At times I wonder how printing professionals think – there seems to be some kind of “magic” involved here and there.

Might it be possible to actually write down a sequence of simple mathematical statements that actually makes sense – at a level that may be followed by a normal human being? ;-)

Being a technician by both profession an heart makes me lean towards numbers as soon as there is something that needs explaining.

Now, to have some sample data; lets assume there are a set of developed 35mm film frames to scan. The special thing here is the fact that the scanner’s output resolution will affect the result – what kind of detail the image will show.

One theoretical fact here is “the Nyqvist frequency“. saying that a “B hertz” signal can be completely determined by “a series of points spaced 1/(2B) seconds apart.”. Now, how would frequencies com into play here? We’re sampling but it isn’t sound-waves… Well, the fact is that you’re still sampling within a space (not frequency space though) and attempting to reconstruct (i.e. print or display) the input with a limited set of samples.

All the theories and formulas mentioned in the Nyqvist article does apply.

This gives that to get a full and true 300 dpi resolution you need to sample at 600 dpi as a pure minimum. A sinus wave – which would be similar to the undulating pattern that appears on sand dunes – with a repetition rate of 300 waves per inch on the negative would then be represented by a series of black and white stripes. Double the sample frequency, i.e. 1200 dpi and you would start to see the undulation, as there might be ONE extra grey level, between the black and white. All this assuming that you manage to synchronize the pattern with the samples – so that one of the samples actually catches the TOP of a wave, and another the BOTTOM. Now, the step up to 2400 dpi adds yet another plausible level, so we have a likely hit ratio of 8 samples over one single full wave. *Compare to how electronics experts select oscilloscopes; the bandwidth needs to be 10 times higher than the highest frequency you intend or need to measure.*

Now, can you tell which kind of ACTUAL resolution and detail you have in your scanned image?

In practice, scanning a 35mm negative at 2400 dpi with an film-capable flatbed will create an image almost 8MP in size. e.g. 3400 by 2250 pixels (7.65MP). Depending on the quality of the scanner the result might contain details to a level that indicates 1500-2000 dpi. Recommended reading: www.filmscanner.info (translated from german by Google).

Lets move on to printing. Given the above image size, it is easy to see that 3400 could be “approximated” to 3300 which is easy to divide by 11; 11*300, so we have enough information to print 300 pixels per inch of paper, for 11 inches. Now note that most consumer printers can’t do that, some Prosumer level ones do – without actually knowing I’d say that just a few professional level printers surpass that.

“Why so” you ask?

Often the case is that 360, 720 or 1440 dots of a certain size may be printed along an axis (vertical or horizontal on the page). Each of the dots printed may have ONE color, which is to be picked out of one of the cartridges you load into the printer. Some printers appear to have capability to vary the size of the dot, which then probably helps to enhance the result – the exact way this is used is beyond my current knowledge, and to some extent not interesting here.

The effect of the above limitation of the printer is that; to actually come close to the color of an image-pixel you need to use e.g. a 6 by 6 matrix of printer-dots – with a certain mix of colors from the cartridges.

This means that a 360 dpi printer prints only at 1/6th of the given resolution – 60 image-pixels per inch.

e.g. 360 dpi, where 6 printer-dots is needed to each image-pixel -> 360/6 = 60… 720 dots per inch -> 120 pixels per inch, 1440 dots per inch -> 240 pixels per inch.

For a printer with 360 dpi resolution, you then might think,

a 60 pixels * 12 inches => 720 pixels wide image would be enough information to print a 12 by 8 inch page (ISO A4 is 11.7 by 8.2″… 120*12 = 1440, 240*12 = 2880).

That about printer capability…. Now, how much detail does a printed image need?

In optics, there is a definition of a “Circle of confusion“, in short “CoC”.

This will be in effect not only as the lens “views” the subject you’re taking a photograph of, but also as you view the print of the photograph – then with your eyeball as the lens.

What is it then?

The answer to “what is a sharp a dot?” and it depends. The farther away from the dot you view it, the larger the “dot” may be.

From this comes that “what is sharp?” can be estimated by the theoretical calculations available for CoC.

Also note that CoC is not only about a simple circle or a dot, it applies to edges and lines in the print – compare to the edges of the “dot”.

Now, I have used inches so far, but need to get into the metric system for some calculations that will be inevitable below… so:

60 pixels per inch is close to 50, which translates easily to 2 dots per mm (50 dots per inch, an inch is ~25mm).

