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: (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° )

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.