Kodak grey card usage

RobC said:

Diapositivo,

Copy this chart and mark on it where you think the speed point is and where you think the exposure will be placed (regardless of what is being metered) and post your guesstimate.
Anyone can try as well and we can see how thick the mud is or if there is a concensus.
You have all the formulas being banded around so them so lets see if you can actaully use them to show what you think they will show.

Click to expand...

So far my understanding is that the spot meter, for any reflectance of the subject and for any level of illuminance, will give you an exposure that is at LogH = -1.0. That is true for ISO 100.
For ISO 200 the exposure will be -1.3 and for ISO 400 -1.6 etc.
The corresponding density, for any reflectance of the subject, for any level of illuminance AND for any ISO should be 1.0 which the density that you see plotted for LogH = -1.0.
I expect films of several makers and different kind to have, for all ISO, almost the same density at the "target grey" point, although the curves will differ when you move toward toe and shoulder.

and where do you think the speed point will be ?

markbarendt said:

That was actually the point I was trying to make back there in response to wiltw:

wiltw's statement is false. One is not inherently more contrasty than the other.

Click to expand...

Mark, I understand that what wiltw meant is that when the sensor originates the image, its response curve being perfectly linear, that will make it "flat" for human vision, or lacking contrast.
A gamma correction has to be applied to render the contrast natural, as the human vision expects it.

Film, instead, is originated, from beginning, with a logarithmic response, it is intrinsically more similar to the human vision.

"Contrast" has many meanings in photography. Sometimes is the gamma stuff, sometimes is the SBR, sometimes is the absolute distance between black point and white point (e.g. a low contrast image (high-key image of a model with pale skin, white pull-over, white trousers, and white cat on her hand) can have a high contrast (the model has some black glasses, and in the background a black vase sits in the shade: the general image is mid-key and is contained within maybe 2 EV but the contrast of the image is relatively high because whitepoint and blackpoint are very far apart).
In the example of wiltw he meant the raw image is very low contrast (true) and the contrast is raised during development (raw conversion).
That's also true for negatives as well, but raw images really are flat!

RobC said:

and where do you think the speed point will be ?

Click to expand...

You find the equations for slide film in Conrad, abovementioned, page 12.
In short, Hg = 0.91 Hm (If I get it right, Conrad calls Hm the speed point).

So as you see the speed point is at 0.14 EV less exposure than speed point.

But you don't care where speed point is with slides, you only care where Hg is, which is the mid-tone exposure the light meter will create on your slide if you follow it "blindly". And you care what "grey point" that corresponds to, which is a tone that will be recognised as a 18% reflectance grey or, if you prefer, center of zone V. (that assume standard development, and normal conditions of vision).

You want to place that white china mug in zone VII? You meter it, you obtain a value (which is always LogHg = -1.0 for ISO 100) and you open 2 EV more (LogHchina mug = -0.4 for 100 ISO).

Remember that with B&W you place your SBR starting from the shadows, at a point which is more or less near speed point, and you "unroll it" toward mid tones and high tones.

With slides if the SBR is high you peg a highlight that must be preserved and well rendered, you place it (2 EV, 2.5 EV, depends on the curve, on the detail, on the image, on the shadows that you need to salvage etc.) and then the SBR will "unroll itself" toward mid-tones and shadows.
This point where you peg your highlight has nothing to do with speed point of a slide film. The highlight is pegged in the highlights, while the speedpoint is the midtones, and very near target grey.

The characteristic curve is quite different and the speed point is calculated in a very different way with slides than with B&W.

here's a little graph for you to help you think about it. I have correctly scaled and overlaid the curve for Kodak T-Max 100 onto the fuji provia 100 curve and marked 3 points 1 of which must be the speed point for provia 100 I think. The red point is the approx speed point for T-Max 100 I think. So do you think the green or the yellow points are the speed point for provia or do you think its somewhere else?
Note: The kodak dev was D76 and the dev time kodak recommend as normal is 6 1/2 minutes so youcan guesstimate where the curves cross. The dev times box is from T-Max 100.
And note that assumming the red and green are the speed points for the two films, then the adjust from speed point would be 4EV and not 3 as I calculated. Please explain that.
And given that a kodak grey card is 2 1/2 stops less than than 100% please explain why -1.0logH is 3 1/3 stops from 100% transmittance. Have you considered that a colour in a slide does not equate via some formula to a density so that 18% grey has no relationship to slide density at all? You can only measure the slide density when you know its an 18% subject reflectance and find out what it is. You can't calculate it.
So once again you must calibrate your exposure if you want it on any specific density or colour. Density only relates to light and dark and not colour unless you are measuring using a colour desitometer which you ain't (so far). And your meter is clueless about colour. It only knows about relative luminance and not colour.

