nikon 5000ed vs minolta 5400

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Alan Browne
Guest

Posted: Thu Dec 09, 2004 1:14 am Post subject: Re: nikon 5000ed vs minolta 5400

Dps wrote:

Quote:

Hi Alan,

That makes sense, so the Drange would be what?

log(2^14.5) - log(2^1.5) = 3.9 which is plausible

For "number quantization levels", what specifically do you mean? the

number of

bits?

"number quantization levels", or L, is the number of different binary
numbers you need and L<=2^b where b is the number of bits, in our case say
b=16 so L<=65536. Now, Dmax and Dmin are the max and min values
respectively of the sampled analog signal. If Dmin=0 and Dmax=65535 (i.e.
65536 numbers), then (Dmax-Dmin)/(L-1)=1. Now, for the particular case of
scanners, I do not know the max and min values, but I can definately tell
you that
log(2^14.5) - log(2^1.5) = (14.5-1.5)*log2=3.9, which may translate:

"The way the signal processing guys have said it to me is that the dynamic
range
coming out of an A/D converter is all the bits less the bottom 1.5 bits."

into (16-1.5)*log2=4.36 but again, I am not sure if this is the way to
calculate it....

It is simply the log compression of the full range, so log(2^16) before
considering noise. (actually log(2^16-1) would probably be right but
insignificant in log terms).

So (Dmax-Dmin)/(L-1) does not make sense to me...

--
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-- e-meil: there's no such thing as a FreeLunch.

Alan Browne
Guest

Posted: Thu Dec 09, 2004 1:51 am Post subject: Re: nikon 5000ed vs minolta 5400

Bruce Graham wrote:

Quote:

"Unattended auto-> scanning" is an oxymoron.

....redundant, but not an oxymoron.

--
-- r.p.e.35mm user resource: http://www.aliasimages.com/rpe35mmur.htm
-- r.p.d.slr-systems: http://www.aliasimages.com/rpdslrsysur.htm
-- [SI] gallery & rulz: http://www.pbase.com/shootin
-- e-meil: there's no such thing as a FreeLunch.

Dps
Guest

Posted: Thu Dec 09, 2004 2:53 pm Post subject: Re: nikon 5000ed vs minolta 5400

Quote:

this is the resolution which tells you how close a number must be to a
quantisation level, in order to make it equal to that level. In the example
I gave you, you have a resolution of 1, so, depending on the roundoff
procedure, you could say that 23000.2 is equal to 23000 and 23000.8 is equal
to 23001. 0.2 and 0.8 are smaller than the resolution, and you have to map
each discrete signal sample to the 'closest' quantisation level.

Regards,

Dimitris

P.S. I am not very good in explaining things, am I? ;-)

Bruce Graham
Guest

Posted: Thu Dec 09, 2004 3:23 pm Post subject: Re: nikon 5000ed vs minolta 5400

In article <bPJtd.26075$Ou1.1640735@weber.videotron.net>,
alan.browne@FreeLunchVideotron.ca says...

Quote:

Bruce Graham wrote:

"Unattended auto-> scanning" is an oxymoron.

...redundant, but not an oxymoron.

you are quite correct - I was waiting for your comment about 2 sec after

posting...I meant that my attempts to scan unattended with a four slide
feeder have been very unproductive.

Alan Browne
Guest

Posted: Fri Dec 10, 2004 4:11 am Post subject: Re: nikon 5000ed vs minolta 5400

Bruce Graham wrote:

Quote:

In article <bPJtd.26075$Ou1.1640735@weber.videotron.net>,
alan.browne@FreeLunchVideotron.ca says...

Bruce Graham wrote:

"Unattended auto-> scanning" is an oxymoron.

...redundant, but not an oxymoron.

you are quite correct - I was waiting for your comment about 2 sec after
posting...I meant that my attempts to scan unattended with a four slide
feeder have been very unproductive.

I was waiting for your *oops*/correction before posting...

