The best UV Blocking-Color for the Digital Negatives ?

[Herzeleid]

Herzeleid test print seems to me to be OVEREXPOSED by about 3 stops

So I recalculate my exposure according to you and it should be 1.5 minutes for traditional cyanotype.
Thanks for you input.

 

[Dan Pavel]

"It has to be the minimum because more exposure will raise ALL THE DENSITY VALUES and the upper tones of the print will be clipped as there will be too much density and not enough light will pass through, to achieve printed highlights."

Dear Luis,
there is an exposure latitude within which the darkest black and the lightest white are still printed correctly. Using ANY exposure within the exposure latitude will allow a script like ChartThrob to make a curve that will correctly map the picture tonalities into the print tonalities. Of course, the curves will be different for different ETs. As long as the exposure of the print is the same as the exposure used to print ChartThrob the final result will be, in theory, the same.
Longer exposure + darker DN = shorter exposure+more transparent DN = CORRECT PRINT. The curve produced by ChartThrob will make, in the case of using a longer exposure, a darkerr DN and the final result will be a correct print, as well, Of course the exposure, once chosen, shouldn't be changed between printing ChartThrob and the final print. All density values (in PS % scale) of the final print won't be raised because the DN will be darker and the same quantity of light will reach the emulsion.
In the case of Mr.Sardar Cyanotype print the max white and max black are still printed correctly in ChartThrob. That means that the ET is still within the exposure latitude and the script will produce a curve that will still map correctly the picture tonalities into the print tonalities as long as he uses for the final print the same ET used to print ChartThrob.

"Mt Herzeleid test print seems to me to be OVEREXPOSED ...Of course the base exposure should be reduced to expand the mid tones and leave only ONE FULL BLACK SQUARE at bottom left."
Yes, it is a bit overexposed. If the exposure is reduced as you recommend the mid tones won't be expanded - they will be only moved to the bottom and the highlights will be expanded with some rows of pure white at the top of the chart.

If the exposure time is chosen in such a way that, while remaining into the exposure latitude, it places the middle grays of the final print in the middle of the ChartThrob chart (to be more clear - if equal nr of "white" and "black" rows appear in the printed chart). the DN will be well balanced and neither the ability of the inkjet printer to print darker or lighter tonalities will be on stress. Wouldn't this be the CORRECT exposure time to choose?
Choosing the minimum exposure time within the exposure latitude could result in a more transparent DN,

Hence my question, which is still valid: why is recommended to choose the minimum exposure time within the exposure latitude?

"It is best in my opinion to use just black." I agree. That's what I am using, as well.

 

[ced]

Jumping in here gentlemen. Not all printers/inks have a good black that can pass for printing a good DNeg.
Sometimes brownish or greenish and sometimes not even dark enough letting the light pass too easily so a decent UV blocking colour is necessary.
Those that can print decent black ink only are the lucky ones.

Serdar I enjoyed reading your interesting thesis pdf. Thanks for putting it out there for those interested like myself.

 

[nmp]

"

Hence my question, which is still valid: why is recommended to choose the minimum exposure time within the exposure latitude?

"It is best in my opinion to use just black." I agree. That's what I am using, as well.

I think the importance of these issues depend largely on the process at hand. For me, I started with what I think is, as it turned out, the most demanding of the processes in terms of making digital negatives: namely, Centennial POP and to a slightly lesser extent, salt prints. These differ from a lot of the other processes particularly the iron based ones in the following ways with regards to digital negatives:

1) Overall exposure times are much longer. Typically they run in the half hour range as opposed to others such as Pt/Pd that are much shorter in the 5-10 minutes scale. These are ball-park figures, depending on the precise chemistry and the power of the exposure unit of course. Longer exposure means you have to wait around longer obviously, but more importantly it also means hotter easel. This creates its own peculiar problem in case of POP. Being gelatin coated, it sticks to the negative as it gets hot and even transfers some of the ink onto itself on separation. So I NEED to keep the exposure at a minimum.

2) Secondly, for printing-out processes like salt prints, because of the self-masking effect, extending the exposure time beyond a certain point only results in more severe clumping up of the shadows to the point of being nearly indistinguishable, resulting in an even steeper curve in that zone which creates its own problems in how it can be plotted on Photoshop Curves.

3) These processes also require only the most UV-opaque of inks to get close to the theoretical Dmin. For POP, so far only HP B9180 with Vivera inks has worked for me. Even then, it has to be colorized in a shade of green. All black is simply not an option. Epson P400 did not work, with or without QTR using several different combinations of PK, MK, and various shades of colors with maximum allowed loadings. I suspect P600 or P800 would have been better, having a different set of inks. The upshot is if the exposure time extended from the minimum, the burden on the ink opacity is even greater. In fact, I would want to calculate the exposure time not on Dmax alone but on (Dmax-Dmin) which signifies the total process range. If I have to choose, I would rather take a hit on the Dmax a little if I can not get the best whites.

:Niranjan.

