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Just saying the logarithm base 2 of 16 is not EV 4 by itself, (except when K=3.3333) it's a minor correction.
Ah Wikipedia where only the truth and accuracy can be found, a bit like the massive development chart
For example people keep writing Ev when it should be Ev. Why? becasue if you use Ev instead of Ev it implies from standard mathematical notation that v is a base number of E (function derived value of E) and it isn't in the formula I used and nor does my meta data explain itself as such. Take care with what you write or some pedant will try and suggest it's wrong.
I'm using the formula which minolta sent to me when I asked them a question about how their meters worked (except they use K =14) and not something I found in the dead sea scrolls.
I'm not going to complicate anything by introducing Bv. It's unecessary in the scope of this topic since the minolta supplied formula I'm using is not using it.
I am not sure I get everything in Connolly and in the text by Stephen but I try to distill this.
Once upon a time, the exposure equation could be expressed as Hg = 10 / ISO.
Hg is the esposure needed to obtain a target grey, and is the exposure suggested by the light meter.
Progress in lens coating led to higher light transmission by the lens. That prompted a recalculation of K which ultimately led to a new formulation of the exposure equation as Hg = 8 / ISO.
This new formulation only differs by 1/3 of a stop from the older. In order not to create havoc and destruction in the photographic industry and community, the old equation was kept as far as negative film is concerned. With negatives, 1/3 of a stop is insignificant, it's all recovered during the printing process, and in any case one can err in the way of overexposure with ease.
With slide film, 1/3 of EV is a significant exposure difference, and it matters. Besides, there's no printing and no way to adjust results. So it was decided, in order to correctly describe the behaviour of the slide film with the new better lenses, to adopt the new equation.
This new value of lens reflectance is possibly the cause that made the K constant to go from 10.64 to 12.50.
Now we consider what happens with the old (still good for negatives) equation and the new (good for slides) equation.
Old equation, Hg = 10 / ISO:
100 ISO -> LogHg -1,0;
200 ISO -> LogHg -1,3;
400 ISO -> LogHg -1,6;
New equation, Hg = 8 / ISO:
100 ISO -> LogHg -1,097; which we round at -1,1;
200 ISO -> LogHg -1,4;
400 ISO -> LogHg -1,7;
The density curve for the new equation is shifted 0.1 exposure (1/3 EV) for the same density.
At each ISO speed for each equation, the Hg density is obviously the same, only exposure changes.
So what is this density, which I think we can define as Dg?
We have to look at the characteristic curve of the film.
Let's take the Provia 100F film curve.
We observe, by sight, that Hg = 1,1 corresponds to around Dg = 1.15.
I underline this is the density for "target grey" at all ISO speeds when following the light meter indication, with the new K = 12.5 constant after the revision.
That corresponds to an opacity of 6.07 which, in turn, corresponds to a tranmission of 0.1647 or around 16,5%, which corresponds very nicely to a lightmeter calibrated with K = 12.5 (Sekonic).
A lightmeter (Minolta-Kenko, or Pentax) calibrated for K = 14 would suggest a slightly inferior exposure, and create a slightly higher density, corrisponding, I do believe, to a transmission of 18% on the slide.
So we have historically three different "target grey":
Old ASA formulation, with K = 10.64, corresponding to the exposure equation Hg = 10 / ISO: a shade of grey darker than 16.5 %; This equation is still used for determination of ISO speed as far as negative film is concerned, but light meters don't use K = 10.64 any more.
New ASA/ISO formulation, with K = 12.5 corresponding to the exposure equation Hg = 8 / ISO: a shade of grey corresponding to 16.5% reflectivity. That's what you get if you shoot slides with a Kenko lightmeter. That's also what you get if you shoot negatives with a Kenko, or a K = 12.5, light meter, because light meters have all adopted for the new higher lens transmission.
A different K = 14, adopted by Minolta (Kenko) and Pentax. That should give you a shade of grey of 18% which exactly matches the middle grey Kodak card and the middle of human vision between black and white.
Comments welcome!
I guess thats why Minolta say it in the formula they use. So what you are saying is you got it wrong when trying to validate it.
And for cd/ft2 you can forget K becasue its = 1. You can't with cd/m2If you are being sincere, I'll continue... You will need to add to your Minolta formula, and accept they are using EV.
Now if you can point out a definition of EV that satisfies you, I can work from that.
This is based on the definitions of the different terms in EV that I found related to the APEX system.