Diapositivo said:

Mark, I understand that what wiltw meant is that when the sensor originates the image its the response curve is linear which makes it "flat" for human vision, or lacking contrast.
A gamma correction has to be applied to render the contrast natural, as the human vision expects it.

Film, instead, is originated, from beginning, with a logarithmic response, it is intrinsically more similar to the human vision.

"Contrast" has many meanings in photography. Sometimes is the gamma stuff, sometimes is the SBR, sometimes is the absolute distance between black point and white point (e.g. a low contrast image (high-key image of a model with pale skin, white pull-over, white trousers, and white cat on her hand) can have a high contrast (the model has some black glasses, and in the background a black vase sits in the shade: the general image is mid-key and is contained within maybe 2 EV but the contrast of the image is relatively high because whitepoint and blackpoint are very far apart).
In the example of wiltw he meant the raw image is very low contrast (true) and the contrast is raised during development (raw conversion).
That's true for negatives, but raw images really are flat!

Click to expand...

Raw data by itself cannot make an image, it's just data.
In order to make an image from raw data you need
A) the raw data
B) a set of instructions that includes, among other things, the contrast specifications
C) software that can use A&B to make a viewable image (and is itself made up of lots of instructions on how to build the raw data into an image.)

The contrast specifications in B come from either the camera settings (default, auto, or user), software (not ISO) defaults, auto, or user settings. These settings are purely arbitrary in practice, they are fully at the whim of the software programmer and user, not inherent for the medium. This includes all the info necessary to appear normal to a human.

The graphing seen in digital software is different than a film curve and can't be compared directly, the straight line is a handle to grab for manipulation not a graphed representation of silver density.

RobC said:

here's a little graph for you to help you think about it. I have correctly scaled and overlaid the curve for Kodak T-Max 100 onto the fuji provia 100 curve and marked 3 points 1 of which must be the speed point for provia 100 I think. The red point is the approx speed point for T-Max 100 I think. So do you think the green or the yellow points are the speed point for provia or do you think its somewhere else?
Note: The kodak dev was D76 and the dev time kodak recommend as normal is 6 1/2 minutes so youcan guesstimate where the curves cross. The dev times box is from T-Max 100.
And note that assumming the red and green are the speed points for the two films, then the adjust from speed point would be 4EV and not 3 as I calculated. Please explain that.
And given that a kodak grey card is 2 1/2 stops less than than 100% please explain why -1.0logH is 3 1/3 stops from 100% transmittance. Have you considered that a colour in a slide does not equate via some formula to a density so that 18% grey has no relationship to slide density at all? You can only measure the slide density when you know its an 18% subject reflectance and find out what it is. You can't calculate it.
So once again you must calibrate your exposure if you want it on any specific density or colour. Density only relates to light and dark and not colour unless you are measuring using a colour desitometer which you ain't (so far). And your meter is clueless about colour. It only knows about relative luminance and not colour.

View attachment 153612

Click to expand...

The red point is probably right. That's a possible speed point for a B&W negative film I imagine.
As you see, for normal development you end up, for light meter recommendation, which is LogH -1.0, with a density of 1, which is where you also find slide density. That might not be a coincidence, but I cannot say I know why. Maybe LogD = 1 is something people at ISO tend to see as middle grey after all. In any case, the target grey is determined by exposure, not by density. Density is the "byproduct" of it. Obviously there is a relation between the two.

Green point is at the beginning of the shoulder and is the point where you might want to place some deep shadows, something like zone I, being around -1.3 LogH darker than target grey point. Last usable shadow (with serious texture) is probably around LogH -2.0 but one should experiment.
Yellow point is around 2 EV above grey point. That's white region. You could place there some china pottery, some porcelain etc.

I really don't follow your logic when you try to peg some light meter value as some distance from the extremes of the range, such as from 100% transmittance. Minolta tells you that your light meter gives you an exposure of H = 0.1 (i.e. LogH = -0.1 on our graph).
They are the maker of that instrument, so trust them.

