Kodak grey card usage

Rob, how would you expose the cow and oak scene, and the geese and oak scene? And why?
Maybe we arrive at the same exposure value through two different "logics".

I think I want to buy some book, or this book, by Dunn. I cannot find the reference any more.

Was it "Exposure manual" by Dunn & Wakefield, Fountain, 1974?
Is there anything else interesting written by Dunn?

I'd like to see if the riding lady sells it in Italy.

I would calibrate my exposure through a simple test I outlined to you way back in the topic. I would know in advance that if I metered a white how much to adjust exposure by to place it on white. And if I needed to put a white card in the subject I would, but its unlikely.
But it's you and benskin who are trying to do it by formulas so now you've been through the learning curve why don't you tell us how you would do it. Its you who has been trying to understand the formulas and asking all the questions and not me.

Diapositivo said:

I think I want to buy some book, or this book, by Dunn. I cannot find the reference any more.

Was it "Exposure manual" by Dunn & Wakefield, Fountain, 1974?
Is there anything else interesting written by Dunn?

I'd like to see if the riding lady sells it in Italy.

Click to expand...

I think there were 4 versions so the year may vary. The later the better.

Diapositivo said:

I mumbled. It really sticks very well.
I apply Conrad abovementioned.

For negative colour film, the speed equation is:

ISO = square root of 2 / Hm. We derive Hm = square root of 2 / ISO.

For an ISO 100 film, Hm = 0.014142.

Now we have the speed point and we must calculate the grey point, Hg.

Hg can be derived from Hm by applying a derivation factor which is [(b * K) / square root of 2].

Conrad uses b = 0.728 and K = 12.5. I will also use K = 14 for comparison.

With K = 12.5 the derivation factor is 6.505382. So Hg is 0.091999 and LogHg is -1.03622.
With K = 14 the derivation factor is 7.206832. So Hg is 0.101919 and LogHg is -0.99174.

The two values are certainly very close apart (we already know they are 1/6 EV) and, as far as the K = 14 is concerned, the value of Hg coincides very nicely with the Hg already derived for slide film, which was -0.99174, absolutely identical!

So the Minolta lightmeter, using K = 14, gives us a result that, for the corresponding shade of grey (18%, actually 17.6%), is the proper one and exactly the same whether we use slide film or negatives.

I wait a confirmation from Stephan that, finally, at last, I applied the speed equations and the derivation of Hg well. (Connolly was confusing, Conrad is better).

The last point remains, in my opinion, to be discussed, and that is how a reflected light meter is bound to be calibrated to a certain shade of grey, chosen by the manufacturer, through the arbitrary factor K.
(To those who say there is no reflected light in the exposure equation, I would shortly answer: there is, and is K, which is chosen after some empirical tests, in order to determine which K matches in a visually satisfactorily way... that certain shade of grey i.e. that certain reflectivity).

Click to expand...

Fabrizio,

For me, the best way to understand how the process is supposed to work is to use the statistical average values. I like to call this the standard model. There are variations which are just as valid, but they should be left until later. It’s also important to understand the theory behind the numbers otherwise there’s nothing to keep you from using unrealistic or misleading values.

Conrad’s value for b (which is q in most equations) is 0.728. The value most associated with q is 0.65. The difference is that Conrad a subject on the lens axis. The standard assumes 12 degrees off axis. The center of the lens is the brightest point and the light falls off as you move toward the outer part of the lens. The use of 12 degrees is for an average value for the lens.

The same standard model principle should be applied to the meter. The calibration values is what is determined as the goal. Too many people mix and match values of K and C until they match a predetermined outcome. My Pentax digital spot meter isn’t going to suddenly change what it reads just because a Sekonic incident meter is in the vicinity.

I’ve explained this point a few times, but a different value of K does not mean there is a different level of exposure at the film plane. A K= 14 also doesn’t necessarily mean it will produce an Ev value different than at K = 12.5. There are two primary reasons for a value of K to be different than 12.5: consideration for zoom lenses with a lower transmission value and spectral sensitivity of the exposure meter’s photo-cell. Remember how Allen Stimson explained how the value of K in the 1961 standard changed because the color temperature of the calibration source changed from 2700k to 4700k. A photo-cell that is under or over sensitive to the calibration color temperature will be off-set by a different value of K.

For a K = 14 that is derived from a compensation for a perceived lower lens transmission, the exposure will suggest a setting that will produce more light entering the lens, but not striking the film. The idea of what Hg should be for a given film speed is fixed. The exposure meter's purpose is to determine a camera setting that will produce that value.

I believe you wrote somewhere that K is arbitrary. You might be thinking of K1 in standard’s equation for K. Connelly uses P for the exposure constant, which is the same as K1. P is the arbitrary chosen value, but in this use, arbitrary just means that it isn’t a physical constant. It was determined through psychophysical testing to determine the most acceptable value.

