[filmscanners] RE: scanner dmax discussion

["Austin Franklin" <austin@darkroom.com>]

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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