Scanning old glass negatives

[kirstenh]

I am doing a lot of historical documents including old postcards and
glass negatives. Has anyone here done glass negatives? A lot of these
are hand dipped. Some are hand cut and not quite square. So far we have
been making contact prints and scanning them. These things are getting
old (like a hundred years old) and the emulsion is starting to flake
off. Each time they are handled may be their last. One problem with a
flatbed scanner is that it focuses on the surface of the glass and not
on the emulsion which is where the picture is. Is there anyone out
there with the same problems ???


Thanks much, K.

 

[CSM1]

I am doing a lot of historical documents including old postcards and
glass negatives. Has anyone here done glass negatives? A lot of these
are hand dipped. Some are hand cut and not quite square. So far we have
been making contact prints and scanning them. These things are getting
old (like a hundred years old) and the emulsion is starting to flake
off. Each time they are handled may be their last. One problem with a
flatbed scanner is that it focuses on the surface of the glass and not
on the emulsion which is where the picture is. Is there anyone out
there with the same problems ???


Thanks much, K.

What about a light box and a camera to photograph the glass negatives? With
the correct copy stand, you can get a very good copy of the glass negative
without touching the emulsion.

I do not claim to know how the scan glass negatives. Much less hundred year
old ones.

You said that the scanner focuses on the surface of the scanner glass, which
is true for a CIS sensor, but you can put the emulsion side of the glass
negative on the scanner glass, if that does not harm the negative. If the
image is reversed or right to left, you can correct in a photo editor
(Photoshop) or even in some scanners twain interface's.

CIS scanners do focus on the surface, but CCD sensors have plenty of depth
of field to handle the thickness of the glass of a glass negative.

If you are using one of the Canon LiDE series scanners, then that is a
problem. Any of the other scanners, with a CCD sensor should work fine.

 

[Don]

CIS scanners do focus on the surface, but CCD sensors have plenty of depth
of field to handle the thickness of the glass of a glass negative.

CCD sensors do have some depth of field but the problem is once you
move the object away from the surface - even a little - you lose
contrast and dynamic range very rapidly (see below).

I would put the glass negative (emulsion down) on the scanner, then
set the scanner to "film" (to turn off the light) and use a diffuse
external light source to illuminate the negative. (I doubt that the
scanner's "film light" in the cover would be enough.)

Another - out of left field - suggestion may be to use a projector and
project the negative focusing to the top of the scanner glass (put a
white sheet of paper to focus, then remove). I haven't actually tried
this, but it should work. Just make sure the projector doesn't shake
as the scanner head moves. As a bonus you would get higher effective
resolution.

OK, back to loss of contrast...

I'm currently wrestling with glossy photos which cause Newton's rings
and sometimes stick to the glass causing gray splotches in the scan.

My solution is to affix the photo to a piece of Plexiglas with
double-sided tape to keep it flat and then use pieces of cardboard
strategically placed around the photo for the Plexiglas to rest on.
That way the whole contraption "floats" just enough above the glass to
avoid Newton's rings and other artifacts. Compared to a regular scan
(photo resting on the glass) there is a noticeable loss of contrast
and dynamic range (e.g. the highlights are darker) but I fix that
later in Photoshop.

Don.

 

[RSD99]

I have done glass negatives ... from the 1940's ... on my Epson Perfection
4870.

The procedure is to place the glass negative emulsion side down on the
scanner's glass bed, and scan exactly like any other negative. Sometimes
careful handling is required, but I have not had any problems with the ones
I have tried. [FWIW: Almost all of them were three-color separations or
three-color "one shot" camera negatives, mostly done on Kodak "Super
Panchro Press Type B" glass plates.

I do *not* think that you will have **any** problems with the emulsion
"flaking off" if it is a standard silver-based emulsion. Unfortunately, I
cannot speak to what kind(s) of problems you might have with older forms of
emulsions ... except that some of them *may* be quite delicate, and you
should probably check with a for-real "Archivist" and learn how to handle
them properly.

