film characteristic curves and log(lux*sec)

I read many previous posts and could not find a clear answer to my question, although I'm sure someone will find it somewhere, so please excuse.

For clarity, I understand logs, fstops, etc.

A film like Fuji Velvia gives the x-axis for its characteristic curve in ABSOLUTE terms.

Velvia 100

https://asset.fujifilm.com/master/emea/files/2020-10/2f3c7f90a0b0c6e605e84f98b7d489c2/films_velvia-100_datasheet_01.pdf

Velvia 50

https://www.ishootfujifilm.com/uploads/VELVIA%2050%20Data%20Guide.pdf

For velvia 100, an absolute exposure of log(lux*sec)= -1 results in a density of 1.

For velvia 50, the exposure resulting in the same density is approximately -0.7, which is approximately 1 stop more of exposure. Which makes perfect sense.

Some films, such as ilford HP5 400, give the x axis as relative log.

In either case, I can figure out the approximate exposure latitude of the film in stops, but what about latitude to over vs under exposure? Where is the anchor point for exposure or density, relative to "correct" exposure?

Based on the table provided in this website, if (log) exposure is given in absolute terms, -1 yields the correct density for middle gray for iso 100 film. -1.3 for 200, etc. Thus, one can figure out how much under or over exposure can occur before the density stops changing linearly. It may be asymmetric, which is why knowing the "anchor point" is useful.

https://www.filmshooterscollective.com/analog-film-photography-blog/a-practical-guide-to-using-film-characteristic-curves-12-25

What about "relative" log exposure? I've seen some kodak reference sheets refer to "log H ref."

Thanks all.

Looking at several different films, it appears that absolute log exposure = - 1 results in a density of 1 for iso 100 SLIDE films. This is the case for kodak e100, fuji velvia 100, and fuji provia 100f. Not the case for color c41 or b&w print films. Maybe because the film's inherent exposure density is different.

So my question still remains regarding a film like hp5, which gives its curve in relative log exposure: if a gray card is exposed properly for the iso marked on box, and we develop according to some preset standard, where do we end up on the curve? Forgive me for ignorance regarding "standard" developing as I send my film out

So my question still remains regarding a film like hp5, which gives its curve in relative log exposure: if a gray card is exposed properly for the iso marked on box, and we develop according to some preset standard, where do we end up on the curve?

I would make the opinion that the easiest way to find this is experimentally, assuming you have the equipment.

To do it theoretically, in a fairly simple manner, assuming that you are an Ansel Adams' Zone System practitioner, I think (not certain) that you can first find the point on the characteristic curve where the b&w negative density just begins to rise above the base+fog level. On the assumption that this represents a Zone I exposure you could then move 4 stops over on the exposure axis to get to Zone V, which is what Adams considered to be the metered point for an 18% gray card.

To do it theoretically, the way I suspect that you want to do it, you need some information from both the film-speed test standard and the exposure meter standard. I don't believe that the relationship is spelled out directly, but it should be possible to work it out. I never have worked it out, but there's a guy on photrio who says it works out to be 3 1/3 stops exposure increase beyond the ISO speed measuring point. Assuming he's right (probably is), use a method similar to the Zone System one above, but only move 3 1/3 stops on the exposure axis. To be a little more correct the b&w negative film speed point is where the film density reaches 0.15 (I think) above base+fog. (Note that the film must also be developed to a specific contrast.)

As a note the published characteristic curves use a log number on the exposure axis. On this scale each full f-stop change is equivalent to 0.30. So a 4 f-stop difference is 4 X 0.30 = 1.20 on the log axis.

One last note: there is a little vagueness in the exposure meter standard. It allows the manufacturer to vary the value slightly to accommodate various things. So two different brand reflective meters might give moderately different results. (Read up on the 'K' value for reflective meters, etc.) This opens a can of worms on the internet with some people saying that the gray card should be 12% instead of 18%, etc., etc. Additionally a gray card will give slightly different readings depending on the angle of the main light source.

If you get into the world of professional color negative films the manufacturers usually spell out some density aim value ranges for a "correct" exposure. (They can do this because the processing is, or should be, tightly controlled.) You can find these on the data sheet. Typically a "red density" value range for both a gray card and for flesh tones.

