FotoDave@aol.com
Tue, 25 May 1999 00:42:33 -0400 (EDT)
Tadeuz,
Thanks for your input. It does stimulate more thinking.
> The question is, why are reflective and transmissive dot area calculated
> differently?
Actually you can use the same. The difference is not between transmittance
and reflectance. The first (simple) one assumes that the densitometer is
calibrated so that Dmin reads 0.
The second one doesn't make that assumption, so Dmin has to be subtracted in
the equation; but even this one is a simplified one because it assumes Dmax
is higher than 2.0 which is true in almost all cases.
However, if you have a Dmax of 2.0 and 3.0, the difference in optical
densities is negligible (sp?) simply because the difference is in the 3rd
decimal place. It is the case where you would find one with a density of .30
and another one has 0.35, for example.
> These processes are
> of very high contrast, so, a D-max of 2, like Dave says, can be
> interpreted as total absense of light.
Yes, the long-range process will require higher Dmax, but as David mentioned,
the output of an imagesetter has a Dmax higher than 4.0.
Your mention of self-masking process is interesting. I think it will affect
the Dmax requirement. I do not doubt that there is a different requirement
for the negatives for gum and for Pt print.
But all these still can't explain why a 2 areas which both measure 50% on a
densitometer can measure different optical densities. From David's last post,
I think that the difference is actually caused by how a particular
densitometer measures dot area.
Dave S
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