I am in the Los Angeles area ... if you have any further questions, you
might try contacting me "off list."

 

[Kennedy McEwen]

Another - out of left field - suggestion may be to use a projector and
project the negative focusing to the top of the scanner glass (put a
white sheet of paper to focus, then remove). I haven't actually tried
this, but it should work.

There isn't a hope in hell of that working with a conventional scanner,
Don, and a simple ray trace will explain why. Ye canny brek the laws o'
physics - light travels in straight lines.

Only the light that passes through the projector lens *and* the flatbed
glass surface *and* the scanner lens can be focussed onto the CCD to
produce the output. Unless both of those lenses are considerably larger
than the scanner glass surface area then only a very small portion of
the image will be focussed onto the CCD - effectively the convolution of
the scanner and projector lens pupils - and that will be pretty small
indeed.
Glass
Scanner G
Lens G
| G
| G
|/SL------I----------------------------------------Projector
x-SL------I----------------------------------------P Lens
|\SL------I----------------------------------------P
^ G
^ G
^ G
CCD G
G
Glass

Only those rays roughly shown above, passing through the points marked
'I' on the flatbed glass can contribute to the image formed on the CCD.
All of the points on the scanner glass marked G have rays which come
from the projector and completely miss the scanner lens, so cannot form
an image. The lenses are fairly small compared to the flatbed width, so
you don't get a very large image area and what you do get is very
unevenly illuminated with a very bright central spot.

A corollary to this method would be if you place a sheet of white card
on your window and project an image onto that for focussing purposes and
then remove the card. Now walk outside and look at the image you see on
your window - you won't see or photograph the image projected, just an
out of focus image of the illuminated projector lens that varies in
intensity as you move your head across the window, changing the relative
alignment of the projector lens with your eye lens.

Short of redirecting every photon incident on the glass towards the
scanner lens, using a fresnel lens, or keeping the white paper on the
glass as a means of scattering the light so that some of it reaches the
scanner lens (although most will be lost) there is no way to get light
form any reasonable image size onto the CCD.

By contrast, if the scanner were a CIS device then there is a good
chance that this technique would work, because the detector itself
subtends the entire scanner width and no scanner lens is necessary, but
there isn't a hope of it working with a conventional CCD scanner.

 

[Don]

There isn't a hope in hell of that working with a conventional scanner,
Don, and a simple ray trace will explain why. Ye canny brek the laws o'
physics - light travels in straight lines.

Hence, the "out of left field' moniker... ;o)

The lenses are fairly small compared to the flatbed width, so
you don't get a very large image area and what you do get is very
unevenly illuminated with a very bright central spot.

I see.

A corollary to this method would be if you place a sheet of white card
on your window and project an image onto that for focussing purposes and
then remove the card. Now walk outside and look at the image you see on
your window - you won't see or photograph the image projected, just an
out of focus image of the illuminated projector lens that varies in
intensity as you move your head across the window, changing the relative
alignment of the projector lens with your eye lens.

That's not exactly the same because there is a distance between the
window pane and the eye. Only if the eye is literally stuck to the
surface of the window pane (or at least the same small distance as
between the scanner lens and the top of the scanner glass) where the
image is focused, would it be the same.

But I take your point. I'm reminded of my 8mm days. A similar effect
may be walking in front of the projection screen, looking at the
projector and expecting to see a clear image. Like you explained
above, the most one may see are image fragments with a bright spot in
the middle.

Followed by stepping on everyone's toes as you try to find your chair
in the dark with severely contracted pupils... ;o)

Short of redirecting every photon incident on the glass towards the
scanner lens, using a fresnel lens, or keeping the white paper on the
glass as a means of scattering the light so that some of it reaches the
scanner lens (although most will be lost) there is no way to get light
form any reasonable image size onto the CCD.

I was just about to suggest that! Is there a ("milky") optical glass
with such properties which could be placed on top of the scanner?

