Digital BW, The Print

> let's see if I got it right: the DR tells us how deep the scanner can see
> into the film,

A higher DR gives that, yes.

> and so higher DR values mean higher ranges of the  analog
> signal strength we get from the CCD.

No.  Typically, the input to the A/D will be +- 3 volts, for a 6V total
swing.  Then each bit represents N millivolts.  For a 10 bit A/D, each bit
is 0.005859375 mv.  For a 14 bit scanner, each bit is 0.0003662109375
mv....BUT...the scale is now shifted, since the CCD is more "sensitive".
Remember, no matter how many bits you have, 2 is always twice 1...and we are
dealing with integer density ratio values.

> On the other hand, bit depth is the
> number of bits that the analog to digital converter (I suppose
> there must be
> one somewhere) uses to map that signal.

As my arithmetic shows above...

> Given that the analog signal is
> continuous, any number of bits we pick will be a limiting factor
> in that it
> will not represent all the analog values (theoretically
> infinite),

I don't know quite what you mean here...we only measure relative values, as
in 2 times, 3 times, 4 times....1000 times...only integer ratio values.

> but this
> is not a problem due to human eye limitations. We have a real limiting
> factor when we use a number of bits which is not sufficient to
> represent the
> ratio between the highest and the lowest analog value (which I
> think is the
> actual definition for DR)

I believe all three statements there are correct.

> so if we have an analog signal ranging
> from 1 to
> 10 (silly numbers, I know) and only use 3 bits (0-7), we lose some
> information, whereas if we use 4 (0-15) we don't. Is this correct?

No, we always lose information...but...what you need to accommodate with the
number of bits is just how small a signal you want to be able to handle...in
other words, that voltage of 1-10 represents a dynamic range (which is an
integer ratio value)...and the number of bits must accommodate that.

> I don't know how scanners are designed, but I guess that once you have an
> analog signal from the CCD (whose range defines the scanner's
> DR), you could
> use different A-D converters: therefore the bit depth (defined as
> the number
> of bits we translate the signal into), would be independent from the DR.
> Does this make sense?

You could design it anyway you wanted...but what you would find, is the
methodology that is currently used, not only make sense to do given the way
CCDs work, and A/Ds work, but it just so happens to fit the integer density
ratio value very very nicely, which is what we need to represent in our data
values.