Stephen

markbarendt said:

I think there were 4 versions so the year may vary. The later the better.

Click to expand...

I agree with Mark. Get the 4th edition.

RobC said:

I would calibrate my exposure through a simple test I outlined to you way back in the topic. I would know in advance that if I metered a white how much to adjust exposure by to place it on white. And if I needed to put a white card in the subject I would, but its unlikely.
But it's you and benskin who are trying to do it by formulas so now you've been through the learning curve why don't you tell us how you would do it. Its you who has been trying to understand the formulas and asking all the questions and not me.

Click to expand...

I know you don't need my hep Rob, but please indulge me in an attempt at translation: EV 13, because that's what experience and prior testing indicates will render the scene as I wish.

That's a fair answer, perhaps a little more exposure than I might guesstimate but since I have never calibrated exposure for provia 100 It would be extremely foolish of me or anyone else to claim it should be any specific exposure based purely on theory. And besides, how much texture you want in the white cow would mean it shouldn't really be white so I expect you would be pretty close.
No theory required except knowing how far from metered value pure white is. And if you want to take that further and do full zone system test for last full textural details in shadows and highlights it could all be done on one test roll of film very easily and that would improve your accuracy when using those adjustments rather than pure white. Then there is no doubt when you are metering.

Since he wants to do it all by theory he'll probably try and use his Minolta shadow and highlight buttons but since he doesn't know the real range of the film he'll not have a clue if they'll be very accurate until he does a test or finds they are not. They can't possibly be accurate for all film types and most likely not spot on for any film type according to taste.

RobC said:

if it makes it any easier and less agricultural for you,

0.10 lxs = EV -4.6438

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0.10 lux = EV -4.64 ISO 100 is the illumination of a scene read by an incident meter with a C=250.
0.10 lux.second is the exposure at the film plane for ISO 100 if the camera setting is as per a reflected light meter with a K=14
Which one did you mean?

an incident meter with C=250 is the most accurate light metering method if you want accurate colour out of the box. You don't need to worry about grey card angles or reflection percentages.

And reflection meters have light incident on them from the luminance of the subject its pointing at.

@Benskin
Stephen, I think I get what you mean and what I mean and how different they are, but yesterday I was away all day (hiking, what a good day is a day spent in the mountains) and today I don't think I have time for a well-explained answer. I promise an answer to #381 shortly.

I would expose the two situations, cow and geese, this way (pre-prepared answer, I have no time now to revise it, please debug!).
RobC, I expected a real answer: EV = answer. Because when you make an exposure, you have to set an aperture and a shutter speed on your camera. You can't feed it theory (whether good or bad or confusing or clear theory).

First, let's resume the scene in front of us:

Case 1:
grass in the sun: EV 14.6;
black horse in the sun: EV 12.3;
white cow in the sun: EV 16.6;
foliage on top of the tree: EV 14.6;
grass in the shade of the tree: EV 11.3;
trunk of the tree in its darkest point: EV 9.3;
the white dog in the shade is EV 13.5.

Case 2:
What if, instead of a cow at EV 16.6, you have some white geese at EV 17.0?

The question is: how do you expose this with slide film.

My pre-prepared answer:
Short answer: cow case: EV 14.6. Geese case: EV 14.9.

Long answer:

First, the cow case.
Your subject brightness range is 16.6 - 9.3 = 7.3 EV. That's beyond the linear part of the curve of a slide. Some parts of the scene will use the shoulder and/or the toe of the film.
Your main problem is not to blow-out the cow.
You evaluate that a cow is a "normal white" subject, not the kind of white that white china, a white shirt, a white sheet of paper would be (90% or so) but white nonetheless. You would place the latter at 2.5 EV above middle grey, but you want to place this cow-white at 2 EV above middle grey.
That means the lower end of your SBR, the tree trunk, will fall 5.3 below grey. That's way into the shoulder. It's OK because that's the way you expose a slide. Some of the tree trunk might gradually shift into black. You can't have everything with slide film. But the rise of density will be gradual and well-managed thanks to the shoulder. The cow at 2 EV above middle grey will entirely reside in the linear part of the film response.

Your light meter gives you EV 16.6 for the cow and that exposure would place the cow at 18% grey. You want it to be white at 2 EV above that. You calculate your exposure: EVcow = 14.6.

That will make the grass fall exactly at middle grey (because the light meter read it at EV 14.6, which is our chosen exposure, so we are actually exposing for having a subject with exposure EV 14.6 to appear the same tone as an 18% grey cat, and that's our grass), the black horse in the sun to fall 2.3 EV below middle grey, and, as said, the oak trunk will fall at 5.3 EV below middle grey.
Your black point will be some even darker spot in the foliage of the tree. That presumably will fall into complete black Dmin. If you have some white cloud, that will go higher than the cow in the curve, and set a higher white point. Else, the brightest part of the cow will be your white point.
The scene will be natural: the cow will appear white and will be within the linear response of the film, with full texture, the grass middle grey, the shaded parts will gradually fall into black, there will be "snap".