The reason why they can say that is that they calibrated their instrument, through many tests, so that when it measures an 18% Lambertian cat it gives an H = 0.1 and the cat will be correctly exposed.
Actually, whatever cat (black, white) will be rendered as 18% grey, and will result in an exposure of H = 0.1, but only the 18% cat will be normally exposed. Every other cat will be underexposed (if whiter) or overexposed (if darker). The fact that the only correctly exposed cat is the 18% cat is ample demonstration that your light meter is calibrated for a particulare shade of grey cat, that is an 18% lambertian cat.

So, after many tests, they "hard-wired" in their lightmeter the relation 18% grey -> H = 0.1. (That means there is a K=14 in their equation but that's not relevant, really. What matters is that metering any subject gives on film an exposure of H = 0.1 which is the correct density to represent 18% grey).

Pure white (90% as in Kodak card) is circa 2.5 EV above middle grey (18%).
That 2.5 EV above middle grey is a region where, depending on the slide film you use, you begin losing texture. If it's china bone porcelain that you want to photograph that might be adequate. If it's a tissue, you don't want to lose texture, you would rather use 2.2, or 2.0, above middle grey. That depends on film also. Astia had a better dynamic range than possibly any material around. Velvia has the worse dynamic range around. With a "normal" slide, 2.0 is good for a white with important texture, let's say, and 2.5 is OK for a white object where texture is not important, such as a marble façade or a white painted wall. That IMHO and YMMV.

I don't know what you mean when you say please explain why -1.0logH is 3 1/3 stops from 100% transmittance. Did I ever say that? LogH -1.0 is grey point, target grey, LogHg, is where your spot meter places ANYTHING it measures (at ISO 100). Find this point on the graph. For slide film, it corresponds to LogD = 1. Memorize it. That's where all your reflected light meterings end up being placed by the light meter.

This is a very basic concept. Your reflected light meter will place what it reads at a certain mid-tone, 18%, and Minolta tells you that this 18% is at LogH = -1.0. It is always there. Your reflected light meter (be it spot or average, integrated or hand-hald) always gives you THAT POINT in the film curve. Never one tad to the left or to the right. The lightmeter doesn't know you are measuring china bone, grey cat, black velvet. It sees a certain amount of light. And a reflected light meter gives you the exposure that, on film, gives it LogH = -1.0. It's as easy as that. You always end up with 18% grey (or something that your mind, working on average like the mind of a large sample of human minds, constructs as an 18% grey).

If it's china bone, you end up with an 18% china bone (badly underexposed). If it's black velvet, you end up with an 18% black velvet (not black at all, badly overexposed). Your reflected light meter ALWAYS gives you an image where the metered area is 18% grey.

I don't know why you want to see your light meter working differently from how it does. Even when you use negative film, your light meter always gives you 18% grey. It never gives you "speed point"!


For B&W material, I might be saying some nonsense, and I do request correction if I am wrong. What I think at 01:14 at night is what follows:

If, for your black and white negative, you want to place something on your speed point, you must know where your speed point is relative to Hg and then do the math.

You derive your LogHm, which let's imagine being -2.3 (from the graph you posted). This is as you know the exposure of your speed point.
LogHg which is the exposure the light meter gives you, the target grey, is, as we know, -1.0.

In order to place this reading at your M point you must subtract LogH 1.3 or 4.3 EV from the metered indication (add EV). So if you point your lightmeter at some deep shadow where you want to preserve detail, and you read EV9, you place it exactly at EV13.3 and you will have placed it at your speed point.
If your speed point is the lower boundary of your Zone I, and you want to place your shadow at Zone I, you place it 4 EV below light meter reading, in this case at EV 13.

But that depends on how you develop, and on the film curve, and it is naturally crucial how you choose the zone to measure (not easy to deal, visually, with Zone I, I presume). Who cares, you can place zone V, or zone VI, or zone IV, of whatever thing you clearly see that must occupy a certain zone in your negative. The rest will fall accordingly.

You can measure the exact speed point of your final material (given film and development), and determine the distance there is between metered point (target grey) and speed point, if that is what you need (which would be a sensible way to operate with B&W I presume, because you have a point where to place the shadows and than you "roll over" the rest of your SBR up the curve).

