On Friday, July 11, 2003, at 07:37 AM, Austin Franklin wrote:
> Hi David,
>
>> Hmm. Since the density range is defined logarithmically, you only get
>> the
>> same density range if the noise in the CCD output signal is
>> 1/256th of 6V or
>> greater. If the noise is less, you get a larger dynamic range
>> with the extra
>> bits.
>
> Agreed, but why does it matter if it's logarithmic or not?
>
> Regards,
>
> Austin
>
Austin,
I really don't want to get into a big discussion about this, but the
logarithmic
issue is of the utmost importance. It makes ALL the difference.
I thought you just agreed to my little example:
> Now do the 3 bit mapping:
> a) increment by 2: 2,4,6,8,...14,16
> density range = log(16) - log(2) = 1.2 - 0.3 = 0.9
> b) go half way: 1,2,3,4,...,7,8
> density range = log(8) - log(1) = 0.9 - 0 = 0.9
>
> The range is exactly the same both ways. The logarithmic nature
> of density throws off our intuitive feel for how the numbers work.
If you left out the log you'd get (a) 16-2 = 14 and (b) 8-1 = 7 and
incorrectly believe that the ranges were different (and as intensity
ranges, yes, they are different, but as density ranges they are
identical). This is very counter-intuitive, but nonetheless how the
physics and math works.
Roy
-
Roy Harrington
roy@harrington.com
Black & White Photo Gallery
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