So that is a dot size of 0.5mm, which we then need to compare to the Circle of Confusion – the size of which in turn depends on the viewing distance…

… and here we dive into some intricate mathematics from the world of optics. ;-)

To get a formula to use I have checked the wikipedia article on Circle of confusion (below). There one can read that CoC needs to be “0.025 cm for viewing at 40 to 50 cm”(0.0025 cm <=> 0.00025 meter). This means that the angle made up of ONE DOT should be equal to: a = 2*arctan( 0.00025 / 2 / 0.50 ) => a= ~1/28°, i.e. around 1/30°, or two arc seconds (2/60°)

So, by some algebraic substitution back and forth we get:

1/30° = 2 * arctan( CoC / 2 / ViewingDistance ) <=>

tan( 1/30° ) / 2 = CoC / 2 / ViewingDistance <=>

ViewingDistance = CoC / tan( 1/30° )

Check:

0.025 / tan ( 1/30° ) => 50 [cm]

…

ViewingDistance = CoC / tan( 1/30° )

with a 0.5mm CoC (50 dpi from above) gives:

ViewingDistance = 0.5 / tan( 1/30° )

That is MINIMUM ViewingDistance

ViewingDistance = 859.43.. mm <=> ViewingDistance= 1 meter.

So, with a true 50 dpi print you need only back off 50 cm more, for a total of 100 cm (1 meter, 40”) as viewing distance.

References:

http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem

http://en.wikipedia.org/wiki/Circle_of_confusion

http://en.wikipedia.org/wiki/Depth_of_field

http://en.wikipedia.org/wiki/Airy_disk

http://en.wikipedia.org/wiki/Point_spread_function

2013-01-29 (Tuesday, week 05) at 06:16

Great thoughts on this. I’m a little confused though, what is your conclusion? I worked in a Photo lab for a while and do a lot of graphic design now. I found that at a certain point, the number of pixels no longer matters because your eyes can’t make out a difference. Typically, when using Photoshop, the tipping point was 300 DPI, with the pixels (i.e. 2500×2500) having little to do with the end quality. We often printed posters on photo paper 24 inches by 30 inches (as an example), and at 300 DPI, the resolution was great, though we could go up to say 800 just to be safe. Get down to the common 72 DPI that many customers brought in…it was trouble even if the pixels were at 2500×2500.

2013-01-30 (Wednesday, week 05) at 21:44

The conclusion is to come from the maths at the end, I have yet to fill it in ;-)

The plain truth is; any “DPI” rating for a digital image has nothing to do with how large you actually can print the contents of the file.

Printed “dots per inch” comes from available pixels in the file and how much you “stretch” those, nothing else. A 6MP image is 3000 by 2000 pixels (assuming 3:2 side ratio), a print with 300 pixels plastered over each inch makes the print be 10 inches wide (approximately 250mm).

Whether that looks good or not depends a lot more on the actual output parameters of the *printer driver*, which paper you print onto, whether you have profiled your printer correctly and have been using the benefit of a calibrated display for the editing.

I’d say that if you have a good image, with decent sharpness pixel-wise, you will be able to print 120 pixels per inch and still have quite nice results.

Additionally: Depending on

CircleofConfusion calculations (as start on above) will allow even lower pixels per print-inch – if the viewer backs off to a reasonable distance, based on the CoC size and calculation.2013-04-15 (Monday, week 16) at 21:02

More…

http://www.digitalphotopro.com/technique/workflow/the-right-resolution.html

2013-12-28 (Saturday, week 52) at 18:38

FWIW; I consider the megapixel gauntlet to be “over”.

Already a good 3MP photo can be printed on A4 (297x210mm, 11.7×8″) without anybody’s eyes getting sore, presuming a reasonable viewing distance. You will only need 200+DPI prints only as somebody picks out a loupe for close examination.

An experienced photographer fills the frame with his subject, with some forethought and luck – so has no need to crop more than for correction of composition.

3MP: e.g. 2100x1400px.

Printed 250mm wide, 250/2100 => 0.119mm per pixel, 8.4 pixels per mm.

Close limit for viewing:

ViewingDistance = CoC / tan( 1/30° )

If the print is perfectly clear that means CoC = 0.12mm, which then suits the formula above, and we get VD=0.12 / tan( 1/30 ) => VD = 204 mm

2014-03-13 (Thursday, week 11) at 00:05

The “resolution” of human vision.

http://petapixel.com/2014/03/12/answering-unanswerable-whats-resolution-human-eye/

2014-08-12 (Tuesday, week 33) at 19:30

Text copied here from https://www.flickr.com/groups/panoramas/discuss/72157645973145079/72157646391956832/"quailty" depends on which distance you intend to (let others) view them from, together with how big you intend to print.

The width of one very sharp dot in your image (ref: Circle of Confusion) should be 1/30° at the very limit on how close you can stand

without seeing ‘pixels’, but I’d say you’d need to be at double that distance to have a "margin".With a wide image that distance might be "too close" to actually be able to perceive the full view.