I am still wondering why an 18% grey is rendered in a slide with a 0.1 density which, when projected, gives a 10% transparency. I would expect an 18% transparency.

I was thinking about it today and I came up with an hypothesis, which I propose for discussion.

When we are in sunlight we have a much brighter light than what comes out from a slide projector.
In sunlight, we might not have so dark a "black point", but our eye see, somewhere, a black point.
The white point is, though, much brighter.

When we are in a projection room the black point is still fairly dark, but the white point is not so brightly white. Black point and white point are less distant than in sunlight.

The eye adjusts, and because it adjusts, we preserve the relation between black, grey, white. We will always see middle grey at midway between blackpoint and whitepoint.

But blackpoint is black, whitepoint is - in comparison to sunlight - whitish. We have a compression of brightness range.
Just because our vision is adaptive, middle grey must fall half way between black and white. If white is moved toward grey of a certain amount, so grey must be moved toward black of a certain amount.

That might be the reason why a density of 10% is perceived, in projection, as an 18% grey.

Not that I am very much convinced myself, though.

Diapositivo said:

I am still wondering why an 18% grey is rendered in a slide with a 0.1 density which, when projected, gives a 10% transparency. I would expect an 18% transparency.

I was thinking about it today and I came up with an hypothesis, which I propose for discussion.

When we are in sunlight we have a much brighter light than what comes out from a slide projector.
In sunlight, we might not have so dark a "black point", but our eye see, somewhere, a black point.
The white point is, though, much brighter.

When we are in a projection room the black point is still fairly dark, but the white point is not so brightly white. Black point and white point are less distant than in sunlight.

The eye adjusts, and because it adjusts, we preserve the relation between black, grey, white. We will always see middle grey at midway between blackpoint and whitepoint.

But blackpoint is black, whitepoint is - in comparison to sunlight - whitish. We have a compression of brightness range.
Just because our vision is adaptive, middle grey must fall half way between black and white. If white is moved toward grey of a certain amount, so grey must be moved toward black of a certain amount.

That might be the reason why a density of 10% is perceived, in projection, as an 18% grey.

Not that I am very much convinced myself, though.

Click to expand...

If the scene is purely reflective and front lit (because otherwise the 18% reflective is meaningless) then a pure white subject would have a density about 0.25 on the slide or about 55% transmission. I think lowering the 18% to 10% transmission because it's difficult to make the slide completely transparent and still retain decent contrast in that range. You would notice that at 0.25 density is the start of the toe where contrast is lowered.

Diapositivo said:

I don't know what you mean when you say please explain why -1.0logH is 3 1/3 stops from 100% transmittance. Did I ever say that? LogH -1.0 is grey point, target grey, LogHg, is where your spot meter places ANYTHING it measures (at ISO 100). Find this point on the graph. For slide film, it corresponds to LogD = 1. Memorize it. That's where all your reflected light meterings end up being placed by the light meter.

Click to expand...

LogH -1.0 is 3 1/3 stops from LogH 0 which is what I told you all the way back at post 215 but becasue you didn't read it we're now at post 361

But you haven't seen the problem yet. LogH 0 can not be the speed point becasue its a different exposure than for B&W film and how would you meter know to use a different exposure for slide film. It doesn't. The green point must be the speed point. And if green point is speed for slide film the B*S/K which gives a 3EV adjsutment from speed to place exposure at -1.0 can't be right because the adjustment from speed point to -1.0 is 4EV on the graph (just for thats a 1.2 LogH shift which is 4 stops. So once again. and I remind you its you bringing all the formula kiddie stuff into the argument with the aid of benskin, where does does the 4 stop shift from speed point to -1.0 come from? I can only account for 3 stops with the lightmeter formula.
And the meter can't be shifting exposure one way for B&W film and another way for slide film so they must both being adjusted by the formula in the same direction and that direction increases exposure by reducing EV and your supposition can't be right unless you can account for the extra stop.Why 4 stops shift when the formula calculates it as 3 stops shift to put it on middle of curve. And the density at -1.0 logH is wrong for B+W film which is more likely to be 0.7logD.

