What does this characteristic curve mean?

Discussion in 'Exposure Discussion' started by Doc W, Jun 13, 2018.

  1. Doc W

    Doc W Subscriber

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    Just when I thought I was getting a handle on understanding characteristic curves, I came across this one (scroll down to the second page):

    http://bergger.com/media/wysiwyg/Fiches_techniques/BERGGER_PANCRO_400_DATASHEET_01_2017.pdf

    Here are my questions. First, I understand the vertical axis - that is simply what the densitometer sees, but what about the horizontal? It is labelled "relative log exposure" but is in whole numbers, rather than stops (i.e., 0.3, 0.6, etc.).

    Second, what does this curve tell us? The toe looks a little odd, starting so high.

    Any light you can shed is most appreciated. Try to keep it really simple. I have great difficulty with these sorts of things.
     
  2. Andrew O'Neill

    Andrew O'Neill Subscriber

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    It's a short toe.... so, more contrast in the shadows... which I prefer.
     
  3. Bill Burk

    Bill Burk Subscriber

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    Whole numbers are logarithmic which represent 1, 10, 100, 1000, 10000. They call the exposure "relative" so that you don't get too hung up on trying to figure out what the actual exposure is.

    Each stop is 0.3 logarithmically, so what you see is that from toe to shoulder there's 3 whole numbers. 10 stops of exposure from toe to shoulder is a good long range of exposure that you will not likely use up because average scenes have about 7 stops.
     
  4. Chan Tran

    Chan Tran Member

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    Most of the characteristic curves I have seen the horizontal axis are in Log of exposure in Lux.Sec. But the one you link to only show log of relative exposure so there is no unit and thus there is no speed information.
     
  5. OP
    OP
    Doc W

    Doc W Subscriber

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    Here is what I get from this, so far.

    The fbf of the film is about halfway between 0 and 0.5, thus about 0.25, as measured by a densitometer. My experience with this film is that it has a very high fbf, so I get that. As exposure increases, it begins to rise, at about 1.5 logH on the horizontal axis. But I don't know what those integers on the x axis mean. 1.5 what? 1 to 7 of what? That's my main problem.
     
  6. Bill Burk

    Bill Burk Subscriber

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    If they had labeled the meter candle seconds of exposure, you would know.

    If they had developed the film to ASA/ISO contrast (approximately 0.62) you could look where the curve crosses 0.1 B+F and look up to where I have 400 on this scale.

    As it stands though, line up 400 near where the toe has risen 0.1 and that’ll be close.

    upload_2018-6-13_12-31-25.png
     
    Last edited: Jun 13, 2018
  7. Kino

    Kino Subscriber

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    I had to re-write this, as I wasn't paying attention to the actual question!

    The 0.70 gamma on the chart is about 0.05 gamma higher than "normal" (0.65), but the stock appears to have a very long straight-line portion of the curve, so it should be a very long-scale producing negative.

    The base+fog is about normal for grey-based negative stocks used in motion picture work.

    The horizontal axis is a density scale. Each measured density of the ORIGINAL test wedge that is being printed onto the negative is located along the bottom of the scale and projected upwards as as straight line.

    This wedge is then printed to the negative stock using some known light value and it is processed at the noted time/temp and chemical mix.

    The density results from each of these patches on the wedge that has been printed to the negative, is measured by a densitometer (status M usually) and is noted on the vertical axis and a line is drawn horizontally. Where the two meet, a single point is drawn.

    Then a curve is drawn through all the projected points of the wedge to give you the characteristic curve.

    As Bill notes, the exposure value was not disclosed, so only gamma relative to time/temperature and chemistry is available.

    Typically, the steeper the straight-line portion of the curve, the higher the relative contrast for that time/temperature and chemistry combo, the more horizontal the straight-line portion of the curve, the lower the relative contrast.

    Sharp toe (bottom) or shoulder (top) curves denote a film that tends to suddenly exhibit reciprocity failure and can "clip" hard on shadows or highlights.

    Gradual toe or shoulder curves show gradual reciprocity failure and tend to compress gracefully without sudden clipping of values.
     
