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View Full Version : On the effect of shutter speed on image contrast for R5 Monobath film-development



Lenrick
7-Mar-2016, 06:33
Disclaimer: If you are looking for a well-throughout, well-executed and well-analyzed test done according to industry standards written in industry jargon, I’m afraid you have to look elsewhere.

I wanted to test if it was possible to increase or decrease the contrast in New55 Atomic-X film developed in New55 R5 Monobath. I’m not sure what I can do with agitation or time emerged in liquid, but now I tested exposure time. I took two exposures of a still life setup, one with short exposure time and one with long exposure time and compared the negatives.

The two exposures were done back to back on the same equipment and in the same lighting condition (no filters). My incident-light meter read EV 9 which gave me the two exposures as:
1) f/5.6 and 1/15 s
2) f/64 and 8 s, which I compensated to 50 s due to reciprocity.
But I really messed up the second (the long) exposure since 8 s should be adjusted to about 37-38 s according to the reciprocity information provided by New55. My bad, but I think we can still get some information from this test. But remember the long exposure is overexposed compared to the short exposure. The negatives were developed simultaneously in the same tank, and later scanned in an old Agfa Duoscan T2500.

Figure 1 is a photo of the scene captured on my iPhone, the gray floor is quite close to a gray card. The statue made of gypsum and is positioned on a sheet of A3 typing paper.

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Figure 2 shows the TIF-scans of the negatives seen as they were exported from scanner, a RAW format would probably be better but I think this is ok for now.

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As you can see in Figure 2, the development is in uneven over the negatives (there is a “pouring-pattern”), and since the negatives were places in the tank face to face different parts of the two negative have been hit by the pouring liquid. I use a Combi Plan tank, which is not suited for R5 Monbath unless you can plunge the film directly into the open tank in total darkness (as recommended by New55). I don’t have total darkness anywhere outside my changing bag so I’m limited to using the narrow ports. The pixel intensity line plots show that the bright side of the nose is saturated in both exposures and the floor boards have very different intensity in the two images. I’m not sure how much the scanner is to blame for the saturation; perhaps a better scanner can penetrate the dense parts of the negatives. You can also see that the non-exposed areas of the negatives (the areas hidden in the film holder grooves) are completely black while there is some intensity in the very dark areas between the floor boards.

Now in hindsight I realize that it is almost pointless to compare images with such different depth of field. There is almost no texture in the floor boards in the short exposure image but plenty of texture in the floor boards in the long exposure images. I should of course use neutral-density filter to adjust the exposure time. I should also have found a better subject with both dark and bright regions.

In Figure 2 I have also marked four small areas of interest in the two images. “SE” is short for short exposure and “LE” for long exposure. The pixel intensity valued for the eight areas are:

___SE | LE
1: 220 | 225
2: 108 | 145
3: 027 | 057
4: 015 | 006


The histograms can be seen in Figure 3.

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The area with the highest intensity while still having some texture (position 1 in both images) are very close in intensity (a difference of only 5 units), even though one image is overexposed. The two mid-gray areas (position 2 and 3) have both increased with about 30-40 units due to the overexposure in the long exposure. The very dark shadow (between the floor boards in position 4), actually have lower intensity in the overexposed image. This is, however, most likely due to the increased depth of field in the overexposed image. The floor board gaps are smeared into gray goo in the short exposure image while being clear and distinct in the long exposure image.


It is hard to say how the long exposure time affected the contrast, but it seems the contrast increased quite a bit in the darker areas while still maintaining acceptable in the brighter areas. It seems that the very bright areas are affected much less by overexposure than the darker areas. Is this finding true for all overexposed negatives developed in R5 Monobath, independent of exposure time (shutter speed)?