When I do full 5400 dpi ICE scans I also work on previous scans in parallel.
Usually the scanner provides work faster than I can do my cropping, color
corrections, USM save,
re-size-USM-reload-save-re-size-USM-save-reload-re-size-USM save (etc.) loop.
Sometimes I'll load a final four when I go to bed, but it's not really worth it.

Cheers,
Alan.

--
-- r.p.e.35mm user resource: http://www.aliasimages.com/rpe35mmur.htm
-- r.p.d.slr-systems: http://www.aliasimages.com/rpdslrsysur.htm
-- [SI] gallery & rulz: http://www.pbase.com/shootin
-- e-meil: there's no such thing as a FreeLunch.

Alan Browne
Guest

Posted: Fri Dec 10, 2004 4:16 am Post subject: Re: nikon 5000ed vs minolta 5400

Dps wrote:

Quote:

It is simply the log compression of the full range, so log(2^16) before
considering noise. (actually log(2^16-1) would probably be right but
insignificant in log terms).

So (Dmax-Dmin)/(L-1) does not make sense to me...

this is the resolution which tells you how close a number must be to a
quantisation level, in order to make it equal to that level. In the example
I gave you, you have a resolution of 1, so, depending on the roundoff
procedure, you could say that 23000.2 is equal to 23000 and 23000.8 is equal
to 23001. 0.2 and 0.8 are smaller than the resolution, and you have to map
each discrete signal sample to the 'closest' quantisation level.

That I know, but consider what we're trying to find out... what is the Drange?

If we follow your equation we get:

Dmax-Dmin/(L-1) = (Log(2^16)- Log (2^1.5)) / Log (2^16-1) = 0.9 ... and that
don't dog don't hunt...

--
-- r.p.e.35mm user resource: http://www.aliasimages.com/rpe35mmur.htm
-- r.p.d.slr-systems: http://www.aliasimages.com/rpdslrsysur.htm
-- [SI] gallery & rulz: http://www.pbase.com/shootin
-- e-meil: there's no such thing as a FreeLunch.

Dps
Guest

Posted: Fri Dec 10, 2004 1:30 pm Post subject: Re: nikon 5000ed vs minolta 5400

Hi Alan,

Quote:

It is not my equation, it comes right out of every single introductory
handbook to DSP. DRange=Dmax-Dmin. Resolution=DRange/(L-1). And yes, I am
obviously not good at explaining things ;-D

Dimitris

Alan Browne
Guest

Posted: Fri Dec 10, 2004 9:26 pm Post subject: Re: nikon 5000ed vs minolta 5400

Dps wrote:

Quote:

Hi Alan,

That I know, but consider what we're trying to find out... what is the

Drange?

If we follow your equation we get:

Dmax-Dmin/(L-1) = (Log(2^16)- Log (2^1.5)) / Log (2^16-1) = 0.9 ... and

that

don't dog don't hunt...

It is not my equation, it comes right out of every single introductory
handbook to DSP. DRange=Dmax-Dmin. Resolution=DRange/(L-1). And yes, I am
obviously not good at explaining things ;-D

It is "your" equation in this thread, you presented it.

And oops. What did I miss? Now you're saying DRange is Dmax-Dmin... so it is
as I stated at the start: log(2^14.5) or log(2^16) - log (2^1.5). Which is
possibly higher than the Dmax of the film... so 16 bit scanners should be
getting more Drange than the film can provide.

Cheers,
Alan

--
-- r.p.e.35mm user resource: http://www.aliasimages.com/rpe35mmur.htm
-- r.p.d.slr-systems: http://www.aliasimages.com/rpdslrsysur.htm
-- [SI] gallery & rulz: http://www.pbase.com/shootin
-- e-meil: there's no such thing as a FreeLunch.

Dps
Guest

Posted: Fri Dec 10, 2004 9:42 pm Post subject: Re: nikon 5000ed vs minolta 5400

Quote:

It is "your" equation in this thread, you presented it.

