


I'm trying to see how to take our word problem and turn it into a formula. The word problem: "Because his light meter can be configured to display exposure values in terms of f/stop for a given flash, Rafal wants to find the f/stop indication that is required for a reading taken above the step wedge, which would give a threshold exposure to his film under step 28 (density 2.6) for a 320 ISO film."
I'll admit, instead of trying to do the algebra, I just measured my sensitometer with flashmeter (which gave, without anything in the path  f/16 at ISO 100), and knowing that I usually have a 6 stop ND filter for 400 ISO film, the answer is likely to be somewhere in the vicinity of f/2, f/1.4 or f/1.0...
His formula didn't get him there?

Originally Posted by Chan Tran
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
So this formula is the issue? Was this incorrect because it labels the terms lux.sec while you believe it should be lux ?

Originally Posted by freemp
Rafal used the following formula to calculate the EV value he expects to read on his flash meter: Numberofstopstorightoff/1.0 = log _{2} (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/Exposur...ing_conditions
Freemp, I'm trying to understand the discrepancy you are referring to. As Bill has just pointed out, based on the earlier suggestion from Chan, the formula I was using gave me an exposure value, hence in lux s, and not just lux, which would express illuminance. Bear in mind that an electronic flash controls the illuminance and the duration of the output, hence the use of an exposure metric (rather than illuminance), lux s, for my calculation. What is confusing, however, that the very same light meter can be also used to measure only the illuminance, when not in flash mode.
I'm not sure which part of the Wikipedia article you are referring to, but assuming that you meant the equation E=2.5*2^{EV}, 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 numberofstopstorightoff/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?
It is quite possible that I may have made a mistake in my thinking, so please let me know if you think I got this wrong. Even though that experiment was a success, it wouldn't be a surprise to me if the logic was somehow flawed—thank you for chiming in.
Last edited by Rafal Lukawiecki; 05282013 at 01:31 PM. Click to view previous post history.

Originally Posted by Rafal Lukawiecki
I'm not sure which part of the Wikipedia article you are referring to, but assuming that you meant the equation E=2.5*2^{EV}, 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 numberofstopstorightoff/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?
I think it is not possible to just put luminous exposure [lxs] values for illuminance [lx] and vice versa just because it results in something that appears to be reasonable. I completely agree about how you get to the luminous exposure value (1.13 lxs) required to make step 28 of your step tablet appear as the speed point on your film. With the next step, you're trying to get an EV (unitless exposure value) by using a formula that actually requires an illuminance [lx] with your luminous exposure value (1.13 lxs) instead. From my understanding, the result of this calculation is not what you would like it to be. However, if your experiments were proving the results to be correct, there is not much left to argue. Except maybe, that the results, even though wrong, were by chance close to the results of the correct calculation or at least still in some reasonable range.
Anyway  Since I'm just trying to do exactly the same thing here (exposing a film through a step tablet to do the sensitometry), I was just searching for more information about how to get to a flash meter reading that tells me that the film will be properly exposed.

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 incidentlight 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:
N^{2} / t = (E ∙ S) / C
so:
N^{2} / 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 ∙ N^{2}
Therefore, plugging the aperture fstop 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 fnumber (N):
N = 2^{AV/2
}
For example, 5.6⅓ is fiveandathirdofastop to the left of f1/0, so 2^{5.33333/2} is 6.3496... or f/6.3. Plugging this equivalence between fnumber and AV into the E x t = 2.5 x N^{2} we get:
E ∙ t = 2.5 x (2^{AV/2})^{2} = 2.5 x 2^{AV}
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 fnumber 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 2^{EV}
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 Wikipediaquoted formula has a unit of illuminance, while the righthand side has 2 tothepowerof a unit of exposure. Still, AV is only the aperture, but EV includes the metric of time, as it is defined as:
EV = log_{2} (N^{2} / t)
and so, in terms of AV:
EV = log_{2} (2^{AV} / 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 ^ (log_{2} (2^{AV} / t)) = 2.5 x 2^{AV} / t, and if we multiply both sides by t, we get back to luminous exposure formula:
E ∙ t = 2.5 x 2^{AV
}
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 fnumber my flash meter should show, bearing in mind that my flash meter only shows fnumbers in flash mode, and it had no lx∙s display, which was the reason for these calculations. So:
AV = log_{2}(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.
Last edited by Rafal Lukawiecki; 05292013 at 10:01 AM. Click to view previous post history.
Reason: Clarified the last two formulas, tidied up lx symbol when referring to lux.

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Originally Posted by Rafal Lukawiecki
... 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.
That was the missing piece. Thank you for clarifying. I was mistaken by thinking you were trying to calculate an EV. Sorry for the fuss.
Anyway, I think there is little flaw in the actual calculation of the AV, since the equation log_{2} 1.13/2.5 = 1.15 assumes ISO 100. If I understood right, it should be log_{2} (1.13 * 320 / 250) = +0.53 instead because you also calculated the required exposure for ISO 320 before.