Now I would like to find a better subject and capture 3 images and with the aid of neutral-density filters keep the aperture constant.
1) One image at normal exposure and short exposure time
2) One image at normal exposure and long exposure
3) One overexposed image at short exposure time

Questions, suggestions or other feedback?

ic-racer
7-Mar-2016, 07:43
Exposure time does not affect film contrast unless the time is very short, like 1/10,000 or less or very long, like 10 seconds or more. If you discover something, let us know. If you only want to expose only a portion of the tonal range of you scene, then you can under-expose to remove the unwanted low values. This will give you a low contrast negative with no shadow detail. Common jargon for that technique is 'push processing.'


Make note of the sections "Family of Curves" and "Exposure Latitude"
http://motion.kodak.com/KodakGCG/uploadedfiles/motion/US_plugins_acrobat_en_motion_education_sensitometry_workbook.pdf

Ken Lee
7-Mar-2016, 10:43
Exposure time does not affect film contrast unless the time is very short, like 1/10,000 or less or very long, like 10 seconds or more.

What happens to contrast at those extremes please ?

Jody_S
7-Mar-2016, 10:52
Any exposure done under conditions of reciprocity failure will necessarily have a different contrast than one done in 'normal' conditions (which results in over-exposing the highlights, and under-exposing the shadows). No matter the film, but obviously worse with some films that are notoriously bad for reciprocity failure.

Simple logic: if you meter a scene and discover the highlights require a 2s correction for RF, and the shadows require 10m correction. You meter for a midtone and end up adding 12s to your metered time. You have over-exposed the highlights by 10s, and under-exposed the shadows by 9m48s (for neutral grey). This will mean your scene with 5 stops between shadows and highlights will be exposed so as to have 8 stops.

Ted R
9-Mar-2016, 09:02
There is a lot to learn about photography. I learned a lot by reading books from Kodak, Ilford and technical books written for black and white photographers. There are important and fundamental principals involved. Perhaps some study of the basic texts on the meaning of film density, the purpose and effects of development changes, and the concept of exposure variables would be appropriate? Look for technical books on black and white photography. Here is one suggestion http://www.amazon.com/Negative-Ansel-Adams-Photography-Book/dp/0821221868/ref=sr_1_5?ie=UTF8&qid=1457539534&sr=8-5&keywords=ansel+adams+book

Ken Lee
9-Mar-2016, 10:42
Any exposure done under conditions of reciprocity failure will necessarily have a different contrast than one done in 'normal' conditions (which results in over-exposing the highlights, and under-exposing the shadows). No matter the film, but obviously worse with some films that are notoriously bad for reciprocity failure.

Simple logic: if you meter a scene and discover the highlights require a 2s correction for RF, and the shadows require 10m correction. You meter for a midtone and end up adding 12s to your metered time. You have over-exposed the highlights by 10s, and under-exposed the shadows by 9m48s (for neutral grey). This will mean your scene with 5 stops between shadows and highlights will be exposed so as to have 8 stops.

I never thought about it that way.

It's common practice to meter and expose for the shadows - and to determine compensation for reciprocity failure after determining exposure.

Your analysis exposes a flaw in that approach: how do we solve the problem ?

I use development by inspection, so perhaps during development I am compensating unknowingly - but I rarely adjust development by more than +/- 1.

koraks
9-Mar-2016, 10:59
You solve it by contraction: decrease the EI and develop shorter to decrease contrast. Af the same time, this makes the reciprocity issue bigger. It then follows that at some point, you run out of options to deal with the situation and you have to resort to other measures - open up the aperture and/or use different film.

Michael R
9-Mar-2016, 12:30
I never thought about it that way.

It's common practice to meter and expose for the shadows - and to determine compensation for reciprocity failure after determining exposure.

Your analysis exposes a flaw in that approach: how do we solve the problem ?

I use development by inspection, so perhaps during development I am compensating unknowingly - but I rarely adjust development by more than +/- 1.

It is to some extent an unavoidable problem when we make long exposure reciprocity adjustments based on the metered exposure time rather than actual subject luminances. It's a shortcoming of typical reciprocity tables/graphs. Long exposures generally cause an increase in contrast (a consequence of reciprocity failure to begin with). However current films perform quite a bit better than older films.