Now, let's make the case of the geese.
They are whiter than our cow. We consider them to be very near 90% reflectance, very near a white sheet of paper. That would be 2.5 above middle grey. The geese are 2.4 EV above grey, and they are measure 0.4 EV above the cow. But that's a region of your film where you are probably in the toe (depending on your material). The texture will be in danger. The geese are near you, you want to see the plumage. There are very subtle tones of "darker extreme white" that you want to preserve if you want to maintain the texture in the geese plumage.
So you don't want to place the geese at 2.4 above mid-grey where they would "belong" for their reflectance. You want to place them let's say at "only" 2.1 EV above middle grey, so that you protect texture. It's the same reasoning about fotographing flour or sugar that I was making in another thread. They are 0.4 above the previous case with the cow in reflectivity (because they measure EV 17.0 and the cow EV 16.6], but we place them only 0.1 above the cow in our final image.

[Passing note: An incident light meter would give you an exposure EV 14.6, that would place the mid-tone grass at mid-tone, and the geese 2.4 EV above middle grey, i.e. when you follow the incident metering they would inevitably fall "where they belong for their reflectance in respect to 18% middle grey reflectance", which in this case is in a zone of your film where texture could be jeopardized.]

So we place the geese at 2.1 above middle grey. Middle grey of the geese would be EV 17.0 (which is what our light meter tells us) and we set an exposure of EVgeese 14.9, and that will set the geese where we want them on the curve, 2.1 above middle grey. [The geese will fall at LogHgeese = -0.3, because middle grey always falls at Logg = -1.0 with ISO 100].

That gives an exposure which is 0.3 EV darker than the previous cow case. That doesn't surprise us. If we had followed the same logic of the cow, and had placed the geese at +2 EV above middle grey, our exposure would be 0.4 EV darker than the cow case, because the geese are 0.4 EV brighter.
But we place the goose 0.1 higher on the curve than the cow, so we close 0.3 EV in comparison to the cow case.

The grass will now fall 0.3 EV below middle grey, but that's still an acceptable grass.
The black horse will fall 2.6 EV below middle grey, still a very acceptable black horse, with texture.
The tree trunk will fall 5.6 below middle grey (its middle grey would have been at EV 9.3). So we see that we are stressing a bit the capability of our film, here. Will the trunk at 5.6 EV below middle grey be rendered as a black spot? No (look at the film curve), but certainly very dark indeed. But the geese are near us and relatively prominent, and blowing out them would spoil our image. A trunk in the shade with faintly perceivable details will not break our image. So we go for an exposure of EV 14.9, protect the geese, and hope for the best regarding the tree trunk.

If this was a commissioned and paid work, we would bracket. Else I wouldn't because film, whatever they say, is not cheap, unless this is a shot that has a particular value or significance for me.

If this was a black & white negative situation, as usual with black & white you would just expose with the thumb rule and you would get it.

If you have an old-style camera with the aperture command on the lens body, remember that you can vary continuously the aperture, you are not bound to respect the "clicks" of your lens.

So for the cow that would be e.g. 1/125s and 60% of travel between f/11 and f/16. You can obviously set f/9.5 if you only can vary aperture in half stops and you will set f/10 if you can vary aperture in thirds of stop. That will not make a great difference. But I would do all the "reasoning" in tenth of EV, personally. In any case the same logic applies if you think in thirds of EV and if you set your light meter in thirds of EV.

Exposing at EV 13 would push the geese at 4 EV above middle grey (remember they were metered EV 17.0) and frankly I don't think you would find any decent texture in that region of a slide film, that would place LogHgeese = +0.2, look in our graph where that is. But who knows? YMMV.

Please debug everything I am open to any kind of contribution.

Like I've been saying all along the thread. YOU MUST CALIBRATE EXPOSURE by doing a very simple film exposure test so you know exactly how far shadow and and highlight separation, and pure white and black are from metered value. The theory DORS NOT cover this. You can look at the manufacturers H&D curve but that is only a guide since you may be using a different developer and there will be some varaition. And it will certainly vary from film type to film type.
You must get get to know your own "system" before you can place exposure exactly where you want it. No theory can tell you that. Taking a light reading is an objective process. Placing an exposure is a subjective process based on experience and testing.

If you want accurate colour get an incident meter and pray becasue even with an incident meter it will be off a tad or more depending on conditions.

Diapositivo said:

Exposing at EV 13 would push the geese at 4 EV above middle grey (remember they were metered EV 17.0) and frankly I don't think you would find any decent texture in that region of a slide film, that would place LogHgeese = +0.2, look in our graph where that is. But who knows? YMMV.