OK, OK, I just said it is n't mine, as I cannot claim its invention ;-)

Quote:

And oops. What did I miss? Now you're saying DRange is Dmax-Dmin... so
it is
as I stated at the start: log(2^14.5) or log(2^16) - log (2^1.5). Which
is
possibly higher than the Dmax of the film... so 16 bit scanners should be
getting more Drange than the film can provide.

That's what I've been saying in the first place!!! check my first post
(8/12)!!!! BUT all I wanted to say (also check my first post) is that there
is no Dmax or Drange or whatever defined form *FILM*. These are only
meaningful in an A/D conversion, there is no A/D conversion when shooting
film whatsoever!!!!. There is no quantisation when shooting film, because
film exposure is an analog process.

Regards,

dimitris

Alan Browne
Guest

Posted: Fri Dec 10, 2004 11:40 pm Post subject: Re: nikon 5000ed vs minolta 5400

Dps wrote:

Quote:

It is "your" equation in this thread, you presented it.

OK, OK, I just said it is n't mine, as I cannot claim its invention ;-)

And oops. What did I miss? Now you're saying DRange is Dmax-Dmin... so

it is

as I stated at the start: log(2^14.5) or log(2^16) - log (2^1.5). Which

is

possibly higher than the Dmax of the film... so 16 bit scanners should be
getting more Drange than the film can provide.

That's what I've been saying in the first place!!! check my first post
(8/12)!!!! BUT all I wanted to say (also check my first post) is that there
is no Dmax or Drange or whatever defined form *FILM*. These are only
meaningful in an A/D conversion, there is no A/D conversion when shooting
film whatsoever!!!!. There is no quantisation when shooting film, because
film exposure is an analog process.

Dynamic range is quantifiable regardless of a signal being digital or analog.
Ask an electrical engineer. Or here: http://www.jeffrowland.com/tectalk6.htm

Film is not an "analog", it is an image. ("Analog" means, "by analogy", such as
a voltage representing a temperature sensor and a meter indicating that voltage
as a temperature instead of as a voltage.).

The confusion (as much mine as anyone's) is that the term Dmax for film means
maximum density, and this has a figure of 4.0. A film burned clear would have a
Dmin approaching 0 and a signal on the A/D would be at or near maximum.
Conversely, the densest area would have a signal at the A/D approaching 0, but
necessarilly would contain noise. The range between the two can be construed as
the Dynamic range of the film.

Cheers,
Alan.

--
-- r.p.e.35mm user resource: http://www.aliasimages.com/rpe35mmur.htm
-- r.p.d.slr-systems: http://www.aliasimages.com/rpdslrsysur.htm
-- [SI] gallery & rulz: http://www.pbase.com/shootin
-- e-meil: there's no such thing as a FreeLunch.

Alex Tutubalin
Guest

Posted: Sat Dec 11, 2004 1:26 pm Post subject: Re: nikon 5000ed vs minolta 5400

mark wrote:

Quote:

I am deciding between two scanners: the nikon coolscan 5000ED vs. the
minolta scan elite 5400. Both have a 4.8 Dynamic range and share

Mark,

I've used both SE5400 and 5000ED in parallel for several months.
After that, I've decided to keep Nikon and sell Minolta.

Nikon scanner is
* sharper! (esp. visible on Velvia scans)
* faster
* can scan entire roll (with SA-30 adapter)
* film loading is faster and confortable

On the other side, Minolta can resolve more fine details but
it is not needed for 35-mm scans (I've never print photos larger
than 16x20 from 35-mm)

Alex Tutubalin,
Moscow, Russia

Kennedy McEwen
Guest

Posted: Sat Dec 11, 2004 8:46 pm Post subject: Re: nikon 5000ed vs minolta 5400

In article <B4mud.31618$bD6.758876@wagner.videotron.net>, Alan Browne
<alan.browne@FreeLunchVideotron.ca> writes

Quote:

Dynamic range is quantifiable regardless of a signal being digital or
analog. Ask an electrical engineer. Or here:
http://www.jeffrowland.com/tectalk6.htm

Of course, and the term existed before the concept of an analogue to

digital convertor was even invented.