Too much info?

The answer is simple - Ilford is too bone-idle to properly calibrate its vague curves in Lux-seconds. Logarithmic or otherwise.

Longer answer:

If Ilford's developing of HP5plus conforms to the ISO standard, and if it is truly 400 ISO, then the 'speed point' falls at 0.1D above base+fog on the Y axis, and at - 2.7 LOG Lux.seconds on the X axis.

Of course the scale and line thickness of Ilford's near-useless published curve makes exact placement of the speed point near impossible. But if you say 1 on Ilford's Relative Log E scale is actually minus 2.7 Log Lux.seconds, you won't be far off.

Therefore you need to add Ilford's relative figures to - 3.7 to get the approximate absolute Log Lux.second value. Making the mid-grey value fall at about 2.1 on the X axis of Ilford's rubbishy chart.

P.S. The relationship between the ISO speed-point (@ 0.1D above B+F) and ISO film speed is:

ISO speed = 0.8/reference Lux.second exposure

NB. Working in linear Lux.seconds, not the Logarithmic value.

Conversely the speed-point exposure is got from:

0.8/ISO speed

Again, using the linear Lux.second value, which must be converted to/from the conventional H&D chart Log E value.

...the 'speed point' falls at 0.1D above base+fog...

Right! Not at 0.15 like I said.

Thanks for the thoughtful reply. How you determined that hp5's speed point is has an absolute (log) exposure value of - 2.7?

From there, i understand that we're making a relative value of 1 = - 2.7. And if the absolute exposure for iso 400 for middle gray is about - 1.8 or- 1.7, then yes, that's 1 or 1.1 x units from the speed point. So about 3.5 stops.

How you determined that hp5's speed point is has an absolute (log) exposure value of - 2.7?

It isn't obvious from the characteristic curve, or whatever. You have to know how the ISO method works, meaning that you looked it up previously.

In the ISO speed standard (for b&w general purpose neg film) the arithmetic film speed is, as rodeo_joe says in post #5, 0.8 divided by the quantity of light needed to achieve the speed point. So, 0.8/0.002 = 400, right? If you take the log of 0.002, it's -2.7

FWIW the speeds are rounded off according to rules which is why you never see "in-between" film speeds.

Graphically, the ISO speed for B&W film is determined thus -

Where m denotes the 'speed point' and Hm is the exposure in Lux.seconds required to acheive m density.

The ISO speed is then 0.8/Hm and rounded to the nearest preferred value.

That the X axis scale is in Log units can be a cause of confusion, but the maths is all fairly straightforward.

(BTW, the money-grubbing ISO would have you pay about 40 Swiss Francs for the above trivial piece of information, if you were to purchase the relevant ISO document. So much for 'freedom' of information.)

Got it. Thank you. I've seen another formula for iso (may be used in color neg film? Or maybe differing standards over time?)

Iso=10/H(mid) where H(mid) is the exposure that gives properly exposed middle gray. Thus for iso 100, H is 0.1 so log H is -1.

The formula based on speed point is new to me. Thanks for sharing.

I've seen another formula for iso (may be used in color neg film? Or maybe differing standards over time?)

Hi, the formula you listed sounds like that used for color reversal (slide) films.

In the case of negative films (both b&w and color) the test methods are based on a film speed point where density is only slightly above base + fog. But for color reversal films they look at the entire sensitometric curve, then select a film speed point that is roughly midway in the curve. So obviously the log exposure point is far away from that just barely producing an exposure.

If you want to try this on the Velvia curve (in the data sheet that you linked) here's the general method: find a first point on the curve where film density is 0.20 higher than the minimum density. Note the log exposure at that point. Next, draw a line, starting from that point on the curve tangent to the curve in the high density range. If the tangent point has film density greater than 2.0 (above minimum density), which Velvia does, find the point where the line crosses the film density value of 2.0 + minimum density. Note the log exposure at that point. Next, average the two log exposures, (first_log + second_log)/2, then use the result in your formula.

Color neg film uses yet another method. It's similar to that for b&w negative film. It finds exposure points for each layer that is 0.15 density above base + fog, then "weights" the color layers to get a single exposure value. The speed formula is somewhat similar to the b&w method (I don't wanna give it without explaining the weighted method more clearly).

