Approximate ISO-EI speed test with in-camera (LF) contacting
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)
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.
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.
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 Stephen Benskin; 05-26-2013 at 02:10 AM. Click to view previous post history.
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?
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.
Sponsored Ad. (Subscribers to APUG have the option to remove this ad.)
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).
Actually it's (1/10^0.3)*0.08 = 0.040
Originally Posted by Michael R 1974
if H = (1/10^0.3) * Illuminance
then Illuminance = H / (1/10^0.3)
Illuminance = 0.08 / 0.5
What am I missing here?
Somehow you arrived at a greater amount of light given a certain amount of light blocked by a significant density.
Can't be right.
A-ha. You figured out how much more light would be required to make the exposure the same given the density you just introduced.
Originally Posted by Bill Burk