Quote:

Film is not an "analog", it is an image. ("Analog" means, "by
analogy", such as a voltage representing a temperature sensor and a
meter indicating that voltage as a temperature instead of as a voltage.).

However, Alan, by that definition (which I partially disagree with in

any case) film certain is an analogy of the scene, and thus analogue.

My disagreement with your definition becomes apparent when you consider
that the temperature can also be represented by a series of digital
numbers on a meter just as well - the numbers are an analogy of the
temperature. Thus, your definition leads to the immediate paradox that
digital is analogue - and thus, using the established "proof by
contradiction" method, your definition must be false.

In general terms, analogue means "continuous" whilst digital means
"discrete". A digital representation of the signal can only indicate
discrete values, whilst an analogue representation can indicate all
values with infinitesimal discrimination. The digital representation
can only indicate the signal to the noise floor if there are sufficient
discrete steps, however the analogue representation *always* indicates
the signal into the noise floor.

Quote:

why the DRange of the scanner *MUST* be significantly higher than the
DRange of the film it is scanning. This relates to perception and gamma
as much as it does to the difference between analogue and digital.

Film reproduces the luminance changes in the shadows by increasing the
number and size of silver grains or dye clouds per unit area. This is
continuous, and thus effectively analogue - even at the quantum level it
remains analogue with the presence of atoms in a unit area being
probablistic.

If the DRange of the scanner simply matched the DRange of the film then
the Dmax on the film would represent a count of 0 from the scanner,
whilst the Dmin on the film would be represented by a number close to
(2^n)-1. The next darkest level on the scanner from this Dmin
representation would be a count of one less, which would be a virtually
indistinguishable visual change of density on the film. However the
next lightest level from Dmax which the scanner could represent would be
1, due to the discrete nature of the digital data. This would represent
DMax-0.3 on the film, and would be visibly discrete and lighter than the
Dmax. For example, film Dmax is generally around 3 - 3.6, which should
require no more than 10-12-bits, however a 1 bit change in the shadows
of such a linearly encoded digital image is clearly visible. In short,
an ADC which has a DRange equal to the film is inadequate to
discriminate the shadow information.

The problem this throws up is that the equations that both you and
Dimitris have been debating relate to *linear* representations of
signals, whilst our perception of density is very non-linear. It is
well known that your vision is more sensitive to fine changes of
luminance in the shadows than it is in the highlights of the scene,
hence the use of gamma encoding to minimise the bit depth used to
describe the image digitally. This means that the dynamic range
required to describe the shadows on the image is much higher than
dynamic range necessary to describe the highlights - however, your
equations relate to a uniform dynamic range in a linear system which, at
best, indicates an *average* dynamic range. Since perception of
discrete steps is the driver for quantisation level, your equations for
Drange should be applied in perceptual space, the space in which the
discrimination of discrete steps is equal throughout the range, ie.
perceptually linear. The results can then be transformed via the
inverse gamma space to voltage, luminance and digital linear space to
determine the Drange necessary to describe the shadows without
posterisation, and thus the minimum Drange necessary on the scanner.
--
Kennedy
Yes, Socrates himself is particularly missed;
A lovely little thinker, but a bugger when he's pissed.
Python Philosophers (replace 'nospam' with 'kennedym' when replying)

Alan Browne
Guest

Posted: Sun Dec 12, 2004 4:48 am Post subject: Re: nikon 5000ed vs minolta 5400

Kennedy McEwen wrote:

Quote:

An image of a scene is an image. We can look at it, project it, scan, replicate
it, etc. We don't need to 'convert' its meaning in any way.

Quote:

The 'digital' temp meter has made a translation in information type. It samples
the voltage, converts to digital and then discrete logic or s/w converts it into
numbers.

IAC, as previously ranted, I simply do not like the term analog for film as I
believe it had come up as some sort of counter for the term digital as applied
to cameras. Did anyone call film "analog" before digital? We certainly called
analog circuitry analog circuitry long before the great rise of digital systems.
(Most computers in my industry were certainly dominated by their analog
interfaces and sensors over the relative simplicity of the digital computer
part).... it never entered anyone's mind to call film "analog"... it was a
recording of an image and that image was (post dev.) identifiable as such on the
same film on which it was shot.

