Hi Roy,
I was talking about your context so we are discussing the same thing. You
have already got a response from Vincent which puts that case in terms of
resolution, here's my quick take from the dynamic range point of view - the
two arguments are otherwise essentially the same.
The important thing I think to remember about DyR is as always, the
definition, and what defines the minimum signal, the MDS, in that
definition. The DyR is the range from the max signal down to the MDS, not
down to some "zero" figure. So in the case you are discussing, and if we
call black the low end of the range as you have, we can establish the DyR
as follows.
We know it is max signal / MDS. What is the MDS? It is the minimum signal
that can be detected ABOVE whatever corresponds to background or 'zero
level'. In this case, and with all digital step limited situations, the
MDS is step 1, not step 0. It is the first, lowest signal you have any
chance of discerning above the zero signal level, which in this case is
pure black.
So in the first case, the DyR is : max/MDS = (4096 steps) / (1 step), and
in the second 256/1. i,e, DyR 4096 vs 256.
Looking at it another way, with an 8 bit file, the bottom step is the same
level as step 16 was in the 12-bit case. So when you converted from the
12-bit to the 8-bit, you lost the 16 lowest steps and combined them all
into 1, the lowest level of the 8-bit situation. In that conversion you
lost the 16 lowest shades of gray, permanently. So all that info is gone
and your MDS is now 16 times larger, and correspondingly your DyR has
diminished by the same amount, 16 times.
If you then converted back to 12-bit, you can't regain those bottom 16
shades, so your picture is permanently degraded. Despite the 12-bit
digitisation which implies DyR of 4096, the actual image bottom step- the
new minimum discernable signal above black MDS - is actually step 16 of
your 4096 and so the DyR is now 4096/16 = 256, the same as the 8-bit
case. This must be so because the information content is exactly the same
in both cases.
Does this make sense?
Julian
At 17:14 01/09/02, Roy wrote:
>I'm curious whether we're talking about two different things or that you
>disagree with what I was actually talking about.
>
>It think that your post (in response to Austin) was talking specifically
>about scanner output. In other words the phrase "the number of bits LIMITS
>the dynamic range" was in the context meaning "the number of bits in a
>scanner LIMITS the dynamic range of that scanner". In this context I
>entirely agree -- a 16-bit scanner has more dynamic range than an 8-bit
>scanner.
>
>My 8-bit versus 16-bit comment was in a very different context. I was
>talking about a 16-bit Photoshop that was ready to be printed. Thus
>value 0 was the max black and value 65535 was the max white. At this
>time the file was converted to 8-bit such that value 0 represents the
>same max black as 0 in the 16-bit file, and value 255 in 8-bit file
>represents the same max white as 65535 in the 16-bit file. So both
>files represent the same black to white range. In this context I
>say the 8-bit file and the 16-bit file have the same dynamic range
>because they represent the same tonal range on a output print. The
>endpoints are the same only real difference is how many levels are in
>between.
>
>So, is there disagreement? If so I'd like to know why and how you
>look at it.
>
>Thanks,
>Roy
>
>
>Roy Harrington
>roy@harrington.com
>Black & White Photography Gallery
>
>
>
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