    Last edited: Jun 13, 2018
  8. pentaxuser

    pentaxuser Subscriber

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    I too noted that the Bergger 400 characteristic curve was based on a Gamma of 0.7 which is higher than that associated with Kodak or Ilford film curves, although I couldn't see what the Gamma /CI is for Ilford HP5+ . I wanted to compare this with Ilford HP5+ curve to see if I could draw meaningful conclusions as to the differences between the two films.

    HP5+ uses relative log exposure as well and seems to cover the same number of stops on the straight line portion but whereas HP5+ starts to rise after 1, Bergger rises at a later point but seemingly more steeply. I don't have the computer skills to attach both curves but for educational purposes could those who know about these matters and about reading curves explain what meaningful conclusions can be drawn from the two films?

    I have used HP5+ simply as an example for the purposes of being able to draw correct conclusions. I stress this post is driven by a desire to improve my ability to draw meaningful conclusions from characteristic curves and is not a backdoor attempt to turn this into a Bergger v Ilford thread

    Thanks

    pentaxuser
     
  9. Kino

    Kino Subscriber

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    Gamma is a function of the filmstock's reaction to level of exposure, time/temperature and developing agents. The baseline developer, D76, is a Hydroquinone/Metol developer with about equal parts of each developing agent. HQ is a hard/very active developing agent and Metol/Elon is a soft, low energy developer.

    Potentially, if the Bergger was processed in a developer that had lower amounts of HQ than Metol/Elon, the curve could be closer to .65 at the same time/temp.

    The Illford HP5+ datasheet shows it being developed in Illford ID-11 developer, which is almost identical to D-76 BUT it has a bit more Metol/Elon; about 3 grams to 2 in D76 (from what I can find) and should produce a slightly lower contrast/density image.

    If you developed Bergger 400 in ID-ll, it MIGHT be slightly closer to .65, I have no experience with this stock.

    If you want to compare two filmstocks, you have to establish a time/temp/developer/agitation standard and process both identically in the same manner.

    You can conclude that HP5+ in ID-ll @ 7 1⁄2min. at 68°F (20°C) with intermittent agitation appears to have a lower base+fog level, lower contrast and lower total density range than Bergger Panchro 400 at it's incompletely stated development time/temp in Stock D76. Not trying to be fatuous, but it's not apples to apples unless you use identical exposure/time/temp/developer/agitation standards for both stocks.

    You can, however, generalize and state from the published curves that it appears that Bergger is more contrasty and dense than HP5 at their manufacturer's indicated time/temp/developer/agitation standards as per their datasheets...

    Hope this makes sense...
     
  10. Kino

    Kino Subscriber

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    Also wanted to add, when you typically see a large base+fog, and it is NOT a result of the film base dye for antihallation properties pushing the entire density range upward, then it indicates to me the film has been developed at a higher temperature than is optimum.

    If the Illford is clear base film and the Bergger is gray base film, then you can subtract the density of the dye from the Bergger plot for a better comparison.
     
  11. Bill Burk

    Bill Burk Subscriber

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    B0DBACD9-0433-4EB9-97E8-A5765C2DA0C4.jpeg
    Here is a partial graph and from it, using my new delta-x overlay, an estimate of the meter candle seconds of exposure which the factory spec applied to the film.
     
  12. Photo Engineer

    Photo Engineer Subscriber

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    The true speed of the emulsion is the inflection point of the curve from fog. This can be seen in the graph to the left of the notation delta x. The label in the graph "speed point" is defined by an ANSI statement, whereas the former is defined by physics and will be constant in all cases.

    PE
     
  13. iandvaag

    iandvaag Subscriber

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    TL;DR Relative log(exposure) means that the numbers on the x-axis have no absolute meaning, but the difference between two numbers is meaningful. The horizontal axis of the characteristic curve linked by the OP is expressed in "units" of log10(exposure). These aren't really units, just like per cent (%) isn't really a unit, it's just a different way of expressing a value. If you take the difference between two x-axis numbers, and want to convert this difference into "units" of "stops of light", simply take the difference in units of relative log(exposure) and divide by 0.3.

    There's a lot of fantastic answers above, but I thought I'd throw my hat into the ring as well. First off, if you are at all interested in learning more about sensitometry, I highly suggest the Kodak Sensitometry Workbook. It is very tractable, and it gives lots of examples.