I'm just saying without K, you can't say 16 cd/m2 is 4 on its own. It makes some assumption of K, which if you would like, puts a 1 in the denominator in where K would be used in the definition of luminance that is used in EV. Now the formula I found has B / KN where N is 0.3 and 3.3333 for K times 0.3 is effectively putting 1 in the denominator.
what happens if the develpment time is a tad out or another manufacturers developer is used or maybe the developer is a tad less active than it was. That is going to mean your target won't be hit. And then yuor lenses most likely only have 1/3 stop detents so you can't set exposure accurately even if your meter works in 1/10ths of a stop and most camera meters aren't that accurate anyway.
In short all this level of theoretical mathematics can not be matched in the field so its a waste of time unless you are designing a light meter. Your best option is always to do practical tests to iron out discrepancies in your meters accuracy . your technique and development and to visually judge if the neg colour is what you want it to be.
It's about exposure and not density.
OK I begin asking questions, one by one, about Connelly and about some sensitometry in general. That might be boring, or interesting for those reading the thread, and certainly interesting for me.
I have a problem with Hg and HR in Connolly.
Your explanation seems clearer.
If I get you right, HR is the speed point determined as a quadratic mean of the two exposure point F and S.
Hg is where the light meter places the middle grey. This is determined "empirically" and is outside of all our equations, so far. We take it as an exogenous datum.
Given the derivation of HR which is the "speed point" for slide film, the speed equation for reversal is now S = 10 / HR.
But HR is not the exposure that gives middle grey.
Following the indication of the light meter, I expose for Hg and obtain a certain density Dg.
Hg / HR = 0.8. Or, more simply, Hg is 1/3 EV less exposure than HR.
If all of this is right, I just throw Connolly in the dustbin and proceed from this firm ground.
My fundamental questions are:
1) Which is the target grey that the light meter creates on my slide?
2) Which is the corresponding density of this grey, Dg?
The answer to question 2) might be:
first I find HR using speed equation;
second I derive Hg;
third I plot Dg on the film curve.
For a ISO 100 reversal film, S = 10 / HR so 100 = 10 / HR, so HR = 0.1 so LogH(0.1; 10) is -1.
Now I subtract 1/3 EV from this exposure in order to obtain Hg, which is -1,1.
I stop here and ask for confirmation of what's wrong in what written above.
I had no feedback to my text #238 so I obstinately go on with the reasoning:
If we now observe the charactheristic curve of Provia 100 ISO, the exposure of -1.1 corresponds to a density of around 1,15 by eyesight.
That corresponds to an opacity of 6.07 which, in turn, corresponds to a tranmission of 0.1647 or around 16,5%.
The density at middle grey at all ISO is - given the same exposure equation - always the same. For every level of light, for every film speed, middle grey must be the same middle grey so I expect the same density with small variations.
If we trust this source: http://dpanswers.com/content/tech_kfactor.php this 16.5% grey corresponds very closely to a lightmeter calibrated with K = 12.5 (15,7%, Sekonic should use this K).
So far, I am concluding that it all sticks very well, the circle is closed as I expected:
A light meter with K = 12.5 is calibrated to correctly render a 16% grey. When using the light meter, we obtain an exposure value, Hg, which is 1/3 of EV closer than the speed point of a slide film which is HR and, given ISO = 10 / HR, corresponds to an exposure of -1.0. So my lightmeter gives me an exposure of -1.1 which, in turn, gives me back my 16% middle grey on the projection screen.
Don't tell me there's something wrong because I would be very
And I even think that by using a light meter calibrated with K = 14, which would correspond to a 17.6% grey, I would actually get that grey on my slide.
Yes there are little variations due to small imperfections of the process, but the result will be "satisfactorily" where expected if all elements of the process are well executed.
Here's something else to think about. P/q = K. So 8.11/.65 = 12.5. What if the lens had a lower transmittance and q equals 0.58 instead of 0.65? 8.11/.58 = 14. This is basically K. So a K of 14 will produce a camera setting that will let more light into the camera, but if the lens has a q of 0.58, the same amount of light will hit the film plane as a lens with a q or 0.65 and a K of 12.5. The idea is to create the same exposure at the film plane.
Thank you for your reply. The reason I ask is that I actually use non-TTL averaging reflected light metering (6-degree 'view') and have used that system since the 1970s when I got my first 6-degree meter. I wanted to read the text to see how my method of reflected meter exposure index calibration fits in with the others.Sorry, my bad. It's from the appendix in Dunn's Exposure Manual. It appears to be based from a graph in the paper Safety Factors.