Please debug everything I am open to any kind of contribution.

Click to expand...

Can't fault your logic. But you are taking a picture of a cow and a goose. I'm taking a picture of a dog and a tree. Blowing out the highlights a bit don't bother me in this situation.

Rob proposed that I was overexposing a bit, so his EV would likely be closer to yours.

But truth be told, before I started thinking "analytically" I was thinking of just using the reading off the sunlit grass. Or using an incident meter. But what I did was split the difference between the two grass readings. Maybe it's been too long since I've shot transparency and I'm in too much of a color neg mindset.

BrianShaw said:

Can't fault your logic. But you are taking a picture of a cow and a goose. I'm taking a picture of a dog and a tree. Blowing out the highlights a bit don't bother me in this situation.

Rob proposed that I was overexposing a bit, so his EV would likely be closer to yours.

But truth be told, before I started thinking "analytically" I was thinking of just using the reading off the sunlit grass. Or using an incident meter. But what I did was split the difference between the two grass readings. Maybe it's been too long since I've shot transparency and I'm in too much of a color neg mindset.

Click to expand...

I think that, with slides, EV 13 would in any case bring, in that situation, an overexposed picture. The grass would end up 1.6 EV over middle grey, and probably the entire scene would have a "milky" quality.

I'm sure with negative film that would be the right exposure. With negative, the highlights never are a problem. The problem is, in case, the lack of density in the deep shadows. So it is appropriate to expose more than the normal light metering and have a "safety buffer" in the shadows, the possibility to "dig deep" in the shadows.

An incident metering would give an acceptable picture with slide-cow, and with negative film. I remain convinced that an incident reading, EV 14.6, would jeopardize geese texture in the geese case, which is more taxing for the dynamic range of the film. This is to show the thumb rule that incident light meters makers give: the reading is good for almost everything, but for very white and very dark subjects, do compensate a bit bringing the placement some 0.5 EV toward the grey.

Actually the rule of thumbs should be: with slide film, incident reading almost always works, but do reduce 0.5 EV if you have pure white subjects where you want texture. With negative film, incident reading almosto always work, but do open 0.5 EV if you have very dark subject parts where you want texture.
(incident reading is plain always fine with slides, if you don't have a very white subject, and is plain always fine with negatives, if you don't have a very black subject).

These situations are also why I mistrust averaging different incident readings when using slides. Ultimately, in difficult situations, one must decide where he wants to put the highlights exactly, and averaging two incident readings does not give an exact placement on the curve. Reading the value of the spot meter gives higher probabilities of an "exact" placement where one wants the subject to be.

Anyway, what I was trying to convey is that all the reasoning in using a reflected spot metering is dependent on where the instrument places the reading, which mid-tone. The value at which each subject is measured (EV 17 for geese, or EV 9.3 for the tree trunk) is the reading which would make that subject mid-grey.
The calculation that follows in order to place it makes very important to know where exactly this mid-grey is.

If the mid-grey is 8% or 18% makes more than 1 EV of difference in the final choice, and that makes a world of difference with slides, especially in those situations (nocturne pictures of monuments comes to my mind) where the maximum exploitment of the limited slide dynamic range is important. Even 12% - 18% makes 0.5 EV and this is a "fixed mistake" that I would absolutely like to avoid when using slide film. Not without reason spot light meters specify that their instruments are calibrated for 18% grey.

I think that people using only black & white can "afford" not using the spot meter precisely (misuderstanding its behaviour) because the "slack" that negative film allows both in exposure, and then in the printing process, allows for compensation. One can even never really realize how improperly it is using its spot light meter (e.g. being convinced it gives a value of 8% rather than 18%) and be happy in any case with the final results, after the print.

With slides one can certainly not afford being more than 1 EV off at every shot (thinking light meter places at 8%), and IMHO also cannot afford being 0.5 EV at every shot (thinking light meter places at 12%). Understanding where the light meter places the subject becomes crucial.
Always IMHO.

Stephen Benskin said:

I believe you wrote somewhere that K is arbitrary. You might be thinking of K1 in standard’s equation for K. Connelly uses P for the exposure constant, which is the same as K1. P is the arbitrary chosen value, but in this use, arbitrary just means that it isn’t a physical constant. It was determined through psychophysical testing to determine the most acceptable value.

Click to expand...

Mmmmh, yes. K for me sometimes, and wrongly, was the "overall adjusting factor" that will make, after examination by a sufficiently large number of people, the best conversion from metered light reflected by the subject to an exposure value to be set in the camera. I have no problem in calling it K1 or P.

