the mean log luminance of a statistically average scene is approximately 0.95 log units below the highlight log luminance (edge of detail in white) and 1.25 log units above the shadow log luminance (edge of detail in black), and that this mean luminance is assumed to be the luminance metered, directly or indirectly, for exposure determination. These values result in the mean luminance correlating with a Lambertian scene reflectance of 12% for 100% highlight reflectance."

I agree for a statistically average scene, the mean reflectance is 8%. Jack Dunn uses 9% reflectance in his book Exposure Manual, most probably because he rounded the luminance range down to 7 stops. Eight percent reflectance can be considered the real world average reflectance. If this sounds fairly low for photography, it is, for contrary to what most people believe, the average reflectance for photography isn’t the mean of the values of the real world subject, but the mean of the values of the real world subject measured within the camera. From a photographic perspective, the mean of the camera image is the value that really matters, and the fundamental difference between the real world luminance range and the camera’s illuminance range is the camera image includes flare (Graph 1).

View attachment 152349

Take another look at Holm's mean log luminance of a statistically average scene. It falls 0.95 log units below the highlight log luminance and 1.25 above the shadow log luminance. Add one stop flare and that makes the shadow log luminance fall only 0.95 below the mean. 0.95 above and 0.95 below.

Stephen,this subject is difficult enough to get our heads around without the variable 'flare'.Let's make that the next step but let's not introduce it now. peace.

 

[Stephen Benskin]

Stephen,this subject is difficult enough to get our heads around without the variable 'flare'.Let's make that the next step but let's not introduce it now. peace.

Ralph, I apologize for offering answers. Please continue with the pet theories and confirmation bias.

 

[RobC]

last we argued about flare you accused me of using a setup where I didn't get any and therby implying I wasn't playing by the rules. Must be my wonderful zeiss lenses which don't have any worth taking into consideration.

 

[Stephen Benskin]

the film speed in the Kodak chart is 125. Not valid as formulas based on 100 speed film.

Please adjust.

Which formulas? If you are referring to the Ev value derived from the meter, then yes, it's based on 100 speed. Otherwise, please explain.

 

[RalphLambrecht]

Ralph, if you are referring to "The Brightness Scale of Exterior Scenes and the Computation of Correct Photographic Exposure," I don't remember anything about 18% reflectance in that paper (although it's been more than a few years since I've thoroughly read the paper). In fact, I remember finding it interesting that it's not mentioned in any of Jones' seminal papers. I believe the perceived importance of 18% gray is primarily promoted from popular photographic writings.

The 18% gray card is more of a visual reference than an exposure device; however, it should be noted that the instructions that comes with the Kodak Neutral Gray Card says that if the card is used for metering an exterior scene, the exposure should be increased by 1/2 stop. This would indicate an effective 12% reflectance (to be clear, exposure meters are NOT calibrated to reflectance).

Rob, I've written rather extensively about K. There's a 20 plus page document available at http://64.165.113.140/content/benskin/. It's titled "Defining K." One of the problems of deriving information from the standards is that the purpose of the standard isn't about theory. All the research and supporting arguments for what's in a particular standard comes from scientific papers. The one exception to this is in the appendixes in the standard for general purpose photographic exposure meters (ANSI PH3.49 - 1971). The standard makes it clear that the appendixes are not part of the standard. The various sections that explain K are almost verbatim from the previously published paper Re-evaluation of Factors Affecting Manual or Automatic Control of Camera Exposure by James F. Scudder, C.N. Nelson, and Allen Stimson (also available at the above link). Here is an excerpt from the paper's abstract:

"The film exposure level maintained by an automatic control in a camera depends primarily on the film speed but several other variables can manifest substantial influence. The effects of field-luminance distribution and spectral sensitivity, as well as sensitometric, optical and photometric relations are expressed analytically and the equations for camera exposure are derived. The resulting constant relates the ASA standard film speed to the preferred exposure for an area in an average scene having the average luminance indicated by the meter. This constant, when combined with nine variables which are a function of camera design, meter design, and scene structure, provides an equation that is simplified by substituting empirical values for all but three parameters. The exposure constant is expressed as a function of the lens transmission, spectral characteristics of the detector and the discrimination of the field luminance measurement."



I've attempted to explain this to you a number of times in the past. By evaluating the variables in the equation for K, it's purpose can be better understood. I did a breakdown of the variables in Defining K that might be useful. Rob, I hope to adequately explain that while you got the math correct, your interpretation and application of K is in error. There is a lot to unpack and I'm still trying to determine an approach and how much effort I want to put into it. But something to think about for starters, you determined correctly that the ratio of C and K produces the effective average reflectance of the reflective meter (12%), but then mistakenly conclude meter's are calibrated to 8%. And yes, the 8% correlation is just a coincidence.

