nikon 5000ed vs minolta 5400

  • Thread starter Thread starter mark
  • Start date Start date
Kennedy said:
As I pointed out, the Minolta is not designed for speed or bulk
scanning. That is Nikon's realm - at the moment.
Click to expand...

Yes, that was one of the plusses for Nikon when I made my
recent purchase of the Minolta 5400. The Nikon add-on slide
handler in particular so that a whole roll of slides could be scanned
unattended (as opposed to the Minolta's maximum of four slides
for unattended auto-scanning).

Mike
 
mark said:
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
other similarities, but in terms of use, image capture, resolution,
software, reliability and end results which is the right one to pick?
What should be my other considerations when deciding between these two
scanners. At first I was sold on the Minolta 5400, but I have read
several less than inspiring reviews. While the Minolta model has an
edge over the nikon by offering a claimed 5400 ppi, I have recently
read a less than enthusiastic review in PC Magazine regarding the 5400
(rating it worst among dedicated and flat bed scanners) and I have
read several user reports complaining about mechanical malfunctions
from users. In CNET, the review rated the Minolta 5400 7 out of 10
with the following remarks: "While the Dimage 5400 produces very good
scans, with excellent dynamic range and shadow detail, it isn't
capable of the sharpness we saw from the Nikon Coolscan V. As you can
tell, I am a little confused in deciding which model to pick. Any
advice would be greatly appreciated.

Thanks.

Mark
Click to expand...

This has to be the most comprehensive and informative single thread
I've seen on this topic. You're a lucky guy. :)

false_dmitrii
 
Watch it, or you'll suffer the consequences. Your chair's getting
comfier already....

false_dmitrii
 
Watch it, or you'll suffer the consequences. Your chair's getting
comfier already....

false_dmitrii
 
Maciej said:
Mark, I am not an expert on theory of these things but I understand that
4.8Dmax is _theoretical_ value, determined by the number of bits in the
A/D converter. The real world values are probably closer to 3.2-3.2.
My guess would be that they are similar for both scanners.

Anyone out there who can verify/correct this statement?
Click to expand...

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.

For a 16 bit A/D, manufacturer claimed Dmax: log(2^16) = 4.82
real world Dmax: log(2^14.5) = 4.4

Then, what does film hold? I often hear that the best slide film has a max Dmax
of 4.0...

....

Cheers,
Alan
 
Mark, I am not an expert on theory of these things but I understand that
4.8Dmax is _theoretical_ value, determined by the number of bits in the
A/D converter. The real world values are probably closer to 3.2-3.2.
My guess would be that they are similar for both scanners.

Anyone out there who can verify/correct this statement?
Click to expand...

Yes, I think you are quite right. I have only seen one advert that
specified the optical density range (as well as the theoretical range)
and the scanner didn't do that well anyway.

-- Steven
 
The Nikon add-on slide
handler in particular so that a whole roll of slides could be scanned
unattended (as opposed to the Minolta's maximum of four slides
for unattended auto-scanning).
Click to expand...
My Canon is like your Minolta (4 slides at a time). "Unattended auto-
scanning" is an oxymoron.
 
Then, what does film hold? I often hear that the best slide film has a max Dmax
of 4.0...
Click to expand...

dynamic range=Dmax-Dmin, I believe you confuse dynamic range with Dmax

AFAIK the dynamic range is related to the quantisation in a DSP procedure...
The quantisation step (or resolution) = (Dmax-Dmin)/(L-1) where L is the
number of quantisation levels. There is no DSP procedure when shooting film
.....

-- dimitris
 
Dps said:
max Dmax



dynamic range=Dmax-Dmin, I believe you confuse dynamic range with Dmax

AFAIK the dynamic range is related to the quantisation in a DSP procedure...
The quantisation step (or resolution) = (Dmax-Dmin)/(L-1) where L is the
number of quantisation levels. There is no DSP procedure when shooting film
Click to expand...

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?


Cheers,
Alan
 
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?
Click to expand...
"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....

Regards,

Dimitris
 
Dps said:
Hi Alan,



number of


"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....
Click to expand...

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...
 
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...
Click to expand...

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? ;-)
 
...redundant, but not an oxymoron.
Click to expand...
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.
 
Bruce said:
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.
Click to expand...

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.
 
Dps said:
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.
Click to expand...


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...
 
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...
Click to expand...

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
 
Dps said:
Hi Alan,




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
Click to expand...

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

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



it is



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.
Click to expand...

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.
 
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