Quote:

In general terms, analogue means "continuous" whilst digital means
"discrete".

I agree with that, but only as that is the nature of most analog signals. The
signal however remains some abstraction (as a voltage, current, frequency) of
something else it represents. A film image is undeniably an image.

Quote:

The confusion (as much mine as anyone's) is that the term Dmax for
film means maximum density, and this has a figure of 4.0. A film
burned clear would have a Dmin approaching 0 and a signal on the A/D
would be at or near maximum. Conversely, the densest area would have a
signal at the A/D approaching 0, but necessarilly would contain
noise. The range between the two can be construed as the Dynamic
range of the film.

What seems to be missing in this entire thread is any consideration of
why the DRange of the scanner *MUST* be significantly higher than the
DRange of the film it is scanning. This relates to perception and gamma
as much as it does to the difference between analogue and digital.

Film reproduces the luminance changes in the shadows by increasing the
number and size of silver grains or dye clouds per unit area. This is
continuous, and thus effectively analogue - even at the quantum level it
remains analogue with the presence of atoms in a unit area being
probablistic.

If the DRange of the scanner simply matched the DRange of the film then
the Dmax on the film would represent a count of 0 from the scanner,
whilst the Dmin on the film would be represented by a number close to
(2^n)-1. The next darkest level on the scanner from this Dmin
representation would be a count of one less, which would be a virtually
indistinguishable visual change of density on the film. However the
next lightest level from Dmax which the scanner could represent would be
1, due to the discrete nature of the digital data. This would represent
DMax-0.3 on the film, and would be visibly discrete and lighter than the
Dmax. For example, film Dmax is generally around 3 - 3.6, which should
require no more than 10-12-bits, however a 1 bit change in the shadows
of such a linearly encoded digital image is clearly visible. In short,
an ADC which has a DRange equal to the film is inadequate to
discriminate the shadow information.

Agree. I don't see how having more bits gives much improvement in the shaddows,
however, as a minute change in the light going through, with the high resolution
of the low order bits will raise the detector value from very low to some what
high. eg: a slight change in the light level will result in several bits worth
of information. (eg: go from values close to 0 to values in excess of 32 or 64,
quite quickly.)

Quote:

The problem this throws up is that the equations that both you and
Dimitris have been debating relate to *linear* representations of
signals, whilst our perception of density is very non-linear. It is
well known that your vision is more sensitive to fine changes of
luminance in the shadows than it is in the highlights of the scene,
hence the use of gamma encoding to minimise the bit depth used to
describe the image digitally. This means that the dynamic range
required to describe the shadows on the image is much higher than
dynamic range necessary to describe the highlights - however, your
equations relate to a uniform dynamic range in a linear system which, at
best, indicates an *average* dynamic range. Since perception of
discrete steps is the driver for quantisation level, your equations for
Drange should be applied in perceptual space, the space in which the
discrimination of discrete steps is equal throughout the range, ie.
perceptually linear. The results can then be transformed via the
inverse gamma space to voltage, luminance and digital linear space to
determine the Drange necessary to describe the shadows without
posterisation, and thus the minimum Drange necessary on the scanner.

I'm beginning to see the light on this one. Thanks. Is it possible, then, to
map the gamma curve over the film to A/D response and thereby determine the
minimum number of bits required to span the response of the film?

Cheers,
Alan

--
-- r.p.e.35mm user resource: http://www.aliasimages.com/rpe35mmur.htm
-- r.p.d.slr-systems: http://www.aliasimages.com/rpdslrsysur.htm
-- [SI] gallery & rulz: http://www.pbase.com/shootin
-- e-meil: there's no such thing as a FreeLunch.