Yes, that's the paper I was referring to;have to dig it out and read it again.

 

[Stephen Benskin]

lets go at this from a different angle.

We'll asume that film will capture 7 1/3 stops of subject range and print exactly onto B+W paper full a full range of tones from pure black to white.

Now if we meter from an 8% card, at what density would you expect if to appear on the film curve? And at what density when metering from a 12% card and at what density when metering from an 18% card.

I just want densities and NO FORMULAS. A very simple question which does not require a contrived complex answer.

First of all, why do you get to use formulas but keep wanting to restrict others from using them?

Secondly, if you are metering a target with a single tone, that value will fall at 8/ISO. The density would depend on the curve shape, but the exposure will always fall at the same point. That's how the meter / film speed combination works. There's a ratio between the speed point and the metered exposure point. You need to plug your idea in and see if it will produced the expected results. I think your understanding will change.

last we argued about flare you accused me of using a setup where I didn't get any and therby implying I wasn't playing by the rules. Must be my wonderful zeiss lenses which don't have any worth taking into consideration.

Wow. Simply wow. I believe you are referring to how I've proven Zone System testing is essentially flare free. This is because you basically need a range to produce flare and shooting a target with a single tone doesn't have a range. All lenses have flare and many pet theories go wrong because they don't factor it in. The film is exposed by the camera image and that includes flare which increases the shadow exposure and reduces the effective subject luminance range.

I remember now. I was pointing out how the statistical average Luminance range of 2.20 is determined. That it doesn't incorporate specular highlights and accent black. That paper LER doesn't incorporate the whole curve which then has room for those values. Flare plays a part as the 2.20 log subject Luminance range is reduced to an optical image of approximately a 1.80 - 1.90 illuminance range.

 

[Stephen Benskin]

https://en.wikipedia.org/wiki/Light_meter

This is the formula I am using
View attachment 152387

Bill, that equation is the key to understanding how K works. We can reduce it down a little more if we use a shutter speed that is the reciprocal of the film speed. Now, place K to the side for the moment and what we have is the relationship of the subject Luminance with the aperture. Using Sunny 16, N^2 and Lg should be equal. Lg should be 256 cd/ft^2 or 2755 cd/m^2. But due to the light loss in an optical system, the amount of light entering the lens needs to be higher. This is the primary function of K. It adjusts for the light loss in an optical system. 256*12.5 = 3200. 3200*.65 / 256 = 8.1 (P) or the value of Eg at the film plane. Reminder: 0.65 is the value for q which is the factor that takes into account the light transmission characteristics of the lens system.

Of course P (denoted as K`1 in the K equation) is the answer only when the variables r, p, and R in the K equation equal unity which I will be covering later. Hint: It's usually these variables that produce values of K other than 12.5.

 

[markbarendt]

... if you are metering a target with a single tone, that value will fall at 8/ISO. The density would depend on the curve shape, but the exposure will always fall at the same point. That's how the meter / film speed combination works. There's a ratio between the speed point and the metered exposure point.

That is the magic for me, the relationship between the film speed point and the meter reading is a known.

Because of that relationship, with a bit of experimentation or math we can very accurately find the relationship between any known tone, like a gray card or white card or whatever, and the speed point of the film.

The relationship, the distance, between meter reading and film speed is fixed to the shadow end of the curve, not in the middle. The distance from measured tone to print highlight is variable.

A pull moves a measured mid-tone relatively darker in a print, a push moves a measured mid-tone lighter in print. IMO re-centering the mid-tones is the goal of the EI changes that are suggested with a push or pull.

 

[RobC]

So we are standing outside in bright sunshine with a scene in front of us with a white wall in it which we want to retain as fully white on the negative and which we want to print as just fully white on G2 paper.
The scene has some shadows in it too. We take meter readings with our trusty spot meter and find that the wall is 5 stops brighter than the shadows we want to retain good detail in.

So we now whip out our trusty 18% reflectance grey card. We know that in the relative scale we are about to photograph that the wall will be reflecting approx 90% of the light falling on it. Nothing reflects 100% and pure bright white around 90% of the light falling on it, so within a 1/5 of a stop of 100%. We know our grey card will reflect 18% of the light falling on it. We can work out that the shadows we want full detail in will be reflecting approx 3% of light falling on them because if the wall is 90% reflectance then 5 stops less will be 3% reflectance.
We also know that we are expecting to capture a negative with 7 stops of range from black to white with the white wall having a density on film of 1.3log above fb+fog. In other words a pretty normal full scale neg capturing a full 7 stops of subject brightness range.

