Steven, I don't think I can go along with this part, "print tones need to be lighter..." I don't have that edition, so can't confirm what it says, but I really think you may be misreading something.
Originally Posted by Stephen Benskin
Regarding the two tone-reproduction charts you attached, I don't believe that either of those responses can make a successful general purpose print (but they may be ok for copy work). My reasoning: if you look at reproduction of white, the "perfect white" in the subject (a hypothetical 100% diffuse reflector with equivalent reflection density = 0) is being mapped onto paper white (density ~ 0.10, the lightest photo papers get).
What this means is that there is no headroom for specular reflection. I think it is pretty crucial to allow specular objects to print lighter than a "pure white" in the scene. If you don't do this, shiny silver jewelry doesn't look like silver anymore, flesh highlights don't look shiny, etc. Everything has the look of being "clipped" to white rather than being shiny.
When you redo the tonal reproduction to allow for some specularity (on your upper right quadrant, map the scene beyond density = 0 to at least -0.20 and preferably to -0.30 (equiv to 1 f-stop brighter than "pure white" in the scene); this gets mapped to the white paper base.) When you do this, unfortunately everything else on the print has to become darker; the result is that almost all of the tones on the paper end up having a higher density than the equivalent subject density.
I don't know if I'm being clear on this. Anyway, this is not just academic, it's also from experience using studio-type portrait work. Anyone who has a high-quality print including a Macbeth Colorchecker (or other gray scale) and a densitometer can cross check; I think you'll find that the 2nd and 3rd lightest patches are similar (perhaps print is even a touch lighter), but going darker the print will have progressively higher densities.
Thanks Stephan. In the 4th edition chapter 19 Tone and Color Reproduction page 554.
Mark Barendt, Ignacio, CO
"We do not see things the way they are. We see things the way we are." Anaïs Nin
There’s a lot not covered. The quote in post #45, is only the basic conclusion. A good section of the chapter goes on to support the statement. It even discusses how the eyes response to different levels of illuminance and different luminance ranges.
Now, here’s the example of a referred print tone reproduction curve from The Theory of the Photographic Process.
Notice the luminance range of 2.20. This is considered the statistically average luminance range. It also doesn’t include specular highlights and accent or cavity blacks. More on this later.
From the text’s discussion of the above example, “The tone-reproduction curve in Figure 22.2 is an average of the curves obtained for the first-choice prints in the experiments of Jones and Nelson, and Jones and Condit, who made and studied several thousand black-and-white prints of approximately 170 outdoor scenes. Fifteen to twenty prints differing in contrast, density, and tone reproduction curve shape were produced for each scene. The prints were viewed with an illumination of 100 footcandles and judged for subjective quality by a number of observers.”
“For scenes having unusually long log luminance ranges (2.5 – 3.0), the tone reproduction curves for the first choice prints were slightly to the left of the curve shown in Figure 22.1 (above). For scenes having unusually short log luminance ranges (1.5 – 1.9), the corresponding curves were slightly to the right of this curve. Scenes of normal log luminance ranges (2.0 – 2.4) gave curves that were essentially identical with the curve shown in the figure.”
Slightly later in the chapter, “Consequently, the characteristics of the human eye rather than those of the photographic paper appear responsible for the necessity of making reflection-type black and white prints about 0.25 less dense than the hypothetical print, represented by a 45 degree line, that would reproduce the luminance ratios or log luminance differences of the original scene…The eye is apparently able to compensate adequately when the luminance differences are about one-fiftieth those of the original scene, and thus make the photograph look like a sunlit scene, but not when the luminance differences are one-hundredth of those of the original scene. The slope of 1.15 required in the middletone region of the curve is also probably related to the characteristics of the eye. The reproduction gradient in the highlight region is sacrificed in order to fulfill the need for a middletone gradient greater than 1.0.”
I hope it's becoming apparent why I constantly encourage people to understand theory. Chapter 20 in The Theory of the Photographic Process 3rd edition is entitled The interpretation of Sensitometric Results. After all, what good are curves if you don't know how to read them?
Last edited by Stephen Benskin; 03-07-2013 at 07:49 PM. Click to view previous post history.
Apologies, I should have said "Stephen."
Originally Posted by Stephen Benskin
I don't think you addressed this point, which is important (IMO). I suspect that you concluded this from an inappropriate use of the reference lines used in various graphs. If you want them to represent actual density equivalents the line would have to pass through a point of 100% diffuse reflectance (pure white) on both axes. For the printing paper, this means a density value of zero, not paper-white. Most of the graphs seem to pass the reference line through paper-white, with a density value ~ 0.10.
If you were to shift these reference lines (use a straight edge, hold it parallel to the existing reference, then shift it over to cross zero density) you'll probably find that the print density values are essentially NOT lighter than original subject density equivalents.
