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There are cases where you want to remove the paper texture when you scan. An often mentioned technique is to do two scans, rotating the paper 180° between the two. Then in an image editor you load both images, rotate one 180° so that they overlap again, and take advantage of the different shadows to remove them ("Lighten only" mode or else).

At least, this is the theory. But if I try this on my Epson V200 the two picture don't overlap. If I scan a detached sheet of squared paper, vertical lines (in portrait mode) are shifted, and I cannot have them all overlap across the paper: if they overlap around the center, they won't on the edges, and vice-versa, and there is about as much as 1mm of discrepancy across an A4 sheet. On the other hand horizontal lines can be made to overlap across the whole height of the paper.

For instance this is the difference between the upright scan and the rotated scan (rotated again in the editor), if I optimize the overlap on the edges:

enter image description here

And this is the same if I optimize the overlap at the center:

enter image description here

(The small white arrows point at the lines from the rotated scan)

In other words at 6OOPPI the center line is about 15px to the side, so if I overlap with the rotated picture there is a 30px discrepancy.

In addition I find that the horizontal resolution is about 610PPI (vertical is 599.6).

Any clues for the cause of the problem?

  • Common problem with home-grade scanner (actually, seems to be inherent to CCD scanners?)
  • Scanner is old technology
  • Scanner is old, period
  • Scanner is broken/out of specs

My surprise is that the scan head transport (timing belt/stepper motor) is still accurate, but whatever happens in the scanning head (I thought there were no moving parts there, according to this answer is not. Some web pages I have come across mention in passing that the scanner could be re-calibrated?

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  • \$\begingroup\$ Most of my documents scanning is with a camera, often my phone. Sometimes a stills camera with on-camera flash. I suppose if that wasn’t good enough, I could set up strobes \$\endgroup\$ Feb 21, 2022 at 4:30
  • \$\begingroup\$ If you have Photoshop you can resize the width or length separately. Works great if the dpi difference between them is consistent. From the menu: Image->Image-size \$\endgroup\$
    – doug
    Feb 21, 2022 at 4:56
  • \$\begingroup\$ @Doug Yes, but this isn't even the problem. The image is distorted: what is exactly halfway between the edges on the paper isn't exactly in the middle of the scan. \$\endgroup\$
    – xenoid
    Feb 21, 2022 at 7:28
  • \$\begingroup\$ Probably you're aware, but here's a relevant question on removing paper texture from scanned images (the answers include the 180-scan-trick, but also others): photo.stackexchange.com/q/23445/9161 \$\endgroup\$ Feb 21, 2022 at 12:34
  • \$\begingroup\$ @SaaruLindestøkke Yes, I know these techniques... \$\endgroup\$
    – xenoid
    Feb 21, 2022 at 13:05

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A household flatbed scanner head doesn't have a CCD the full width of the scan. (I suspect, most of them don't). Instead, it has a fairly wide angle lens and a narrow mirror (or two) which concentrates the image on a much smaller linear sensor.

I should have one in pieces to show somewhere, but meanwhile a schematics could be found in this article:

Flatbed scanner head diagram

Consequently, it has all the problems of wide-angle (or any) lenses, including distortion. And this distortion is non-linear, in the sense that the edges may be stretched relatively more than the centre. Like cameras, some scanners might compensate for it, some don't; in any case, the horizontal (cross-wise) resolution is only approximate.

At the same time, lengthwise resolution can be made highly accurate and linear. It's just a question of good mechanics, and not expensive at that. Lengthwise distortions are typically the missed or repeated lines due to some random glitches, and they are rare.

This can be easily observed by scanning something round, such as a CD. It never turns out the same pixel dimensions across and along, and scaling one dimension to match the other will not make it perfectly round (try with a circular mask).

Unfortunately, I don't know of a distortion correction filter that works only in one dimension. It can be improvised without great difficulty though. I never needed accuracy greater than what linear scaling gives, but lining it up like you do may require it.

P.S. If the paper is white, why not just blow it beyond highlights?

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  • \$\begingroup\$ Thanks for the explanation. As explained in the OP, I'm trying to use a technique that requires a decent overlap between two scans. Btw, after taking some measurements, I was able to generate a displace map for Gimp that greatly reduces the distortion (not perfect, but much better). \$\endgroup\$
    – xenoid
    Feb 21, 2022 at 7:38
  • \$\begingroup\$ According to your explanation, the distortion should be more or less symmetrical. But may be they moved the lens a bit so that smaller scans in the top left corner are more accurate. Last question: how can I tell if a given scanner model uses a lens or not (or how wide/distorting the lens can be)? \$\endgroup\$
    – xenoid
    Feb 21, 2022 at 7:47
  • \$\begingroup\$ @xenoid, I don't know how to tell, but all those 3 or 4 I pulled apart had a lens. But I haven't laid my hands on bigger professional ones. The lens angle is quite wide (>90°); you can estimate it by checking the width of the scanning head; your V200 (like mine V100) seems to have two mirrors and the light is "folded" twice, so roughly double the head width (i.e. along the scan) vs page width. \$\endgroup\$
    – Zeus
    Feb 21, 2022 at 8:25
  • \$\begingroup\$ Having a lens brings all its potential issues like misalignment, so perfect correction may not be possible. But calculating even a perfect one may not be trivial: we don't know how much is distorted optically and how much (and how) is corrected digitally (like in many modern cameras). If you think about it, the more suitable lens for this would be fisheye... \$\endgroup\$
    – Zeus
    Feb 21, 2022 at 8:26
  • \$\begingroup\$ Given the output I doubt that there is a correction don in the scanner. Manufacturer's scanner software could apply a correction (but since I use Linux this isn't a solution anyway). \$\endgroup\$
    – xenoid
    Feb 21, 2022 at 8:30

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