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Re: [Digital BW] Bit depth, was Minolta DiMAGE ScanMulti PRO

Re: [Digital BW] Bit depth, was Minolta DiMAGE ScanMulti PRO

2001-09-27 by Austin Franklin

> I
> think I get why
> with low bit justification higher bit depth leads to higher DR, thus a
> larger histo, but I don't get why the histo doesn't get larger when it's
> high bit justified.

Higher bit depth, whether high bit justified data, or low bit justified
data, can represent/hold a higher dynamic range.  High bit justified data
will show a wider span on the histogram than low bit justified data, even
though they are "the same" data.  Remember, high bit justified data has
space between the valid data points...and low bit data does not have any, it
uses contiguous values (if the original scan was able to provide contiguous
values that is).

> If a person doesn't know how their data is being
> justified, can they make a meaningful assessment of the histogram? Ooooh,
> maybe that's why you want a 16 bit histo....

Well, you should always take the data and set the setpoints and spread it
out...that will eliminate any issues with high bit justification or low bit
justification.

> PS, when you do discuss the stuff below maybe you can work this
> in too. I've
> heard it said over and again, including from Andrew Rodney, that
> DR and bit
> depth are totally unrelated. When people say that, do they mean in all
> instances other than at point of capture?

I don't know why they say this, because it's simply wrong!  Yes, you CAN
represent any dynamic range with any number of bits 2 or more...BUT you
don't get the intermediate values (tones).  Can you show me where Andrew
Rodney states this.  I would like to read exactly what he as to say.  Ed
Hamrick agrees with me on this, BTW.

> Is it that they just don't
> understand how capture devices work, but they are correct in all other
> instances? If you are correct about the relationship, and I trust you are,
> why are so many otherwise knowledgeable people mistaken?

Well, I don't know that they are hardware engineers...they may be very
experienced in imaging, but just don't understand how the hardware works.  I
don't actually know that they are truly knowledgeable.  The knowledgeable
people I know, agree with me ;-)

Re: [Digital BW] Bit depth, was Minolta DiMAGEScanMulti PRO

2001-09-28 by Todd Flashner

>> I've
>> heard it said over and again, including from Andrew Rodney, that
>> DR and bit
>> depth are totally unrelated. When people say that, do they mean in all
>> instances other than at point of capture?
 
> I don't know why they say this, because it's simply wrong!  Yes, you CAN
> represent any dynamic range with any number of bits 2 or more...BUT you
> don't get the intermediate values (tones).

Right, but doesn't that just show you CAN have one independent of the other?
That IS the point, no? You've always maintained that you can't get to a high
DR w/o sufficient bit depth, but what you offer above, to my mind, violates
that. I understand high DR with low BD won't give us imagers what we want or
need, but on a fundamental level.....

>  Can you show me where Andrew
> Rodney states this.  I would like to read exactly what he as to say.  Ed
> Hamrick agrees with me on this, BTW.

Alright, I've asked similar questions on other lists in the past. But most
people who've responded aren't engineers, so they probably look at it
differently. I really think the step by step approach of: voltage to CCD, to
front end, to A/D converter, to tones assigned, etc., which you've taken, is
the only way to approach this subject in a meaningful way. The other
answers, while illuminating, are more of a random thought approach which
leaves room for too many gaps. ;-)

I will try to collect a few responses to show you a sampling of some of
those answers. Did you see the one I posted a couple of days ago from Dan
Margulis (he spoke of the extra digits on the thermometer, and related it to
noise)? If not I'll send it to you off list -
not that it is terribly relevant to this.

But, in short, most people are of the mind that bit depth and dynamic range
are independent qualities, or that's how I've interpreted them. I'll venture
they may be right when considering the attributes of a file already before
us, but not realize it's a little different when we are discussing the
elements of a capture device.

Your comments are welcome.

********

From Andrew Rodney:


Todd asks:
> First off, when in 16 bit mode in PS does the histogram indicate how the
> file maps into 16 bit space, or an 8 bit space?

Andrew answers:
In a way both. The numbers in the Histogram (0 to 255) remain the same.
Technically that¹s not really correct as there are a whole lot more numbers
in a high but file. But consider that Photoshop wisely treats 8 bit files as
one thing, ANY file more than 8 bits (10, 12, 14, 16) bits as another. The
Histogram would be a mess if we had numbers for 12 bit verses 14 bit verses
16 bit. What Adobe did was just leave everything the same as far as numbers
are concerned. The Histogram you see is accurate to the bit depth you are
currently in. Turn off cache for Histogram in preferences.