An important detail with respect to the color films is that they must be tested with the correct color of light. For daylight balanced film it calls for "ISO sensitometric daylight (D55)." So if you do shooting tests midday in a shaded area, this will likely have excessive bluish light (a higher color temperature), meaning that the blue-sensitive layer of the film is relatively overexposed (compared to the other two). So this would tend to rock the boat, so to speak, with respect to any critical film speed measuring tests.

I've seen another formula for iso (may be used in color neg film? Or maybe differing standards over time?)

The formula and graph I gave for B&W film speed determination has been the same since it was developed - sorry - as an ASA/BS standard, way back when the Swiss Gnomes of the ISO were in short pants (or Lederhosen?). If it's changed at all since then it's in the most insignificant of wording.

The formula and methodology for determining B&W ISO film speed today are essentially identical to those of the ASA standard proposed, and almost universally accepted, circa 1963.

The methodology for development is embodied in the graph - i.e. the film must receive development such that an exposure of 20 x Hm results in a density increase of 0.8D above density m. In other words 0.9D above B+F at Hm + 1.3 Log E.

Any variance from the above is not in accordance with the ISO standard for determining B&W film speed. But other and different standards exist for colour negative and colour reversal film. Especially for reversal films, where the minimum exposure results in a close-to-maximum density and vice versa. Plus the gamma of colour reversal film is close to 2, while that of B&W negative film is normally in the range of 0.6 to 0.7, and that of colour negative film is 0.5 or less.

The family of T-max curves - amalgamated from Kodak's datasheets. Incidentally showing that 'push' processing hardly moves the curves leftwards at all, and therefore also hardly alters the film speed.

Ektachrome 200 datasheet curve.

KODAK Gold curves by layer. Blue-sensitive/Yellow-forming, Green-sensitive/Magenta-forming, and Red-sensitive/Cyan-forming.

The formula and graph I gave for B&W film speed determination has been the same since it was developed - sorry - as an ASA/BS standard, way back when the Swiss Gnomes of the ISO were in short pants (or Lederhosen?). If it's changed at all since then it's in the most insignificant of wording.

 

The formula and methodology for determining B&W ISO film speed today are essentially identical to those of the ASA standard proposed, and almost universally accepted, circa 1963.

 

The methodology for development is embodied in the graph - i.e. the film must receive development such that an exposure of 20 x Hm results in a density increase of 0.8D above density m. In other words 0.9D above B+F at Hm + 1.3 Log E.

 

Any variance from the above is not in accordance with the ISO standard for determining B&W film speed. But other and different standards exist for colour negative and colour reversal film. Especially for reversal films, where the minimum exposure results in a close-to-maximum density and vice versa. Plus the gamma of colour reversal film is close to 2, while that of B&W negative film is normally in the range of 0.6 to 0.7, and that of colour negative film is 0.5 or less.

 

[ATTACH=full]1436630[/ATTACH]

The family of T-max curves - amalgamated from Kodak's datasheets. Incidentally showing that 'push' processing hardly moves the curves leftwards at all, and therefore also hardly alters the film speed.

 

[ATTACH=full]1436631[/ATTACH]

Ektachrome 200 datasheet curve.

 

[ATTACH=full]1436629[/ATTACH]

KODAK Gold curves by layer. Blue-sensitive/Yellow-forming, Green-sensitive/Magenta-forming, and Red-sensitive/Cyan-forming.

This is good information. Much easier to understand than my Kodak Sensitometry book...

I JUST Love it when people talk TECH!

I always think that I am understanding it, but 5 minutes later I don't.

Perhaps it's just age, I do still remember the order of the carpal bones even 50 years later

(example "Never Lower Tilly's Pants, Grandmother Might Come Home"

Note that as described, for black and white negative films you follow the rules to determine the development time, and then determine the ISO speed from the resulting curve.

For C41 and E6, the development chemistry and timing is standardized. You just draw the curve and find the ISO value.

I am not sure about black and white reversal films.

It does seem that the same film, developed for negative or positive, has a different ISO value.

It does seem that the same film, developed for negative or positive, has a different ISO value.

Taking the B&W film criteria and comparing them to the 200 ISO Ektachrome curve(s), it's not that far apart.

