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# Thread: Hiding in Plain Sight

1. It's in the Journal of Photographic Science, Vol. 16 1968 page 188...

Talking about the arithmetic speed where the first way is sensitometric testing and the 0.1 speed point... "In the second way, Sa is determined by a photographic test, and is equated to the reciprocal of the effective exposure time when a satisfactory photograph is taken with a camera set at f/16, on a clear sunny day with a solar altitude of about 40 degrees."
...
"It is evident that the two methods of film speed assessment determine in effect the ratio..."

"k1 = Hg/HM"

---
So...

It almost seems like you can say "The arithmetic film speeds were setup that way to keep Sunny 16 valid."

Then perhaps indirectly since by this time there is a tie-in to the sensitometrically determined speed point...

(Another thread) "The metering point was chosen to align the average scene to agree with Sunny 16 results"

And what I kind of find funny: The ratio for Sunny 16 seems to solve to K but why? It is an "incident" light scenario - shouldn't it relate to C?

2. Originally Posted by Bill Burk
Talking about the arithmetic speed where the first way is sensitometric testing and the 0.1 speed point... "In the second way, Sa is determined by a photographic test, and is equated to the reciprocal of the effective exposure time when a satisfactory photograph is taken with a camera set at f/16, on a clear sunny day with a solar altitude of about 40 degrees."
...
"It is evident that the two methods of film speed assessment determine in effect the ratio..."

"k1 = Hg/HM"
You could even say that the sensitometric speed point is about the characteristics of the film curve (lower gradient limits with B&W negative) and is represented by Hm, and the photographic testing (think psychophysical testing) sets the value of the speed constant.

So...

It almost seems like you can say "The arithmetic film speeds were setup that way to keep Sunny 16 valid."

Then perhaps indirectly since by this time there is a tie-in to the sensitometrically determined speed point...

(Another thread) "The metering point was chosen to align the average scene to agree with Sunny 16 results"

And what I kind of find funny: The ratio for Sunny 16 seems to solve to K but why? It is an "incident" light scenario - shouldn't it relate to C?
You're presenting some interesting questions here. You should flesh them out.

3. Originally Posted by Bill Burk
And what I kind of find funny: The ratio for Sunny 16 seems to solve to K but why? It is an "incident" light scenario - shouldn't it relate to C?
Bill, I've been hoping you expand on this point because it looks like you see something. I'm not sure exactly what. Sunny 16 is about the the average illuminance which has to do with C, and K has to do with the average luminance which is a product of the average illuminance. They are interconnected.

4. Originally Posted by Stephen Benskin
Bill, I've been hoping you expand on this point because it looks like you see something. I'm not sure exactly what. Sunny 16 is about the the average illuminance which has to do with C, and K has to do with the average luminance which is a product of the average illuminance. They are interconnected.
OK it makes sense that Sunny 16 should describe average illumination which would be a constant related to the incident meter, and the incident meter constant is C.

And that amount of light falls on an average scene, which has a reflectance that might be 12% or 18% or somewhere in that vicinity. When that average % of light comes back it relates to the constant K.

Is it then oversimplifying to say that from the 0.1 speed point K is the distance to the average reflected scene. And C is the distance to the average sunlight falling upon it?

5. Originally Posted by Bill Burk
Is it then oversimplifying to say that from the 0.1 speed point K is the distance to the average reflected scene. And C is the distance to the average sunlight falling upon it?
The speed point really doesn't factor into the exposure meter constants, and I'm not sure what "distance" means in this context. I can see that you are working out how everything is connected and it all does fit. It will eventually all click together if you keep playing with it.

Both K and C are constants that factor in the elements of the camera's optical system and elements of the exposure meter. K and the exposure equation's constant, q, are closely related and have many of the same variables. The only difference is that q doesn't contain the variables for the exposure meter.

The excerpt below is Appendix C from the 1971 ANSI Exposure Meter standard. You should be familiar with the values and equations.

6. I'm thinking this needs a new graphic diagram.

A diagram could be used to illustrate the 18% and 12% controversy.

Another diagram could show the metered points and K and C. I can't always follow along with math formulae unless they are explained step by step. And I think with diagrams, the concepts could be revealed even more clearly

7. Bill, I'm not sure how to do a diagram to illustrate K and C. I might be able to go step-by-step though. As for 12% and 18%, I think I might have a way of looking at it.

The first graph is keyed to 12% reflectance. You can see that the relationship between the speed point and the metered exposure is 10x (1.0 logs). 18% reflectance falls 0.15 log-H further up the curve.

The second graph is keyed to 18% reflectance. The shadow exposure, including flare, now falls 1/2 stop further down on the curve.

8. The 18% graph is somewhat of a cheat. It supposes the same luminance distribution as the 12% model. Of course, this would lead to a different film speed / metered exposure ratio between the two. But what if 18% was middle grey and had the same luminance distribution as the 12% model? The 18% curve would look identical to the 12% one above. One big difference. The math doesn't work for an 18% grey model.

9. I have another theory about how the 18% gray mystique came about.

Kodak make the card to use in print press calibration, four color process, to reveal when a press ink balance is correct (neutral gray). If you make printing ink too dark, it will be harder to judge neutral. So a lighter tone was selected for the purpose of showing color balance.

Then when the idea to use a gray card as an average subject "proxy" - like using the palm of your hand - the idea was given out as "use a known shade of gray, like this 18% gray card". Well the mistake that point forward is that the known shade was not adjusted into the exposure calculation to place it at the known place.

Everybody knew using the palm of your hand was brighter, so you always open up. Probably because the first time you forget to open up your picture is far from expected.

But the gray card was pretty close, so the discrepancy didn't cause much grief.

10. What you say sounds possible. No one can know for sure where or how it all began. One of the problems with the question of 18% gray is that there multiple ways it is used. Of which two have to do directly with photographic exposure. Is it the average reflectance and are exposure meters calibrated to it?

I believe the use of 18% gray in printing and two dimensional art has it's origins with Munsell.

Munsell's steps are based on perceived equally spaced steps between black and white with the middle step being 18% gray. Kodak's Gray Card Plus has a white and a black area included. According to their information, the white has a 90% reflectance (RD 0.046), and the black has a 3% reflectance (RD 1.52). The mean is (1.52-0.046) / 2 = 0.739 Reflection Density or 18% reflectance. So 18% can easily work as the average for a two dimensional subject.

And metering off the hand? Here's a spectral reflectance graph of Caucasian skin. Looks like a high level of infrared reflectance. I guess metering off the hand is more a rule of thumb. sorry.

C and K weren't always the values they are today either.

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