My version of the maths behind it is here.

Links to background data is also there.

Now with that done, you have to select the type of paper to print on; Canvas, Matte, Lustre and Glossy are just a start – telling the surface type. For the choice of either Matte or Canvas you will be limiting the possible color output too – the gamut on these are limited. Lustre and Glossy come very close to "photographs" when printed on e.g. a Epson 2880/3880s.

How about framing, paper thickness, quality (fine art paper!?)…

So, now you have chosen the paper type; you need to have a color profile for the combination of printer and paper. Calibrate!

To see what you’re actually going to print you need photo editing software that understands that profile. And that software requires a calibrated display – or else it will not know what it presents, and more importantly you do not see the same color that the printer will put on the paper. Calibrate!

2014-09-13 (Saturday, week 37) at 14:15

Raw text, saved here from elsewhere – will be edited as time allows:

—

The “PPI” doesn’t matter much, the sharpness of your image does though – as that determnines the Circle of Confusion the subject makes up.

THAT is the the basis for at which “PPI” to print the image. If the size of the CoC covers as much as 1/30 degree of the view you MAY see “pixels” – and due to this the viewing distance matters, in the same way as a long lens shows you only part of your subject if you’re too close to it.

If you view a SHARP image – printed at actual 50 PPI and viewed from the correct distance, as given by the above – you will not see any pixels.

—

Please note that my use of “PPI” has nothing to do with the printer’s “DPI” numbers – above I refer to actual PRINTED ppi – i.e. image pixels per inch as put on paper.

Printer DPI are almost always a multiple of some strange number that depends on the hardware and mechanics used when the printer was built. It may be a multiple of 180, e.g. x2: 360, x4: 720, x8: 1440, x16: 2880.

These “printer resolutions” are used to put a SINGLE CARTRIDGE COLOR on the paper and belong hidden within the technical sides of the printer used and nowhere else.

“Draft” modes are QUICK-PRINTING of approximated images,likely at an intermediate / low printer resolution.

—

A laser printer must have a larger portion (if not the entire image) available as it is about to be printed (I’d say: due to limitations in the laser printing technology), hence need the actual amount of RAM for an entire page at the used resolution.

e.g. en.wikipedia.org/wiki/Laser_printing

Note also en.wikipedia.org/wiki/Laser_printing#Anti-counterfeiting_… – which may make a color laser printer consume more yellow than it might need otherwise.

—

My comments about PPI and the print PPI above – in practice:

An experiment, take it backwards:

1. Print or Draw a small square on a sheet of paper (a postit!), the actual size is not important, just make it a precise measure and rather small – say 1/12 inch <-> 2mm per side or some such. Also try to add another one near it, as close to HALF THE SIZE as you can make it.

2. Measure the square’s width (same as height) and write it down

3. Tape the paper with the square onto a wall at comfortable height for your eyes.

4. Walk away from the paper to a distance where you cannot longer tell that the larger square square IS a square – ponder on the smaller square; is it a dot? Walk away further to make it a dot.

5. For me, with current old glasses, the larger one is "not a square" at a 2m distance (just below 7′), that is 2000mm – also making the smaller one close to "hard to see".

6. Now; at a viewing distance (=ViewingDistance) of 2000mm I’d say my "Circle of confusion" (=CoC) is 1mm (the width of the smaller square)

7. The formula for shortest viewing distance, given by "1/30 degree" above is:

ViewingDistance = CoC / tan(1/30°)

so does this match the numbers here?

note: the formula comes from basic maths below and is discussed further here.CoC = ViewingDistance * tan( 1/30° ) =>

2000 * tan (1/30) = 1.163552966 —> that is ->

CoC is just above 1mm (mm, because 2000 is mm)

So, what does this imply?

1mm is 1/25 of an inch (25.4mm exactly), this means than a CoC of 1mm allows a fair view at a distance of 2m –>

a print at 25 PPI viewed from 2m distance is “OK”

I’m not saying “good” – take note! – Print at 50 PPI and you have a better view…

Now it may be discussed where the view passes into “good”

Maths behind the formula:

( v– click on this to view the image)

Image linked from and also shown at http://www.cimt.plymouth.ac.uk/projects/mepres/step-up/sect4/ind...

2015-03-12 (Thursday, week 11) at 21:51

https://www.flickr.com/groups/756291@N24/discuss/72157651310945105/72157648993485033

2015-07-08 (Wednesday, week 28) at 20:20

More of the same, another POV:

http://photo.stackexchange.com/questions/456/is-there-a-general-formula-for-image-size-vs-print-size