Approximate ISO-EI speed test with in-camera (LF) contacting

Discussion in 'Exposure Discussion' started by michael_r, May 24, 2013.

  1. michael_r

    michael_r Subscriber

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    1. Shortcut: I’m blanking here – is there a series of formulas to convert a meter reading (for example 1s@f/8, ISO 100) to H in lux-seconds (or millilux-seconds)?

    2. A long way (in-camera contacted wedge) which will include flare:

    (i) Use long lens and/or avoid wide apertures to eliminate falloff
    (ii) Mask lens to substantially eliminate lens/camera flare
    (iii) Point camera at white card target (set to infinity focus)
    (iv) Meter card at film ISO speed
    (v) Give 5 (arbitrary) stops more exposure so that wedge step with D 1.5 corresponds to metered (middle grey) exposure on film
    (vi) Develop film, measure densities: If no flare was present and development was correct, exposure from wedge D 2.5 should correspond to film net D 0.1 and exposure from wedge D 1.2 should correspond to film net D 0.9

    Obvious problem is flare. Other problems? eg: meter calibration (12%, 18% etc.)?

    3. A longer way (no wedge) with no flare (same as roll film methods):

    (i) Mask lens to substantially eliminate lens/camera flare
    (ii) Point camera at white card target (set to infinity focus)
    (iii) Meter card at film ISO speed
    (iv) Expose sheet at 3 1/3 stops less exposure than metered
    (v) Expose second sheet at 1 stop more than metered
    (vi) Develop film, measure densities: If development was correct, film net Ds should be 0.1 and 0.9 respectively

    Same meter calibration issues as in test (2)

    How bad is all this on a scale of 1-10, 1 being great, 10 being absolute failure?

    (My lack of sensitometer is why I continue to contemplate these procedures)
     
  2. Stephen Benskin

    Stephen Benskin Member

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    Michael, just a couple of small points.

    1. It's reciprocal. All metered exposure is 8 * 1/ISO. For 100 speed film, metered exposure is 0.080 lxs.

    2. There will be no flare with contacting. Density at one stop over metered exposure is dependent on curve shape and development. No point of density other than the speed point can be accurately assumed.
     
  3. michael_r

    michael_r Subscriber

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    Stephen, re your second point - apologies there shouldn't have been a reference to flare in the second scenario. I had originally also typed up a third scenario in which a wedge was photographed in various ways (which I abandoned as I don't like those methods) and I must have screwed up copying/pasting sub-bullets along the way. Too late to edit it unfortunately.

    Regarding the formula, I have to think about that some more. I didn't think that formula could be used on it's own for what I was trying to figure out. Thought it would be more involved.
     
  4. Stephen Benskin

    Stephen Benskin Member

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    Maybe you're thinking of how to determine the value of the illuminance on the step tablet and the value of the transmitted light. The equation to determine the amount of light transmitted through a unit of density is to multiply the illuminance by the reciprocal of the antilog of the step tablet density:

    H = 1/10^density * illuminance
     
    Last edited by a moderator: May 26, 2013
  5. michael_r

    michael_r Subscriber

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    Can I replace H in the above equation with H-metered from post #2? For a 100 ISO film and a step tablet density of say 0.3, we'd have:

    Illuminance at film = (0.08)(10^0.3)

    I haven't worked it all through with the proper units but I assume somehow things cancel out and you're left with lux?

    ~0.16 ?

    Thanks

    Edit: Part of what I'm trying to get at here is that while contacting is the best way to plot a curve, most people need a reasonable way to do it without a sensitometer. For example - the current thread in the B&W film forum with the 0.6 gamma etc.
     
  6. Bill Burk

    Bill Burk Subscriber

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    The units pass through the formula. Lux in results in Lux out. Likewise Lux-Seconds in results in Lux-Seconds out.

    Stephen's 0.08 is "Lux-Seconds" because it is an amount of exposure given to the film.

    You might be trying to find a suitable "Lux" as an amount of light that you will then expose the film to for a suitable time to arrive at the same exposure. (So you can take the time out of the equation, but somewhere you are going to apply time).
     
  7. Stephen Benskin

    Stephen Benskin Member

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    Actually it's (1/10^0.3)*0.08 = 0.040
     
  8. michael_r

    michael_r Subscriber

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    Wait,

    if H = (1/10^0.3) * Illuminance
    then Illuminance = H / (1/10^0.3)

    Illuminance = 0.08 / 0.5

    = 0.16

    What am I missing here?
     
  9. Bill Burk

    Bill Burk Subscriber

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    Somehow you arrived at a greater amount of light given a certain amount of light blocked by a significant density.

    Can't be right.
     
  10. Bill Burk

    Bill Burk Subscriber

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    A-ha. You figured out how much more light would be required to make the exposure the same given the density you just introduced.
     
  11. Stephen Benskin

    Stephen Benskin Member

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    The first example didn't use the reciprocal of the anitlog. You only used the anitlog.
     
  12. michael_r

    michael_r Subscriber

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    Whoa!!! :D

    I think I must have read Stephen's post (#4) incorrectly and put the brackets in the wrong place or something? Is this formula correct or not:

    H= 1/10^D * I

    H = I / 10^D

    or is it

    H = 10^D * I

    Sorry about this, guys.
     
  13. Stephen Benskin

    Stephen Benskin Member

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    H= 1/10^D * I
     
  14. michael_r

    michael_r Subscriber

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    What the heck is going on here?

    H = 0.08

    1/10^0.3 = 0.5

    0.08 = 0.5 * I

    I = 0.08 / 0.5

    I = 0.16

    This can't be. I'm losing my mind.
     
  15. Stephen Benskin

    Stephen Benskin Member

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    H = 0.08 * 0.5 = 0.04
     
  16. michael_r

    michael_r Subscriber

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    So my first substitution was wrong (H=0.08). I wasn't using the equation properly.

    Need to rethink this.

    I'm not understanding what this formula does. H is exposure in lux-s. Then on the right side of the equation we have I in lux, and then the other term which is the reciprocal anti-log of D. Somehow multiplying I (lux) by the reciprocal anti-log of D gets us to lux-s? How?
     
    Last edited by a moderator: May 27, 2013
  17. Stephen Benskin

    Stephen Benskin Member

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    Both are in lux sec. Don't over think it. Light falling on the step tablet, the density of the step tablet, and light that passed through.

    Use this for a reference.

    Sensitometric exposure for 125 speed.jpg
     
    Last edited by a moderator: May 27, 2013
  18. michael_r

    michael_r Subscriber

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    Apologies I was thinking of I as intensity, not illuminance. Amazing :sad: