Photographic resolution and scans.

I’ve heard it said that every time you use an equation you lose half the audience. I’m going to take that risk : In photography there are a lot of equations which come up in the form 1/x + 1/y = 1/z , and one of those is for recorded resolution. 1/Lens-Resolution + 1/Recording-Resolution = 1/ImageResolution. It’s also a manifestation of the law of diminishing returns. But why do I think it is worth a post ?

First: there is a limit on the smallest detail that lens can resolve in an image: one test to get an indication of this is to look patterns of parallel black and white lines and see how fine the lines can be before they blend into a grey mush.

Second: However many Pixels you have, the digitization process can’t put in detail which wasn’t captured by the lens. There will always be some loss in the process - (or, if you prefer, to record as much detail as the lens could resolve the sensor would need to have infinite resolution). Increasing the sensor resolution will reduce the loss but each successive increase produces smaller and smaller benefits (and remember that if the number of lines the sensor can resolve doubles, the number of Pixels quadruples).

That equation says if a lens can resolve x pairs of lines per unit of distance (it doesn’t matter if it’s lines per mm or lines per image width), and the sensor records ½x , the net resolution is 1/3x ; if the sensor records x, net resolution is 1/2x ;  2x line pairs at the sensor gives a net resolution of 2/3x , go up to 4x and the result is 4/5x, 8 times lens resolution at the sensor gives 8/9 of the lens detail in the output. You can see the progression – but I’ve just described a 16 fold increase in linear resolution, or a 256 fold increase in pixels – like going from a basic 240x320 Pixel QVGA webcam to 20 Mega pixels – (in 2010) that’s the realm of professional equipment – but improving detail recorded detail by a factor of less than 3. Of course that would only be true if the image being digitized were the same – the pro camera will have a lens which resolves more detail (thousands of lines over the image width, against hundreds for a web cam lens). Changing whichever component has lower res will have a bigger impact than changing the higher res one. There’s no point in making a web cam where the lens has many times the resolving power of the sensor, or mounting a lens on a pro camera with much less resolution than its sensor.

I’ve known this for years, the upgrades to my digital SLR cameras have increased the Pixels but the images only show a fraction more detail – although it is easier to see with my best lenses. But I’ve recently been going over the problem with scanned images.
I made the picture above in 2003, printed it on roll paper using my A4 printer and it has been on my wall ever since. But it was shot on film and transferred to CD at the lab – the JPG files are quite low resolution. The border makes up about 1/3 of the height and the actual picture is roughly 14cm / 5½” tall – and covered by 1100 pixels. (The border is a useful trick for making the aspect ratio of the picture a bit squarer so the print isn’t so long. Instead of being 7700 x 1100 – a 7:1 aspect ratio, it is 8000x1400 a 5.7:1 ratio).  I’ve been thinking about doing a new version, the picture can be cropped less at the top & bottom as well and I can re-visit the ideal size of border; but I can fix a couple of other things the sepia toning is excessive and there is a stitching error (look at the legs at the landward end of the pier). 

The final result would go into silverlight deep zoom and I’m already looking at 13” / 330mm rolls of paper for my new super-A3 size paper. (The current print is 4 feet /1.2M long. The exact size of an A3 print would depend on the cropping and the border – with a border like the original the aspect ratio would be 3.8 : 1 – so the print wouldn’t longer just taller.) But I want to wring the maximum possible detail from the negatives. It’s not simply a question of smearing the same detail over more pixels – printing software can do that, so the even current image on bigger paper won’t look pixelated. I want something which rewards looking closer.  I can scan the prints which came back with the film, I have two negative scanners (one of which was the subject of the video I just made)  and I can set my 14 Megapixel camera up as a slide copier. Which will give the best results – is it the one with the most pixels ? No. Using the camera gave the most pixels, but the results weren’t great. But the question turns out to be much more complex than I expected, because no two digitisations produce the same range of tones, and they all have different levels of noise – noise can be processed out in software but at the price of some detail. Subtle details can be lost through a lack of contrast rather than a lack of resolution or swamped in the noise.

I spent some time trying to get examples of how each looked and gave it up as impractical – different details rendered better in different scans, and trying to find a single piece of the picture which shows both the good and bad from the different scans proved to be impossible – especially since the panorama software handles the overlapping sections differently in different sets of scans, so one might be comparing the fuzzy edge of a frame in one result and the sharp part of an overlapping frame in another (there was a flock of birds flying round the collapsed central ballroom and they appear – or don’t – depending on the whim of the software; the original used some 3rd party software and I’ve had 3 versions of Microsoft software since). In short – the more time I spent trying to be objective the less conclusive the results became, which print looks best isn’t necessarily the one with the most detail. As I said before “Because my experience has been bad I don’t scan much, and because I don’t scan much I won’t spend the money to get a better experience.”. So the key might be to stop wasting time scanning my own negatives and send them to a professional scanning service.

update: Fixed a bunch of typos.