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[filmscanners] RE: Density vs Dynamic range



> You are talking about noise, say in the shadows...where it
>becomes quite non-linear (increases)...  If that's what you're saying, I
>agree.

I am embarrassed to say it; but I am not sure if that is what I am saying or
not.  I am prepared to say and accept that the dynamic range of a density
range may not be lineal (e.g., the segments described may not all be of
equal size); but I do not know if this is empirically or theoretically the
actual case in engineering conceptualizations and principles.  I am also
willing to say that noise increases in an non-lineal manner in the shadow
portions of the density range such that (a) it is not equally distributed
across the entire density range and (b) it may not even be equally
distributed within the shadow portions of the density range from one defined
segment to another at that end of the spectrum.  But I am willing to admit
that I do not know what I am taking about and could be way off base in these
assumptions.

What I guess I am trying to say in that paragraph is that within the upper
and the lower ends of the density range or spectrum there may be empirically
discernable differences in noise levels making for shades of white or shades
of black which can be said to analytically exist but which are not
practically distinguishable so as to yield useful differentiated information
even if they do empirically provide small levels of differentiated
information when measured by some measurement device but not when perceived
by the human eye. Is this at all clear; does it make any sense?



-----Original Message-----
From: filmscanners_owner@halftone.co.uk
[mailto:filmscanners_owner@halftone.co.uk]On Behalf Of Austin Franklin
Sent: Monday, June 10, 2002 12:32 PM
To: laurie@advancenet.net
Subject: [filmscanners] RE: Density vs Dynamic range


Hi Laurie,

> >It does, you are right...but only to the point of reaching the
> noise level.
> >Once you have resolved down to the noise, resolving any further simply
> >doesn't give you any more information.
>
> Here is where I tend to get myself in trouble; but here it goes.  As a
> practical empirical matter of pragamtic significance, I agree and in my
> comments have sort of taken that as a given eventhough logically,
> analytically, and theoretically you could extend the descrete segments
> beyond the noiseless portions of the density range into the noise.  If one
> did so, it would be meaningless from an empirical and practical
> standpoint.

Correct.

> That is my thinking and why I could not visualize noise in the equation (I
> suppose one could just define as a matter of definition the
> density range as
> being only the noiseless portion of the density spectrum ignoring that
> portion which lies below; but I guess I am reluctant to do this for
> analytical reasons of defining a phenomenon like the density range which
> often extends beyond the clean, noiseless, visable - taken
> metaphorically -
> elements of that phenomenon). But with respect to my conceptual
> understanding as it is, am I incorrect about this?

Hum.  I believe I understand what you are saying, I just have to think about
it a bit.  You are talking about noise, say in the shadows...where it
becomes quite non-linear (increases)...  If that's what you're saying, I
agree.

> >Simply put, dynamic range is the number of steps within the
> density range.
> >I believe we have no problem with the definition of density
> range...right?
>
> That is my generic understanding of what you have been saying,
> and one with
> which I understand and agree.  To carry it further, I also
> understand you to
> be saying in general terms that the bit depth (number of bits) has no
> bearing on the size of the density range, but it does determine the number
> of steps or segments that that density range is divided into (not
> necessarily the size of any of those steps or segments).

Bingo!

> Or to put it
> another way, the bit depth describes the structure of the dynamic
> range and
> not the size of the density range.  Am I off base or close to what your
> understanding is.

Nope, right on the dot!

Regards,

Austin

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