But I see your point about most of the light being lost as it seems to
relate to my current struggle:

I'm now wrestling with glossy photos which not only literally stick to
the scanner glass (causing gray "splotches") but also cause Newton's
rings. I "solved" this by affixing the photo to a piece of Plexiglas
with double sided tape to keep it flat and then raising the whole
contraption with strategically placed pieces of cardboard around the
photo to support the Plexiglas. Although the scanner does have enough
depth of field I do notice a quite radical reduction in contrast which
I have to correct for later.

You explanation above (serendipitously) seems to suggest why.
Apparently, having a photo "float" fractions of a mm above the scanner
glass seems to lose a lot of light too. However, after I correct the
image later I don't really see any loss of detail (subjective
estimate). Is there (theoretically and objectively) any loss of detail
by having the photo "float" just above the glass?

By contrast, if the scanner were a CIS device then there is a good
chance that this technique would work, because the detector itself
subtends the entire scanner width and no scanner lens is necessary, but
there isn't a hope of it working with a conventional CCD scanner.

It would be an interesting CIS test, although as I mentioned last
time, there is still a problem of the scanner and the projector moving
because - unlike an object on the scanner glass - the two are
uncoupled and move independently as the scanner assembly shuttles back
and fort.

Don.

 

[Kennedy McEwen]

That's not exactly the same because there is a distance between the
window pane and the eye.

There is a distance between the scanner lens and glass bed - otherwise
the lens pupil would have to extend the full width of the glass. So it
is just the same situation as described above.

But I take your point. I'm reminded of my 8mm days. A similar effect
may be walking in front of the projection screen, looking at the
projector and expecting to see a clear image. Like you explained
above, the most one may see are image fragments with a bright spot in
the middle.

Exactly - the glass does virtually nothing, excepting some minor
scattering and linear shifts of the non-orthogonal rays, to the light
coming from the projector.

Is there a ("milky") optical glass
with such properties which could be placed on top of the scanner?

A fine ground glass screen can work, which is effectively all that
Anti-Newton-Ring glass is. Even so, that will only allow some of the
rays from all of the image to be scattered into the lens pupil, but
there will still be light fall off towards the edges, or a "hot spot" in
the middle, just as you get with back projection systems if viewed on
axis. A fine fresnel lens will correct this, if you use the correct
focal length, just as the viewing screen on an SLR camera uses ground
glass for image forming and a fresnel lens for illumination flattening.
In this context though, quite an expensive proposition.

Apparently, having a photo "float" fractions of a mm above the scanner
glass seems to lose a lot of light too. However, after I correct the
image later I don't really see any loss of detail (subjective
estimate). Is there (theoretically and objectively) any loss of detail
by having the photo "float" just above the glass?

Light loss in your case is because you are increasing the distance from
the light source. For a point source this is an inverse square law, so
double the distance and you quarter the illumination. Being an extended
source, the fall off is less radical, but how much depends on the light
source geometry.

Loss of detail should only be due to defocus and reduced signal to
noise.

It would be an interesting CIS test, although as I mentioned last
time, there is still a problem of the scanner and the projector moving
because - unlike an object on the scanner glass - the two are
uncoupled and move independently as the scanner assembly shuttles back
and fort.

I doubt that this would be significant given the actual exposure times
for each CCD line - but it depends on the size of the projected image.

 

[Don]

A fine ground glass screen can work, which is effectively all that
Anti-Newton-Ring glass is. Even so, that will only allow some of the
rays from all of the image to be scattered into the lens pupil, but
there will still be light fall off towards the edges, or a "hot spot" in
the middle, just as you get with back projection systems if viewed on
axis. A fine fresnel lens will correct this, if you use the correct
focal length, just as the viewing screen on an SLR camera uses ground
glass for image forming and a fresnel lens for illumination flattening.
In this context though, quite an expensive proposition.

What's a fresnel lens? (I meant to ask last time but I forgot.)

Light loss in your case is because you are increasing the distance from
the light source. For a point source this is an inverse square law, so
double the distance and you quarter the illumination. Being an extended
source, the fall off is less radical, but how much depends on the light
source geometry.

