P is the important constant in this. It's the same as K1 (not to be confused with K) from the exposure meter standard. This is where Hg will fall for a given film speed. 8/100 = 0.080 for 100 speed film. HR in Connelly and currently Hm in the current standard is determined. This as well as HF and HS can be known points of density. The speed equation will produce a film speed value that will want to put the exposure Hg 1/3 stop to the left of HR.
Do you refer to equation [15], page 190, Connolly? As it is, it poses H
R = H
g.
I don't think you do, because you tell me H
g is 1/3 EV to the left of H
R.
Connolly proceeds by saying: "It is to be noted that film manufacturers are re-assessing...and may require a change...". Do you mean that they have actually reassessed and the change was granted?
That would lead us to the difference of 1/3 EV between H
g and H
R.
I suppose you are referrring to this.
Is it correct to state that, in this case, we have a K = 11.6? (It is equation (16) immediately below, the scan is a bit confused).
This paper only deals with reflected exposure meters, and there are two types, integrated and spot. For spot, the Hg will be from anything you point at. For an integrated meter for a statistically average scene and basically the calibration Luminance, it's 12%. This comes from the calibration Luminance value as compared to the calibration Illuminance value. I don't use incident meter and haven't spent nearly as much time on them. I've only read The Incident Light Method of Exposure Determination maybe two times and that was over 10 years ago (although I'm going through it now).
As far as I understand, spot and average reflected light meters work exactly the same. Spot, pointed at anything, will give me a value H
g which will give me a certain shade of grey whichever the brightness of the object. There is only one shade of grey for which H
g is the correct exposure, factory determined "target grey". If the object I point the spot meter at is whiter than target grey, the slide will be underexposed (and make the object appear target grey). If the object I point the spot meter at is darker, the exposure will be overexposed (and appear target grey).
An integrated light meter has the advantage of being able to calculate the light transmission exactly, but the "average" scene that it sees (the determined "target grey", that is, supposing target grey is an average of the world reflectance) is the result of the same kind of arbitrary choice that the exposure maker makes with a spot light meter. Basically the light meter wants to "guess" the reflectance of the target, and the guess that minimizes exposure mistakes is the average.
There are two averages here: the average reflectance of the world, and the average reflectance in the scene. If they coincide, the exposure is correct. If the scene is let's say high-key, no cigar.
I quote Ralph Lambrecht in another thread: "
it is my understanding that the 18% reflection value is based on a Kodak field study, conducted by Condit in the 1940s, where he took readings of over 100 scenes around Rochester, determining an 18% reflection average and a 7 1/3 avg SBR;". I have read something similar as well. I would be very surprised if human vision placed "middle" tone in a place that doesn't correspond with "middle" tone in the general world. And I would be generally surprised if "target grey" for separate spot light meters was different from "statistically average scene" for integrated light meters. That would cause a lot of head scratching and, frankly, would make no sense from a practical point of view.
In my view either all reflected light meters of the same maker are calibrated around 12%, or they are all calibrated around 18%, or whatever same shade of grey (be it "on a spot" or "average in the scene").
One question you need to ask is how are the values perceived on the transparency. 18% and 12% are only half a stop difference, then there's how the transparency is viewed. You might find this interesting and I think the chart will answer some of your questions. The attached is from Jack Holm's paper Exposure Speed Relations and Tone Reproduction.
That's very interesting, thanks very much.
That also leaves me perplexed.
I understand that there must be some "compression" of tones when printing a slide on paper. That will be dealt by the printer but I do understand one might "optimize" exposure in advance for the printing process, just like one might optimize it, let's say, for the scanning process.
What raises my surprise is the difference there is between the theoretical exposure for middle grey and the suggested exposure, when optimization is for the projection on a screen.
The more I study this, the more I get confused, really.
Half a stop is the difference between a properly exposed white marble façade and a burned white marble façade. To me, whether the meter is calibrated to 12% or 18% makes a world of difference. And I have always considered it calibrated to 18% obtaining, with my Minolta spot meter, a very precise and expected placement of high lights (and when I went one day with my Gossen Sixtino, a very defective and useless piece of equipment, to take pictures I basically spoiled several shots of Villa Pamphili because the walls came out washed up, lacking in detail in a very unnice way).
EDIT: considering that burning the highlights is what spoils a slide, it might be that the table suggests, basically, an ample "margin of safety", a marked underexposure of slides, counting on the adjustment of human vision after the first slide. But they must all be underexposed in the same way, or it wouldn't work.
And besides, a nocturne picture with this underexposure would become sad to look at, considering the amount of scene that would disappear into D
max like into a black hole.