RobC said:

LogH -1.0 is 3 1/3 stops from LogH 0 which is what I told you all the way back at post 215 but becasue you didn't read it we're now at post 361

But you haven't seen the problem yet. LogH 0 can not be the speed point becasue its a different exposure than for B&W film and how would you meter know to use a different exposure for slide film. It doesn't. The green point must be the speed point. And if green point is speed for slide film the B*S/K which gives a 3EV adjsutment from speed to place exposure at -1.0 can't be right because the adjustment from speed point to -1.0 is 4EV on the graph (just for thats a 1.2 LogH shift which is 4 stops. So once again. and I remind you its you bringing all the formula kiddie stuff into the argument with the aid of benskin, where does does the 4 stop shift from speed point to -1.0 come from? I can only account for 3 stops with the lightmeter formula.
And the meter can't be shifting exposure one way for B&W film and another way for slide film so they must both being adjusted by the formula in the same direction and that direction increases exposure by reducing EV and your supposition can't be right unless you can account for the extra stop.Why 4 stops shift when the formula calculates it as 3 stops shift to put it on middle of curve. And the density at -1.0 logH is wrong for B+W film which is more likely to be 0.7logD.

Click to expand...

The way b&w and reversal color both work with the same exposure meter is by using the speed constant and knowing the relationship between the speed point and Hg. Like I've written multiple times.

Fabrizio said he wasn't that familiar with B&W. The ratio between Hm and Hg is 1.0 or 10 times or 3 1/3 stops. The average shadow falls approximate one stop below that or 4 1/3 stops (1.30) from Hg. I posted a graph that illustrates how the average scene falls, as well as shown the math, as well as made papers available that explain it.

Stephen Benskin said:

View attachment 153627

Click to expand...

Very nice chart!
Is that also from Dunn?

Chan Tran said:

If the scene is purely reflective and front lit (because otherwise the 18% reflective is meaningless) then a pure white subject would have a density about 0.25 on the slide or about 55% transmission. I think lowering the 18% to 10% transmission because it's difficult to make the slide completely transparent and still retain decent contrast in that range. You would notice that at 0.25 density is the start of the toe where contrast is lowered.

Click to expand...

Can you explain what you mean for "purely reflective and front lit"?
If we exclude mirrors and polished surfaces or metallic surfaces, photographic subjects have a surface that we can approximate - if I am right - with a Lambertian surface, i.e. a surface which reflects lights in all directions in the same way.
For instance, when the sun is shining (or is not shining) on the snow, or on the grass, we see the snow, or the grass, with the same brighness even when we change our angle of vision.
A cat appears to us of the same tone when we make a round around him.

Diapositivo said:

Very nice chart!
Is that also from Dunn?

Click to expand...

It is. Thought you might like it.

RobC said:

LogH -1.0 is 3 1/3 stops from LogH 0 which is what I told you all the way back at post 215 but becasue you didn't read it we're now at post 361

But you haven't seen the problem yet. LogH 0 can not be the speed point becasue its a different exposure than for B&W film and how would you meter know to use a different exposure for slide film. It doesn't. The green point must be the speed point. And if green point is speed for slide film the B*S/K which gives a 3EV adjsutment from speed to place exposure at -1.0 can't be right because the adjustment from speed point to -1.0 is 4EV on the graph (just for thats a 1.2 LogH shift which is 4 stops. So once again. and I remind you its you bringing all the formula kiddie stuff into the argument with the aid of benskin, where does does the 4 stop shift from speed point to -1.0 come from? I can only account for 3 stops with the lightmeter formula.
And the meter can't be shifting exposure one way for B&W film and another way for slide film so they must both being adjusted by the formula in the same direction and that direction increases exposure by reducing EV and your supposition can't be right unless you can account for the extra stop.Why 4 stops shift when the formula calculates it as 3 stops shift to put it on middle of curve. And the density at -1.0 logH is wrong for B+W film which is more likely to be 0.7logD.

Click to expand...

It is not clear at all what you want to convey, frankly. The meter doesn't know where the speed point is. The meter only knows where the target grey point would be if you follow its advice. Sounds like a dialogue between auditively challenged persons!

Let's make a real-life example. It's a glorious afternoon of a summer day and you are on a meadow with some sparse oak trees with a large umbrella of foliage. There are a black horse and a white cow pasturing in the sun. There is a white shepherd dog quietly sitting under the shade of one tree.

The sun is fairly high and is 45° behind your right shoulder. There are almost no clouds in the sky.