 

[Dan Pavel]

Yes, if the UV-opacity of the black ink is not high enough for what a certain process requires then using the most blocking color helps. In such a case choosing the minimum exposure time in order not to stain the highlights is logical. It is a decision depending on the process/inks used..
I am using a P600 and with pd/pb, VDB, Gum and Cyanotype the black ink is sufficiently opaque to allow some exposure latitude.
Probably the recommendation to choose the minimum exposure that gives max. black comes from the time when the opacity of the inkjet inks was not that good. With modern inks, capable of giving an exposure latitude with certain processes, choosing the best ET instead seams, IMO, more logical.

 

[LUIS GUEVARA]

So I recalculate my exposure according to you and it should be 1.5 minutes for traditional cyanotype.
Thanks for you input.

Dear Serdarr .
I said it "seems" to be overexposed by about 3 stops by looking at a webpage image. That is all I can say by looking at a web reproduction of a test . I cannot tell exactly by what amount it needs to be adjusted but it is obvious that it needs less exposure . However if you are also going to change to the "Best color", you will have to start anew and determine the Minimum exposure time for that color. The aim is to achieve Dmax in the lowest square , hopefully coresponding to the negative Dmin, (0 % negative ink ) and paper White in the Highest square ( 100 % negative ink ) so that you get the maximum amount of tones in between. In practice Dmax Black at 2% and Paper White at 98% is common accepted practice .

Regarding the need to use color ink to achieve sufficient "Spectral Density " to secure paper white I have to say that that was the case when inkjet printers used dye inks that didn't provide sufficient density. Modern inkjet printers use ( or have the option to use ) Pigment inks and nowadays we have the opposite requirement , that is , the printer produces too much density and , therefore negatives created with them when printed on alternative emulsions, ,achieve paper white TOO SOON , reducing the amount of Print midtones between Paper white and Max Black in the chemical print. Today it is generally required to reduce MaxInk deposition ( K value in QTR or Max Optical Density on Epson driver ) for that reason. See attached image showing how to do that when using color ink on an epson printer driver. Either use Black ink only or reduce the Max Optical density as shown, to expand the tonal scale.

Regarding self masking it is something that happens to some Print Out emulsions , particularly Salted Paper, but not to all of them and for sure it makes Paper Black point determinations much more difficult , if not impossible, That is why the use of POP method BY INSPECTION is considered ANTAGONIC to a "step by step", rigorous, computerized method , where adjustments are done in the Computer side rather than in the Printing side that is normally kept as a constant.

IN OTHER WORDS , WHEN MAKING COMPUTER DIGITAL NEGATIVES ONE HAS TO USE A STANDARD CHEMICAL (DEVELOPING OUT ) PRINTING METHOD AND NOT MAKE CHANGES TO IT DURING PRINTING OR YOU WILL NEVER ACHIEVE CALIBRATION OR , MUCH LESS , PREDICTABILITY. POP out methodology is adequate to adjust tonality by inspection when using less than properly developed Film negatives that cannot be modified , therefore you adjust the printing to suit the negative. In digital we do the opposite , we adjust our negative to suit the Print Emulsion.

 

[Herzeleid]

Dear Luis,

There is no way of estimating how many stops a chartthrob or similar chart is over or under exposed. It is not like a stouffer tablet, it cannot be measured.
Thank you for restarting a year old thread, and thank you for pointing out what I would have done differently eight years ago.

 

[LUIS GUEVARA]

Dear Luis,

There is no way of estimating how many stops a chartthrob or similar chart is over or under exposed. It is not like a stouffer tablet, it cannot be measured.....

Dear Serdar.
Hibryd printing is a very current subject but one full of misunderstandings and misinformation. Particularly when using automated methods , like charthrob.

However the concepts are the same ; The computer Monitor Black point is Zero Binary and in a Positive print it becomes 100% ink coverage or Max Black. The Monitor White point is 255 binary and in a Positive print it becomes Zero Percent (0 % ) ink coverage or paper white. Due to the fact that you are creating a NEGATIVE instead of a POSITIVE, these values will become reversed when printing so Monitor Black equals digital negative zero percent ink or clear base. Monitor White becomes of course 100% ink coverage.

When a digital negative is printed to a Translucent base such as pictorico OH no matter what system you use when generating a digital negative for calibration , the monitor Black point will become a zero ink coverage step (0 % ) in the negative. if chartrob does not map monitor black point to zero percent ink in the negative something is wrong in the process or in the operator understanding of the process. if it does map correctly then that clear square of zero ink becomes the basis for determining minimum exposure time for maximum chemical print black. the rest of the 255 squares of the negative should have increasing densities just like a stouffer step wedge does.The analog print should of course have decreasing densities from maxBlack towards Paper white, hopefully of uniform density change.

Each binary step correspond to doubling the density value of the previous step and that can be used to estimate how far off the exposure is from the target value after printing to your chosen chemical process. Colored steps however have ACTINIC values and their blocking power would be different than black blocking power because they act as a color filter and its equivalent "density" has to be determined with an UV densitometer or a Colorimeter but the printed negative chart has to have ZERO INK where it Maps to MaxBlack. and 100% ink where it maps to Paper White.