Kennedy McEwen
Guest

Posted: Sun Dec 12, 2004 6:52 am Post subject: Re: nikon 5000ed vs minolta 5400

In article <UGLud.34374$bD6.1552342@wagner.videotron.net>, Alan Browne
<alan.browne@FreeLunchVideotron.ca> writes

Quote:

Kennedy McEwen wrote:

Film is not an "analog", it is an image. ("Analog" means, "by
analogy", such as a voltage representing a temperature sensor and a
meter indicating that voltage as a temperature instead of as a voltage.).

However, Alan, by that definition (which I partially disagree with in
any case) film certain is an analogy of the scene, and thus analogue.

An image of a scene is an image. We can look at it, project it, scan,
replicate it, etc. We don't need to 'convert' its meaning in any way.

But what is recorded on film is an analogy of that scene, comprising a

mass of grains and/or dye clouds to represent the luminance levels in
the original scene. The fact that this can be viewed as an image by
projecting light through it is no different from the voltage recording
of a sound wave, which can just as readily be used to reproduce a
further analogy of the original sound be feeding it to a speaker.

Quote:

Only in the same way as the digital scan of the image, whether or not
film forms an intermediary step.

Quote:

It samples the voltage, converts to digital and then discrete logic or
s/w converts it into numbers.

Just as happens with the scanner or digital camera.

Quote:

IAC, as previously ranted, I simply do not like the term analog for
film as I believe it had come up as some sort of counter for the term
digital as applied to cameras. Did anyone call film "analog" before
digital? We certainly called analog circuitry analog circuitry long
before the great rise of digital systems.

Actually, we didn't. In fact, even systems that we would now consider
to be digital in nature, such as the original PCM system proposed by
Alec Reeves in 1941 were not described as digital for some considerable
time, and neither were any that we would now call analogue so described.
Even Claude Shannon's 1948 paper on information theory, whilst refering
to systems of decimal-digits and binary-digits (and inventing the term
bit) failed to discriminate between analogue and digital encoding
schemes, referring to continuous and discrete. Even the original
publications on CCDs by Boyle & Smith or Amelio in 1970, as a device
which quantised what we would now term analogue signals in time, but not
level, did not use language to discriminate digital and analogue
techniques. The use of the terms analogue and digital first appear
during the development of the computer, to distinguish between computing
systems which used continuous analogue computations and those which used
discrete numerical calculations. The transfer of that terminology to
describe electrical systems only appears in the late 1960's when such
computational processing could be applied to real signals.

Quote:

as digital, film is an analogue representation of the image. The issue
being that it is effectively a continuous rather than a discrete record
of the luminance it each point in the scene. The fact that we didn't
understand the distinction at the time of the original invention doesn't
mean the distinction didn't exist. Darwin wasn't aware of DNA when he
recognised the principle of evolution, but we now know that it is random
mutations of DNA that is responsible for it.

Quote:

In general terms, analogue means "continuous" whilst digital means
"discrete".

I agree with that, but only as that is the nature of most analog
signals.

It is the nature of *all* analogue systems. That is what distinguishes
digital from analogue. We have become used to consider digital as only
being binary in nature, with only two discrete levels, however there are
multi-level digital systems as well, with 3, 4, 5 or more discrete
levels of digital data. Indeed, if you ever used a 56K modem in your
computer then its operation at the highest data rate relied on such
multi-level digital signals.

Quote:

The signal however remains some abstraction (as a voltage, current,
frequency) of something else it represents. A film image is undeniably
an image.

It is still an abstraction in a different medium from the original scene
and does not contain all of the information of the original scene, only
a limited selection that camera lens can reproduce and the film record.
It is inherently analogue in nature.

Quote:

I don't see how having more bits gives much improvement in the
shaddows,

Because your eye is more sensitive to changes in the shadows.

Quote:

analogue signal then you will not be able to represent those small
changes at low levels. You need more bits than the analogue Drange. In
addition to this, you need to digitise into the noise floor to
adequately represent the noise itself.