Your spotmeter, set for ISO 100, pointed at the following various elements describes this scene:

grass in the sun: EV 14.6;
black horse in the sun: EV 12.3;
white cow in the sun: EV 16.6;
foliage on top of the tree: EV 14.6;
grass in the shade of the tree: EV 11.3;
trunk of the tree in its darkest point: EV 9.3;
the white dog in the shade is EV 13.5.

You want the depict this scene with the slide film of which you posted the chart with the green dot, ISO 100.
You want the scene to appear in natural tones: the black horse to be recognised as a black horse, the white cow as a white cow, the grass as a natural tone of grass.

How do you expose such a scene? And, especially, why? What is the "algorithm" leading to your choice?

What if, instead of a cow at EV 16.6, you have some white geese at EV 17.0?

markbarendt said:

Raw data by itself cannot make an image, it's just data.
In order to make an image from raw data you need
A) the raw data
B) a set of instructions that includes, among other things, the contrast specifications
C) software that can use A&B to make a viewable image (and is itself made up of lots of instructions on how to build the raw data into an image.)

The contrast specifications in B come from either the camera settings (default, auto, or user), software (not ISO) defaults, auto, or user settings. These settings are purely arbitrary in practice, they are fully at the whim of the software programmer and user, not inherent for the medium. This includes all the info necessary to appear normal to a human.

The graphing seen in digital software is different than a film curve and can't be compared directly, the straight line is a handle to grab for manipulation not a graphed representation of silver density.

Click to expand...

Mark, it is my understanding that in the raw you already have a meaning of "contrast", which is the difference between the darkest point and the brightest point, let's call it "total contrast" or "SBR" (it's not, it's more like Dmin - Dmax difference, the difference between the darkest pixel and the brightest pixel). This is a given of the image after capture.

But the contrast curve is very flat. You have to apply an S-curve correction in order to arrive to an image which distributes the contrast in a way that human vsion sees as normal contrast, i.e. that has "midtone contrast" like we perceive it. That' a "redistribution of total contrast", the total contrast always remains the same, it's only the "midtone contrast" that change.

The contrast settings (high contrast, low contrast, default contrast) correspond to different S-curve corrections applied when you make the raw transformation into a final image (an in-camera JPEG). The setting "high contrast" will yield an image with higher midtone contrast than the "default" or "low contrast" settings.

The idea that a contrast curve is only a redistribution of total contrast in order to increase or decrease midtone contrast is explained here:
http://www.cambridgeincolour.com/tutorials/photoshop-curves.htm

You can read it with interest because it is something that exactly applies to analogue photography. I never printed but I assume the different gradations of paper do just that, apply different S-curves to the negative.

Diapositive said:

Mark, it is my understanding that in the raw you already have a meaning of "contrast",

Click to expand...

No.
The tone reproduction information is kept separate from the raw data, separate folders in the same file box. Software programs like PS or LR can use the internal instructions that the camera wrote (as shot) or external instruction sets applied in our computers. In either situation the defaults are very reasonable for human viewing. Pick up any new digital camera, point and shoot, and you'll get a workable result.

Any missing piece of information from the instruction set results in a corrupt file which means getting no image or an obvious failure.

https://en.wikipedia.org/wiki/Raw_image_format see processing.

2Ev-3

markbarendt said:

No.
The tone reproduction information is kept separate from the raw data, separate folders in the same file box. Software programs like PS or LR can use the internal instructions that the camera wrote (as shot) or external instruction sets applied in our computers. In either situation the defaults are very reasonable for human viewing. Pick up any new digital camera, point and shoot, and you'll get a workable result.

Any missing piece of information from the instruction set results in a corrupt file which means getting no image or an obvious failure.

https://en.wikipedia.org/wiki/Raw_image_format see processing.

Click to expand...

Mark, I don't think you got the digital technology right. You probably don't nurture a great interest for it

Ultimately an image file is a set of pixel values. All "sidecar information" that a raw file can contain (there are many) is just a sidecar information, containing e.g. the EV measured at the time, the shutter release, the aperture, the ISO setting, and the S-curve used to develop the JPEG preview which is often incorporated in the raw file, the name of the photographer, the GPS coordinates and whatever else. The raw file, as you say, has lots of meta-information about the image. The raw data is the raw image.