In practice it doesn't work out too badly for a few reasons. 1) Since most Zone System users apply reciprocity adjustments to their Zone III metered exposures, the contrast effect is typically minimal throughout the normal "textural" range. Instead the reciprocity effects are confined mostly to the lowest light levels (ie below Zone III). For an example of this, refer to Howard Bond's reciprocity tests. Since he was targeting a constant Zone III for his exposure adjustment factors, he found there was virtually no minus development adjustment required. 2) Flare counteracts the effects of reciprocity failure. 3) The effects of (1) and (2) combine with other slop in the process so that we don't notice. We compensate unknowingly in printing/editing. When it comes to making negatives, we don't have quite the control and precision we think we do, no matter how much testing, calibrating and fiddling we do. Fortunately it doesn't matter because there is enough latitude in the process and materials that we can work around these inaccuracies (usually without realizing it) and make great prints.

Now, I don't think any of this explains anything in the original post, but it is an interesting question nonetheless. It highlights the fact that low light/long exposure reciprocity failure depends on the intensity of light hitting the film, not the exposure time per se.

David Schaller
9-Mar-2016, 12:39
I never thought about it that way.

It's common practice to meter and expose for the shadows - and to determine compensation for reciprocity failure after determining exposure.

Your analysis exposes a flaw in that approach: how do we solve the problem ?

I use development by inspection, so perhaps during development I am compensating unknowingly - but I rarely adjust development by more than +/- 1.

I'm with Jody. I do it, when I remember(!), by metering for the shadow, and applying the reciprocity correction right away, to place the shadow. So, for example, if the meter says the shadow is EV 8 at a film speed of 100, and my aperture is f32, it would be 4 seconds so to compensate that would be 8 seconds. Then I divide the corrected time by 4 to place the shadow in zone III, and my exposure is 2 seconds. Then I see where the highlight fell, and apply N- whatever is necessary. So in many situations I am doing fairly big N minuses to tame the highlights, and to have adequate shadow exposure. If it's more than N-3, I'll place the shadow in Zone IV, since there will be a loss of film speed with reduced development.

Of course these are different ways of talking about the same phenomena. Really this has turned into a discussion of compensating for reciprocity failure, especially when the shadows and the highlights are on opposite sides of 1 second.

koraks
9-Mar-2016, 14:58
> It highlights the fact that low light/long exposure reciprocity failure depends on the intensity of light hitting the film, not the exposure time per se.
Isn't it a matter of what happens at the toe of the film curve? Threshold sensitivity and the non-linear bit at the start of the curve?

Michael R
9-Mar-2016, 21:25
The reciprocity behaviour of an emulsion depends on light intensity. An emulsion actually functions most efficiently within a small intensity range. From a practical perspective there is a much larger intensity range within which the emulsion functions close enough to its peak efficiency that the deviations are immaterial. This is the intensity range within which we say the reciprocity law holds for the film. Outside that range, reciprocity fails. The effects on the characteristic curve depend on how far outside the reciprocity range different parts of the scene are when the light hits the film. With respect to the resultant characteristic curve, generally what you could say is that low intensity reciprocity failure has the effect of making the toe shorter/steeper. But - flare has the opposite effect. So it is difficult to know what you're getting.

koraks
10-Mar-2016, 08:41
Good point; there was a flaw in my thinking. The problem with reciprocity failure in combination with the film characteristic curve is, as I understand it now, that the curve describes the densities that are produced (given the development regime used for establishing the curve) if various amounts of light hit the film in a timeframe that is long enough and also short enough for reciprocity to hold true. Reading a bit further on it, it appears that a grain will only form a latent image (or become developable) if sufficient photons hit it; apparently a grain requires something like 4 photons to achieve this state. The compound effect of the photons hitting the grain will result in a stable state of the grain allowing it to be reduced during development. However, if there is too much time between the arrival of the photons, the effect doesn't compound; the grain reverts to its unexposed state after only one or two photons hit it and some time passes.

