[filmscanners] RE: IV ED dynamic range... DYNAMIC RANGE!

["Austin Franklin" <darkroom@ix.netcom.com>]

Hi Todd,

> First off (Dmin) I was hoping to see how you'd respond to Julian's point
> that MDS is not always defined by noise.

I've always said minimum discernable signal isn't always defined by noise,
but in the case here, it is.  The ISO proposal clearly says that, what you
call MDS, is Dmax, and that Dmax is based on noise.

> Your eq is based upon noise, but
> all definitions reference min disc signal. Where are you when the min disc
> signal is not defined by noise?

I don't quite understand what you're asking.  I did transpose Dmax and Dmin
in my original post, sorry for the mistake.  I believe I posted an immediate
correction...before I left the house ;-)

> > 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.
>
> I don't get how you can have a range of 5-0 when noise is 1. How
> do you get
> a range below noise?

You can have noise even in negative voltages.  It doesn't matter whether
it's -12 volts, with 1V of noise, or +12 volts, with 1V of noise...

> Doesn't noise limit your range on the low end, ie
> define your MDS, just as saturation/clipping does on the high end?

No, noise doesn't limit the range at all on the low end.  If you're talking
A/C, then all you get is an amplitude.  If you're talking D/C, then you can
go as low as you want...

Think of noise this way.  If you have a meter that is only good to 1V, when
your meter gives you 3V, you will be measuring a signal that has a value of
~between 2.5V and 3.5V.  I know what you're going to ask...what about when
you are measuring 0V...

> Seems to me in your examples above you'd have two choices for
> each scenario:
>
> > A RANGE of 0-5V (which could be stated as a range of 5V) with 1V noise
>
> This could be ISO = DR = Dmax - Dmin.... = 5 - 1 = 4, which I
> *think* is how
> Julian would approach it.

Yeah, but that's wrong.  The minimum signal is still 0...not 1.

> or Austin = DR = range/noise.... = (5 - 1)/1 = 4 which would appear to
> account for noise twice, though when noise is assigned a value of 1 it
> doesn't affect the final DR value (while conceptually it's way different).

No, your minimum signal is STILL 0, not 1...so it's (5-0)/1 or 5.

> > A RANGE of 0-5V with 1/2V noise
>
> This could be ISO DR = Dmax - Dmin.... = 5 - .5 = 4.5

No, you are subtracting non-log numbers, and subtraction only works with
logarithmic numbers, as far as the dynamic range equation goes.  So, it is
actually:

5 / .5 or 10.

> or, Austin DR =(5 - .5)/.5 = 9

You're changing the terms here.  I said the minimum discernable signal is
.5, but that the minimum signal level is 0.  Just take it that the range is
capped at 5V on the high end, and 0V on the low end.

> Obviously, I must be mistaken about my assumption of noise and MDS...

You're over complicating it...

> But the greater surprise is your definition of Dmax as a range.

Actually, it's Dmin that's the overall range.  It's the saturation point of
the CCD, clipping point as you will...

> For your
> interpretation to make sense, your Dmax would need to be the entirety of
> what Julian considers to be dynamic range.

I believe this is correct, meaning I believe Julian believes that density
range and dynamic range are the same, which is incorrect.

> For instance, if you were
> determining the the DR of a frame of film, isn't Dmax just the max density
> of the film?

Not quite...it's the max density minus the minimum density.  If the film has
a max density of 1.8 and a minimum of .2, the Dmax, as you are using it
here, is 1.6.  YES, I know it's using the same term to mean two different
things (dMax that is), and I don't like it...what people have been assuming,
is that when they say Dmax in that sense, that the Dmin is 0.

> By your definition Dmax would be the entire density range of
> the film?

The range from the lowest density to the highest density, yes.

> So is DR effectively = density range / noise?

Yes.

> If it were that it would be pretty easy to say so. Never seen it in any of
> the sources cited.

Er, that's exactly what I've said at least a thousand times.  If you convert
the equation above to log, it's the same as what ISO says...since division
in regular numbers is subtraction in log math.

So, in log form, your equation above is:

DR = density range - noise

But...with transmissive density measurements, the darkest is the noise...not
the lightest, as noise is the highest density number, and saturation is the
lowest density number...hence, why ISO has it stated as Dmax - Dmin and in
log math it all works out correctly.

BTW, I just wanted to post a SIMPLE statement of correction to some
misinformation posted, not get DRAGGED back into this discussion.

Regards,

Austin

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