Loss of detail should only be due to defocus and reduced signal to
noise.

As I mentioned before, since I have a conventional CCD scanner there
is sufficient depth of field so I don't see any focusing issues.

I was only concerned with loss of detail but, if I understand
correctly, that shouldn't really be a problem. I need to boost
contrast somewhat in postprocessing to "cheer up" the highlights which
are dulled due to the loss of light, but other than that I'm not
really losing anything, am I?

Don.

 

[Kennedy McEwen]

On Wed, 19 Jan 2005 18:36:59 +0000, Kennedy McEwen

What's a fresnel lens? (I meant to ask last time but I forgot.)

A thin lens - one which is designed without the normal glass bulk. I am
sure you have seen them as cheap lightweight reading magnifiers etc.

Lens power is a consequence of the surface curvature, which refracts the
light according to Snell's law and the angle of incidence due to that
curvature, and the internal medium has no effect since light continues
to travel in a straight line through it to be refracted again at the
rear surface, which is usually at a different angle. If this was the
same angle as the front surface, then the lens has no power and the
light is refracted back to its original path - which is why you can see
an undistorted image through a flat sheet of glass, or a meniscus lens
at the correct viewing distance.

In a conventional lens the curvature of the surface requires significant
bulk in the lens body because the surface is continuous. However, since
only the curvature is important in the lens function, the surface can be
designed with discontinuities so that the bulk of the lens is
eliminated. If the discontinuities are perfectly perpendicular to the
light path then the Fresnel lens will perform exactly as well as the
conventional lens, but this can only be achieved for one particular
point, usually on the focal plane. At all other points, the
discontinuities in the surface cause scattering and loss of performance.

Some lens designs combine the fresnel approach with conventional lenses
and another of Fresnel's developments, the zone plate, to produce what
are known as binary or diffractive optics, where the discontinuities
occur exactly on the zone plate transitions, so contributing to the lens
performance rather than detracting from it.

As I mentioned before, since I have a conventional CCD scanner there
is sufficient depth of field so I don't see any focusing issues.

Nevertheless, they will be there, however minor they are. There is only
one point of focus, everywhere else is just an approximation.

I was only concerned with loss of detail but, if I understand
correctly, that shouldn't really be a problem. I need to boost
contrast somewhat in postprocessing to "cheer up" the highlights which
are dulled due to the loss of light, but other than that I'm not
really losing anything, am I?

You will possibly be losing some detail due to defocus and, whilst you
don't notice any, without knowledge of the scanner design it is
impossible to determine whether any loss is significant or not.

 

[tom]

Hello, the best kind of scanner for glass negs with peeling emulsions
is the kind with a "film drawer" such as the Microtek 1800f or the old
Agfas. In this arrangement the negative is scanned with the emulsion
up, so the emulsion never is touched. Unfortunately there is newton
ring problems when the glass plate touches the glass bed of the film
drawer, so I have to apply a little tape on each side or corner of the
base side of the plate to act as a spacer to lift the plate slightly up
off the glass.

Top issue for scanning old plates is dmax, which is the range of dark
to light values that a scanner can see through. As far as I know, the
Artixscan 1800f, which is available for about $1,000, has the greatest
depth of any scanner under $10,000, and it is better than the Artixscan
2500 or any of the Epsons. If you have the money, the best scanner for
glass plates is the Creo Eversmart Supreme 2, which is about $28,000.
Used Creos, Fujis, and other high end scanners are occasionally
available used for about $5000, but if you are buying used scanners you
need to excersise extraordinary precaution to protect yourself if the
deal goes bad, and you need to learn how to dismantle and repair the
scanner yourself. Most scanner factories now only offer repair service
for the warranty period, and after that you need to pay to exchange
your old scanner for a new one. Parts and service are unavailable for
Microtek, Epson, etc., after the warranty period. Buyer beware.