> More specifically, when I capture a raw unmanipulated HDR file from my Leaf
> scanner, which I believe scans in 14 bit, and bring it into 16 bit space in
> Photoshop, the tones in the histogram are bunched up.
> 
That¹s because Leaf provides you untonede (linear) raw data. Newer scanners
can provide high bit files that are toned. The Leaf is pretty old stuff so
at the time, it was about the only scanner that even provided you high bit
data but they just gave it to you untoned. You can pull the histogram and
make it look nice and lovely and you¹ve just toned it. Since it¹s high bit,
no damage done. Once you convert to 8 bits, you¹ll see a lovely resulting
Histogram with no data loss.

> Would it look any different if the same film were captured on a similar device
> that captures at a true 16 bit level?
> 
If said scanner also provided you untoned data, no. If it toned the data, it
could look a whole lot different. With my Imacon Flextight, I can tone and
capture 16 bit data. The software allows me to pull curves and set endpoints
and then provides me 16 bit scans. Your Leaf doesn¹t provide that option so
you get that dark looking file with the Histogram pushed to one side.

> Or, Does it indicate that the dynamic range of the film scanned was just not
> as large as that possible even in an 8 bit space?
> 
Dynamic range has nothing to do with bit depth! They are completely
different spec¹s. You can have a scanner with 16 bits per color and a
dynamic range of 3.3 and you can have a scanner with 12 bits and a dynamic
range of 3.8. Bit depth is the number of steps. Dynamic range is the height
of the star case. You can have a staircase that¹s 20 feet high and have 40
steps. You can have a staircase that¹s 30 feet high and have 30 steps.

> Is it a function of the
> dynamic range of film itself, the dynamic range of the capture device, or
> the bit depth of the capture device relative to the space?
> 
Film has a dynamic range, so do scanners and digital cameras. It¹s the
device that plays a role here.

> Basically, I am confused overall as to how the histogram works in 16 bit
> mode, in that anything over 8 bits is handled as 16 bit. Thus, does it
> display 10 bit data differently relative to say, 14 bit, or any other
> possible comparison of that sort?
> 
No, Photoshop deals with 8 bits pre color in one fashion and then everything
higher (what I like to call ³High Bit²) as 16 bits pre color. At least
that¹s what you see in the mode  menu. There is simply no reason to break
down a high bit file into categories.

Andrew Rodney 

******

>At 05:15 PM 7/26/01 -0400, Todd Flash wrote:
>>I'm rather confused about the interaction/relationship of these concepts:
>>Dynamic Range, Bit Depth, and the Histogram.
>>
>>First off, when in 16 bit mode in PS does the histogram indicate how the
>>file maps into 16 bit space, or an 8 bit space?
>>
>>More specifically, when I capture a raw unmanipulated HDR file from my Leaf
>>scanner, which I believe scans in 14 bit, and bring it into 16 bit space in
>>Photoshop, the tones in the histogram are bunched up. I know, I haven't set
>>endpoints or gamma yet, but is this showing that the dynamic range of the 14
>>bit data is not as large as that which is allowable in a 16 bit space? Would
>>it look any different if the same film were captured on a similar device
>>that captures at a true 16 bit level?


Todd, as I recall, our buddy Austin claims that the
Leaf can do perfectly fine scans over a wide range
of exposures.  This seemed odd to me.  But I did say,
at the time, that this could only be true if the A/D
was using only a small portion of its available input
range.  So in a way -- your observation about bunched-
up histograms makes some sense.

Rafe Bustin

********

On Fri, 27 Jul 2001, tflash wrote:

