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Old 11-21-2013   #48
Photo_Smith
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Quote:
Originally Posted by RichC View Post
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.
If you're not interested in printing then fine but that still doesn't change the fact that you can't approximate the diffraction of 4x5 by adding a 3x multiplier because the focal length is 3x.
Diffraction is one of the min causes of degradation of the image in other words the spread function of the light.
It is dependant on the size of your output though so I'm not sure how you are going to duck out of that one.
Simple fact diffraction will act like a blurring filter over your sensor/film and will greatly lower your 'theoretical' lp/mm figure.

Quote:
Originally Posted by RichC View Post
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.
If we are talking science then what Nikon say in their literature is not worthy of discussion.

Quote:
Originally Posted by RichC View Post
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).
Sure, I'm more pointing out when you say the D800E has xx resolution it doesn't-just clarification.

Quote:
Originally Posted by RichC View Post
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) + (1/s)]

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.
You might like to look at your figures again if you think a drum scanner tops out at 80 lpm I think and I'll have to find out it depends on the aperture of the scan it can resolve grain and further to that I'm not sure of your maths either.

Quote:
Originally Posted by RichC View Post
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!
When arguing the ultimate difference in resolution of two systems you need to think about those systems as a whole, price shouldn't come into it!
Since we are talking about resolution I presumed (falsely) that that resolution would have a purpose-large display.

Quote:
Originally Posted by RichC View Post
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 .

By reading my post more carefully...?
I read your post, possibly I didn't understand it certainly I don't understand why you would need resolution if it's not for large display?
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