    Good question. When I have done film testing and produced my own "Hurter-Driffield curve" aka "H-D curve" aka "D - log(E) curve" aka "D - log(H)" aka "characteristic curve", I always express the x-axis in terms of relative exposure. Why do I do this? Because I don't know what the absolute exposure is!

    Consider a Stouffer step wedge: a piece of film with 21 steps of different opacities, where each step is progressively denser by 0.15 "density units". Since you know that each step is 0.15 denser than the previous step, you can simply place this wedge atop the film I want to test, and expose the film under the enlarger. What you have done is provide 21 different exposures to the film, in increments of 0.15 log(H) "units". You don't actually need to know the exact exposure that you gave the film in lux*sec (i.e the illuminance produced by the enlarger bulb in lux multiplied by the exposure time in seconds) to produce a D - log(H) curve. You can simply plot "relative log(H)" on the x-axis and arbitrarily choose the exposure received by the film under the densest portion of the step wedge as "zero relative log(H)". This is how I suspect the Bergger Pancro 400 curve you linked to was produced.

    † Note: you can measure the density of the step wedge with a densitometer and plot that density on the x-axis.

    Density is a unitless quantity. Density is log10(opacity) and opacity is the ratio of incident illuminance to transmitted illuminance. Opacity is unitless since it is a ratio of two quantities having the same units (the units cancel.)

    Absolute exposure has units. Exposure is illuminance (often expressed in lux or millilux) multiplied by time (expressed in seconds), so the typical dimensions of exposure are lux*sec.

    A number given in "relative exposure" has no absolute meaning. If you take a number given in relative exposure and add some constant, you will obtain absolute exposure. But if the people presenting the D - log(H) curve did not measure the absolute exposure they gave the film, this constant is not known! Even so, a D - log(H) curve expressed in terms of relative log(H) still has all the useful information contained in a characteristic curve, except for information regarding the speed of the film (ISO/ASA). So, a number on the x-axis in such a curve is not meaningful by itself, but the difference between two numbers on the x-axis is meaningful! For example, the difference between two points on the curve can give information about exposure latitude. The ratio of a difference in density to a difference in relative exposure for two data points on the curve (i.e. the slope between two points) can give information about the gamma, contrast index (CI), and average gradient.

    You can think of the relative log(exposure) axis as meaning that we don't know where to "fix" the curve along the x-axis. You can slide the entire curve left and right along the x-axis (i.e. horizontal translation), the curve shape doesn't change, so most of the useful information is retained. Note the sliding the curve along the x-axis doesn't involve scaling the x-axis. This is what I mean when I say that the absolute x-value at some point along the curve has no meaning, but the difference between two points is meaningful.

    As stated in a few posts above, if you are more comfortable in stops (more accurately "steps"), a change of 0.3 relative log(H) is equivalent to 1 stop. Once again, a "stop" of light has no absolute meaning, it is a relative term. If you measure a scene with a light meter, and then the sun comes out from behind a cloud and the scene is suddenly illuminated by twice the number of photons, your meter will indicate a decrease in EV by 1 stop, thus ensuring the film receives the same exposure as before the change in amount of light in the scene. A doubling of the illuminance is equal to a increase of illuminance of 1 stop. Now consider a light meter that reads in units of log(H). A doubling of the illuminance would show an illuminance increase of 0.3 "log(H) units" on such a meter.
    ______________________________
    ~A brief aside~
    Here's a question: why does a "stop" equal 0.3 of a log(H)? It seems a bit arbitrary. I tried to make this as clear as possible, but there's probably a better way to explain this.
    First of all define exposure:
    (1) H = E*t
    where H is exposure, E is illuminance, and t is time.

    The exposure (in lux) can be related to EV at ISO 100 by the following formula:
    (2a) E = 2.5*2EV
    It can be shown that the change in illuminance in units of EV, or "stops", is related to the change in illuminance by the expression (2b). This is the mathematical representation of what I described in words in the paragraph immediately preceding this aside.
    (2b) ΔE = 2ΔEV
    or, alternatively
    (2c) ΔEV = log2(ΔE)
    where ΔE is the change in illuminance and ΔEV is the change in illuminance in "units" of exposure value.