Kodak's graph doesn't seem to go quite to Dmax; so determining the density drop (rather than rise) for a minimum exposure 'speed point' is a bit of a guess. But given the higher gamma - about 3 times that of negative film - Dmax minus 0.3D comes at around -2.4 Log E, which is the same as it should be for a 200 ISO B&W negative film.

Panatomic-X negative is ISO 32, but reversal processed it is 64.

There is a Tri-X reversal movie film:

https://www.kodak.com/content/products-brochures/Film/TRI-X-7266-Technical-Data-EN.pdf

that is ISO 200. There are no development suggestions for negatives, but one might expect

it to be 400 like other Tri-X.

Regarding the characteristic curve, I do wonder why they like different scales on x and y axis.

It is much easier to judge the slope when the scales are the same.

The amount of exposing energy is given in logarithmic notation, otherwise, the graph paper length becomes inconvenient as to length. Most graphs plot in ½ stop increments. The test exposure strip likely covers 21 steps. To plot on graph paper using decimal notation requires a sheet of graph paper many yards long. Even if the plot only covers 14 exposures, that’s 8,192 times the 1st exposure.

Using log notation base 10, 0 represents 1: 1 represents 10: 2 represents 100: 3 represents 1,000: 4 represents 10,000. Thus, using log notation makes the graph more easily interpreted.

As to slope: Use a protractor and find the angle of the slope. Typical pictorial film has a angle of about 38⁰. Next, we find the tan of this angle = 0.8. This reveals the contrast. If the angle is 45⁰ the tan = 1.

Now the f-stop is a 2X change in light energy. The value 2 in log notation base 10 is 10 elevated to the 0.30 power. For convenience we just say 1 f-stop = 30 units of density change. However. that is true if the slope angle is 45⁰ (gamma 1). That is likely too contrasity, so films typically have a slope of about 38⁰ gamma 0.8. For this contrast, 1 f-stop change is 0.3 X .8 = 0.24 density units change per f-stop.

Nobody said this stuff is easy.

Also, for color negative film, it was designed to produce prints on paper using an enlarger. The enlarger is equipped with cyan, magenta, and yellow filters plus lens aperture and time of exposure. All are used to control the exposure film to paper.

Because the cyan filter is a poor red controller (crosstalk), its use is avoided. Thus, the system is biased so the cyan filter is seldom deployed. That dictates that the red exposing energy be controlled not by the cyan filter but by exposure time and or lens aperture. To achieve this, the lens aperture is stopped down to obtain the correct red exposure. The green and blue light on the easel are now in excess. Magenta filters are deployed to adjust down the green exposure and yellow filters to control the blue exposure. This is the reason that, for a color negative film, the red emulsion is the one measured for ISO (speed).

(snip)

 

Using log notation base 10, 0 represents 1: 1 represents 10: 2 represents 100: 3 represents 1,000: 4 represents 10,000. Thus, using log notation makes the graph more easily interpreted.

 

As to slope: Use a protractor and find the angle of the slope. Typical pictorial film has a angle of about 38⁰. Next, we find the tan of this angle = 0.8. This reveals the contrast. If the angle is 45⁰ the tan = 1.

(snip)

If you look at the graph in this one:

https://www.kodak.com/content/products-brochures/Film/TRI-X-7266-Technical-Data-EN.pdf

the graph is square, but there are three log10 units on the vertical axis, and six on the horizontal axis.

The actual curve on the graph only goes from -3 to 0 on the horizontal axis, so they could have expanded the graph.

Otherwise, the slope (in log units) is twice what it should be if you use a protractor.

Now, in this one I believe it is close to 2:1, but many graphs aren't so convenient, maybe 1.5:1 or something else.

So, not only do you measure the slope with a protractor, but also the scale change.

And you can't compare graphs, side by side, with different scale ratio.

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.

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.

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!

Why make it so hard to read the graphs!

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! :confused:

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!!!!!!!!!!

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 [uSER=4774668]@alan_marcus|2[/uSER] has stated (that 18% is the anchor point). I found at least 1 other thread on photo.net in which [uSER=4774668]@alan_marcus|2[/uSER] and [uSER=2403817]@rodeo_joe|1[/uSER] 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