[filmscanners] Re: scanner dmax discussion

["Karl Schulmeisters" <karlsch@earthlink.net>]

>"> and "1", means "0.5 to 1.0", so if you see the center of
>> that range as the "meaning" of the reported value, a 1-bit system is only
>> representing the range 0.25 to 0.75.
>
>I think that is what's incorrect, as the meaning IS the range, not the
>center of the range"

Except Austin, that's not how these devices are calibrated or tested.  The
universal (I have yet to see another one used but I am willing to be
corrected) approach is to use a step tablet with density increments of 1/3
of a stop and the 'dynamic range' is measured by seeing where the scanner
ceases to differentiate between two adjacent steps.

Assuming that is step 'n',  and that maps to 'all bits on', then it is
ambiguous as to whether 'All but LSD bit on' means
a) the range from 'all bits on -2' to 'all bits on -1'
or
b) the range from 'all bits on -1' to 'all bits on'
or
c) the range of density values that map to 1/2 way between 'all bits on -1'
to 1/2 way between 'all bits on'.

Since it is ambiguous, the most consistent way to do this is to assume that
each discrete measurement represents the center of a value range, +/-(0.5 x
the full value range).

Karl

----- Original Message -----
From: "Austin Franklin" <austin@darkroom.com>
To: <karlsch@earthlink.net>
Sent: Friday, July 11, 2003 8:28 PM
Subject: [filmscanners] RE: scanner dmax discussion


Hi David,

> >>>>>>>>>>>>
> Well, hum.  Even if 8 bits wasn't enough, and it really needed
> 16, does the
> density range actually increase?
> <<<<<<<<<<<<
>
> Yes, the measurable density
> range increases as well, by definition.

By what definition?

> Since density range is
> defined as a
> dynamic range sort of thing

Well, no, density range and dynamic range don't have to be the same thing.
You can have any density range represented by any dynamic range.

> (i.e. <full range>/<step size>), the data that
> comes out has a wider/narrower range depending on how many
> significant bits
> got thrown out.

I, obviously, disagree with that.

> That's the confusion in scanner specs: "density range" is
> _defined_ to be identical to "dynamic range".

The confusion I've seen in scanner specs is they claim the number of bits
the A/D uses is the density range/dynamic range, without taking into
consideration that some of the bits may be "into the noise".

> >>>>>>
>   Keep in mind, that every value is +- 1/2
> the bit value.  So, the bottom and top number still CAN represent
> the lowest
> value (as well as the top number), and therefore, the density
> range that is
> "seen" can very well be identical.
> <<<<<<<
>
> This is what I now believe to be incorrect.

It's a statement of how this works though...  Keep in mind, the output from
the CCD will be identical, and have identical noise...no matter how many
bits you are using.

> Here's what I see as the right way of looking at it: Each value
> measured by
> the scanner corresponds to a range of densities:

That is correct...and keep that in mind ;-)

> in a 1-bit system, "0"
> means "0 to 0.5",

I disagree with that, unless you specifically design the system to work that
way, but when measuring using an A/D, it doesn't work that way, unless noise
is from -.5 to +.5...  All the steps will be of equal range.  Check the
coding of some of the A/Ds...I'll see if I can dig up some of my old Analog
Devices documents on this.

> and "1", means "0.5 to 1.0", so if you see the center of
> that range as the "meaning" of the reported value, a 1-bit system is only
> representing the range 0.25 to 0.75.

I think that is what's incorrect, as the meaning IS the range, not the
center of the range.  Keep in mind this is purely a representative system,
and given that, you have to take into account what is actually being
represented.

> To bash this nail a bit more: in a 1-bit system, any density in
> the range 0
> to 0.5 will be reported as 0. So when you look at a measured
> value of zero,
> the best you can say about it is that it's "0.25 with an error
> tolerance of
> +/- 0.25". It's quite problematic to look at a measured value of
> 0 and claim
> it means "0" with an error tolerance of +0.5/-0.0", which is what you are
> doing in the above.

What you look at 0 as saying is that it represents a value of 0-.5V (using
your example that I disagree with, as stated above)...and the value can be
anywhere within that range.

> As you increase the number of bits, the "center values" of the extreme
> values (0 and 2^n -1) get closer and closer to 0 and 1 respectively.

Agreed, but 0 can STILL represent a value of 0 in either case, which is my
point.

> Again this makes sense from a measurement standpoint, since a
> measurement should be seen as the center value of the range of physical
> measurements that could return that value.

Why?  Why not say that in reality it is a range, because it is?

> Arbitrarily declaring
> two of the
> measured values to have different meanings is problematic.

I'm not seeing what you're saying here...

If you want to take this off-list, please feel free to email me at
austin@darkroom.com.

Regards,

Austin

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