    The relative log(exposure) axis refers to log10(H). Substitute the expression (1) for H:
    (3) log10(H) = log10(E*t)
    By the product rule of logarithms, we can write:
    (4) log10(H) = log10(E) + log10(t)
    Now substitute expression (2a) for E into the first term on the right-hand side of (4):
    (5) log10(H) = log10(2.5*2EV) + log10(t)
    Simplify:
    (6) log10(H) = (EV)*log10(2) +log10(2.5)+ log10(t)
    Now, log10(2) is approximately 0.3, and the remaining terms on the right hand are just constants (let's add the constants together call the sum C.) So we can write our final expression:
    (7) log10(H) = (EV)*0.3 + C

    So, you can convert an exposure measured in EV at ISO 100 to a log10(H) by multiplying by 0.3 and adding some constant.
    So that's where the 0.3 comes from! A change of base from log10 to log2 involves multiplying by a factor of log10(2).
    ______________________________

    ~An example of a practical piece of information that you can glean from such a curve~
    So, as Bill said, if you consider that the straight line portion of the curve goes from 2 to 5 relative log(H), this represents a range of 5 - 2 = 3 relative log(H). Convert to stops if that's more convenient:
    (3 relative log(H))*(1 stop / 0.3 relative log(H)) = 10 stops.
    Now consider that you have measured your scene with a spot meter. The brightest spot in the scene measures as having 7 stops light more than the darkest spot in the scene. Thus, your exposure latitude is:
    10 stops - 7 stops = 3 stops

    I hope all of this clarifies rather than confuses. I also hope this can give you a better intuition of sensitometry and maybe even helps you make better photos!
     
    Last edited: Jun 14, 2018
  14. pentaxuser

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    So my conclusion from the characteristic curves shown for HP5+ and Bergger, would be on the range of stops and exposure latitude only, that the Bergger film and HP5+ are about the same. Is this correct?

    On fbf, Bergger has a higher one than HP5+ but there are no meaningful( in the practical sense of that word) conclusions to be drawn. When it comes to actual use in a compare and contrast mode, this difference in the two fbf can be ignored or does it have any practical consequences in terms of use.

    Thanks

    pentaxuser
     
  15. MattKing

    MattKing Subscriber

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    Excellent post Ian.
    It is amazing that such a detailed understanding of slopes and curves would come from someone in a location with the topography of Saskatchewan! :whistling:
     
  16. jonasfj

    jonasfj Member

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    The characteristic curve is not really a description of the film stock, rather of a particular negative, which is the film stock in combination with development.

    When determining the ISO of a film according to the ANSI standard, the curve should have a specified slope. I do not know exactly how they came up with the definitions and methods, but the practical intent is that 10 stops of exposure should fit within the useful part of the curve. Note, that at box ISO and developed according to the standard, all films have approximately the same slope (same contrast) of the characteristic curves.

    The slope of the curve can be modified by changing the development time (keeping all other parameters constant). Longer time => higher contrast and vice versa.

    The latter is very useful, because it makes it possible to take the contrast of the scene into account when exposing and developing the negative (zone system). High contrast scene => Reduce ISO and development time

    I find It difficult to extract any practically useful information out of characteristic curves in technical sheets. The reason is that the final result depends on your entire workflow and process. Exposure, development (developer and method), printing (enlarger diffusive/condenser, paper, filters etc.).

    The best (easiest vs result) way I have come across to get negatives that are easy to print is to follow this method:

    http://www.halfhill.com/speed1.html

    Or to just shoot at half the box speed! ;-)
     
  17. Saganich

    Saganich Subscriber

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    This thread should be saved for reference, well done.
     
  18. Bill Burk

    Bill Burk Subscriber

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    One thing I noticed... the characteristic curve in the factory spec. looks very steep (looks like a high contrast curve) but that’s because the x-axis is compressed in their graph. It’s actually pretty close to normal development.
     
  19. iandvaag

    iandvaag Subscriber

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    Thanks Matt! I've also thought that it's a bit peculiar that I have such a strong interest in stereo photography given my location on the flat prairie landscape! :D
     
  20. Stephen Benskin

    Stephen Benskin Member

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    Thank you Bill.:smile: We're generally used to seeing a scale with a nice 1:1 ratio (3.0 up and 3.0 over) where a gradient of 1.00 would be 45 degrees. This one is over 2:1.
     
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