> Thank you Andrew and Maris. You guys seem to be in accord, and what you
> expressed was in agreement with my assumptions.
> 
> Rafe, you seem to feel other wise, so I would like to address this to you.
>  
> >>> More specifically, when I capture a raw unmanipulated HDR file from my
Leaf
> >>> scanner, which I believe scans in 14 bit, and bring it into 16 bit space
in
> >>> Photoshop, the tones in the histogram are bunched up. I know, I haven't
set
> >>> endpoints or gamma yet, but is this showing that the dynamic range of the
14
> >>> bit data is not as large as that which is allowable in a 16 bit space?
Would
> >>> it look any different if the same film were captured on a similar device
> >>> that captures at a true 16 bit level?
> > 
> > 
> > Todd, as I recall, our buddy Austin claims that the
> > Leaf can do perfectly fine scans over a wide range
> > of exposures.  This seemed odd to me.  But I did say,
> > at the time, that this could only be true if the A/D
> > was using only a small portion of its available input
> > range.  So in a way -- your observation about bunched-
> > up histograms makes some sense.
> 
> Could you say more about what makes sense?
> 
> This question does in fact stem from a difference of opinion between myself
> and Austin. He is of the mind that the histogram of the raw (linear) data
> appears the way it does because it is 14-bit capture mapped into/displayed
> in 16-bit space. I am of the opinion that it is due to the dynamic range of
> the data, whether it be due to the DR of the film, or the scanner. He
> believes the DR of the data is in part a function of the bit depth. He
> referenced the way higher bit A/D converters allows scanner manufacturers to
> make extended DR claims.
> 
> At least that's the way I interpreted his statements.


No, I'm not trying to take sides in your dispute
with Austin.  As you recall, Austin claims (and
cites the Leaf manual as proof) that the Leaf
can render perfectly fine scans over a wide
range of exposures at the scanner-driver level
(on the Leaf, these exposures would be measured
in milliseconds per scan line, I presume.)  This
didn't make a lot of sense to me.

As a (former) designer of instrumentation
circuitry, my goal was generally to set
analog gains so that one used as much of
the A/D input-voltage range as possible,
while also ensuring that over-range and
under-range conditions would not occur.

To may way of thinking, the only way
that Austin's claim could be true was
if the Leaf was only using a small part
of the available A/D input voltage
range on any given scan.  And, voila --
this is what your "narrow" histograms
seem to show.

I'm only trying to put "2 and 2 together"
here, that's all.

Rafe Bustin

(Austin, I still never did understand what Rafe was saying here. Too
technical for me. Could you expound on this in more simple terms?)

*********

Dynamic range is a separate concept from bit-depth. Think of bit depth as
the number of data points that can exist between dMin and dMax. In Photoshop
Levels, dMin and dMax will always be 255_0. In 16-bit mode, the Levels
theoretically should be 65,535_0, but since that would be impractical to
display on screen, the distributions are scaled to 255_0.

If the levels in your histogram (14bit-16bit) are bunched up or do not fill
up the entire range (255_0), it is caused by your capture settings and/or
the dynamic range of the captured image.

Nathan

RE: [Digital BW] Bit depth, was Minolta DiMAGEScanMulti PRO

2001-09-28 by Austin Franklin

> >> I've
> >> heard it said over and again, including from Andrew Rodney, that
> >> DR and bit
> >> depth are totally unrelated. When people say that, do they mean in all
> >> instances other than at point of capture?
>
> > I don't know why they say this, because it's simply wrong!  Yes, you CAN
> > represent any dynamic range with any number of bits 2 or more...BUT you
> > don't get the intermediate values (tones).
>
> Right, but doesn't that just show you CAN have one independent of
> the other?
> That IS the point, no?

I don't know what their point is, but scanners/CCDs/ADs work the way I said
they do.

> You've always maintained that you can't
> get to a high
> DR w/o sufficient bit depth, but what you offer above, to my
> mind, violates
> that. I understand high DR with low BD won't give us imagers what
> we want or
> need, but on a fundamental level.....

IMO, it's a silly exercise, that has absolutely no substance or meaning.
What is the point of it?  It's not the way CCDs work, it's not the way
density works...and it makes no sense to attempt to encode things
differently.  When you are encoding density ratio values, you use integer
values...and you also need to have just so many integer values to be able to
represent every ratio value in the dynamic range.  It's just so simple.

Sometimes things just naturally work well and fit together a particular way,
and unless there is a compelling reason to do it differently, why bother?

Also, if you do remap the dynamic range, you then have to know what the
mapping is.  If you map it to a lower bit depth, then you lose data.

I really don't get the big deal and why this is even an issue to anyone.  It
only comes about because of a lack of understanding that density is a ratio,
and ratios are integer values.  Once you understand that this is a ratio
issue, and how ratios are represented, it all just makes sense.  You know
that a ratio of 1.5 to 1 is the exact same as 3:2?  You can represent ANY
ratio value in integers.

