Digital BW, The Print

OK so let me first of all say that you guys know a LOT more about this than
me and that I am simply fumbling around in the dark trying to learn about
luminance and illuminance (excuse the pun) with a very dodgy recollection of
logarithmic mathematics. So forgive me if I don't use the right terminology,
if I am overly exhaustive in outlining my thoughts or if I tangle myself up
in logic knots here.

In short, Paul, I think you are one step ahead of me here in your answer and
that what you are doing is a defined mixture of the two step process I saw
happening - bear with me.

I understand that there is a direct relationship between luminance (display)
and illuminance (prints).  Gamma (contrast) defines the rate of change of
luminance (or illuminance) as we move from normalised pixel values of 0 to
1.  That is, in a simplified form, luminance = (normalised pixel
value)^gamma.  So even if a printer (and paper and ink combination) were
capable of the same tonal range as a display, unless it used the same gamma
setting the prints would look very different from the screen.  In other
words, because we work in a non-coloursynced workflow world (where
differences in gamma would be adjusted for) we must ensure that any printer
linearisation (the process of ensuring that the log of each side of this
equation is a straight line) uses the same gamma as our display. As soon as
we have a different gamma we depart dramatically from "what you see is what
you get".

Now our printers are not capable of the same tonal range as our monitors.
We don't get a bright white and we don't get dark blacks.  Even so, ideally,
we would still have a printer setup that faithfully portrays the various
in-range luminances that we see on screen - that is, the in-range tonal
range would be reproduced with the SAME gamma and mid point.  If not then we
morph the entire image and begin to lose precise control of dealing with
"out-of-gamut" shades.  Out of range pixel values (highlights and,
particularly, deep blacks) can't be reproduced by the printer.  In this
scenario, the out of range blacks would simply be reproduced as black as
possible.  This compression can be simulated in PS with a curve whereby the
end points on the input scale are clipped to constant output points (using
the 0-100% scale, lift the highlight point up then right and the black point
down then left, in each case keeping the input equal to output).  Note the
overall image is not radically changed, gamma is the same (the slope of the
PS curve has not changed), except at the two ends of the spectrum (where
clipping occurs). 

This clipping is not generally satisfactory and we would normally be
prepared to make controlled adjustments to the IMAGE in order that we get
good relative tonal display in the displayed or printed image.  We would
most likely take this hard edged inverted Z curve and smooth it to a nice S
curve around the same mid.  We can decide over what range we are prepared to
accommodate tonal changes to "open-up" the shadows and highlights (limiting
the range, say, by anchoring the 25/25 and 75/75 points for example).

If I can define (and I must be able to) where my printer tonal range starts
and stops in the little output gray scale in PS's curves function then I
have precise control over how the image tonal range is compressed into the
printer tonal range.

What I am not sure about is how programmes like QTR really do their
linearization.  I suspect, though, that they linearize such that rate of
change in -log(luminance) [guess I should have the negative because a print
is illuminated rather luminates] for a change in log(normalised pixel value)
is constant from dmin to dmax.  In other words, gamma is different from the
display gamma.  Because the tonal range is compressed this gamma is by
definition lower than display gamma and most likely the mid-grey point has
shifted. (Here is where I admit fully I need input from people like Roy.)

Paul, I think what you do is in effect a hybrid approach whereby you are in
effect creating the S curve and defining placement of the mid point in one
go.  Does this make sense?  I am not sure where the 50% point should be
EXCEPT to say that it should be the same as the display or workspace.

I would advocate, I think (!), a more rigorous two-step process.  (A) Define
the two end points of our printer's tonal range and ensure that the image's
tonal range is compressed with a PS curve (of whatever shape a person may
desire) before the image is sent to the printer.  (B) Calibrate or linearize
the printer such that it faithfully reproduces its tonal range with the same
gamma as our display gamma.