So because we don't really understand exposure we think a kodak grey card will solve it for us. We place our grey card against the white wall and meter it. We find its 2 1/2 stops less bright than the white wall. Thats odd, a kodak grey card is supposed to be middle grey but its only 2 1/2 stops less than the white wall and the middle of our film curve would be 3 1/2 stops less than the white wall. But being a bit dim we believe that a kodak grey is the middle and we should expose using the meter reading given.

That would result in the kodak grey card being placed in the middle of the film curve. But since it's really only 2 1/2 stops less bright than the wall it would actually be under exposed by a whole stop and at the same time our shadows would be underexposed by a stop and our white wall too. In fact everything would be underexposed by one whole stop becasue of our mis-understanding of how bright a kodak grey card really is. To get the full scale of 7 stops on our neg we would need to meter the grey card and open up 1 stop from the reading so that the grey card would get the correct negative density and also the shadows and the white wall. YES, open up one whole stop from what we meter our Kodak 18% grey card at.

But there is one big assumption being made in the above and that is, our spot meter will place whatever we meter in the middle of the film curve. How do we know that is true? We don't unless we look at the formula and values a meter actually uses. So we do that and find that our reflection meter actually measures the light and then takes 8% of that reading and uses it to give us the Exposure Value (Ev) we should use.

Well 8% is a tad over 3 1/2 stops less than 90% so that would put the exposure in the middle of the film curve which is what we want if a kodak grey card was really the middle of a 7 stop range. But since it isn't, we have two choices, we either meter the grey card and open up 1 stop or we get a card which has same reflectance %age as our meter exposure reduction factor reduces the reading by. i.e. an 8% reflectance card.

Life would be so much easier if all the writings and books explaining exposure stated that a kodak grey card is 2 1/2 stops less than 90% reflectance and that a modern reflection meter is calibrated to use an exposure based on 3 1/2 stops less than 90% to place the exposure in the middle of the film curve.

Now I'm fully expecting to be told I'm wrong by those who insist on going to the bottom and looking up themselves. But not one of them is capable of explaining where the flaw they think exists in my practical approach actually is. Could be becasue there isn't one
For example flare:

 

[markbarendt]

Life would be so much easier if all the writings and books explaining exposure stated that a kodak grey card is 2 1/2 stops less than 90% reflectance and that a modern reflection meter is calibrated to use an exposure based on 3 1/2 stops less than 90% to place the exposure in the middle of the film curve.

The problem I see here is that the gray card doesn’t have a fixed relationship to the printable highlights, the fixed relationship is with the printable shadows.

 

[Stephen Benskin]

Reflectance uses the theoretical Lambertian surface which means values can easily be over 100% Reflectance. Jack Holm, "That the mean log luminance of a statistically average scene is approximately 0.95 log units below the highlight log luminance (edge of detail in white) and 1.25 log units above the shadow log luminance (edge of detail in black), and that this mean luminance is assumed to be the luminance metered, directly or indirectly, for exposure determination. These values result in the mean luminance correlating with a Lambertian scene reflectance of 12% for 100% highlight reflectance or 14% for 128% highlight reflectance."

 

[Diapositivo]

So we are standing outside in bright sunshine with a scene in front of us with a white wall in it which we want to retain as fully white on the negative and which we want to print as just fully white on G2 paper.
The scene has some shadows in it too. We take meter readings with our trusty spot meter and find that the wall is 5 stops brighter than the shadows we want to retain good detail in.

So we now whip out our trusty 18% reflectance grey card. We know that in the relative scale we are about to photograph that the wall will be reflecting approx 90% of the light falling on it. Nothing reflects 100% and pure bright white around 90% of the light falling on it, so within a 1/5 of a stop of 100%. We know our grey card will reflect 18% of the light falling on it. We can work out that the shadows we want full detail in will be reflecting approx 3% of light falling on them because if the wall is 90% reflectance then 5 stops less will be 3% reflectance.
We also know that we are expecting to capture a negative with 7 stops of range from black to white with the white wall having a density on film of 1.3log above fb+fog. In other words a pretty normal full scale neg capturing a full 7 stops of subject brightness range.