> Did you see the one I posted a couple of days ago from Dan
> Margulis (he spoke of the extra digits on the thermometer, and
> related it to
> noise)?

Yes, I saw that...but that relates to some things, and not others.  I don't
recall what his exact point was...

> But, in short, most people are of the mind that bit depth and
> dynamic range
> are independent qualities, or that's how I've interpreted them.

Well, they just aren't.  You absolutely do need so many bits to represent an
entire range of integer density values over a particular dynamic range.  It
just works that way.  It's like saying counting and numbers are independent.

> I'll venture
> they may be right when considering the attributes of a file already before
> us, but not realize it's a little different when we are discussing the
> elements of a capture device.

I am specifically talking only about image capture devices.  I don't know
what you mean by the attributes of a file already before us.


> From Andrew Rodney:
>
>
> Todd asks:
> > First off, when in 16 bit mode in PS does the histogram indicate how the
> > file maps into 16 bit space, or an 8 bit space?
>
> Andrew answers:
> In a way both. The numbers in the Histogram (0 to 255) remain the same.

That is correct.

> Technically that\ufffds not really correct as there are a whole lot
> more numbers
> in a high but file.

That doesn't matter.  The histogram only uses the top 8 bits, so as he said
above, the histogram remains the EXACT same.

> The Histogram you see is accurate to the bit depth you are
> currently in.

I believe that is simply because the data is always high bit justified.

> > More specifically, when I capture a raw unmanipulated HDR file
> from my Leaf
> > scanner, which I believe scans in 14 bit, and bring it into 16
> bit space in
> > Photoshop, the tones in the histogram are bunched up.
> >
> That\ufffds because Leaf provides you untonede (linear) raw data.
> Newer scanners
> can provide high bit files that are toned.

Toning has nothing to do with it.  Toned data is NOT raw data.  Toning data
also does NOT really spread it out, setting the setpoints spreads it out,
whether you chose to tone it or not.  He is talking about something other
than raw data.  You CAN get both high bit raw data, and high bit toned data
out of newer scanners.

> > Or, Does it indicate that the dynamic range of the film scanned
> was just not
> > as large as that possible even in an 8 bit space?
> >
> Dynamic range has nothing to do with bit depth! They are completely
> different spec\ufffds.

As you know, I know that is completely incorrect.

> You can have a scanner with 16 bits per color and a
> dynamic range of 3.3

That is in fact correct, only by virtue of the fact that the number of bits
is NOT limiting the dynamic range.

> and you can have a scanner with 12 bits and a dynamic
> range of 3.8.

Absolutely wrong.  A dynamic range of 3.8 specifies integer ratio values
from 1 to 6,309, which means you need a number of bits that can hold these
numbers, and 12 bits can only hold 4,096 different values.

> Bit depth is the number of steps. Dynamic range is
> the height
> of the star case. You can have a staircase that\ufffds 20 feet high and have 40
> steps. You can have a staircase that\ufffds 30 feet high and have 30 steps.

He is right, but contradicts his example above!  If you have a bit depth of
12, as he states above, you only have 4,096 steps.  If your dynamic range is
3.8, which is 10**3.8 or 6,309...you don't have enough steps to hold it.
Pass this along to him and see what he has to say.

The only way to put a dynamic range of 3.8 into 12 bits is to lose data
values, period.


I snipped Rafe's comments, since I believe that is all covered in our recent
discussions...and we can get back to that if they still confuse you.

> *********
>
> Dynamic range is a separate concept from bit-depth.

Yes, they are separate concepts, BUT, YOU NEED A MINIMUM NUMBER OF BITS TO
CONTAIN A PARTICULAR DYNAMIC RANGE.  In other words, the number of bits does
limit the maximum dynamic range you can represent if you want to include all
intermediate integer density ratio values.

> Think of bit depth as
> the number of data points that can exist between dMin and dMax.

That's true.

> In Photoshop
> Levels, dMin and dMax will always be 255_0.

Yes, but that's only the top 8 bits if the file is of higher bit depth, so
it is not a 1:1 representation of the actual data

> In 16-bit mode, the Levels
> theoretically should be 65,535_0, but since that would be impractical to
> display on screen, the distributions are scaled to 255_0.

Correct.