Thinking about it, it seems that flare would provide a number of stray photons across the entire image area, assisting in the directional (actually image forming) photons to expose the individual grains in low-light areas - a kind of in-situ preflashing, in a way. Although the analogy of preflashing doesn't hold up on closer scrutiny, as preflashing just lifts the black point to an overall base density so that the actual image is lifted up on the curve, partly or completely past the toe of the curve, so it exposes several grains fully. In contrast, flare in a situation of low light where it moderates reciprocity failure, mainly plays a role on the level of individual grains by exposing them partly.

Ken Lee
10-Mar-2016, 09:21
However, if there is too much time between the arrival of the photons, the effect doesn't compound; the grain reverts to its unexposed state after only one or two photons hit it and some time passes.

Excellent.

Does all of this tell us anything about photographs made in the early days of photography, using very slow emulsion and long exposures ?

Were early photographers always encountering reciprocity failure, which imposed on their images a certain non-linear curve, as it were ?

koraks
10-Mar-2016, 09:39
I wonder and I couldn't say. I'm not sure how the physics are different when you compare small-particle emulsions such as wp collodion emulsions with the larger grained and aniline dye sensitized ones. I can imagine that the smaller particle emulsions could requirer fewer photons per particle to activate them and therefore suffer less from reciprocity, but simply require a lot more photons for a given area (more small particles per area than larger particles in more modern emulsions) in order to create sufficient density for reproduction. Again, it would be necessary to compare what happens on the level of single particles vs larger areas. But I'm just speculating now.

Michael R
10-Mar-2016, 09:42
Good point; there was a flaw in my thinking. The problem with reciprocity failure in combination with the film characteristic curve is, as I understand it now, that the curve describes the densities that are produced (given the development regime used for establishing the curve) if various amounts of light hit the film in a timeframe that is long enough and also short enough for reciprocity to hold true. Reading a bit further on it, it appears that a grain will only form a latent image (or become developable) if sufficient photons hit it; apparently a grain requires something like 4 photons to achieve this state. The compound effect of the photons hitting the grain will result in a stable state of the grain allowing it to be reduced during development. However, if there is too much time between the arrival of the photons, the effect doesn't compound; the grain reverts to its unexposed state after only one or two photons hit it and some time passes.

This is essentially the theory. See "Gurney-Mott". An analogous effect occurs when light intensity is very high combined with a short exposure duration (high intensity reciprocity failure). You have a kind of "choke" effect in that case, causing the latent image to form more slowly than the exposure equation (reciprocity) predicts.


Thinking about it, it seems that flare would provide a number of stray photons across the entire image area, assisting in the directional (actually image forming) photons to expose the individual grains in low-light areas - a kind of in-situ preflashing, in a way. Although the analogy of preflashing doesn't hold up on closer scrutiny, as preflashing just lifts the black point to an overall base density so that the actual image is lifted up on the curve, partly or completely past the toe of the curve, so it exposes several grains fully. In contrast, flare in a situation of low light where it moderates reciprocity failure, mainly plays a role on the level of individual grains by exposing them partly.

Actually the analogy of flare to in-situ pre-flashing/pre-exposure as you explained it does hold up well. Flare and pre-exposure are both non-image forming light, and have the same effects. The only difference is that pre-exposure is "controlled" whereas flare is a largely uncontrolled variable.

koraks
10-Mar-2016, 15:24
But there's one inconsistency in the flare/pre-exposure argument: pre-exposure happens at some point in time and then has an effect that endures relatively long; you can pre-expose/flash minutes or perhaps even hours (days? Not sure how long it lasts) before making the actual exposure. Flare under conditions of very low light needs to occur at about the same time when the main image photons hit the film - otherwise the individual grains would revert back to a fully unexposed state. In addition, preflashing constitutes a pretty large photon bbardment as you generally preflash to zone II or so. It seems to me that that's generally a lot more light than the few photons that reach the film under low light levels where reciprocity failure occurs. Or am I missing an essential element?