I gave for granted, so far, that the world at large considered b (q) as something fixed, and that all the adjustement due to the statistical observation by the sufficiently large number of people was meant to derive a K that gave the best result (the best match between light measured and exposure). If b (q) is a calculated factor, the result of an equation, and not something that is the result of statistical analysis, I thought, then the variation due to the result of statistical perception must end up "all in K" and the b*K in Conrad would end up being:

b fixed, result of an equation, something calculated;
K derived through empirical observation of perceived best matching result, something "observed" with statistical analysis.

But one might say that the result of the statistical observation ends up to influence both b and K. So it is their product, b*K, which is the result of observation, the overall matching factor.

No problem with that (although one would ask why so much theory about calculating b. In the end, is the observed b*K that would matter, not b nor K, and one would also ask why light meter producers give a figure for K and not for b!). But we should now ask Mr. Minolta what does he call K (which is = 14) in his answer to RobC's question and in the instruction booklet.

I suspect that Mr. Minolta calls K what Conrad calls K (different from what I called K, but same if we keep b as exogenous, calculated) and in that case I suspect that also for Mr. Minolta what Conrad calls b must be the same value as in Conrad, because Mr. Minolta tells us that his exposure always gives Hg = 0.1.

We should keep in mind that when Mr. Minolta tells us, that is:
1) He uses K = 14
2) His light meters give to LogHg a value of -1.0
3) He calibrates his light meter for a mid-tone of 18%
he tells us a lot of stuff.

The three elements all seem to nicely sum up if b = 0.728 (just like in Conrad), K = 14 (K as in Conrad) and mid-tone is 18% because that gives a LogHg a value of -1.0 both for slide and for negatives.

(If, instead, Mr. Minolta called K what Conrad calls b*K, then he wouldn't be true when he tells us that his instruments create a density of 0.1 on film).

They also sum up if we consider that these values create on the film the exact same exposure that an incident light meter with a flat disk and a C = 250 would give (following the logic given by Conrad, not that I approve of his considerations immediately after that).

Of course the same value of K, or of K*b, or of (K*b)/10, or the same value of whatever the "overal adjusting factor is", would be obtained by a reflected light meter producer also if incident light meters never had come to existance, and also if the C calibration factor never had been conceived by human mind.

The reflected light meter producer would arrive to the same conclusion, the same calibration, by the sole statistical analysis of best perceived match between reflected light and exposure values.

On the other hand, an incident light meter producer would arrive to a calibration of its incident meter by statistical observation of the best match between incident light and exposure value, even if reflected light meters never had been invented.

But, if the sample of people is significative and well chosen, the two sets of observations must converge toward the same result. Human vision is the same when it observes pictures, it doesn't know if the light metering was made with incident or reflected light meters.

you're still making an awful lot of assumptions. You must test your film and calibnrate your exposure.

Diapositivo said:

Mmmmh, yes. K for me sometimes, and wrongly, was the "overall adjusting factor" that will make, after examination by a sufficiently large number of people, the best conversion from metered light reflected by the subject to an exposure value to be set in the camera. I have no problem in calling it K1 or P.

I gave for granted, so far, that the world at large considered b (q) as something fixed, and that all the adjustement due to the statistical observation by the sufficiently large number of people was meant to derive a K that gave the best result (the best match between light measured and exposure). If b (q) is a calculated factor, the result of an equation, and not something that is the result of statistical analysis, I thought, then the variation due to the result of statistical perception must end up "all in K" and the b*K in Conrad would end up being:

b fixed, result of an equation, something calculated;
K derived through empirical observation of perceived best matching result, something "observed" with statistical analysis.

But one might say that the result of the statistical observation ends up to influence both b and K. So it is their product, b*K, which is the result of observation, the overall matching factor.

No problem with that (although one would ask why so much theory about calculating b. In the end, is the observed b*K that would matter, not b nor K, and one would also ask why light meter producers give a figure for K and not for b!). But we should now ask Mr. Minolta what does he call K (which is = 14) in his answer to RobC's question and in the instruction booklet.

I suspect that Mr. Minolta calls K what Conrad calls K (different from what I called K, but same if we keep b as exogenous, calculated) and in that case I suspect that also for Mr. Minolta what Conrad calls b must be the same value as in Conrad, because Mr. Minolta tells us that his exposure always gives Hg = 0.1.

We should keep in mind that when Mr. Minolta tells us, that is:
1) He uses K = 14
2) His light meters give to LogHg a value of -1.0
3) He calibrates his light meter for a mid-tone of 18%
he tells us a lot of stuff.

The three elements all seem to nicely sum up if b = 0.728 (just like in Conrad), K = 14 (K as in Conrad) and mid-tone is 18% because that gives a LogHg a value of -1.0 both for slide and for negatives.

(If, instead, Mr. Minolta called K what Conrad calls b*K, then he wouldn't be true when he tells us that his instruments create a density of 0.1 on film).