Even if the second part is done automatically I need to know what my two
tonal range end points are in terms of the 0-255 scale - or perhaps more
easily the 0-100% scale - used by PS in its curves function in order to
accurately compress the image tonal range to the printer tonal range before
sending it to print.  There must be a mathematical relationship here.  Take
an input pixel of 85% for example.  Dark grey.  Annoyingly PS has this
around the other way to normalised pixel values. Luminance = (normalised
pixel value)^2.2 = 0.15^2.2 = 0.0154. Density = -log10(luminance). So this
equates to a density of 1.81. Is my maths right?  Of course we can go in
reverse, beginning with our density readings from our linearized step
wedges.  

Have I almost answered my own question?  My mathematics is rusty!!! If I
know the density value then what is the formula for deriving luminance and
then getting back to a normalised pixel value which I can then reinterpret
into PS's 0-100% (or 0-255) curves scale?

(A dMax of 1.65 (EEM) equates to a normalised pixel value of around 0.175 (I
am simply guessing numbers here because I can't do the maths above!) which
is equal to 82.5% on PS's 0-100% scale.  So the max output I can have is
82%.  Let's assume the min input I can have is 5%.  It is not coincidental
that if you do this straight line "curve" and toggle it on and off and
compare it to toggling a soft proof of QTR's EEM curve that you see similar
tonal range adjustments.  I suspect that most of us are really interested in
soft-proofing not for proofing the change in hue (is this the right term?
tint) but for previewing relative and absolute tonal range.)

So I would be grateful if someone could help me with my maths problem and
also for input on the printer gamma point.

Cheers

Steve


> From: Paul Roark <paul.roark@...>
> Reply-To: <DigitalBlackandWhiteThePrint@yahoogroups.com>
> Date: Tue, 30 Nov 2004 18:19:11 -0800
> To: <DigitalBlackandWhiteThePrint@yahoogroups.com>
> Subject: RE: [Digital BW] Tonal range and linearization
> 
> 
> Steve,
> 
> I have some test strips at http://home1.gte.net/res09aij/Test-files.htm that
> show some relationships I find helpful.  This is just my approach, however.
> I've never seen a real scientific discussion of the theoretically "correct"
> way to "linearize" a gray scale.  Just what the midtone density should be is
> not 100% clear to me.
> 
> However, you noted the range of 0.04 for the paper white to 1.65 for the
> 100% black.  That just happens to be a typical EEM range.  So, reading a
> 0.04 EEM paper white I get a Lab L reading of 97.  Taking a reading of the
> 1.65 black with my X-Rite, I get L = 16.7.  The committee that made the Lab
> scale did a tremendous amount of work on how we see.  So, I take their scale
> as rather persuasive.  So, if you average the EEM black and white Lab
> Luminosity values you get about L = 56.9.  You'll see on my test strips that
> I put 50% at 0.61 visual density.  This is about Lab L=56.9.  (I use the
> readouts of the X-Rite for these equivalents.  I think there is a formula,
> but I don't have it.)
> 
> Perhaps not by coincidence, this was also about the average PiezoBW 50% I
> measured some years ago when I was still printing with that system, and I
> wanted my files to be able to print on either system.
> 
> When I profile other papers with a higher dmax, I don't change the 50% spot,
> because I think (perhaps incorrectly) that the midtone density is what most
> people see.  That seems to determine how bright or dark the overall image
> is.  A great black and bright white just give more dynamic range.  Our
> ability to see into those 2+ black densities, however, is limited.  So, I
> use 0.61 for my target 50% density for all papers.
> 
> On the other hand, the Kodak standard gray card is L = 50.  I think one
> could also argue that we ought to standardize on that as a uniform 50%
> density no matter what the paper white and ink black happen to be.  If I
> were starting from scratch, I'd be tempted to do just that.  I think using
> the L=50 midpoint might make the average monitor closer to the print.  L=50
> is about density 0.74 on my X-Rite system.
> 
> I'm not sure this is what you were looking for, but this is what I do.
> 
> Paul
> www.PaulRoark.com