If you are scanning old plates, you will be much better off to do it in
16 bit rather than 8 bit mode. Feel free to contact me off-list, i
scan glass plates for a living.
Tom Robinson
http://historicphotoarchive.com

 

[Don]

A thin lens - one which is designed without the normal glass bulk. I am
sure you have seen them as cheap lightweight reading magnifiers etc.
...

Thanks as always, Kennedy!

You will possibly be losing some detail due to defocus and, whilst you
don't notice any, without knowledge of the scanner design it is
impossible to determine whether any loss is significant or not.

That's OK. I ran tests by comparing a normal (contact) scan with a one
where I elevated the photo very slightly - just enough to eliminate
Newton's rings. The elevated scan had a bit dull highlights which, at
first blush, looked like loss of some detail. But after I adjusted the
contrast and then compared the two I did not really see any noticeable
difference. There may be some but taking all things into consideration
it's certainly acceptable.

For example, by affixing the photo to a piece of Plexiglas with
double-sided tape and so keeping it very flat, I've eliminated
distortions present in the regular, contact scan. Namely, in spite of
it being held down by the scanner lid, the photos are not really
totally flat. I wasn't even aware of this until the test. So, on
balance, that compensates for any minute loss of detail. Besides, the
number of my glossy prints (where this is needed) is quite small.

One other interesting thing I noticed on some photos was the "color
shine" on leading and trailing edges. This became really apparent when
I scanned B&W photos (I scanned them in color.)

On some prints (particularly the ones on thicker paper) the leading
edge had a distinct blue "shine" while the trailing edge had a red
"shine". It's less than 1 mm in absolute terms but magnified in
Photoshop it's quite visible. I then turned the picture around and the
same thing happened indicating there was nothing on the photo.

Why is that? Could it be like the red and blue shift observed in
astronomy? The speed of the scanner assembly is not exactly on par
with celestial objects moving away or towards us but, then again, one
notices the pitch shift of the police car siren when it passes by and
sound is considerable slower than light.

Or is this shift caused by filters in the scanner being at different
angles?

Anyway, I've just finished my last picture yesterday so it's back to
torturing myself with Kodachromes next... ;o)

Don.

 

[Kennedy McEwen]

One other interesting thing I noticed on some photos was the "color
shine" on leading and trailing edges. This became really apparent when
I scanned B&W photos (I scanned them in color.)

On some prints (particularly the ones on thicker paper) the leading
edge had a distinct blue "shine" while the trailing edge had a red
"shine". It's less than 1 mm in absolute terms but magnified in
Photoshop it's quite visible. I then turned the picture around and the
same thing happened indicating there was nothing on the photo.

Why is that? Could it be like the red and blue shift observed in
astronomy? The speed of the scanner assembly is not exactly on par
with celestial objects moving away or towards us but, then again, one
notices the pitch shift of the police car siren when it passes by and
sound is considerable slower than light.

In both of the comparison examples this is the doppler effect. I can
assure you that you are NOT seeing doppler effects on your scanned
images.

Or is this shift caused by filters in the scanner being at different
angles?

More likely to be due to uneven spectral distribution on the scanner
lamp.

 

[Don]

In both of the comparison examples this is the doppler effect. I can
assure you that you are NOT seeing doppler effects on your scanned
images.

I didn't really think so either, but I've been "ambushed" so many
times in this scanning adventure that I "don't trust" optics anymore
and I'm willing to entertain anything at this point.

More likely to be due to uneven spectral distribution on the scanner
lamp.

It's not only in spots but along the whole length of the photo. And
then only on the leading and trailing edges. If it were a spectral
problem with illumination shouldn't it be observable throughout the
photo?

Also, the heavier the paper the more pronounced the effect. On thin
paper I can barely see it even at maximum magnification.

Anyway, it's not really a problem, but I was just curious.

Don.

 

[Kennedy McEwen]

It's not only in spots but along the whole length of the photo. And
then only on the leading and trailing edges. If it were a spectral
problem with illumination shouldn't it be observable throughout the
photo?