They also sum up if we consider that these values create on the film the exact same exposure that an incident light meter with a flat disk and a C = 250 would give (following the logic given by Conrad, not that I approve of his considerations immediately after that).

Of course the same value of K, or of K*b, or of (K*b)/10, or the same value of whatever the "overal adjusting factor is", would be obtained by a reflected light meter producer also if incident light meters never had come to existance, and also if the C calibration factor never had been conceived by human mind.

The reflected light meter producer would arrive to the same conclusion, the same calibration, by the sole statistical analysis of best perceived match between reflected light and exposure values.

On the other hand, an incident light meter producer would arrive to a calibration of its incident meter by statistical observation of the best match between incident light and exposure value, even if reflected light meters never had been invented.

But, if the sample of people is significant and well chosen, the two sets of observations must converge toward the same result. Human vision is the same when it observes pictures, it doesn't know if the light metering was made with incident or reflected light meters.

Click to expand...

Fabrizio,

There's a relationship between the constants, P, q, and K. P/q = K. We can also use it as K*q = P. For K = 14 and q = 0.72, P would have to equal 10 and that wouldn't work with b&w. Also, why adjust the reversal speed constant if the value of P is also going to change?

As for the equation for q, most of the same variables are also in the equation for K. Add L/A2 to the equation and you have the long version of the exposure equation.

Stephen Benskin said:

Also, why adjust the reversal speed constant if the value of P is also going to change?

Click to expand...

And, in a previous post, #199, page 8 (I'm re-reading all the thread):

Stephen Benskin said:

Don't forget the ratio from the speed point to the metered exposure point. This and the film speed sets where the exposure should fall.

Click to expand...

For me the exposure should fall at Hg for middle grey. The metered exposure point is the exposure point for the middle grey.

You seem to say that the exposure point is, in fact, not Hg, and that it should fall somewhere else, in a calculated spot which takes into account speed point and metered exposure point.

Is this the reason why my calculation never matches yours?

To me, "metered exposure point" and Hg and "where exposure should fall" are, when using a reflected light meter aimed at a mid-tone subject, exactly the same thing. And they are the same exposure for slide film and for negative film.

Would you please explain what additional steps would be applied to arrive, from metered exposure point, to "where the exposure should fall", both in the slide case and the negative case?

Stephen Benskin said:

Diapositivo, here is a quick answer for you.


The speed of B&W negative and color reversal film are determined at critical points. With B&W negative film it has to do with the limiting gradient of the film curve. This defines the minimum exposure which can yield a quality print. When the contrast parameters of the ISO standard are maintained, this point falls approximately one stop to the left of the exposure for the fixed density speed point (Hm where the density equals 0.10 over Fb+f). To calculate B&W film speed Hm is plugged into the equation S = 0.80/Hm where 0.80 is known as the speed constant. Using this equation, the value of S is 1/3 stops slower than what Hm would otherwise indicate. This is difference is to adjust for the color temperature of the sensitometric exposure; however, the idea that the film speed value can be different from the speed point is important. To calculate what the Hm should be for any speed, the equation is 0.80 / film speed = Hm.

With reversal color film, the speed point is found in the “middle” of the curve. Please refer to the example. Point HF is located on the curve 0.2 above the minimum density. The exposure value HR is the geometric mean between HF and HS. The speed equation is S = 10 / HR. Please note that the nomenclature in this example is outdated.

Sometime in the 70s or early 80s the speed equation changed from 8/HR to 10/HR thus increasing the speed of reversal film by 1/3 stop. The way the speed point was determined didn’t change, just the speed constant. It’s all about the relationship between the speed point and Hg. In this example, we know where HR was determined and we know where Hg should fall. At one time this was the same point with reversal color film, but it was decided to reduce the overall exposure. Now Hg falls 1/3 stop below HR.

The exposure meter wants to place the exposure at Hg. This relationship can be determined with the equations Hg/Hm for B&W film and Hg/HR for reversal color film. For B&W this value is 10. For color reversal it’s 0.8. In the previous incarnation it was 1 which means the speed point for reversal color film was the same point as where the exposure wanted to place Hg. The equation can also use the speed constants and the value for Eg.

Eg can be thought of as the value of exposure striking the film plane if the shutter speed is at one second. From this Hg can be determined with the equation Eg * 1/ISO = Hg. Eg as a constant is referred to as K1 in the exposure meter standard or in some papers as P. So the equation can be reduced to Hg = 8/ISO.

8/100 = 0.080 lxs 8/125 = 0.064 lxs 8/400 = 0.020 lxs

If it’s determined there needs to be a change where Hg should fall on the film plane, the meter doesn’t need to be adjusted, nor the method of determination the speed point. All that needs to be changed is the speed constant. This will change the speed number and consequently placement of Hg by the exposure meter.

