Re: [Leica] Re: dMax and Dyanmic range - scanners

["Michael Abbott" <lists@mabot.com>]

> What scanner manufacturers *actually* mean by Dmax is that the A/D
> converters in their machines capture so many bits of information.
>
> For example, a scanner which can capture 12 bits is capable of
representing
> 4096 distinct shades of gray (ignoring color for the moment). It's
> *theoretical* dynamic range is therefore
>
>     log10(4096) - log10(1)
>
> Which is 3 and a bit (sorry don't have slide rule handy)

it's 3.6. For n bits your formula gives

   TDR  =  log10(2^n)  =  n log10(2)  =  0.3 n

This seems to work for the few scanners I know of. Do you know if it is a
universal practice? I always assumed they worked out the number of bits
needed after managing to make a device with whatever dynamic range.


I'm confused about what your formula means. It says that th. dyn. range in
*stops* = number of bits. This means that for n stops of range, you divide
each stop into 2^n/n regions:
   2-stop range (0.6 of density) ==> 2 different greys per stop.
   3            (0.9)               ~3
   10           (3.0)                100
   13           (3.9)                630
I thought that visually we needed a constant number of greys-per-stop in
order not to see the steps. That would suggest a theoretical dynamic range
for an n-bit number of

   TDR  =  2^n 0.3 / #vis-greys-per-stop
or
   n  proportional  log(TDR)

although that just doesn't seem right. Can anyone see what I've
mis-understood? (This will probably be embarassing!)


Confusedly yours,
Michael


(With apologies to all the non-mathematicians here. Less OT than lawyers
though.)


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