What I don't get in your reasoning (possibly because I use slide film) is this logical jump between the SBR (or Subject Luminance Range) in your subject and in your film.
If the interval between the highest luminance and the lowest luminance of your subject is 5EV, I expect your film to have 5 EV of range.

If the light hitting the subject is not uniform, so that the overall luminance range is 7 EV (5EV due to the different reflectivity, and 2 EV given to the different illumination, that gives 7 EV of total range, taking the hightlight in "full light" and the darkes part in "full shade" as the extremes) than I expect those two extremes to have the same 7 EV range on film.

I don't understand how do you expect - if I understand you well - to obtain a 7 EV range in your film from a 5 EV range scene. That's probably due to a specific way you develop your negative? Am I right in assuming that, given a certain standard development process, the two ranges should coincide? That's the way I suppose slide film should work. I don't reckon I find more, or less range unless the scene has a range which is wider than what the slide can record, and I will fatally end up with blocked shadows or burned highlights.

But if the scene is, let's say, 3 EV of total inner contrast (summing up difference of reflectivity and difference of illumination) how can I obtain a film with a range higher or lower than 3 EV?

PS I understand that flare reduces somehow the EV range on film especially if the EV range on the scene is pretty high. I don't understand having an EV range on film larger than on scene.

 

[Bill Burk]

well there you go. I consistently get my zone V on 0.6 above fb+fog.

As you say, you develop to lower gradient, I think 0.50 or thereabouts, so that would be about where your Zone V lands. That's fine.

 

[Bill Burk]

We can work out that the shadows we want full detail in will be reflecting approx 3% of light falling on them because if the wall is 90% reflectance then 5 stops less will be 3% reflectance.

It might be helpful to think of what the shadow of a typical photograph is. Not trying to make any point, just bringing something helpful to the discussion...

Something we want detail in, could reflect a low middle tone like 8% of the light that falls on it... but it is in shade. With two stops less light to reflect back to us. So what might be Zone IV if it were in open daylight, we might want to place on Zone II

 

[RobC]

What I don't get in your reasoning (possibly because I use slide film) is this logical jump between the SBR (or Subject Luminance Range) in your subject and in your film.
If the interval between the highest luminance and the lowest luminance of your subject is 5EV, I expect your film to have 5 EV of range.

If the light hitting the subject is not uniform, so that the overall luminance range is 7 EV (5EV due to the different reflectivity, and 2 EV given to the different illumination, that gives 7 EV of total range, taking the hightlight in "full light" and the darkes part in "full shade" as the extremes) than I expect those two extremes to have the same 7 EV range on film.

I don't understand how do you expect - if I understand you well - to obtain a 7 EV range in your film from a 5 EV range scene. That's probably due to a specific way you develop your negative? Am I right in assuming that, given a certain standard development process, the two ranges should coincide? That's the way I suppose slide film should work. I don't reckon I find more, or less range unless the scene has a range which is wider than what the slide can record, and I will fatally end up with blocked shadows or burned highlights.

But if the scene is, let's say, 3 EV of total inner contrast (summing up difference of reflectivity and difference of illumination) how can I obtain a film with a range higher or lower than 3 EV?

PS I understand that flare reduces somehow the EV range on film especially if the EV range on the scene is pretty high. I don't understand having an EV range on film larger than on scene.

You are assuming I would expose for the shadows but I don't and nor does the example I gave. It uses the actual meter reading or meter reading opened up one stop. i.e. expose for middle of curve or one stop more than middle of curve.

Slide film accepts a shorter range than 7 stops, more like 5 stops (but not all slide films are the same). A kodak 18% grey card is well suited to slide film. It is not well suited to negative film which accepts 7 1/3 stops of range which should be printable straigh to G2 paper without doing somersaults.

 

[Bill Burk]

Reflectance uses the theoretical Lambertian surface which means values can easily be over 100% Reflectance. Jack Holm, "That the mean log luminance of a statistically average scene is approximately 0.95 log units below the highlight log luminance (edge of detail in white) and 1.25 log units above the shadow log luminance (edge of detail in black), and that this mean luminance is assumed to be the luminance metered, directly or indirectly, for exposure determination. These values result in the mean luminance correlating with a Lambertian scene reflectance of 12% for 100% highlight reflectance

Any chance you could draw for 100 ISO, for example the metered point would be 0.008 mcs arithmetic, -2.1 Log mcs for 100 ISO.

Now what exact exposure does the light meter want to put at the metered point at the film plane for 100 ISO film? Maybe that short answer will clear up a lot of doubt about things like what density the film would be at that point.