Here's an example of how those reference lines can mislead: if you were to use the existing reference lines, which pass through paper-white, you might photograph a gray card (measured density ~ 0.74) and print it to produce the same density (~0.74), but someone else concludes that the print is lighter than the original because the reference line is 0.10 units higher (~ 0.84 density).
As I mentioned before, a print should be able to accomodate some specular reflections, so a scene white should be printed a slight amount darker than paper-white. When this is done, everything on the print gets a bit darker. Here's an example of such a graph, showing clearly that nearly all print densities are equal to or higher than the original scene density equivalents. It's from Digital Color Managment - Encoding Solutions (Giorgianni, 1998).
In this graph, the reference line should be a legitimate comparison for density. The section where the two lines run together is roughly where flesh tones of a light-skinned person would be. To the right of that, the curve is getting flat; this gives specular highlights some "separation", albeit compressed. (Without this compression, the rest of the print would get much too dark.) My most experience is in portrait work, where I would say this is a very good general-purpose tonal response (my work experience confirms this.) But I don't know how well it would hold up for landscapes, and the like.
(Gosh, it sure takes a long time for such a small amount of writing.)
I pretty much agree with everything else being said.
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We have computers nowadays, and it's easy to fit a cubic spline or somesuch to the toe-area to get accurate measurements of its gradients. So I'd like to compute speeds based on the fractional gradient method directly.
I'm familiar with the fractional gradient method of Gmin = 0.30*Gbar(1.5), where Gbar is the mean gradient over a log-E range of 1.5.
But I question that 1.5. Page 440 of The Theory of the Photographic Process (3rd ed, Mees-James) states:
The scene chosen was an average one giving an image illuminance range on the negative material of 32, which is very nearly the statistical average of a large number of scenes when photographed with a camera system having average flare characteristics.
Note the 32. Log-base-2 of 32 is 5, so this is a 5-stop range on the negative. 5*.3 = 1.5, so this is the source of the log-E range of 1.5.
Note the "average flare". That was for cameras made in 1935-1945, which had uncoated lenses. It seems to me that modern multicoated lenses have less flare, and thus the log-E range of 1.5 should be higher nowadays.
Stephen has said that after accounting for flare, the average range is about 1.8. So if we were to use the fractional gradient method with modern cameras, should we change the formula to this?: Gmin = 0.30*Gbar(1.8)
If so, do you think the 0.30 would also change?
Nelson touched on this in Safety Factors in Camera Exposure. Lens coating shifts it back 0.05 log-H units. This simply introduces a small safety factor. In fact, I believe this might be what is at least partially responsible for the ASA / ANSI standard's note in the forward about a built in 1/3 stop safety factor.
Originally Posted by albada
From Safety Factors, "As point out by Jones and Condit, however, the shadow point should coincide with the speed point only when the flare factor is 4. When the flare factor is 2.5. the deepest shadow can be placed about 0.05 to the left of the speed point because a lower slope on the toe of the curve becomes usable when the shadow contrast in the camera image is increased by the reduction in camera flare. Consequently, the "first-excellent" point in Fig 3 is considered to lie 0.05 to the left of the speed point. The "first excellent" point, therefore, lies 0.37 in log-H units to the left of the shadow point, c, representing the exposure obtained from the use of the exposure meter and the ASA exposure index. The interval of 0.37 is the logarithm of the safety factor. These calculations, therefore, lead to the conclusion that the safety factor is 2.35" (for the fractional gradient method)
So basically, yes I believe the factional gradient value could be lower with coated lenses. You can also see that 1.80 was being used when considering the change from fractional gradient to the fixed density / Delta-X method.
Last edited by Stephen Benskin; 03-08-2013 at 06:27 PM. Click to view previous post history.
May I go back to the Delta X Criteion paper. Being confused as ever, reading and re-reading. (Having a headache)
I have now various curve for various films. To use the DeltaX I take the curve where the ISO triangle matches most closely and basically subtract the Delta X value. Being 0.296
Next I take the rest of the curve measure using the same ISO triangle length 1.30 log-H but with the higher Delta which touches the curve to find my Delta X Criterion from the table.
This will then show all speed point for the different curve.
The various ΔX values are based on the value of ΔD. There's a table with the values. Say you have a ΔD of 0.86, the ΔX value is then 0.268. The fractional gradient point is 0.268 log-H units to the left of the fixed density of 0.10 over Fb+f. Take the log-H at that point and plug it into the equation 0.4 / Hfg.
Originally Posted by AndreasT
I thought something looked familiar. But "I'd like to hold off judgment on a thing like that until all the facts are in." - General Buck Turgidson.