...Old ASA formulation, with K = 10.64
P is the important constant in this. It's the same as K1 (not to be confused with K) from the exposure meter standard. This is where Hg will fall for a given film speed. 8/100 = 0.080 for 100 speed film. HR in Connelly and currently Hm in the current standard is determined. This as well as HF and HS can be known points of density. The speed equation will produce a film speed value that will want to put the exposure Hg 1/3 stop to the left of HR.
This paper only deals with reflected exposure meters, and there are two types, integrated and spot. For spot, the Hg will be from anything you point at. For an integrated meter for a statistically average scene and basically the calibration Luminance, it's 12%. This comes from the calibration Luminance value as compared to the calibration Illuminance value. I don't use incident meter and haven't spent nearly as much time on them. I've only read The Incident Light Method of Exposure Determination maybe two times and that was over 10 years ago (although I'm going through it now).
One question you need to ask is how are the values perceived on the transparency. 18% and 12% are only half a stop difference, then there's how the transparency is viewed. You might find this interesting and I think the chart will answer some of your questions. The attached is from Jack Holm's paper Exposure Speed Relations and Tone Reproduction.
Now there's some fun. Because with K = 10.64 using cd/m2 then K in Imperial units would have been really close to 1
Do you refer to equation [15], page 190, Connolly? As it is, it poses HR = Hg.
I don't think you do, because you tell me Hg is 1/3 EV to the left of HR.
Connolly proceeds by saying: "It is to be noted that film manufacturers are re-assessing...and may require a change...". Do you mean that they have actually reassessed and the change was granted?
That would lead us to the difference of 1/3 EV between Hg and HR.
I suppose you are referrring to this.
Is it correct to state that, in this case, we have a K = 11.6? (It is equation (16) immediately below, the scan is a bit confused).
As far as I understand, spot and average reflected light meters work exactly the same. Spot, pointed at anything, will give me a value Hg which will give me a certain shade of grey whichever the brightness of the object. There is only one shade of grey for which Hg is the correct exposure, factory determined "target grey". If the object I point the spot meter at is whiter than target grey, the slide will be underexposed (and make the object appear target grey). If the object I point the spot meter at is darker, the exposure will be overexposed (and appear target grey).
An integrated light meter has the advantage of being able to calculate the light transmission exactly, but the "average" scene that it sees (the determined "target grey", that is, supposing target grey is an average of the world reflectance) is the result of the same kind of arbitrary choice that the exposure maker makes with a spot light meter. Basically the light meter wants to "guess" the reflectance of the target, and the guess that minimizes exposure mistakes is the average.
There are two averages here: the average reflectance of the world, and the average reflectance in the scene. If they coincide, the exposure is correct. If the scene is let's say high-key, no cigar.
I quote Ralph Lambrecht in another thread: "it is my understanding that the 18% reflection value is based on a Kodak field study, conducted by Condit in the 1940s, where he took readings of over 100 scenes around Rochester, determining an 18% reflection average and a 7 1/3 avg SBR;". I have read something similar as well. I would be very surprised if human vision placed "middle" tone in a place that doesn't correspond with "middle" tone in the general world. And I would be generally surprised if "target grey" for separate spot light meters was different from "statistically average scene" for integrated light meters. That would cause a lot of head scratching and, frankly, would make no sense from a practical point of view.
In my view either all reflected light meters of the same maker are calibrated around 12%, or they are all calibrated around 18%, or whatever same shade of grey (be it "on a spot" or "average in the scene").
That's very interesting, thanks very much.
That also leaves me perplexed.
I understand that there must be some "compression" of tones when printing a slide on paper. That will be dealt by the printer but I do understand one might "optimize" exposure in advance for the printing process, just like one might optimize it, let's say, for the scanning process.
What raises my surprise is the difference there is between the theoretical exposure for middle grey and the suggested exposure, when optimization is for the projection on a screen.
The more I study this, the more I get confused, really.
Half a stop is the difference between a properly exposed white marble façade and a burned white marble façade. To me, whether the meter is calibrated to 12% or 18% makes a world of difference. And I have always considered it calibrated to 18% obtaining, with my Minolta spot meter, a very precise and expected placement of high lights (and when I went one day with my Gossen Sixtino, a very defective and useless piece of equipment, to take pictures I basically spoiled several shots of Villa Pamphili because the walls came out washed up, lacking in detail in a very unnice way).
EDIT: considering that burning the highlights is what spoils a slide, it might be that the table suggests, basically, an ample "margin of safety", a marked underexposure of slides, counting on the adjustment of human vision after the first slide. But they must all be underexposed in the same way, or it wouldn't work.
And besides, a nocturne picture with this underexposure would become sad to look at, considering the amount of scene that would disappear into Dmax like into a black hole.
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