Click to expand...

I am re-reading this text in the light of what I have learned so far during the thread.

"This defines the minimum exposure which can yield a quality print. When the contrast parameters of the ISO standard are maintained, this point falls approximately one stop to the left of the exposure for the fixed density speed point (Hm where the density equals 0.10 over Fb+f)."
Do you mean: ... falls approximately one third of one stop to the left ...
If the density is 0.1 over fog, one stop to the left of that is inside the fog.

"Sometime in the 70s or early 80s the speed equation changed from 8/HR to 10/HR thus increasing the speed of reversal film by 1/3 stop. The way the speed point was determined didn’t change, just the speed constant"
I don't get that. If the speed is increased, I endup with a different box speed on the package. The same material that previously was rated ASA 100 is after the change rated ASA 125. That change made, all the rest should remain equal. The new Hg should naturally be 1/3 of EV closer than the old EV for the same material.

"It’s all about the relationship between the speed point and Hg. In this example, we know where HR was determined and we know where Hg should fall. At one time this was the same point with reversal color film, but it was decided to reduce the overall exposure. Now Hg falls 1/3 stop below HR.

The exposure meter wants to place the exposure at Hg. This relationship can be determined with the equations Hg/Hm for B&W film and Hg/HR for reversal color film. For B&W this value is 10. For color reversal it’s 0.8. In the previous incarnation it was 1 which means the speed point for reversal color film was the same point as where the exposure wanted to place Hg. The equation can also use the speed constants and the value for Eg"

We come to the meet here.
Case 1
Do you mean a lightmeter gives me an Hg which would have been the not just the "exposure point" but also the "speed point" with the old rating. But given that there is a new rating of film speed, I can't use the exposure value from the light meter "as is" (we call this Hg) but I should apply a correction factor, which is 0.8? But then, why is the slide film not rated with an ASA value that allows me to use, "as is" the exposure value given by the instrument? Am I really supposed to read ISO 100 and expose a slide film as ISO 125 because I am supposed to know that the speed equation was changed? (I don't think that is the case).

Case 2
Or is it the other way round? The speed point was raised 1/3 EV but not the ASA speed rating. Only raising the speed point, without changing the speed equation, would produce an exposure of 1/3 EV closer than with the previous rating. BUT, compensatively, the speed equation is changed from 8/HR to 10/HR. That implies there is now a compensative factor 0.8 in order to make the exposure fall where it used to fall, and the ISO speed remains the same even if the speed point is defined differently.
The "speed point" for slide film doesn't coincide anymore with the "exposure point" (no problem with that. That also happens in negative film, where the speed point lies at a distance relative to exposure point). But the "exposure point", thanks to the 0.8 correction factor, falls again where the light meter wants to place it, at Hg.
In this case we have this situation:
The light meter gives the "exposure point", or Hg, which is valid for both slide film and negative film, at ISO rating.

Knowing that, I still don't understand why I should need to care at all about the existence of a speed point somewhere, and of speed equation.

All that I need to know is what my ISO speed is and that the light meter will give me mid-tone, i.e. Hg.

When using slide film, I know that I have let's say 2.3 useable texture above mid-tone (the exact amount depends on the slide film, and on where I cut "useable").
When using negative film I know that I have let's say 4.3 useable texture below mid-tone when using standard development (the exact amount depends on the negative film, and on where I cut "useable").

Is Case 1 you mean, or Case 2? Or case 3?

Diapositivo said:

I am re-reading this text in the light of what I have learned so far during the thread.

"This defines the minimum exposure which can yield a quality print. When the contrast parameters of the ISO standard are maintained, this point falls approximately one stop to the left of the exposure for the fixed density speed point (Hm where the density equals 0.10 over Fb+f)."
Do you mean: ... falls approximately one third of one stop to the left ...
If the density is 0.1 over fog, one stop to the left of that is inside the fog.

"Sometime in the 70s or early 80s the speed equation changed from 8/HR to 10/HR thus increasing the speed of reversal film by 1/3 stop. The way the speed point was determined didn’t change, just the speed constant"
I don't get that. If the speed is increased, I endup with a different box speed on the package. The same material that previously was rated ASA 100 is after the change rated ASA 125. That change made, all the rest should remain equal. The new Hg should naturally be 1/3 of EV closer than the old EV for the same material.

"It’s all about the relationship between the speed point and Hg. In this example, we know where HR was determined and we know where Hg should fall. At one time this was the same point with reversal color film, but it was decided to reduce the overall exposure. Now Hg falls 1/3 stop below HR.

