# Comparison of Reflectance from 18% and 12% Zone Models

Discussion in 'B&W: Film, Paper, Chemistry' started by Stephen Benskin, Dec 26, 2011.

1. ### Stephen BenskinMember

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This is kind of interesting. I've put together three exposure models comparing the subject reflection densities and Reflectances for Zone V = 18%, Zone V = 12%, and the standard model of exposure. As anyone familiar with the Zone System knows, Zone VIII falls three stops above the metered exposure point (Zone V). Zone I falls four stops under the metered exposure point for a total subject luminance range of 7 stops (2.10 logs). The standard model of exposure is slightly different as it doesn't base exposure from the metered exposure point but from 100% Reflectance with a 7 1/3 stop (2.20 logs) subject luminance range.

If Zone V equals 18% Reflectance, what would Zone VIII equal? How about when Zone V equals 12%? And there are other implications depending on the model used.

A little fun and a little food for thought.

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2. ### Bill BurkSubscriber

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The numbers at 18% are not pretty - but that's what Ansel Adams wanted.

At 12%, the percent reflectances are convenient and percentages progress through sensible numbers, like 95% for the white side of the card.

The standard model seems to have a clean series of relative densities.

3. ### Stephen BenskinMember

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So, you don't have a problem with a Zone VIII as having a Reflectance of 144%? Doesn't that sound somewhat unrealistic to you?

4. ### michael_rSubscriber

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I'm 150% sure zone VIII doesn't have a 144% reflectance

5. ### Bill BurkSubscriber

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I don't have a problem with it because it might assume a long range subject - reflectances of a part of the scene in shadow plus another part of the scene in open daylight - so a natural scene could exceed 100%

6. ### Stephen BenskinMember

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I know you are just being funny, but think about it. It's the math. 18% is a reflection density of 0.74. Add three stops and that equals an RD of -0.18 or 144% Reflectance for Zone VIII.

This is simply putting numbers to the Zones. If not 144% then what?

From Bill Burke
This is for a normal 7 stop scene when Zone V is at 18%.

7. ### Bill BurkSubscriber

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I just measured, in bright daylight, from top-hat black hole to white wall is 7 2/3 stops.

A reflection grayscale I made only measures about 4 stops. It has "less than ideal" characteristics.

But I believe even a reflective target with ideal characteristics, from 0 - 100% would be less than 7 stops.

8. ### Stephen BenskinMember

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From 100% - 0.8% Reflectance is seven stops. I guess technically the range could be infinite as you would never reach zero if you are always dividing by 2.

Remember, the use of Reflectance in relation to luminance usually involves a Lambertain surface which is a perfect diffuser. 100% Reflectance doesn't mean 100% reflection - (illuminance x reflectance) / pi = luminance.

Scenes with Reflectances of 100% and above incorporate non Lambertain surfaces and specular reflections. There's also relative Reflectance like when shooting a scene with multiple lighting conditions, like the cave and the beach. You have a a normal scene in bright sunlight and then a normal scene in dark shadows. If you are shooting the inside of the cave, the exterior scene will appear to have Reflectances easily over 100%.

However, that still doesn't answer of Zone VIII being at 144% when Zone V is considered to be metered at 18% gray.

9. ### Bill BurkSubscriber

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Well... isn't Zone V 12% gray? After all... Zone V is what the meter reads and that's supposedly 12-13%

10. ### Stephen BenskinMember

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Yes it is!!! Nice catch. This is just one way to prove it's not 18% using very little math. Zone VIII isn't supposed to have specular reflections. So, it shouldn't be over 100%. As Zone VIII is three stops over Zone V and if the meter "reads" 18%, then Zone VIII is over 100%; therefore, the meter can't be reading 18%. Another photographic myth bites the dust.

I almost worked myself into a corner as exposure meters don't "see" Reflectance, but it's possible to extrapolate an equivalent number based on the reflection meter's and incident meter's calibration. This is also the beginning of a more precise way to determine exactly what a meter "sees". Apply the average illuminance of Daylight to the different reflection densities (basically log Reflectance), plug them into the camera exposure equation, and see which one produces the correct exposure at the metered exposure point for a given EI.

And for the record, it's 12%.

11. ### michael_rSubscriber

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I never used % reflectances to do anything anyway, but out of curiosity, where did the 18% myth come from then? And why does/did even Kodak sell "18% grey" cards?

12. ### DREW WILEYMember

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18% is the midpoint for averaging meter use, and esp applicable to color film. But how many stops
of range with a black and white film the Zone Systems applies to either side of this really depends on the specific film and developer. I once took an entire stack of gray cards from several mfg and
read them on a high-end spectrophotometer. None of them were 18%, and all of them varied with
the exact wavelength of light because none were truly gray.

