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Velvia 50 - 14 stops


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<p>Edward, the step wedge test on dpreview you mention gives the 40D over 9 stops.</p>

<p>In your own experience, how much wider is Ektar than the 40D? ( Please answer directly without twisting words if possible "using a number of stops" in the answer; or answer "I don't know").</p>

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<p><em>Yes you based your point (density 0.7) using calculations based on YOUR interpretation of contrast building. The figure 0.7 BTW is very close to mid grey certainly higher than open shadow on a sunny day. Why start so high up the curve? do you think no detail is recorded between the inertia point and 0.7?<br /></em></p>

<p>In fact, I did not draw the sketch in this thread to scale. It was intended strictly to illustrate the problem with Mauro's interpretation of his test. I presume you refer to the green curve (to which my witness line extends). It really doesn't matter which color you choose because there is no difference other than the value for DMin - the curves have the same shape in the toe area. The density value of 0.7 is a coincidence, not by design.</p>

<p>ISO sensitivity measurement starts at 0.10 density units above DMin*. Since the DMin value for the green line is approximately 0.60 density units, it is perfectly reasonable to measure from 0.70 density units for both sensitivity and dynamic range. The slope at that point is approximately 20%, which is also a reasonable figure for the minimum usable contrast. So it appears we agree in fact if not in principle. We also seem to agree that DMin is not a reasonable starting place.</p>

<p>* ISO sensitivity is calculated from the amount of light it takes to increase the density from 0.10 units above DMin to 0.9 units above DMin.</p><div>00TQoj-136827584.jpg.372ec7355273fb365143b01f7976b84c.jpg</div>

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<p>Is anyone considering the orange mask for Ektar? Dmin gets affected by that; unless Kodak did some test with a special emulsion sans the orange mask.</p>

<p>Given the presence of the orange mask, a corner obtained by making the linear region intersect with a line of zero slope and height of Dmin is acceptable as the starting point for the DR calculation. I can't see any argument with that.</p>

<p>As for finding out Dmax; Kodak's curve is quite useless. I mean, it is so blatantly useless as to be a waste of time for them to even publish. Who cares about a half of a characteristic curve? This could be because their wedge ran out steps or whatever, but it is sloppy methodology nonetheless.</p>

<p>Unfortunately, without an accurate Dmax corner, an accurate dynamic range derivation isn't possible; but we have to assume that there isn't much curve left after the max on the curves primarily because if there were appreciable length to the curve beyond what's shown, Kodak would want to shout it out from the proverbial rooftops and would presumably devise a test to measure it.</p>

<p>So let's say we'll add a stop of DR (that doubles the range, by the way) to what can be derived from the curve. Edward's lower corner derivation is quite generous; a corner derived by the actual zero slope line intersecting the linear line gives you barely -2.1 or so as the lower corner. The curve stops at about 1.15 or thereabouts. So you get 3.25/log2 which is about 10.8 stops. Add a stop, you get 11.8 stops, say 12 stops at most.</p>

<p>Mauro's test doesn't show 12 stops, it shows somewhere between 10 and 11. Besides, Mauro and I have a difference in what we call a range: for say 0 - 10, my definition of range is 10, Mauro's is 11. Actually, 10 is correct by all conventions, but what is right is immaterial. When Mauro says Ektar has 12-14 stops of DR, it is 11-13 by my definition. A generous derivation from the curve puts it at 12, bang in the middle of Mauro's 11-13.</p>

<p>Where's the argument?</p>

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<p>The generally accepted definition of DMin is the density of the base plus fog. The relatively high DMin for each color is consistent with the presence of the orange mask when the curve was drawn, as is the relative position of the three curves to each other.</p>
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<p>That is exactly my point. The orange mask increases the absolute value of Dmin, so a measurement of 0.7 or even 0.9 is not incongruent with Dmin for a channel. Zero slope (or nearly zero) is therefore good enough to find a corner, without getting into Dmin values. IOW, Edward, my point of view is that your methodology is quite OK for the lower bound and would be for the upper bound as well, were it not for Kodak's publishing a useless chart that doesn't show the shoulder.</p>
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<p><em><strong>* ISO sensitivity is calculated from the amount of light it takes to increase the density from 0.10 units above DMin to 0.9 units above DMin.</strong></em><br /> No its not, that is incorrect even for B&W part there but inaccurate. Colour neg uses a different calculation. Colour neg has a mask, the calculation for colour neg factors out the masking and averages the 3 plots so your modus is incorrect. Why not just use the ISO standard?<br /> People have already done the maths for this, there are tests just because you are unaware of them doesn't mean you make your own inaccurate ones valid.<br>

A<em><strong> generous derivation from the curve puts it at 12, bang in the middle of Mauro's 11-13 Where's the argument?</strong></em><br>

The argument Edward is in you berating him for his tests based on inaccurate methodology, you then go on to put forward YOUR inaccurate method -both of you are wrong.<br>

That is all I'm pointing out. Dynamic Range has no test per se but recordable density range does, I think in practice they are similar- why not use those tests.<br>

The ISO have a perfectly adequate set of workable standards- no need for your methods</p>

 

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<p><strong><em>As for finding out Dmax; Kodak's curve is quite useless. I mean, it is so blatantly useless as to be a waste of time for them to even publish. Who cares about a half of a characteristic curve? This could be because their wedge ran out steps or whatever, but it is sloppy methodology nonetheless.</em></strong><br>

