I think most of your logic is flawed. Some glaring mistakes you might want to look into is the role of diffraction, and calculating the differences in diffraction over different formats and even different sensors (pixel size) and how they change data.
For instance a 6x7 film with standard lens won't become diffraction limited until after ƒ22 given a 12" wide print.
I disagree...
This is only true if aiming to make prints of the same size from all formats. As I've stated, I'm not interested in print size here, just the amount of information recorded on a negative or by a sensor.
Also you seem to forget (or be unaware) that the megapixel count of a camera is not a measure of it's resolution, that would be it's pixel density and size.
Which is precisely what I said. If you go to, say, Nikon's website, they talk about "a resolution of 36 MP" for the Nikon D800 - so, for better or worse, "resolution" is now used to mean "number of pixels". To reiterate, I am not using it in this sense.
Further to that a 12mp sensor can't resolve 12mp of detail with a Bayer sensor.
Of course it can't - I state that too! This is precisely why I use measured resolving powers (in lp/mm) from actual real-world tests. These values tell us how exactly how much detail is resolved (assuming the test results are trustworthy).
I don't know where you get the figure of 80% degradation with drum scans, or the idea that film needs to be scanned.
A scanner is an optical system with a lens, sensor, etc., and as another link in the imaging system it will inevitably lead to loss of data. How did I get 80%? The resolving power of a drum scanner is about 90 lp/mm (measured - Google for confirmation). To work out the loss in resolving power, the resolutions of film and the scanner are combined. The following formula is empirical (i.e. it closely matches experimental results but is not based on theory):
1/
R = 1/√[(1/
f)² + (1/
s)²]
where
R is the resolution of the scan (
not the scanner),
f is that of film (70 lp/mm) and
s is that of the scanner (90 lp/mm). Plugging in these values gives
1/
R = 1/0.0173
R ~ 55 lp/mm
So, scan resolution = 55/70 = 80% that of film.
I don't know where you get the idea that film needs to be scanned.
Of course you don't have to scan colour film - if you don't, ignore that step. Most people do scan today, and I certainly do. In fact, I can't imagine where I can get large (> 20 inch) colour darkroom prints today, and dread to think of the price!
If we output to print what effect does mathematical dithering have on inkjets?
Which is why I barely touched on printing. As the post stated, I primarily wanted to know how much detail film captures compared with digital - ending up with a digital image file, not a print. I'm not dismissing printing - it's important to me, and I always print my photographs, aiming to get prints on gallery walls. But printing is irrelevant to my calculations .
So many basic errors, mis-calculations and false assumptions in your OP that really I don't know where to start.
By reading my post more carefully...?
😉