# film characteristic curves and log(lux*sec)

Discussion in 'Film and Processing' started by weasel_bar, Aug 13, 2022.

1. Huh?
The slope, gamma, CI, or whatever you care to call it, is got by dividing the density increase (or decrease) by the log E increase that causes it. Those figures are read off from the Y and X axes.
No protractor is necessary, nor indeed very useful.

2. Well, yes, but it is not easy to do.

The tic marks are every log10 unit. Tic marks every 0.1 log10 unit would make it much easier.

But even so, if the x and y (or B and H) scales are the same, then we can almost estimate the slope looking at it, or use a protractor.

It is not so hard to judge a slope of 1.0 when the scales are the same. And then a little more, or a little less, or a lot less.

And even worse, most of the time there really isn't a straight section. There are parts that are less curvy, but not quite straight enough.

Why make it so hard to read the graphs!

3. Because most users don't care that much?

There are a few free computer programs that can trace a graph curve semi-automatically and extract the raw data from it, usually in CSV format. I use one of those and create a spreadsheet - if I'm that interested. The spreadsheet then makes it very easy to determine an average slope, sectional slope, etc. etc.

('Grafula' is one such program, if you're interested)

But, let's face it. Film enthusiasts today, using stale film and kitchen sink processing, are hardly going to notice if their gamma/CI is plus or minus 0.1 away from what it should be. And how would they know? Because I bet densitometer ownership is less than 1 in a thousand; and who wants a collection of stepwedge pictures?

As for the many push-processors that think a 'thicker' negative is a better negative - Urghh! Last edited: Aug 19, 2022
4. I am sitting at work not working. So I figured out the source of the formula I cited above log (H-midgray) = log (10/ISO).
It comes from Hm, the speed point, and proper developing, as described by rodeo joe, placing mid-gray at 0.75D above baseline+fog, as described by alan marcus, and an assumption of linearity in the density/exposure curve between log(Hm) and log(H-midgray)

H(mid-gray): exposure needed for 0.75D above baseline+fog
Hm: exposure needed for 0.1D
If film properly developed, 1.3 log exposures above log(Hm) yields density of 0.9
Therefore, the slope of the curve, if linear, from log(Hm) to log(H-midgray) = (0.9-0.1) / 1.3 = 0.6
Density of 0.75 is 0.65 units above density at Hm (0.1).
0.65 occurs at 1.1 log exposures above Hm: 0.6*1.1 = 0.65

Hm= 0.8/ISO ==> log Hm = log 0.8 - log iso
Log(H-midgray) = log Hm + 1.1
= log 0.8 - log ISO + 1.1
= -0.1 - log ISO + 1.1
= 1 - log ISO =
log (10/ISO)

So Log (H-midgray) is 1 - log(iso) = log(10/ISO) if the response curve is linear between the speed point and proper exposure for mid-gray

QED!!!!!!!!!!

5. The above works if 0.75 is indeed the correct TRANSMISSION density on a NEGATIVE for mid-gray. I think this would be correct if 18% transmission in a negative results in 18% reflectance in the printed image, as @alan_marcus|2 has stated (that 18% is the anchor point). I found at least 1 other thread on photo.net in which @alan_marcus|2 and @rodeo_joe|1 argue this point vociferously.
18% Gray Card

I did not come up with the equation for the relationship between iso and proper exposure for mid-gray. I found relationship stated on the website below. I only tried to figure out why it might make sense. It does, if the density of 0.75 is right, and we take the definition of speed point and use a gamma of 0.6.
A Practical Guide to Using Film Characteristic Curves | Film Shooters Collective

6. Oops. I made a typo. I meant:
"Therefore, the slope of the curve, if linear, from log(Hm) to log(exposure where density 0.8 above Hm) = (0.9-0.1) / 1.3 = 0.6."