Digital BW, The Print

I thought I'd through in my $0.02 to clarify the issue a little bit. IMHO, Austin is pretty much on track.

BD and DR are related. Put simply, one is base-2 while the other is base-10 (all these math courses are finally paying off!). In addition, BD refers to individual colors (usually) which means that you have to multiply by 3 to get the number of colors possible (using RGB images) while DR refers the shades of grey between white and black. Assuming a constant DR for each color, they can be converted as demonstrated bellow.

To convert BD to DR, you can simply take the log of BD to the power 2 (since it is base-2). For example, an 8 BD would equal to a DR of 2.4. The calculations are as follows: log( 2^8 ) => log(256) => 2.408.

I hope this helps.

Regards,

Jean-René (JR) Geoffrion, P.Eng., C.M.C., M.B.A.
Chicago, IL
jr@...

----- Original Message ----- 

From: Austin Franklin 
To: DigitalBlackandWhiteThePrint@yahoogroups.com 
Sent: Friday, September 28, 2001 5:38 PM
Subject: [Digital BW] Dynamic range and bit depth. How they relate.


There have been claims made that dynamic range and bit depth are not
related.  That is not true.  They absolutely ARE related, and this explains
why.

What these claim perhaps mean, is that because a certain scanner uses, say,
14 bits, that the dynamic range of that scanner is not necessarily 4.2.
That is an absolutely true statement, but that does not mean they are not
related.

It take just so many bits to represent a range of values.  This goes as
follows:

8 bits can represent whole number values from 0-255
12 bits can represent whole number values from 0-4,095
14 bits can represent whole number values from 0-16,383
16 bits can represent whole number values from 0-65,535

Density values are ratio values, like 1:1, 1000:1 35673:1 etc.  They are
positive integer values, simply because they are always related to 1.  These
values have a range, and in this case, since they are related to 1, they are
a dynamic range...the minimum discernable value is 1, and the maximum is
what ever it is.

When a scanner says it has a (measured) dynamic range of 3.6, that means it
can discriminate from density ratio values of 1:1 all the way up to 10 to
the 3.6th power, or 3,981.  That means that every ratio from 1:1 to 3,981:1
needs to be accommodated. That includes all intermediate values, 2:1,
3:1...186:1...2,999:1 etc.  That means we need to have a sufficient number
of bits to be able to hold all whole numbers from 1 to 3,981.

Since 11 bits can represent whole number values from 0-2,047, that would
mean we would not be able to hold all the values we wanted to hold.  That
would mean we would need to use 12 bits to hold the numbers from 1 to 3,981.

We could, of course, use more bits than we need...and that does not increase
the dynamic range of the scanner though.

So, the conclusion is that dynamic range and number of bits ARE DIRECTLY
related, in that you need a minimum number of bits to represent a particular
dynamic range...but just because a scanner uses a particular number of bits,
that does not mean that it can actually produce a dynamic range that
uses/requires all those bits.

Austin



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