camera scanning b/w negatives - prints are much too flat

[Ed_Ingold]

ust because a color space and document are encoded as 8-bits per color, that absolutely doesn't mean there are 256 steps there. As the POSTERIZE examples prove!

The histogram abscissa represents 8 bits of data. There are precisely 256 increments. In its simplest form, the ordinate of the histogram represents the number of pixels at each density value (0 through 255). The ordinate is usually normalized so that the value is the percent of the total number of pixels, in order to make an informed comparison regardless of the resolution of the camera. The basis for normalization for a particular image doesn't change when you manipulate the image. If the image is resampled, the basis and all data points remain proportional to the initial resolution.

 

If you were to add the heights of each bar, you would reach the same number before and after truncating the curve, regardless of any gaps which ensue. What has changed is the density each data point represents, and the fact that some intermediate values no longer exist. The existence of a gap means that at very least, the density increment is twice what it was before adjustment. That's still a pretty small ratio, nearly invisible unless you are viewing a large area of neutral density with a very low gradient, such as blue sky.

[digitaldog]

The histogram abscissa represents 8 bits of data. There are precisely 256 increments.

There can be and there are often not 256 increments. I've provided a means you can see ONE increment. Or 12, or 64 or anything between 1 and 256. And when you see combs within, there isn't 256. Missing data; data loss. Again, the Posterize command will make this clear to you once you actually try it! Anything else you state is a dismissal of these facts as explained to the OP who asked why he's got 'vertical lines" in his Histogram. Some of us know how to produce edits such, an 8-bit per color Histogram seen after editing has no such vertical lines, no data loss on this encoding.

[Ed_Ingold]

Missing is not equivalent to lost in this instance. The gaps occur because existing data has been redistributed. The histogram always has 256 values in the abcissa, but there may be fewer than 256 discrete density values in the when the data is truncated and expanded to spread over the entire histogram. Think of them as "stretch marks."

[digitaldog]

Missing is not equivalent to lost in this instance. The gaps occur because existing data has been redistributed. The histogram always has 256 values in the abcissa, but there may be fewer than 256 discrete density values in the when the data is truncated and expanded to spread over the entire histogram. Think of them as "stretch marks."

And the redistributed data is exactly what causes this data loss that degrades the image. But those experts who've been vetted make the facts quite clear if you can understand the bits below (no pun intended) about Tonal Expression and Tonal Expansion; and their EFFECT on the data. Something not to worry about with high bit edits.

http://digitaldog.net/files/TonalExpansion.pdf

 

http://digitaldog.net/files/TonalExpansion.pdf

Edited December 9, 2019 by digitaldog

[paddler4]

I think the two of you actually agree. As Ed wrote, redistributing data does not cause data loss; it simply changes the integer values of the pixels that have been changed. Rodney wrote precisely the same thing, in different words, in the link in post 54. This is on p. 28:

 

A different type of image degradation occurs when you expand a tonal range. You don't lose any data [emphasis added], but you stretch the data that's there over a broader tonal range, and hence run the danger of losing the illusion of a continuous gradation.

[digitaldog]

I think the two of you actually agree. As Ed wrote, redistributing data does not cause data loss; it simply changes the integer values of the pixels that have been changed. Rodney wrote precisely the same thing, in different words, in the link in post 54. This is on p. 28:

Perhaps, perhaps semantics. Here's what I know from testing: applying identical edits on a high bit vs. 8-bit per color document, the edits on the 8-bit per color document produces 3408 fewer unique device values. If that isn't data loss, something else is getting lost.

[Ed_Ingold]

Data is lost if it cannot be unilaterally restored to its original state. To that extent, information below the new black point and above the new white point are lost because they are lumped into buckets at the new endpoints without regard to their previous values (density, color). Between these boundaries the data is stretched, opening gaps between density values. That is because the density values are integers and cannot assume intermediate values. If you multiply every density value by two, for example, every other value becomes blank. This transformation is largely reversible. If you collapse the results, you restore the original contour and appearance with minimal interpolation.

 

Unfortunately there's not a convenient tool to do that in Photoshop, but it's well within the capability of numerical analysis. For simplicity, the OP's image is B&W, and can be regarded as a having single channel.

[digitaldog]

Data is lost if it cannot be unilaterally restored to its original state.

Then what I saw and reported is some kind of data loss. Unless you can instruct me how to reconstruct 3408 unique colors (device colors) from the gradient short of rebuilding it which isn't a fix (or the original edit would be required again). Again, the two edits that were required (only based on the mapping of the Histogram) were identical. The only difference was the bit depth upon which the edits were made. One, the 8-bit per color document has less unique color values. They got lost somewhere. They didn't when the edit was applied in high bit.