Only if it had vertical edges throughout the photo. Only the leading
and trailing edges are vertical *and* illuminated by the lamp is a
different position from the flat edges, such as the side edges are.

 

[Kennedy McEwen]

It's not only in spots but along the whole length of the photo. And
then only on the leading and trailing edges. If it were a spectral
problem with illumination shouldn't it be observable throughout the
photo?

Also, the heavier the paper the more pronounced the effect. On thin
paper I can barely see it even at maximum magnification.

Anyway, it's not really a problem, but I was just curious.

Don.

Actually, after thinking about this a little more, the answer is fairly
obvious - don't know why I didn't notice it before. ;-)

It is a similar effect to what you are seeing with the light fall off
when you raise the image off the glass. The reason one edge is red and
the other blue is because each coloured line of the CCD is sampled in
exactly the same position relative to the edge, but the light source has
moved closer or further away from it, depending on the edge and the
scanner design, thus illuminating it more or less than the other colour.
It is only when the subject is close to flat that this movement of the
light source between sampling each of the RGB channels cancels out. With
the vertical edges of the print, the light is actually at a grazing
incidence and the edge will not obey Lambert's Law as closely under
these circumstances. So any change in light position will cause an
effective change in reflected brightness.

Lambert's Law states that the light reflected from a perfectly diffuse
surface varies as the cosine of the angle of incidence. Since the area
of the surface subtended in a given field of view is proportional to the
cosine of the angle that the surface is viewed from, the net effect is
that perfectly diffuse surfaces appear to have the same brightness no
matter what angle they are illuminated at or viewed from. So a Kodak
grey card looks just as grey if you hold it at 45deg to the light and
view it from 45deg in another axis as it does if you hold it
perpendicular to the light source and view it from any other angle. Real
surfaces however tend to depart quite significantly from Llambert's Law
for grazing angles of illumination. So that same Kodak gray card looks
quite reflective and light when held such that the light source is
viewed as reflecting from the card at grazing angles.

Reminds me of an incident as a child when the physics teacher tried to
explain that white surfaces reflected light while black surfaces didn't
reflect any light. Well that isn't right, I protested, because I can't
see anything written on your blackboard when the sun shines in the
window! The simple explanation I was given at the time was that the
blackboard wasn't a perfect black surface, but I wasn't to get the full
explanation, involving Llambert's Law and departure from it, for several
years. It is actually very difficult to create a good black surface
that obeys Llambert's Law to very low angels of incidence.

You should be able to reduce, if not eliminate, the problem by placing a
black surface behind the print since the departure from Llambert's Law
is most significant when viewing and illumination angles are close to
180deg apart - so most of the problem is probably caused by light
reflecting back off the background surface, off the vertical edge and
back into the scanner. This would be an interesting test to perform to
confirm that this indeed the cause of the not really problem. ;-)

 

[Don]

Actually, after thinking about this a little more, the answer is fairly
obvious - don't know why I didn't notice it before. ;-)

That's the spark I was waiting for! ;o)

It is a similar effect to what you are seeing with the light fall off
when you raise the image off the glass. The reason one edge is red and
the other blue is because each coloured line of the CCD is sampled in
exactly the same position relative to the edge, but the light source has
moved closer or further away from it, depending on the edge and the
scanner design, thus illuminating it more or less than the other colour.
It is only when the subject is close to flat that this movement of the
light source between sampling each of the RGB channels cancels out. With
the vertical edges of the print, the light is actually at a grazing
incidence and the edge will not obey Lambert's Law as closely under
these circumstances. So any change in light position will cause an
effective change in reflected brightness.

Let's make sure we're talking about the same orientation because I
would call the edges horizontal, not vertical. So, just to make sure:

It's the edges which are parallel to the lamp and the CCD row. In
other words, the first few rows of pixels scanned by the CCDs have a
blue cast, and the last few rows scanned are reddish.

In other words, looking at the image afterwards, the top edge has a
blue cast, and the bottom edge the red. Are we talking about the same
edges?