barnacle
11-Mar-2016, 10:16
But the difference there is the amount of light, surely? If it takes (say) four photons to arrive within a certain time before the latent image is irrevocably imprinted, then those four photons can arrive at any time within the time limit as stray flare, but might only expose to zone '0'; just enough that any *more* will imprint an image.

Preflashing to zone 2 is going to be a lot more photons; enough that the exposure is as permanent as any other.

Neil

Michael R
11-Mar-2016, 11:08
But there's one inconsistency in the flare/pre-exposure argument: pre-exposure happens at some point in time and then has an effect that endures relatively long; you can pre-expose/flash minutes or perhaps even hours (days? Not sure how long it lasts) before making the actual exposure. Flare under conditions of very low light needs to occur at about the same time when the main image photons hit the film - otherwise the individual grains would revert back to a fully unexposed state. In addition, preflashing constitutes a pretty large photon bbardment as you generally preflash to zone II or so. It seems to me that that's generally a lot more light than the few photons that reach the film under low light levels where reciprocity failure occurs. Or am I missing an essential element?

I don't think you're missing anything fundamental. Flare is non-image forming light occurring simultaneously with exposure, while pre-exposure is non-image forming light before exposure. I suppose depending on how far in advance you pre-expose, the latent image keeping properties of the emulsion come into play, but it's probably relatively immaterial. The flare factor is variable depending on the scene. It can be significant, but under most low light conditions which would lead to long exposures, flare may be lower.

Lenrick
14-Mar-2016, 09:21
Quick note: I have to adjust how I use my (newly bought but second hand) spotmeter. As I used it this weekend the negatives seems to come out under exposed by about 2 stops.

In the figure below you can see the scene where I made three exposures, and the “field notes” of the light-metering. The numbers are EV and GC strand for measurement on a gray card.
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Three exposures were made on Atomic-X:
A: f/32 at ¼ s, no filter
B: f/32 at 12 s, neutral density-filter (reciprocity adjustment from 4 s to 12 s)
C: f/32 at 1 s, no filter just “over-exposed”
In the text, I’ll refer to the images by A, B or C from now on.
The negatives were developed/fixed in R5 Monobath following New55’s instruction as close as possible. The negatives were then scanned in a Agfa DuoScan T2500 using VueScan. The raw (.dng) files were imported into Photoshop using Camera Raw where they were inverted, but no other adjustment (everything in Camera Raw import set to 0). In the figure below you can see the histograms from the three exposures; I cropped them to eliminate the contribution from the black unexposed parts of the negatives and the white negative-holder.
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Judging from the histograms alone, image A seems to have a broader range of pixel values (~gray scale) than image B. Perhaps 12 s is too short time to see any reciprocity-induced increase in tonal range? What I find reassuring when playing around with these images in the computer is that they show the behavior you would expect from the zone-system. In image A and B there is clearly tonality in the grass and other areas, but no real texture, as you expect from zone I and II (the toe of the curve). In image C, however, there is texture in those very same areas, since they have moved up into zone III and IV.
In the figure below you can see some pixel values from the 5 positions. The numbers for image A and B are very close, which I interpret as a very good job on the reciprocity table made by the New55 company. The darker areas in image C is remain less affected by the overexposure than to the lighter areas, when compared to image A and B. This is the opposite observations made in the original post of this thread.
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The figure below show some more details from position 1 and position 4 in the form of pixel value line plots. As you can see not only is the pixel intensity value higher in image C, but the contrast seems higher also. (When I write contrast here, I mean intensity variation in two adjacent areas.) Maybe it is just because image C is exposed better for this scene in this light.

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