Interesting new question, Freemp. However, I still think my calculation was correct. Let me explain why.
First of all, the necessary exposure of 1.13 lx∙s above Stouffer already took into account the speed of the film being ISO 320. However, the second conversion, of that exposure in lx∙s into the equivalent fnumber as shown on my flash meter, used the ISO setting at which that flash meter was set during the measurement of the pop, regardless of the actual speed of the film. Since 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 = log_{2}(1.13 / (250/320)) = 0.53
and the expected fnumber, at meter set to ISO 320 should be:
N = 2^{0.53/2} = f/1.2
which while making good sense, brings back some confusion as to the AV numbers (and fstops) effectively expressing a metric of exposure due to the operation of a flash meter—unlike a meter working in continuous light mode, when EV includes time while AV does not.
Incidentally, thanks to you asking this followup question I now have realised that a way to measure and to dial down my flash to less than f/1.0 is to set the meter to a higher ISO setting while measuring, since it cannot display less than f/1.0—assuming that it can actually measure such a small flash output. By the way, my next batch of sensitometry will use an Eseco blue/green sensitometer. Hopefully I won't trade one set of issues for another (spectral sensitivity).

Originally Posted by Rafal Lukawiecki
... 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 = log_{2}(1.13 / (250/320)) = 0.53
and the expected fnumber, at meter set to ISO 320 should be: N = 2^{0.53/2} = f/1.2
With the flash meter at ISO 100 the reading should indeed be what you were calculating. I was assuming the meter was at ISO 320. As you were pointing out correctly, the two AVs are equivalent in consideration of the selected film speed.
Here is another followup 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 fstop) = 4.0 fstops 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.

Originally Posted by freemp
Here is another followup 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 fstop) = 4.0 fstops 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.
That's a good question, Freeamp. The idea was not to measure anything through the Stouffer at all, but to calculate the correct exposure above the Stouffer, which after subtracting the opacity of the tablet at that step would give me as much data for the curve as possible. By virtue of how the ISO speed system has been designed, if you expose black and white film to (0.8 / ISO speed) lx∙s (conveniently, I believe it is the same as mcs) you should see a density of 0.1 over film base+fog. The purpose of my original calculation of the needed exposure was to ensure that step 28 of the 31step Stouffer, which has log density of 2.6, appeared just above film base + fog, indeed 0.1 would be what I would have liked to get there—please see my post #23 for detail.
Exposing for step 16 would not be a good idea, I think, because you could not be sure that you would have enough data to construct the toe, or that you would have as much data as possible for the shoulder, within the limits of the tablet you use. It makes sense to have nothing but fb+fog for the first 2–3 steps but not more, and no less, so you know you have not overexposed the whole tablet (no toe data) and that you have not underexposed it so much that you are missing interesting shoulder data. In fact, even with a correctly exposed tablet you are likely not to have enough data for the full shape of the shoulder for some films, so it may be a good idea to overexpose a second tablet by an additional 23 stops just to get to see that portion of the curve, which is what I am planning to do the next time.
In the end, this worked for me. I am not sure if it was the best way to do it, but it worked better than my previous attempts!

Freeamp, just to add about the results I got with 320TXP, while I was targeting step 28 to get to 0.1 over fb+f, which I assumed had density 2.6 (see post #23), in reality that step had density 2.7, and so I was targeting step 27, actually. Looking back at my results on the sheet closest to ISO normal development time, i.e. developed so that a point n where the log of exposure is 1.3 units greater than the exposure at point m has a density 0.8 greater than the density at point m, I did not manage to hit the 0.1 on step 27, but I seemed to get there around step 21 (on that curve steps 3127 were fb+f, 0.01 on 2625, things really waking up on step 24). However, on the steepest curve (which got the maximum development of 16 minutes) I ended up having 5 steps of fb+f (3127). I conclude that either I had miscalculated my exposure, or that I had a different error in the process, perhaps related to the very short flash exposure duration introducing a reciprocity failure. Overall, I have "wasted" at least 4 steps, therefore I believe I could have exposed the sheets a whole 1.3 stop more.
When I did this test again, for HP5+, I was targeting step 27 again, but this time I have managed it on steps 2425, where I got the ISO speed point density (0.1 over fb+f for the "ISO normal" development). This was much closer, but I was still underexposing by about 0.6 of stop, perhaps again for the reason of the unaccountedfor short flash exposure reciprocity.
I suppose I should post my curves at some stage, but the spreadsheets with the measurements have been posted on the related thread (see post #68). I look forward to your results or other feedback.