> If the levels in your histogram (14bit-16bit) are bunched up or
> do not fill
> up the entire range (255_0), it is caused by your capture settings and/or
> the dynamic range of the captured image.

Well, not exactly.  The number certainly ARE limited by the capture device
or the image, but they can also be bunched up simply because of bit
justification, as we have discussed over the past few days.

Dynamic range and bit depth. How they relate.

2001-09-28 by Austin Franklin

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

Re: [Digital BW] Bit depth, was Minolta DiMAGEScanMulti PRO

2001-09-28 by Todd Flashner

>> But, in short, most people are of the mind that bit depth and
>> dynamic range
>> are independent qualities, or that's how I've interpreted them.
> 
> Well, they just aren't.  You absolutely do need so many bits to represent an
> entire range of integer density values over a particular dynamic range.  It
> just works that way.  It's like saying counting and numbers are independent.

>> I'll venture
>> they may be right when considering the attributes of a file already before
>> us, but not realize it's a little different when we are discussing the
>> elements of a capture device.
> 
> I am specifically talking only about image capture devices.  I don't know
> what you mean by the attributes of a file already before us.

Put aside how the image was captured; I'm talking (for the moment) about a
file of unknown origin (which is what I mean by "a file already before us").
You can have a file with a DR of 4.0 of any bit depth, and you can have a
file of any bit depth with a DR of 1.7. In this sense BD and DR are
independent. I'm just saying that the DR of such an image certainly can be
adjusted independent of it's BD once it's in hand. You can adjust the file's
end points and spread or condense the tones at will without needing to alter
it's bit depth in order to do so.

Now, I can see if you put in the qualifier that it is essential that the
tones in the file also be contiguous integer density ratio values. (which I
take it you are saying is how digital capture devices work) then I can see
how that changes things.

Am I making sense? Am I getting you right?

>> That¹s because Leaf provides you untonede (linear) raw data.
>> Newer scanners
>> can provide high bit files that are toned.
> 
> Toning has nothing to do with it.  Toned data is NOT raw data.  Toning data
> also does NOT really spread it out, setting the setpoints spreads it out,
> whether you chose to tone it or not.  He is talking about something other
> than raw data.  You CAN get both high bit raw data, and high bit toned data
> out of newer scanners.

I think colloquially he's referring to setting endpoints as toning, but I
agree with you that any manipulation of the file makes it something other
than raw. But a lot of scanner operators would love to get toned highbit
data from their scanners. I think he was just speaking to the possibility of
that option.

>> and you can have a scanner with 12 bits and a dynamic
>> range of 3.8.
> 
> Absolutely wrong.  A dynamic range of 3.8 specifies integer ratio values
> from 1 to 6,309, which means you need a number of bits that can hold these
> numbers, and 12 bits can only hold 4,096 different values.
> 
>> Bit depth is the number of steps. Dynamic range is
>> the height
>> of the star case. You can have a staircase that¹s 20 feet high and have 40
>> steps. You can have a staircase that¹s 30 feet high and have 30 steps.
> 
> He is right, but contradicts his example above!  If you have a bit depth of
> 12, as he states above, you only have 4,096 steps.  If your dynamic range is
> 3.8, which is 10**3.8 or 6,309...you don't have enough steps to hold it.
> Pass this along to him and see what he has to say.

Austin, if you'd like to help me somehow put all of this into a coherent
question, or challenge, to put out there to those "experts" who declare that
DR is independent of BD, I'll be happy to post it. I'd love to hear their
responses. OTOH, THEY may very well appreciate the education!

Thanks again,
Todd

Dynamic range and bit depth. How they relate.

2001-09-28 by JR Geoffrion

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@...
Show quoted textHide quoted text
----- 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



[Non-text portions of this message have been removed]

RE: [Digital BW] Dynamic range and bit depth. How they relate.

2001-09-28 by Nij

I think this point is key. To the explanation of everything ;)

Austin, I haven't followed you and Todd and Mike through every twist and
turn of this discussion, but it seems like to me you are discussing the
'optimal' solution... perhaps if you were given the task of designing a
scanner from the ground up - new light source, new CCD, and on through the
system. In other words, as an engineer - you would want to match the noise
characteristics of the light / CCD 'source' to the A/D capabilities (and
probably match the quality of ALL the circuitry all the way through the
system).