Now what would be a useful next step is attempt to define the deltaE differences between the two images. That would tell us how visually different they are. Not easy because in order to do so, they have to have the identical number of Lab values to do such a comparison. And again, there is a boat load of Lab values that disappeared so this ain't an easy analysis we want to produce a colorimetric numeric report.

 

Of course this is all totally moot if and when people produce edits, either at the scan/capture state or after, on high bit data. Which is what the OP was asking about. He should 'fix' the Histogram at the scan stage OR capture as is, in high bit and 'fix' in high bit which apparently isn't possible in Elements.

[Ed_Ingold]

If you can't reconstruct it, it's lost. That said, each pixel in 8-bit RGB has 256^3 possible values (1.7 MM). It might be possible to reconstruct individual 8-bit channels, but you have lost the relative in and out points of each channel.

 

The adjustments needed to render the OP's B&W scan have been treated exhaustively. As with any useful modification, you lose a little, gain a lot in the process.

[andylynn]

Please, everybody, let it go. The question is answered.

[digitaldog]

If you can't reconstruct it, it's lost.

As with any useful modification, you lose a little, gain a lot in the process.

Indeed, seems lost in the example provided.

The OP can gain a lot and lose virtually nothing depending on when and where he applies the correction. Kind of the point of high bit editing.

Edited December 10, 2019 by digitaldog

[PaulWhiting]

I appreciate the work many of you have put in to try and help me, but I'm afraid parts of the discussion are "beyond my paygrade", as the saying goes.

 

Here's another twist to consider:

 

Briefly, my b/w prints result either from a converted color file, or from a scanned b/w negative. Whichever kind of b/w I start with, the prints are flat. But - I have two test pattern prints, one contains a variety of patterns ranging, in 51 steps, from pure black to white. The other test print is a "bullseye", half is also in 51 steps and the other half a smooth fade from black to white. The former is from Northlight Images, maybe the latter too - not sure. I'm considering mixing up a fresh batch of inks.

 

How is that prints from the two sources are flat yet the prints of my test patterns are fine?

[andylynn]

Test patterns usually come in sharp contrasty B/W where the files are set up to occupy the full histogram. Hang on a minute, I’ll see if I can set up an illustration...

[andylynn]

This is your original photo. Look at the graph on the upper right. This is the histogram. The bottom (X) axis represents all the values that could be in a JPG image, from darkest on the left to lightest on the right. The Y axis (the height at any point in the chart) is the number of pixels, relatively speaking, that have that value. See how it's all clustered in the center? That means that your image is all in the middle grays range. There is no black or white in the image, only shades of gray that are mostly in the range of 25%-75% of the available brightness.

 

 

The above shot should look flat and not contrasty. You should be able to see, by comparing the dark areas in the photo to the dark border around it (which is a dark gray but not completely black) that the image lacks strong darks, and by comparing to the light of a blank web page (just fire up a new browser window and drag it over the image so you can compare white space to your photo) that there are no strong brights. If you don't see this - if the image above looks like it has strong contrast - you have to adjust your screen.

 

Here is one possible edit:

 

 

The above image has stronger contrast. It's not meant to be ideal, but it shows how the histogram works.

 

What I did in that one was to just use the Curves function to spread your gray values over more of the histogram. The lower right hand chart is the curve for All (we're not looking at r/g/b values individually because I've simplified it by turning on Black & White). The Curves function (here in its simplest b/w form) processes the image by running the brightness of each pixel through a mathematical translation. The X axis is the input for the translation and the Y axis is the output. See how the little circle on the top (which is a control point) lines up on the X axis at about the same point as the right side of the curve in your original histogram? And on the Y axis you can see that it aligns with white. What this does is to take that spot on the histogram, which is where your brightest grays were, and turn them white. Similarly the control point on the bottom is turning your darkest grays black. Because the line between them is a straight line, all the values between them will be translated linearly (evenly spaced). It could be made into a curve, which I could use to correct the contrast in the darks and lights with more fine control, but that's more advanced than I want to get.

[Ed_Ingold]

Briefly, my b/w prints result either from a converted color file,

If scanning and printing B&W film is challenging, printing B&W from color negatives is an order of magnitude more difficult. Without special treatment, the results are flat even with properly exposed film.

 

The simplest approach in the digital age is to optimize each color separately in Levels, as shown for B&W above. Still in RGB mode, it helps to apply color filters, much as you would do to improve contrast when shooting B&W. Lightroom makes it easy, offering several B&W conversion profiles.

