Julian,
> Julian now replies:
>
> Hmmm. Here is the draft ISO spec, from
> . It is
> entitled
> "Photography — Electronic scanners for photographic images —
> Dynamic range
> measurements." Perhaps there is another ISO spec from which you are
> deriving your beliefs? Perhaps you could post it?
You should know this is the same spec. This spec, when understood, shows
exactly what I've been saying is correct, and shows that your belief is
incorrect. It is merely corroboration I am deriving from it. But of
course, if you don't understand or misinterpret the terms, and the basis for
using them, you simply can't understand what this really means...and
therefore can easily be misinterpreted to mean something it simply doesn't,
which you are doing here.
> -------direct quote from Proposed ISO standard-------------------
>
> 7.2 Scanner dynamic range
>
> The dynamic range is calculated from the Scanner OECF by:
>
> DR = Dmax - Dmin (7.2)
>
> DR = Scanner Dynamic Range
> Dmax = Density where the Signal to noise ratio is 1
> Dmin = Minimum density where the output signal of the luminance OECF
> appears to be unclipped
>
> -------end quote from Proposed ISO standard-------------------
>
> (and OECF is opto-electronic conversion function)
>
> You will notice, it is exactly as I have described it, a RANGE.
I do not see ANYWHERE where it says dynamic range is "a" range. It shows
the RESULT of a calculation WITHIN A RANGE (Dmax), divided by the noise
(Dmin), but the result is NOT "a" range.
It shows (in non-log numbers):
dynamic range = amplitude / noise
and in log math, division is merely subtraction:
Non log example:
1000/10 = 100
log example:
log 1000 = 3 (10**3 = 1000)
log 10 = 1 (10**1 = 10)
3 - 1 = 2
and 10**2 = 100...fancy that.
> It is the
> range between Dmax and Dmin.
No, that's Dmax MINUS Dmin. That's a MINUS sign there, Julian, it's a
mathematical equation. The result IS the dynamic range. It is NOT "the
range BETWEEN", it is the result OF the subtraction of two log
values...which, BTW, is exactly the same a DIVISION if the numbers were NOT
log values. Think about that.
> It is not a resolution, there is no
> mention of
> resolution. Can you tell me then how this says that Dynamic Range is a
> resolution?
Very easily. One of the terms (Dmin) is the noise, as CLEARLY stated
(Signal to Noise Ratio = 1...which means the noise equals the signal), and
the minimum increment of measurability in a system IS noise...and therefore,
you can only measure so many points within the overall range and consider
them discrete points.
The other term (Dmax) is simply the overall amplitude the signal can reach,
which is based on zero being the low bound, and Dmax as being the high
bound. If you divide the overall amplitude by noise, that gives you the
resolution within that overall range with which you can discern.
> Julian:
> > > DYNAMIC RANGE on the other hand, is the smaller range within
> the Density
> > > Range that the scanner can capture AT ONE TIME i.e.
> dynamically i.e in one
> > > scan. It is the instantaneous range the scanner can handle.
>
> Austin:
> >Absolutely not correct. Where on earth did you get that? Please please
> >provide any credible source that says anything to the such. The ISO spec
> >doesn't define dynamic range that way...nor do any of the
> resources I have
> >seen.
>
> On the contrary, the ISO standard states a fairly precise process
> in which
> the Dynamic Range is measured by scanning a single slide in a single
> pass. (They do repeat the same single-scan measurement several times to
> improve accuracy).
So what? You HAVE to scan something to measure the relevant terms...duh!!!
How is that "contrary"? This is called "practical application", as apposed
to "theory". It's a good thing to be able to actually test what the Dynamic
Range of a system is...instead of simply calculate it based on component
specifications.
<snip>
> Austin, I have looked long and hard through archives to find
> where you have
> quoted supporting information for your views, without success. There is
> not a single quote I can find anywhere which contradicts what I
> am saying.
I have ALWAYS contradicted what you're saying. You CLEARLY mistakenly
believe "dynamic range" is "a" range, and it is clearly not. A range
requires two bounding terms (or one with the other being considered 0, or
simply the overall range, as in the amplitude), and dynamic range, as per
the equation you have provided, shows that it is a number, that is in dB,
that is derived from the information available WITHIN a particular range
(DMax is the overall range), and that does not make it "A" range.
A VERY simplified example that I've used time and time again while
contradicting what you have been saying, but you seem to fail to grasp:
A RANGE of 0-5V (which could be stated as a range of 5V) with 1V noise has a
dynamic range of (5-0)/1 or 5.
A RANGE of 0-5V with 1/2V noise has a dynamic range of (5-0)/1/2 or 10.
Note, the RANGE is identical, but the DYNAMIC RANGE is not.
You're absolutely NOT adding ANYTHING new to your argument, you are merely
repeating the exact same things I've shown to be incorrect previously.
Austin
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