You should be able to reduce, if not eliminate, the problem by placing a
black surface behind the print since the departure from Llambert's Law
is most significant when viewing and illumination angles are close to
180deg apart - so most of the problem is probably caused by light
reflecting back off the background surface, off the vertical edge and
back into the scanner. This would be an interesting test to perform to
confirm that this indeed the cause of the not really problem. ;-)

Ah... Houston, we (may?) have a problem! ;o)

The back of the scanner lid is black. What's more, during the time I
was wrestling with glossy photos and Newton's rings, I actually ran
some tests with the lid up. This is so I could better control the
flatness of the photo by - initially - strategically placing small
coins on it. BTW, the effect was observed in both glossy and matte
photos. The only identifiable variable seemed to be paper thickness.

NB: When the lid was up I turned off the lights (!) before scanning.
So there should have been no reflections of any kind around the photo.

Don.

 

[Kennedy McEwen]

On Sun, 23 Jan 2005 23:48:15 +0000, Kennedy McEwen


Let's make sure we're talking about the same orientation because I
would call the edges horizontal, not vertical. So, just to make sure:

It's the edges which are parallel to the lamp and the CCD row. In
other words, the first few rows of pixels scanned by the CCDs have a
blue cast, and the last few rows scanned are reddish.

In other words, looking at the image afterwards, the top edge has a
blue cast, and the bottom edge the red. Are we talking about the same
edges?

Seems to be - those are the edges that I would expect to exhibit this
effect.

....

Ah... Houston, we (may?) have a problem! ;o)

The back of the scanner lid is black.

OK - you are looking at something much more fundamental. A black
background means no light from the back to give a specular reflection at
grazing incidence.

What you have there is just straight forward change in illumination of
one edge relative to the other. The scanner light cannot be in exactly
the same position as the sensor - well it can if a beam splitting
arrangement is used, but not in most flatbed scanners. So it is always
offset from the sensor - either leading or lagging it relative to the
scan head motion. That means that objects which are not flat will cast
shadows, and the size of the shadow will change as the head (and light
source) moves. Usually this just means that one edge is light and the
other dark, but clearly in some positions the edge will transition
between the two. If this transition occurs between the coloured lines
being sampled then one edge will appear blue (because it is in shadow
for green and red) and the other red (because it is in shadow for green
and blue), since the CCD lines are usually in that sequence red, green,
blue.

How much variation you get this will depend on several factors - how
broad the illumination band is, how close it is to the optical axis of
the scanner, how far the scanner head (and light source) move between
sampling each line of the CCD. Best case would be:

* a broad source (so little effect on illumination of the subject when
the source moves). Note that scanning the subject further from the
glass surface will reduce the effective width of the light source,
making illumination variation more likely.
* far off the scanner optical axis (so that vertical edges are well
defined as being in or out of shadow and unlikely to transition between
sampling colour lines). Note that scanning the subject further from the
scanner glass will bring the light closer to the optic axis in angular
terms, making illumination variation more likely.
* short distance between samples of each colour (so there is little
change in lamp position and hence illumination).

I just had a look for this on my Epson flatbed and can only see a very
marginal effect, which I would not have noticed if I had not been
looking for it. I also had a close look at the scanner lighting
arrangement while making a slow (high resolution) pass - sunglasses at
the ready, those lamps are *bright*! Although the scanner lamp is a
fairly narrow tube, the reflector makes this a relatively even broad
source, about 1cm or so. It appears to be mounted at approximately
45deg to the optic axis - assuming the optic axis is perpendicular to
the flatbed. Finally, the optic aperture on the scanner head is only
2-3mm wide, to the scanner head must be moving quite a lot less than
this between sampling the red, green and blue colours of each line of
pixels. Overall, I am not surprised I don't see much of a problem, but
obviously your scanner is different.

Sorry, I really seem to have come at this backwards - from really
complex suggestions through to a much more simple and obvious one!

