Dynamic Range/Latitude Explanation. (film)

[ciaran_mcmenemy]

<p>Hi!<br />From the offset, here's what I <em>think</em> d.r. is. If I'm wrong, point it out. Fix any stupidity :D<br />- see image - <br /><br />I want to see if I'm right or totally wrong. <br />Starting from the principle that negative film has a wider latitude than positive film, I think this is achieved by a roll-off in the response at both ends, for the former. By that I mean as a scene gets darker and darker, there will still be some response. This response gets less and less as the darkness increases, with the behaviour beginning to occur at some amount of stops <em>below </em>the chosen/indicated light level for 'grey' (in the middle of the band). Meaning: even though there is lots of shadow, there is still some silver, so you can represent varying shadows. <br />At the other end of the scale, as the scene gets brighter and brighter, the silver response doesn't quite reach the maximum. It might only be the sun that can be completely white, and this response may occur after some amount of stops <em>above</em> the chosen light level. Meaning: as things get brighter and brighter, the densities don't increase as much, so we can represent varying brightnesses. These two qualities have the effect of <em>compressing</em> the scene onto the film, meaning that very dark things are 'pushed' into the image while very bright things are 'pulled' into the image.<br>

Now, for positives.. I think it works like this: at some point below the chosen exposure (just a couple a' stops), there is no exponential (or is it logarithmic?) roll-off. Instead, it hits the flat, in that anything above a threshold of darkness is completely black. No silver density. On the other side, the response hits the peak, meaning anything too bright is completely white. 100% silver. On top of that, there is a smaller range of stops between the thresholds, meaning that exposure should be careful, especially in contrasty scenes. I have no idea why positive film would work this way, however.<br>

Example: metered 'correct' exposure for a subject is 1/125, with the aperture fixed. Their shadow would take 1/30 (2 stops under), the sky would take 1/1000 (3 stops over), the clouds would take 1/4000 (5 stops over), the shadow from the house nearby would take 1/8 (4 stops under). On negative film: everything is in view, with the clouds and house-shadow being somewhat detailed. On positive film, the person's shadow becomes nearly pure-black, the house-shadow is characterless, the sky is very bright, and the clouds are pure white, no detail. <br>

Now, is that right, or am I re-inventing the wheel by making it square?</p><div></div>

[bebu_lamar]

<p>Both positive and negative film has the shoulder and toe section like you said. Of course they are reverse of each other because one is positive and the other is negative. The reason for the negative film to have more dynamic range or more latitude because the slop is much more gentle. The slope of the typical negative film is only about 0.6 and the positive film is slightly more than 1 to compensate for loss of contrast due to lens flare etc.. <br>

If you scan the negative, correct for the orange mask, invert it into positive you will see it's a very low contrast image. To restore the contrast the printing paper has the slope of about 1.8 which would restore the contrast to about normal. The print actually has less dynamic range than a transparency and thus you can not render all the details in shadows and highlight contained in the negative. You can adjust the exposure used to expose the print to render details in either part of the negative dynamic range but not all of it. The extra dynamic range is then considered as latitude.</p>