13. ### Bill BurkSubscriber

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It's an old concept. Some references that are very old and possibly trace to their respective first editions...

In an old Kodak Data Book, Fifth Edition, 1956 (1947 also 1941, 1943, 1944, 1945) I found "Kodak Neutral Test Card: An accessory designed primarily for use in color photography, but quite useful in determining exposure for copying and other photographic applications. The gray side of the card has 18% reflectance; white side, 90%."

Interesting note: When you use the 90% card you are told to divide your EI by 5. (So EI 400 becomes EI 80.) 90 divided by 5 is 18. Any relation?

1970 (editions 1930, 1937, 1940, 1942, 1944, 1946, 1947, 1954, 1970) LP Clerk's Photography Theory and Practice section 475 Key Tone and Selective Readings. "Exposure errors introduced by subjects of abnormal brightness distribution can be avoided by pointing the meter at an artificial constant tone temporarily incorporated into the scene... If this tone is a medium gray of about 20% reflectivity, the meter reading obtained can be used as if it were an average reflected light reading of an ideal subject.

Wow - 20%, first time I saw that figure.

Now I wonder, with recent advice to take the 18% reading and open up.

And 20% dating to very long ago.

I wonder if the 18% was ideal when there was a safety factor in the ASA ratings of film.

---
Oh and a reading from spotmeter mode on 18% gray card (0.78 reflective density) and a reading from the incident mode on my meter agreed perfectly this afternoon.
---

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15. ### Stephen BenskinMember

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Interesting. In An Interpretation of Current Exposure Meter Technology by Allen Stimson, it says that "the difference [between the exposure meter calibration and the 18% gray card] is due to the angle at which the card is held."

Sorry, small correction, an RD 0.74 equals 18%.

16. ### ic-racerMember

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Without stating the "K" of the meter, you could conceivably point it at just about any value reflectance card and get a correct average exposure.

17. ### Stephen BenskinMember

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Reflectance does come in handy with exposure models like for my program. Well not so much Reflectance but reflection density which is really just log(1/Reflectance). Just playing with some values and a few equations can really make clear how everything fits together.

You can find more on this with a fairly detailed series of uploads titled Defining K: http://www.apug.org/forums/forum37/85217-k-factor-relevant-me-should-i-cancel-out-4.html

I'm not sure if anyone has the answer to the origin of 18% gray. All I can offer is mostly speculation.

First of all, we're dealing with three different concepts. One is the perception of middle gray, the other is the average reflectance, and third is what the exposure meter uses.

The first use of 18% that I know of comes from Munsell. He found that in a 10 step gray scale with visually equal steps, 18% fell in the middle on step 5 (remind anyone of something?). Jones heading JOSA's committee on Colorometry later adjusted this value to 19.6 (I think). So, one use of 18% is psychophysical. It's a good reference card.

The second form of middle gray is from the average range of a scene. The statistically average scene has an luminance range of 2.20 logs or 7 1/3 from 100% R (0.0 RD) to 0.6% R (2.20 RD). The middle is 2.20/2 = 1.10 or 8%. If you round the scene to 7 stops, then the middle is 9%. This conforms with Jack Dunn's Exposure Manual. This can be thought of as the middle gray for the physical scene.

The camera's middle gray is slightly different. Flare reduces the range to by ~ 0.34 or 1 1/4 stops (for a large format lens) to around 1.86. 1.86/2 = 0.93 or 11.7% which can be rounded to 12%. So, 12% can be considered the camera's middle gray.

With the older uncoated lenses, flare was considered to be 2 stops. This makes the camera middle gray closer to 16%.

The final middle is exposure meters. A really quick way to determine what the Reflectance needs to be to produce the same results with a reflection meter as from an incident light meter, simply divide their constants. K for a reflection meter is 1.16. C for an incident meter is 30. 1.16* pi / 30 = 0.12

I'll have to check into it some, but the relationship between the two constants were different in the earlier standards. The problem is I don't think they had the hemisphere for incident meters when the first standards were developed in the mid 1940s, so I believe the value for C was for a flat plate receptor. I believe with this combination, something around 18% is possible.

I'm looking at the ISO 1994 standard, General Purpose Photographic Exposure Meters (Photoelectric Type) - Guide to Product Specification. For the incident constant C, it has two types listed: Cardioid and Cosine. The cardioid (hemispherical) comes with the C = 30 and the cosine(flat plate) comes with a C around 20.3. For the same reflected exposure meter with a K = 1.16, the "reflectance" with the cardioid equals 12% and the reflectance with the cosine equals close to 18%. But this is mostly relative as exposure meters read luminance and not reflectance. The percentage comes from the calibration luminance of the reflection meter compared to the calibration illuminance of the incident meter.