Of course the raw data in itself doesn't give you a workable image, just like an undeveloped and un-printed negative doesn't give you a print. You have to pass it through a chemical process (which is variable), an enlarger, you have do develop some paper etc. The "set of instructions" that you have to use in order to develop a raw file belongs to the raw development process and to the format of the raw file. Whatever information is in the raw file, however you consider it organized internally, is the raw image.

If it was possible to store a "contrast information" independently from the "pixel information", which I don't think it is possible, then that contrast information would belong to the raw information in any case. The raw image has got all what you need in order to have a final image once you "develop it" through a well-known process.

That said, the resulting image has a midtone contrast that result from the curves you apply to it, which are somehow discretionary, just like discretionary is your choice of development, paper etc.

Stephen Benskin said:

Fabrizio said he wasn't that familiar with B&W. The ratio between Hm and Hg is 1.0 or 10 times or 3 1/3 stops. The average shadow falls approximate one stop below that or 4 1/3 stops (1.30) from Hg. I posted a graph that illustrates how the average scene falls, as well as shown the math, as well as made papers available that explain it.

Click to expand...

Now I see. I thought the speed point for B&§W film was the darkest useable part of the film. I see from the graph at page 9, #205, that one can place a shadow point to the left of the speed point.

So the speed point is always 3.3 EV lower exposure that middle grey point for negative film.
Considering that LogHg always is -1.0, LogHm should always be -2.0, or 3 and 1/3 EV below it.

In the graphic by Dunn I can't interpret the X axis, going from 0 to 2.2 and market as Relative Log exposure.
I am used to see a LogH lux-seconds scale with -1.0 corresponding to "statistical average scene and equivalent grey calibration surface for photoelectric reflected-light meters", correspondig because Minolta places that grey at LogH = -1.0.

Diapositivo said:

Now I see. I thought the speed point for B&§W film was the darkest useable part of the film. I see from the graph at page 9, #205, that one can place a shadow point to the left of the speed point.

So the speed point is always 3.3 EV lower exposure that middle grey point for negative film.
Considering that LogHg always is -1.0, LogHm should always be -2.0, or 3 and 1/3 EV below it.

In the graphic by Dunn I can't interpret the X axis, going from 0 to 2.2 and market as Relative Log exposure.
I am used to see a LogH lux-seconds scale with -1.0 corresponding to "statistical average scene and equivalent grey calibration surface for photoelectric reflected-light meters", correspondig because Minolta places that grey at LogH = -1.0.

Click to expand...

10/Hm = Reversal Speed
10/100 = 0.10 lxs

Relative log helps when determining the scene luminance range. 2.20 is 7 1/3 stops.

if it makes it any easier and less agricultural for you,

0.10 lxs = EV -4.6438

Diapositivo said:

Mark, I don't think you got the digital technology right. You probably don't nurture a great interest for it

Ultimately an image file is a set of pixel values. All "sidecar information" that a raw file can contain (there are many) is just a sidecar information, containing e.g. the EV measured at the time, the shutter release, the aperture, the ISO setting, and the S-curve used to develop the JPEG preview which is often incorporated in the raw file, the name of the photographer, the GPS coordinates and whatever else. The raw file, as you say, has lots of meta-information about the image. The raw data is the raw image.

Of course the raw data in itself doesn't give you a workable image, just like an undeveloped and un-printed negative doesn't give you a print. You have to pass it through a chemical process (which is variable), an enlarger, you have do develop some paper etc. The "set of instructions" that you have to use in order to develop a raw file belongs to the raw development process and to the format of the raw file. Whatever information is in the raw file, however you consider it organized internally, is the raw image.

If it was possible to store a "contrast information" independently from the "pixel information", which I don't think it is possible, then that contrast information would belong to the raw information in any case. The raw image has got all what you need in order to have a final image once you "develop it" through a well-known process.

That said, the resulting image has a midtone contrast that result from the curves you apply to it, which are somehow discretionary, just like discretionary is your choice of development, paper etc.

Click to expand...

Created a side bar in the SoapBox for the response to get the digital talk into an appropriate space.

(there was a url link here which no longer exists)

RobC said:

if it makes it any easier and less agricultural for you,

0.10 lxs = EV -4.6438

Click to expand...

0.10 lxs is the exposure at the film plane (in the camera). There are no Evs at the film plane.