However, we all know that 'marketting' and 'costing' come into these
equations. Marketing says "We just got this great new 256bit A/D chip for
$2 - let's stick that in" - where an engineer may well have noise
characteristics that demand a 11 bit A/D [is that possible] - but only have
12 or 14bit in their stock of surplus from last time.

Plus, of course, I would assume that the quality of ADC's can be extremely
variable for the same bit-depth... so more noise possibilities there too.

Best,
Nij


> -----Original Message-----
> From: Austin Franklin [mailto:darkroom@...]
<snip>
Show quoted textHide quoted text
> 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.

RE: [Digital BW] Dynamic range and bit depth. How they relate.

2001-09-29 by Austin Franklin

> which means
> that you have to multiply by 3 to get the number of colors
> possible

It's actually to the third power...as in, 8 bit RGB (8 bits per color) is
((2**8)**3) or 16,777,216, which is the same as 2**24th...

> (using RGB images) while DR refers the shades of grey
> between white and black.

Not necessarily gray, but shades (luminance)...there is a filter over the
sensor for RGB, or different colored lights are used.

> Assuming a constant DR for each color,
> they can be converted as demonstrated bellow.

Interestingly enough, the DR isn't that constant for each color.  It's
reasonably close, but can be off by almost a log factor of .2-.4 (as in 1.5
to 2.5 times the density difference) through the entire range.

Dynamic range and bit depth. How they relate.

2001-09-29 by JR Geoffrion

> which means
> that you have to multiply by 3 to get the number of colors
> possible

It's actually to the third power...as in, 8 bit RGB (8 bits per color) is
((2**8)**3) or 16,777,216, which is the same as 2**24th...

Good catch... I slipped there...

JR Geoffrion
Chicago, IL


[Non-text portions of this message have been removed]

Re: Dynamic range and bit depth. How they relate.

2001-09-29 by Martin Wesley

Austin,

Well stated.

I think some of the misunderstandings arise over confusions about 
what dynamic range is being discussed. Each point in the process has 
its own dynamic range. The original scene, the image on the film, the 
light input to the scanner's sensor(s), the voltage output from the 
sensor(s), the output at the A/D converter, the output from the 
scanner to the computer, the final print itself. Likewise you can 
combine any number of these steps and talk about the dynamic range of 
that segment of the process.

It is important to define exactly what dynamic range is being 
discussed. Which brings us back to the very beginning of the current 
discussion. Logic would assume that when a scanner manufacturer 
claims a certain dynamic range for their product is would be the DR 
for the segment of the process spanning the image input to the 
scanner through what is actually transferred to the computer. In 
practice it appears that the claims made have little to do with the 
function of the scanning process as a whole but rather pick the 
subsection of the process that will give the highest number for 
marketing purposes.
 
Martin Wesley



--- In DigitalBlackandWhiteThePrint@y..., "Austin Franklin" 
<darkroom@i...> wrote:
> 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.
> 
(Snip)

Re: [Digital BW] Re: Dynamic range and bit depth. How they relate.

2001-09-29 by Todd Flashner

> It is important to define exactly what dynamic range is being
> discussed.

I think Martin makes a good point. Though I've never been in this
conversation to know how the scanner manufacturers get away making absurd
claims. I've wondered why some of the PS gurus I admire have stated that BD
and DR are independent, and therefor seemingly in conflict with Austin's
undeniable logic. I'm starting to conclude that some of the differences of
opinion on the subject stem from where the person stands when they look at
the situation. Austin looks at it like an engineer who looks forward. He
needs to figure out how to get a piece of film with a DR of 3.7 into a file
which can be read by an image editing program. For his job, everything he's
stated is absolutely correct. The Photshop gurus on the other hand don't
really care what machinations analog data goes through on it's way to
digital code, they are looking backward. They are concerned with what the
file they just double clicked contains. From their point of view, they know
they've worked on high BD files with low DR, and low BD files of high DR,
and they know through manipulations they can alter the DR of a file at will,
regardless of bit depth.

Er, another light bulb turning on... If what I just stated about the PS
gurus were true, then why is it that those same gurus who proclaim that DR
is not bit depth dependent, are the ones who recommend doing major tonal
corrections (including DR manipulations, like changing end points) recommend
doing them in 16 bit mode rather than 8 bit? I guess we get into matters of
semantics here, where you may technically be able to have a low BD file with
high DR, but from a usability stand point it might suck. Which I think was
what Austin was suggesting.

Todd

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