Has anyone has read the exposure instructions that comes with the Kodak card? It says to open up 1/2 stop when metering the card in an exterior scene and not to adjust the exposure for an interior. In film school we were taught to meter exterior scenes with the dome and interior scenes with the flat plate. Notice how that corresponds to the hemispherical and flat plate receptors with the incident meters?

It's mostly speculative.

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18. ### Bill BurkSubscriber

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Yes, the angle the card was held made a difference in the reading and could be a source of further errors.

I should clarify...
My 18% gray card in the Darkroom Dataguide 1970 (measured 0.78 reflective density)

---
But since averaging meters provide the correct exposure for 90% of all pictures, the purpose of the gray card is to be what the averaging meters are aiming for. They aren't aiming for an average gray that is in the middle of the characteristic curve. They are aiming for a brightness distribution that the average photographer gets when he points his camera at the average picture. Which usually includes a bunch of sky.

19. ### Bill BurkSubscriber

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Ah, but 9% is the middle of the high and low - not the average of the distribution.

20. ### Stephen BenskinMember

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Except it's not about averaging meters. Exposure is about the placement of illuminance on the film curve. I'm suggesting that's where the middle gray is determined. And what makes the camera image different from the original scene? You guessed it - flare. Flare changes how the original scene appears in the camera.

For the average 2.20 log luminance scene, the highlight falls 0.92 log-H above the metered camera exposure. The shadow falls 1.28 below. This imbalance doesn't make sense except when you factor in flare. Add 1 1/3 stops of flare to the shadow and it now falls 0.94 below the metered exposure. Now there's 0.92 log-H above and 0.94 log-H below. As the highlight falls at 100% Reflectance, RD 0.00 and the metered exposure falls 0.92 below, RD 0.92, or 12% Reflectance.

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21. ### Bill BurkSubscriber

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But placing a gray card in front of an averaging meter IS about placing an average scene in front of an averaging meter. You would do this when the scene may be high-key or low-key because you do not want the unusual brightness distribution of the scene to affect your exposure.

Exposure using an averaging meter is about placing the average scene - with a statistically understood brightness distribution - and which has no reason to be in the middle - onto the characteristic curve of the film with the shadow down where it should be and the highlight up where it should be. All without metering the shadow or highlight individually. Just knowing where they usually are in an average picture.

22. ### Bill BurkSubscriber

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You might still be right about the influence of flare. 18% Non-Flare gray card metering might still agree with 12% allowing for Flare in the camera. But I still believe there is no reason to tie the meter's gray to a percentage mathematically. I assert its placement was (as it should be) statistically determined. (Though I have no direct knowledge of it)

23. ### RalphLambrechtSubscriber

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more food for thought

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24. ### Stephen BenskinMember

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Right it's about a single point of exposure representing the entire range of exposure. A gray card is a single tone point. You need to pick a single point of exposure. What about a spot meter? It's calibrated to the same standard.

What I believe you might be thinking about is the bell curve of scene luminance range distribution. That it also tends to equal the exposure mid-point. Would that be the mean equals the average? Happy coincidence? What about unbalanced scenes like snow scenes? I believe that these are two seperate issues though.

I'm problably just not explaining it properly. One of the things I like about participating on APUG is that it can challenge what you know or think you know. This is what happened when I taught a class in photography. You think you understand it until you have to explain it.

How the exposure is placed on the curve is determined by P, as in P*1/ISO (of K1 in the exposure meter standard). Many people tend to misinterpret the paragraph in the exposure meter standard that reads, "This factor K1, has been determined experimentally by psychometrically selecting the "preferred exposure" for scene types, light levels, and camera and meter types covering the ranges normally encountered." Sometimes it is referred to as being an "arbitrary" value, but it's a scientifically arbitrary value. And the psychometrically selected "preferred exposure"? Think Jones and the first excellent print test (and subsequent such tests).

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25. ### Stephen BenskinMember

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Yes, but a computer screen and photographic print are both about the psychophysical middle - Munsell and CIE (CIELAB) . The scene is about the luminance range. Physical verses perceptual.

26. ### Bill BurkSubscriber

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Yes it is all about the bell curve of scene luminance range distribution. Not the actual for an individual shot today, but the average considered when the first light meters were made. Unbalanced scenes are not covered. They end up in bad exposure (unless you use a gray card or incident meter).

It's NOT in the middle. It is a full stop up from the middle. But wherever it is - it is there for coincidental, arbitrary or scientifically chosen arbitrary reasons.

The scale on the Weston Master III progresses
-[.]-[.]-[A]-[^]-[C]-[.]-[O]

Which corresponds to

Zone System
-[II]-[III]-[IV]-[V]-[VI]-[VII]-[VIII]

There are four stops beneath V and three above.