 

[Bill Burk]

Rob, I think I got it.

That formula you are using, takes a value of light and puts it on the film plane as an exposure in meter candle seconds. It goes through an aperture and shutter speed to get there. What arrives is not a percentage of reflection at all.

 

[Bill Burk]

... It's a quantity of light and time.

 

[Diapositivo]

You are assuming I would expose for the shadows but I don't and nor does the example I gave. It uses the actual meter reading or meter reading opened up one stop. i.e. expose for middle of curve or one stop more than middle of curve.

Actually I am not considering exposure. I mean, if the scene has 5 EV of range, the light hitting the film will be 5 EV of range (letting aside flare).

If your film has 9 EV of dynamic range (9 EV of linear response) and your scene has 5 EV of range I expect you will end up with 5 EV of range on film regardless of where you place them, and because you have with negatives some latitude of exposure (you have 4 EV of "slack" in this case), you can place those 5 EV in the film in different places of the linear part of the film curve, but they will always be 5 EV on film, I mean, whether you place them on the lower part of the film curve or on the upper part or centered around the centre of the curve, the brightness capture on film would be no more no less than those 5 EV, that's my reasoning.

It's like a blanket of 5 EV which you move up and down a bed which is 9 EV high. You can place the blanket more near the head, more near the feet, or halfway, but the 5EV of the scene I suppose will create 5 EV on film (if the entire interval is placed somewhere in the linear part of the curve).

 

[RobC]

Reflectance uses the theoretical Lambertian surface which means values can easily be over 100% Reflectance. Jack Holm, "That the mean log luminance of a statistically average scene is approximately 0.95 log units below the highlight log luminance (edge of detail in white) and 1.25 log units above the shadow log luminance (edge of detail in black), and that this mean luminance is assumed to be the luminance metered, directly or indirectly, for exposure determination. These values result in the mean luminance correlating with a Lambertian scene reflectance of 12% for 100% highlight reflectance or 14% for 128% highlight reflectance."
View attachment 152423

Here we go again, half truths, smoke and mirrors.
I don't who wrote it or who agrees with them which you do. If I shine 1024 unts of light at anything it can only reflect at most 100% of 1024 units regardless of what type surface it is unless it's fluorescing. If the numskull that wrote that crap means that if the surface was not a matt surface but high gloss then it would reflect x percent in relation to what the matt surface reflects but not more than 100% of the light incident on it then it might make some sense. But that isn't what you said. And what has Mecury got to do with it or are you another one relying on telepathy.

 

[Stephen Benskin]

Any chance you could draw for 100 ISO, for example the metered point would be 0.008 mcs arithmetic, -2.1 Log mcs for 100 ISO.

Now what exact exposure does the light meter want to put at the metered point at the film plane for 100 ISO film? Maybe that short answer will clear up a lot of doubt about things like what density the film would be at that point.

Hg 8/100 = 0.08 lxs
Hm 0.8/100 = 0.008 lxs

 

[RobC]

Rob, I think I got it.

That formula you are using, takes a value of light and puts it on the film plane as an exposure in meter candle seconds. It goes through an aperture and shutter speed to get there. What arrives is not a percentage of reflection at all.

You got what? I'm not trying to get anything. Its you trying to work out why some different formula or writing work out the same as the standard meter formula. I think the meter formula is the one to use becasue its what my actual meter uses, and yo and benskin seem to think the theory and writings are correct even though my meter doesn't use them. Well it does in that film speed is derived from it.
And if you think flare has anything to do with it then you better have one too:

 

[Mr Bill]

Rob, I think I got it.

That formula you are using, takes a value of light and puts it on the film plane as an exposure in meter candle seconds. It goes through an aperture and shutter speed to get there. What arrives is not a percentage of reflection at all.

Exactly. The exposure meter standard is all about measuring some quantity of light then using the main equation to express this as a combination of shutter speed and lens aperture.

The fine details of the standard(s) include specifics of the light source, tolerances, and that sort of thing.

The actual standard(s) include details of the "exposure meter calibration constant," 'K,' which includes variables Stephen has (somewhat cryptically) mentioned - they include such things as camera lens transmittance, camera flare correction factor, and "effectiveness" of the light source in producing exposure on the film. For example, you would expect different 'K' values if the film has a different spectral sensitivity, or lens/camera combinations have different amounts of flare potential.

A lot of the difficulty with understanding this comes from people trying to make it more complicated than it is, such as the error of saying that an exposure meter is "calibrated for 'x' percent scene reflectance."