The exposure meter wants to place the exposure at Hg. This relationship can be determined with the equations Hg/Hm for B&W film and Hg/HR for reversal color film. For B&W this value is 10. For color reversal it’s 0.8. In the previous incarnation it was 1 which means the speed point for reversal color film was the same point as where the exposure wanted to place Hg. The equation can also use the speed constants and the value for Eg"

We come to the meet here.
Case 1
Do you mean a lightmeter gives me an Hg which would have been the not just the "exposure point" but also the "speed point" with the old rating. But given that there is a new rating of film speed, I can't use the exposure value from the light meter "as is" (we call this Hg) but I should apply a correction factor, which is 0.8? But then, why is the slide film not rated with an ASA value that allows me to use, "as is" the exposure value given by the instrument? Am I really supposed to read ISO 100 and expose a slide film as ISO 125 because I am supposed to know that the speed equation was changed? (I don't think that is the case).

Case 2
Or is it the other way round? The speed point was raised 1/3 EV but not the ASA speed rating. Only raising the speed point, without changing the speed equation, would produce an exposure of 1/3 EV closer than with the previous rating. BUT, compensatively, the speed equation is changed from 8/HR to 10/HR. That implies there is now a compensative factor 0.8 in order to make the exposure fall where it used to fall, and the ISO speed remains the same even if the speed point is defined differently.
The "speed point" for slide film doesn't coincide anymore with the "exposure point" (no problem with that. That also happens in negative film, where the speed point lies at a distance relative to exposure point). But the "exposure point", thanks to the 0.8 correction factor, falls again where the light meter wants to place it, at Hg.
In this case we have this situation:
The light meter gives the "exposure point", or Hg, which is valid for both slide film and negative film, at ISO rating.

Knowing that, I still don't understand why I should need to care at all about the existence of a speed point somewhere, and of speed equation.

All that I need to know is what my ISO speed is and that the light meter will give me mid-tone, i.e. Hg.

When using slide film, I know that I have let's say 2.3 useable texture above mid-tone (the exact amount depends on the slide film, and on where I cut "useable").
When using negative film I know that I have let's say 4.3 useable texture below mid-tone when using standard development (the exact amount depends on the negative film, and on where I cut "useable").

Is Case 1 you mean, or Case 2? Or case 3?

Click to expand...

Pretty much Case 2 without the correction factor.

As for b&w, when the contrast parameters of the ISO standard are met, the shadow will fall at the fractional gradient speed point which is almost one stop below the ISO speed point. This is the Delta-X Criterion. It's introduced in the paper Simple Methods for Approximating the Fractional Gradient Speeds. Average flare tends to raise the shadow exposure to around Hm. Misunderstanding the relationship between Hg and Hm for b&w is why Zone System practitioners tend to produce personal EIs around 1/2 to 1 stop slower than the ISO speed. This is my favorite example on why understanding theory is important.

Bill Burk said:

Or both...
View attachment 153443

My point is that the meter is calibrated to a backlit screen, and that explains why it's literally not calibrated to reflectance.
All the discussion about the calibrated reflectance is us trying to figure out the equivalent.

Click to expand...

In re-reading the thread I saw that I oversaw this text.

My point is that a calibration is only a procedure that you use, in a laboratory, to bring your instrument in a certain state. The procedure must be as far as possible easy, fast, cheap, and reliable.
You know what a mess is simulating in laboratory a certain reflectance of a certain subject. You have to define the light, the properties of the surface, the angles involved etc. and probably that will not be very "repeatable" in different laboratories.
It's much easier to just define a calibration light with certain known characteristics.
So the calibration is made in a laboratory by using a source of light, which is a simulation for the real life world, where you normally have reflected light, although you could also have a light source as your subject (neon street sign, fire, TV set, light-emitting insects...).

The light meter doesn't know whether the light comes from a light source or is reflected by an opaque body, the light meter only sees some light, some photons.

You have a scale in your kitchen, and you use it to weigh potatos. In a lab, calibrating a scale with potatoes is not very practical. You have do define potatoes of a certain volume, a certain density, a certain water content... and they will soon germinate...
You use a pre-certified weight of 1 Kg, in stainless steel. You bring your scale to a certain state (1 kg reads 1 kg on the indicator). That doesn't mean that the scale can only work with steel. The scale doesn't know if it weights steel or potatoes. The light meter doesn't know at all what is in front of it. The "calibration light" is a red herring IMHO.

RobC said:

If he wants to know exactly how his meter works then he can contact its manufacturer technical support and ask them ( like I did with Minolta when they made spot meters)

Click to expand...

In re-reading the thread I came through this and I ask you a favour. Do you still have the text of the question you posed to Minolta, and the entire answer? Can you place it in here?

Besides the wording of the answer I would be interested in knowin the "source" inside Minolta for your answer. Was it cursorily prepared by somebody in the help-desk office or did the answer come, ideally, from some technician at Minolta. The second case would be better.

If you could copy/paste the integral text of the question and answer here, I think it would be interesting for the entire internet and would double APUG accesses within two years