 

[Don]

Seems to be - those are the edges that I would expect to exhibit this
effect.

No, it was just your use of vertical edge that threw me. So better to
make sure we're talking about the same thing.

What you have there is just straight forward change in illumination of
one edge relative to the other. The scanner light cannot be in exactly
the same position as the sensor - well it can if a beam splitting
arrangement is used, but not in most flatbed scanners. So it is always
offset from the sensor - either leading or lagging it relative to the
scan head motion.

So, my initial instinct was correct. I thought that "the filters were
at different angles" meaning in essence what you say; the illumination
source and the CCD can't be in the same place, so there's an angle.

Speaking of which, are there any filters in modern flatbeds? I mean,
in the early days flatbeds made 3 passes to get all colors (causing
misalignment between them). How is that done today i.e. single pass?

I just assumed the white illumination was split into three beams which
passed through each of the filters and then sampled by the CCD array
in succession before moving on to the next line.

I just had a look for this on my Epson flatbed and can only see a very
marginal effect, which I would not have noticed if I had not been
looking for it.

I didn't notice it either until I started scanning B&W photos. After
that I went back to color scans and only then noticed the effect there
as well. Otherwise, I wouldn't have noticed it in color photos either.

The bottom line is that the thinner the paper the less pronounced the
effect.

Sorry, I really seem to have come at this backwards - from really
complex suggestions through to a much more simple and obvious one!

That's quite alright. I myself always favor "Why simple when it can be
complicated" approach! ;o)

Don.

 

[Don]

Seems to be - those are the edges that I would expect to exhibit this
effect.

No, it was just your use of vertical edge that threw me. So better to
make sure we're talking about the same thing.

What you have there is just straight forward change in illumination of
one edge relative to the other. The scanner light cannot be in exactly
the same position as the sensor - well it can if a beam splitting
arrangement is used, but not in most flatbed scanners. So it is always
offset from the sensor - either leading or lagging it relative to the
scan head motion.

So, my initial instinct was correct. I thought that "the filters were
at different angles" meaning in essence what you say; the illumination
source and the CCD can't be in the same place, so there's an angle.

Speaking of which, are there any filters in modern flatbeds? I mean,
in the early days flatbeds made 3 passes to get all colors (causing
misalignment between them). How is that done today i.e. single pass?

I just assumed the white illumination was split into three beams which
passed through each of the filters and then sampled by the CCD array
in succession before moving on to the next line.

I just had a look for this on my Epson flatbed and can only see a very
marginal effect, which I would not have noticed if I had not been
looking for it.

I didn't notice it either until I started scanning B&W photos. After
that I went back to color scans and only then noticed the effect there
as well. Otherwise, I wouldn't have noticed it in color photos either.

The bottom line is that the thinner the paper the less pronounced the
effect.

Sorry, I really seem to have come at this backwards - from really
complex suggestions through to a much more simple and obvious one!

That's quite alright. I myself always favor "Why simple when it can be
complicated" approach! ;o)

Don.

 

[Kennedy McEwen]

Speaking of which, are there any filters in modern flatbeds? I mean,
in the early days flatbeds made 3 passes to get all colors (causing
misalignment between them). How is that done today i.e. single pass?

A tri-linear CCD or, more commonly these days, an offset hex-linear CCD,
or hyperCCD, is used in flatbeds. The filters are built into the CCD
itself - a layer of dye deposited on top of the photocells in each line.

I just assumed the white illumination was split into three beams which
passed through each of the filters and then sampled by the CCD array
in succession before moving on to the next line.

No beam splitting - that would eliminate the problem you are seeing.
Instead each line of the CCD simply samples a different point in the
image, then the head steps to the next position. After a few lines, the
green CCD line gets to the position that the red one was at earlier and
samples the green component of that line of pixels - then a few lines
later and the blue CCD line arrives and samples the data. So the CCD is
sending our red green and blue data all the time, but for different
lines. The scanner driver sorts it all out - as well as correcting for
the positions of odd and even pixels in the hyperCCDs.