The most difficult thing is making some kind of box that will give even illumination. You can flash the film from across the room but that might not always be repeatable, but it certainly is a good way to get an even exposure across the stepwedge.
These little "contact printers" usually go for little money on ebay. You could mount the flash inside one of those.
http://www.ebay.com/itm/Vintage-Air...681?pt=LH_DefaultDomain_0&hash=item2ec1b84559
Opacity of a density of 2.6 is 10-2.6 = 0.002511
Exposure necessary at film plane is 0.8/ISO so 0.8/320 = 0.0025 lux s
Exposure metered above just above the Stouffer needs to be 0.0025/0.002511 = 1.13 lux s
Looking at the earlier given formula to convert a light meter f/stop reading to lux s:
Number-of-stops-to-right-of-f/1.0 = log2 (exp/2.5)
so, log2 1.13/2.5 = -1.15
Make a reading at your box with the flat diffuser. Count the number of stop from f/1.0 for example if it displays 5.6 3 then 5.6 is 5 stops from f/1.0 plus .3 then it's 5.3
The Lux.second is (2^5.3)*2.5=98 lux.sec
Rafal used the following formula to calculate the EV value he expects to read on his flash meter: Number-of-stops-to-right-of-f/1.0 = log2 (exp/2.5)
The formula actually requires an illuminance (without time, unit [lux]) value where he puts an exposure value (with time, unit [lux sec]).
Please compare http://en.wikipedia.org/wiki/Exposure_value#Relationship_of_EV_to_lighting_conditions
I'm not sure which part of the Wikipedia article you are referring to, but assuming that you meant the equation E=2.5*2EV, I can see that it is expressed as illuminance (lux). That article, however, says: Strictly, EV is not a measure of luminance or illuminance; rather, an EV corresponds to a luminance (or illuminance) for which a camera with a given ISO speed would use the indicated EV to obtain the nominally correct exposure. As EV would just be the number-of-stops-to-right-of-f/1.0 assuming ISO 100 and exposure of 1 s (see Table 1 in that article), I assume that it actually is a measure of exposure, and the formula overlooks that so as to make use of it for comparing illuminance levels, regardless of the exposure duration. Are you concerned about that assumed measure of time?
... and it was not being used to calculate EV, which by including a metric of time, indeed, would have required only an illuminance, E, in lux alone! I think the confusion stemmed from the fact that you may have assumed, in your original post #28, that I was calculating EV, while, as the posted formula stated, it was the AV (number of stops to the left of f/1.0) that I obtained from it.
... my flash meter was set to ISO 100 (or it should have been, see my post #26) the equivalent AV that it should have displayed was -1.15, or about f/0.7. Had my meter been set to ISO 320, however, then indeed, the expected AV would be as per your calculation with ISO speed S=320, and the incident light meter constant for flat receptor C=250:AV = log2(1.13 / (250/320)) = 0.53
and the expected f-number, at meter set to ISO 320 should be:N = 20.53/2 = f/1.2
Here is another follow-up question: I do not understand why you need to be able to actually get the calculated (extremely low) AV as a reading on your flash meter. Assuming that your flash meter (as actually most of them) also is calibrated on an 18% gray, I do not even understand how you are able to take a correct measurement on step 28 of the Stouffer at all. Or am I assuming something wrong, again?
Shouldn't it rather be step 16 (which is as far as I know the one at 18% gray), which is not only the value the meter is calibrated for, but also 12 * 0.1 logD (1/3 f-stop) = 4.0 f-stops more transparent? So, if you'd measure step 16, or even better completely without the Stouffer just through the (retracted) dome of the meter (which has approximately the same density as step 16), instead of f/0.7, you could read an easy f/2.8 without any further hassle.
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