[PaulWhiting]

Test patterns usually come in sharp contrasty B/W where the files are set up to occupy the full histogram. /QUOTE]

 

So, in other words, the histogram's author has already done our work. The histogram is complete, it doesn't need any tweaking. Maybe I don't need fresh ink.

Edited December 19, 2019 by PaulWhiting

[PaulWhiting]

Andy,

 

Your explanation of how to tweak the histogram is very clear... thanks. I think that's what I've been doing, yet "the proor is in the print" and my prints have still been flat. Maybe Ed's suggestion to optimize each color separately would help (thanks for jumping in, Ed!).

Edited December 19, 2019 by PaulWhiting

[PaulWhiting]

Ed,

 

Actually I'm not converting color negatives to b/w (I've heard that can get real complicated), I mean converting color files from a camera to b/w. Sorry, perhaps I wasn't clear.

 

(OT, and warning for thread drift): By the way, do you have a stand-alone Lightroom, or do you subscribe, or ... ? I really need to get with the program and learn that app. )

[PaulWhiting]

Wait a minute... I haven't tried optimizing curves for the three colors yet... because that doesn't deal with flat prints from scans of b/w negatives. Keeping that in our discussion tells me I'd like to have one procedure for both scans 1) b/w negatives and 2) conversion from color digital (camera) files.

[Ed_Ingold]

I subscribe to the Creative Cloud Suite. I use Lightroom more than Photoshop, but the latter is best for precision in editing and printing. Besides PS and LR, I use InDesign, Premiere Pro and Acrobat on a daily basis.

 

Contrast and tonality of color conversions from digital images or slides are generally improved with filter emulations (channel manipulation). Lightroom has a comprehensive list of profiles for that purpose, and is completely non-destructive.

[andylynn]

But if the negative you scanned was b&w, you don’t gain from working with channels. You can take a white balance and then turn on b&w mode.

[Ed_Ingold]

But if the negative you scanned was b&w, you don’t gain from working with channels. You can take a white balance and then turn on b&w mode.

Correct, but incomplete ;) If you think it is easy to make an attractive B&W print directly from a color negative, you obviously haven't tried it.

 

Converting color to B&W is the same as B&W to print, except there are three channels instead of one. That said, most scanners have RGB sensors, which do not necessarily have the same response. RGB color can be treated as two color channels, blue and red with a third channel, luminosity. The simplest way to render B&W is to manipulate its luminosity. It is not the only way, however. Printers use CMYK colors. The black channel, used alone, is one way to render B&W, especially if there is more than one shade available. Color can be added to the mix to produce a warmer (redder) or colder (bluer) tonality.

 

This process is comparable to choice of paper in a wet darkroom. Kodabromide has a blue-black tone, often associated with news and forensic photography. By the 1960's, newspapers had shifted to Polycontrast paper, which is more flexible in the darkroom. It is also noticeably warm compared to Kodabromide. The results are comparable to using "process black," where C, M and Y are overlaid to produce black, rather than the black channel alone.

 

Readers might find it educational to make and print images in a CMYK color space, rather than let the printer driver do the conversion. Printers are a lot "smarter" now, but a decade or so ago, I found that CMYK produced much better detail and openness on both inkjet and laser printers for short runs of brochures and CD covers. You will have to use Photoshop rather than Lightroom, which does not have a CMYK option.

Edited December 20, 2019 by Ed_Ingold

[andylynn]

I’m not talking about converting color neg to bw. I’m talking about camera-scanning a bw neg. I think that’s what the OP is trying to do. If you have a bw neg that you’ve camera scanned, getting into channels doesn’t help you, because the neg doesn’t have color information to work with.

 

(But when converting a color neg to a bw image... yeah, you’re best off making a good color positive, saving a file, 16 bit per channel if you can, then doing a good bw conversion on that using your preferred process.)

[Ed_Ingold]

I’m not talking about converting color neg to bw.

Why, then, did you discuss "white balance"?

 

R, G and B (or C, M and Y) have the same density at the same exposure level, whether on a color negative or positive. In either case, using luminance for the B&W conversion without other enhancements will be flat and uninteresting.

[andylynn]

I’m not getting dragged into the mud pit of this thread. I’m just trying to help the OP.

 

You don’t disagree with me. We both understand the following: When you camera scan bw film you can use whatever method you prefer of ignoring or turning off color. When you camera scan color negs you can make it a color positive and then use whatever method you usually prefer of turning a color photo into a good bw. There are as many of them out there as there are software companies that want to sell you something.

 

But the OP wants to know why his prints look flat and I think we’ve given him as much information as is helpful about that.