Freemp, your concern is my concern, too. I'd like to get on top of what is going on, so let me try to work out the equivalences from the actual light meter exposure formula—I hope our resident sensitometry experts, Stephen and Bill, also chime in to check my thinking. First of all, let's start not with EV, which was never part of my original quest, but with the actual f/stop (N, dimensionless) and exposure time (t, in
seconds), the
camera settings, and therefore closely related to exposure at the film plane (H
v) which is expressed in
lx∙s (taking bellows extension, flare, and optics transmittance factors out of the equation as not relevant to our flash sensitometer). We begin with the formula shown in the section of the
Wikipedia article which you mentioned, for
camera settings determination from incident-light measurements, where N and t are those camera settings, E is the illuminance measured by incident light meter (flash meter in our case) in
lx, S is ISO speed number (100), C is the incident meter calibration constant, which for a flat sensor is 250:
N2 / t = (E ∙ S) / C
so:
N2 / t = E (100/250) = E / 2.5
solving for exposure at the film plane, E ∙ t, which is now in
lx∙s:
E ∙ t = 2.5 ∙ N2
Therefore, plugging the aperture f-stop number into this equation gives me luminous exposure in
lx∙s. Since Chan Tran calculated exposure of 98 lux seconds for aperture of 5.6 ⅓, let's see if the above agrees. f/5.6⅓ is about f/6.3. So 2.5 x 6.3
2 is 99.25 lx∙s, pretty close. However, his calculation uses the
number of stops to the left of f/1.0, a measure also called the
Aperture Value (AV). Aperture Value (AV) is a positive number of stops to the right or left of f/1.0, negative if to the left of f/1.0, and it is 0 for f/1.0. There is a relationship between AV and the f-number (N):
N = 2AV/2
For example, 5.6⅓ is five-and-a-third-of-a-stop to the left of f1/0, so 2
5.33333/2 is 6.3496... or f/6.3. Plugging this equivalence between f-number and AV into the E x t = 2.5 x N
2 we get:
E ∙ t = 2.5 x (2AV/2)2 = 2.5 x 2AV
which is exactly the formula that Chan has suggested, except he worded AV as
Count the number of stop from f/1.0, and in his example, E x t = 2.5 x 2
5.3 = 98.49 lux seconds, the difference from the other result attributable to the various roundings of the f-number to f/6.3 and of AV to 5.3.
Interestingly, that last formula is very similar to the one you seemed to be thinking of:
E = 2.5 x 2EV
but the crucial difference is that it is expressed in
EV and not
AV. What's the difference? Time! Having said that, I am uneasy that the left side of that Wikipedia-quoted formula has a unit of illuminance, while the right-hand side has 2 to-the-power-of a unit of exposure. Still, AV is only the aperture, but EV includes the metric of time, as it is defined as:
EV = log2 (N2 / t)
and so, in terms of AV:
EV = log2 (2AV / t)
now, using the formula for illuminance in
lux, which you have been thinking of, and assuming ISO 100 and C=250
E = 2.5 x 2 ^ (log2 (2AV / t)) = 2.5 x 2AV / t, and if we multiply both sides by t, we get back to luminous exposure formula:
E ∙ t = 2.5 x 2AV
So, whichever way we arrive at it, when you use EV we are speaking of exposure, and not illuminance alone, but in my calculations, which followed Chan's formula, I was using AV displayed by the flash light meter, which, confusingly but in accordance with how flash meters seem to operate, was taking into account the duration of the flash exposure while measuring the pop. In other words, given a known exposure E ∙ t, in
lx∙s, we can use the above formula, solved for AV, to tell me what relative f-number my flash meter should show, bearing in mind that my flash meter only shows f-numbers in flash mode, and it had no
lx∙s display, which was the reason for these calculations. So:
AV = log2(E ∙ t / 2.5)
which is the very formula, that I have used and to which you have referred to in your first post. I hope it all now makes it clearer that that formula took luminous exposure, E ∙ t, in
lx∙s, to arrive at AV, 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. Still, I have a suspicion that the above formulas are not really proper equations, but they merely describe equivalences observed in the application of a light meter to obtaining a certain density in the photographic process, as otherwise we would have to have an equivalence of the units of aperture/time (EV, a number that includes seconds) with lux seconds (H
v), which is clearly missing something out—most likely the property of the photographic material as related to the calibration and operation of a light meter.
Freemp, thank you for asking for this clarification, as I now have a better understanding of the relationship of metering to EV and illuminance, which I would not have had if you had not asked. If anything is not clear, please ask again—many thanks.
