This post is a coproduct of a discussion in a different thread

("Again: film vs digital"). I think it deserves a separate thread.

However, the results below are so unintuitive, that it is probable

that some goof have been made; please be watchful when you read this.

In particular, the digital noise-equivalent sensitivity of print film

turns out to be 2 orders of magnitude higher than of the slide film;

do not think this matches reality...

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Conclusions: if one believes the RMS noise numbers provided by

manufacturers, MF digital cameras of today should beat slide film in

formfactor 4x5in as far as noise is concerned. However, they are very

far from matching noise of print film in formfactor 4x5in.

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RMS noise of film measures the variation of film density when averaged

over a HUGE circle, 48um in diameter. Well, this circle is huge when

one thinks in terms of 35mm shots; but film is not a useful media in

the 35mm formfactor. And starting from f-stops close to f/45, the

diffraction pattern becomes close in size to this 48um circle, so this

number is very applicable to LF photography.

How this number, RMS noise, translates to metrics useful in digital

world? The principal measure of digital noise is the full well. So

how to translate RMS to an equivalent full well?

As Roger reads the Kodak definition, it is 1000 * (RMS of variation of

log_base10(density of film)). [This is not how I initially read it,

but it is a reasonable interpretation, and now I believe it is

preferable to my initial reading of "relative variation of the

(linear) density of the film".] See for yourselves in

http://www.kodak.com/US/en/motion/st...ometric6.jhtml
Translating to linear-density space, "RMS number" R corresponds to RMS

variation of linear density of R*log(10). Next, we want to translate

it to linear luminance. So one needs to divide it by the contrast

CONTRAST = | d log Density / d log Exposure |

Contrast of the film depends a lot on the exposure; by Kodak

definition, RMS is measured at log_base10(density) = 1. For slide

film, contrast at this density is about 1.6; for print film, about

1/1.6.

Combining all this together, one gets the luminance S/N ratio

corresponding to RMS number R as

S/N = 1000/log(10)/(R/1.6) ~ 700/R for slide film

S/N = 1000/log(10)/(R*1.6) ~ 270/R for print film

The next question to analyse is to find in which exposition zone is

the density with the log_base10(density) = 1. The curves

density/vs/exposure for 3 channels (R,G,B) are very different. But

usually G is between R and B, and, moreover, green is the visually

most important. So we assume that exposure corresponds to this

density of the green channel.

For typical print film (I looked up Kodak Gold 100), this density

corresponds to the exposure of approx. 0.3% of the max. For typical

slide film (I looked up KODAK PROFESSIONAL ELITE Chrome Extra Color

100 Film), this density corresponds to the exposure of approx. 9% of

the max. Although these numbers will vary with the film, we will use

these numbers in what follows.

Assuming full well of N electrons, the S/N ratio at 9% of the max is

0.3 sqrt(N), at 0.3% of the max it is 0.055 sqrt(N). Comparing with

numbers above, one gets

Equivalent full well = 5.4e6/R^2 for slide film

Equivalent full well = 25e6 /R^2 for print film

Typical values of R for slide film are about 11, for best print films

about 3..5.

Example: Comparing 40MP 36x48mm sensor to an 4x5in Velvia 50 shot

(RMS=9). The scale factor is about 3; so the 48um circle on film

corresponds to diameter=12um circle on digital sensor. This is

approximately 2.6 pixels; applying the formula above, one needs about

26K full well of one pixel to get similar noise at midrange tones.

Such a full well usually corresponds to ISO=200 sensitivity; thus one

needs about 3^2 * 200/50 = 36x smaller exposure time for the shot when

one shots digital vs film.

Example: Do the same with print film (RMS=3, as for Fuji Pro 160S, see

http://www.cacreeks.com/films.htm
). Likewise, one needs full well about 1M to get similar noise in

zone I. Assuming maximum full well of one shot about 60K, one needs

to stack together about 16 shots to get an equivalent noise. On the

other hand, with the scale factor 3, one needs the f-stop 3 times

larger to get an equivalent image on the focal plane. So these 16

shots would have about 1.8 times longer total exposure (assuming

instantaneous sensor readout). If this full well is achieved at

ISO100 (vs ISO160 of the film we compare with), the total exposure

time is going to be about 3x longer for digital.

Enjoy,

Ilya