<strong><em></em></strong><br /> Yep, like most curves they are generated using a 21 (11 stops) test wedge. To be fair to Kodak they are only providing the info needed as defined by standards which use a pre defined exposure range which is deemed to be an 'average scene' somewhere about 7:1 from memory. In some text books they plot B&W curves a lot further, you may be interested to know that density normally builds but then rolls off, the strange thing is it then builds again. There is a name for this effect which escapes me for the moment.<br /> Its an academic point though because at those levels the highlight is slightly de-saturated an effect Kodak call 'tonal crushing' in their literature.<br>

<br /> As a photographer I could use this effect, by placing my subject in shadow,specular highlights in the background drop slightly in tonal value and also colour value making them less distracting.<br /> The best film for this was Agfa Portrait which could take an immense amount of overexposure (but no under) and did this quite beautifully specular highlights 'crushed' to a point where they are only just above diffuse ones meant no burning in during printing.<br /> <br /><br /> <strong><em><br /></em></strong></p>

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<p><em>Edward, my point of view is that your methodology is quite OK for the lower bound and would be for the upper bound as well, were it not for Kodak's publishing a useless chart that doesn't show the shoulder.</em></p>

<p>I agree that data for the upper bound is needed. This should be easy to obtain using a Stouffer T4110 step wedge. I will probably have a go at it once things calm down with end-of-school-year activities (which helps pay the bills). The analysis is the hard part. I think Excel will do Legendre polynomials, otherwise I will have find something that does, or resurrect programs I wrote over 15 years ago.</p>

<p>Mark - I have attached a chart from the Kodak publication on sensitivity showing the determination of B&W film speed from an H-D chart (characteristic curve). Point A is 0.10 density units above DMin. A point on the curve 0.80 density units above this is located, and the differential exposure required is read from the horizontal axis. (Hint - 0.1 plus 0.8 equals 0.9 above DMin).</p>

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<p>Almost there...<br>

What you are missing is the further calculation what you have calculated is point A now we take 1.3 log units to the right (the brightness ratio set by the average scene) the part here is labelled B we then need to do a further calculation to determine the speed.<br>

BUT<br>

this is ONLY relevant for MONO films Ektar is a COLOUR film none of the above works its not relevant, there is another standard for reversal films that is again different mixing up the three is plain daft.</p>

 

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<p><em>What you are missing is the further calculation what you have calculated is point A now we take 1.3 log units to the right (the brightness ratio set by the average scene) the part here is labelled B we then need to do a further calculation to determine the speed.<br /></em></p>

<p>What YOU are missing is that is the incremental 1.3 log(exposure) is the dependent variable. The independent variable is the 0.8 increment in density, which establishes point B according to the function expressed by the curve. This chart demonstrates how the process works, the calculations are trivial.</p>

<p>If you have something positive to contribute to this discussion, do so. If all you can do is parse words and misquote posts, give it a rest. Did I miss something?</p>

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<p><em><strong>What YOU are missing is that is the incremental 1.3 log(exposure) is the dependent variable.</strong></em><br /> Jeeze Edward I'm missing the average brightness 1.3 log exposure is the dependant variable- REALLY ? how many times do I have to explain this to you?<br /> <strong><em>This chart demonstrates how the process works, the calculations are trivial</em></strong><br /> The chart shows you how to find the speed of a MONOCHROME film, and is of no use WRT to this discussion<br /> It is only of use to those testing their own film/development curves- if you use the manufacturers curves to find film speed why bother?-the speed is on the box<br /> What though this discussion YOU have tried to do is prove that your modus for finding DR is of more value than practical tests.<br /> There is no test for DR, certainly yours doesn't fit the bill.<br /> What we need to find is the range of densities that film is capable of recording information a MUCH better method would be to use the films overall latitude.<br /> <strong><em>Did I miss something</em></strong>?<br /> do you really need to ask that?<br /> What is your agenda here? Why push your erroneous contrast driven method? why?<br /> I can only think that you have something missing...<br /> I'm done with you</p>
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<p>Mark, you say: "There is no test for DR, certainly yours doesn't fit the bill."</p>

<p>Allow me to point out that if you can define dynamic range, then merely by definition there has to be a test to measure it and quantify it. I postulate that contrary to your assertion, both Mauro's test and Edward's curve based derivation fit the "bill". And there is nothing to limit the number of different tests to measure DR to just these two methods. Many more tests could fit the bill, all one needs to do is be aware of the accuracy and precision of the test.</p>

<p>The only thing I see is that there was a difference in the interpretation of both Mauro's and Edward's results; so what? Within the margin of error of either (esp. with having to guess at Dmax) they indicate the same rough figure for the DR of Ektar.</p>

<p>Where did the question of measuring the ISO speed of the film, which has little to nothing to do with dynamic range, btw, come from?</p>

<p>On an unrelated note, why on earth do you need Legendre polynomials to find out a simple dynamic range, Edward?</p>

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<p><em>On an unrelated note, why on earth do you need Legendre polynomials to find out a simple dynamic range, Edward?</em></p>

<p>(1) Legendre polynomials are used to interpolate the results and plot them as a smooth curve rather than a polygon. (2) The first derivative of the polynomial expresses the slope as a function of exposure. It is simple and graphic to determine the dynamic range if you plot the slope (as is common practice in radiology), and (3) Because I can ;-)</p>

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