Quote:

The problem this throws up is that the equations that both you and
Dimitris have been debating relate to *linear* representations of
signals, whilst our perception of density is very non-linear. It is
well known that your vision is more sensitive to fine changes of
luminance in the shadows than it is in the highlights of the scene,
hence the use of gamma encoding to minimise the bit depth used to
describe the image digitally. This means that the dynamic range
required to describe the shadows on the image is much higher than
dynamic range necessary to describe the highlights - however, your
equations relate to a uniform dynamic range in a linear system which,
at best, indicates an *average* dynamic range. Since perception of
discrete steps is the driver for quantisation level, your equations
for Drange should be applied in perceptual space, the space in which
the discrimination of discrete steps is equal throughout the range,
ie. perceptually linear. The results can then be transformed via the
inverse gamma space to voltage, luminance and digital linear space to
determine the Drange necessary to describe the shadows without
posterisation, and thus the minimum Drange necessary on the scanner.

I'm beginning to see the light on this one. Thanks. Is it possible,
then, to map the gamma curve over the film to A/D response and thereby
determine the minimum number of bits required to span the response of
the film?

Yes - it is somewhere in the region of 17 to 18-bits of linear encoding

with typical slide film.
--
Kennedy
Yes, Socrates himself is particularly missed;
A lovely little thinker, but a bugger when he's pissed.
Python Philosophers (replace 'nospam' with 'kennedym' when replying)

Alan Browne
Guest

Posted: Sun Dec 12, 2004 9:42 pm Post subject: Re: nikon 5000ed vs minolta 5400

Kennedy McEwen wrote:

Quote:

projecting light through it is no different from the voltage recording
of a sound wave, which can just as readily be used to reproduce a
further analogy of the original sound be feeding it to a speaker.

I really don't want to debate this, but a voltage recording of a soundwave needs
a transducer to allow conversion back to the orignal sound. All I need to look
at filmphoto is the filmphoto.

Quote:

My disagreement with your definition becomes apparent when you
consider that the temperature can also be represented by a series of
digital numbers on a meter just as well - the numbers are an analogy
of the temperature. Thus, your definition leads to the immediate
paradox that digital is analogue - and thus, using the established
"proof by contradiction" method, your definition must be false.

The 'digital' temp meter has made a translation in information type.

Only in the same way as the digital scan of the image, whether or not
film forms an intermediary step.

It samples the voltage, converts to digital and then discrete logic or
s/w converts it into numbers.

Just as happens with the scanner or digital camera.

That is a translation (or conversion). The film, alone, conveys the same
information without translation/conversion. A digital camera makes its A/D
conversion in real time and immediately afterwards destroys the image itself.
The file created is maeningless without interpretation (as is a scan).

Quote:

computational processing could be applied to real signals.

You're going back too far with that one.

Quote:

(Most computers in my industry were certainly dominated by their
analog interfaces and sensors over the relative simplicity of the
digital computer part).... it never entered anyone's mind to call film
"analog"... it was a recording of an image and that image was (post
dev.) identifiable as such on the same film on which it was shot.

Nevertheless, just as we can now refer to the original telegraph system
as digital, film is an analogue representation of the image. The issue
being that it is effectively a continuous rather than a discrete record
of the luminance it each point in the scene. The fact that we didn't
understand the distinction at the time of the original invention doesn't
mean the distinction didn't exist. Darwin wasn't aware of DNA when he
recognised the principle of evolution, but we now know that it is random
mutations of DNA that is responsible for it.

You're stretching you analogies further and further...

Quote:

I don't see how having more bits gives much improvement in the shaddows,

Because your eye is more sensitive to changes in the shadows.

That is now sufficiently clear, thanks.

Quote:

however, as a minute change in the light going through, with the high
resolution of the low order bits will raise the detector value from
very low to some what high. eg: a slight change in the light level
will result in several bits worth of information. (eg: go from values
close to 0 to values in excess of 32 or 64, quite quickly.)

But, if the Drange of the digital data is the same as the Drange of the
analogue signal then you will not be able to represent those small
changes at low levels. You need more bits than the analogue Drange. In
addition to this, you need to digitise into the noise floor to
adequately represent the noise itself.

Got it.

Quote:

Thanks.

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