Kodak grey card usage

[RobC]

if you think it gives middke grey (18%) then what is your problem?

 

[Diapositivo]

which means you must calibrate your exposure if you want accurate colour since you don't know where on the curve for any particular film it will be. From calcs in earlier post I said its 3 stops probably from speed point. And that means as you rightly say it won't be in middle of curve unless its fits manufacturers test curve which most of the time it won't. So how can it be exact middle grey? It can't.
You now seem to be saying meters are calibrated for slide film. Minolta may be better for slide film but I don't think others are.

All lightmeters are calibrated to a defined shade of grey. Minolta and Pentax 18%. Sekonic 16.5%. That's to the best of my knowledge.

Where on the curve the lightmeter places the reading is the reason why I am bombarding Stephan with questions in this thread.
I think for slides it can be determined somewhere around 1.1 or 1.2 density for what I gathered so far.

That must be derivable once you know the speed equation and the ISO speed of the film, and, in the case of slides, as Stephan explained, the offset between HR (speed point) and g (metered point by the reflected light meter, which I intend metered point to give you the k shade of grey the manufacturer plugged in the device).

For a slide film, 3 stops from speed point is WAY too white for most anything different from a white page.

Meters are calibrated for both negatives and slides, although one might argue that Minolta and Pentax, which chose an 18% middle grey, lighter than the 16.5% middle grey of Sekonic, tend to make you close a bit more and so are more thought for slide users because with slides the cliff is on the highlights. You must walk near the cliff, but not overboard.

 

[Diapositivo]

if you think it gives middke grey (18%) then what is your problem?

First, I am not sure it is 18%. Everybody here seems convinced that the calibration grey is less than that (with the exception of Ralph Lambrecht). The more I look at the matter, and the more I convince myself that, instead, it is 18% (for Minolta. Actually, as you showed, they declare that!). Participating to this forum in the last days made me understand some stuff better.

Second, I am not yet sure where, on the film curve, is the corresponding density (or, if you prefer, the corresponding exposure), i.e. where is Hg on the film curve. I need a way to dot, on the film curve of any slide film, the light meter exposure. That will give me where the "cliff" is for highlights. That will prevent those exposure mistakes like the one of the fountain, and, imagine that, without tests! (or at least one makes tests in order to verify that everything is in order).

The speed equation, as said, is the link between the ISO speed and the LogH of Hg in the graph. Then I would read density Dg on the graph. And I would expect the same density for each slide film of whichever ISO speed!

In reality the photographic process is complicated. Placement is the complicated part. But having the right dots on the film curve helps!

 

[Stephen Benskin]

Yes, but that belongs to the quirks of human vision and brain reconstruction of reality.
This link shows other things like False colour (the eye sees colours that don't exist in the image) and Mach Bands (your eyes sees shades of grey that don't exist in reality)

http://www.cambridgeincolour.com/tutorials/cameras-vs-human-eye.htm

Normally though we don't see images of grey squares in white squares. In normal average images (of cats ) we see a "complex" scene, with a black point, a white point, and a grey point in the middle. That's true for 99% of images, I reckon.

You might recall the preferred tone reproduction curves in the Holm example. A projected transparency needs to be more contrasty than one on a light table. I believe this is surround. The determination of Munsell's 18% and CIE's 19.77% were determined under very strict conditions. Ones that normally won't be encountered in real life. Viewing conditions is an important, yet frequently overlooked, aspect of photography.

This is from Holm.

Back to meter calibration and image placement. This is from Holm which I've posted before, but it's a key piece of information.

"That the luminance range of a statistically average scene is 160:1 (log range 2.2), and the resulting exposure range on the image capture medium is 80:1 (log range 1.9), corresponding to a camera flare factor of 2.

That the mean log luminance of a statistically average scene is approximately 0.95 log units below the highlight log luminance (edge of detail in white) and 1.25 log units above the shadow log luminance (edge of detail in black), and that this mean luminance is assumed to be the luminance metered, directly or indirectly, for exposure determination. These values result in the mean luminance correlating with a Lambertian scene reflectance of 12% for 100% highlight reflectance."

The calibration Luminance for the exposure meter is at a value that would equal a 12% reflectance on a sunny day. This value will equal Hg as 8/ISO. For different luminance values, the exposure calculator will adjust the f/stop and shutter speed to equal the same exposure at the film plane as with the calibration luminance. You have your slide film, you know where Hg will fall and the density on the curve corresponding to Hg exposure is known. I saw one of your posts where you believe K = 12.5 corresponds to 16% reflectance. Can you say where that is from?

 

[RobC]

First, I am not sure it is 18%. Everybody here seems convinced that the calibration grey is less than that (with the exception of Ralph Lambrecht). The more I look at the matter, and the more I convince myself that, instead, it is 18% (for Minolta. Actually, as you showed, they declare that!). Participating to this forum in the last days made me understand some stuff better.

Second, I am not yet sure where, on the film curve, is the corresponding density (or, if you prefer, the corresponding exposure), where is Hg. I need a way to dot, on the film curve of any slide film, the light meter exposure. That will give me where the "cliff" is for highlights. That will prevent those exposure mistakes like the one of the fountain, and, imagine that, without tests! (or at least one makes tests in order to verify that everything is in order).

The speed equation, as said, is the link between the ISO speed and the LogH of Hg in the graph. Then I would read density Dg on the graph. And I would expect the same density for each slide film of whichever ISO speed!

In reality the photographic process is complicated. Placement is the complicated part. But having the right dots on the film curve helps!

If you read what minolta say in my quote they are suitably vague for it not to be of any use at all. It is not unfortunate English at all. They say middle of the curve you say you know better and not what the formula says. But have you considered that the Minolta is calibrated for negative film and that most other reflective meters are too so that they won't work for slide film unless you calibrate exposure. They say generally defined as about 18% you say it is 18%. Seems you have all the answers and shouldn't need anymore help.

I hope your subject is not animate as its probably long gone since you've been trying to work out exposure.

 

[Stephen Benskin]

Maybe this will help. It uses the exposure meter calibration Illuminance value and reflectance values down to 8%. Where does the exposure meter calibration Luminance value fall?

 

[Bill Burk]

At the big red arrow?

 

[RobC]

Given that the maths is simple and you have all the formulas, why haven't you been given the answer or worked it out by now?

let us know when you have a definitive answer, I won't hold my breath while I'm waiting.

 

[Bill Burk]

...(with the exception of Ralph Lambrecht)...

I believe Ralph Lambrecht is working on finding the statistics that represent the average reflectance of HIS photography.

 

[BrianShaw]

What I want to know is if anyone's taken any pictures lately. I just pointed my meter and read the dial. It worked great!

 

[Stephen Benskin]

At the big red arrow?

12%. A2*K = 3200 when A=16 and K = 12.5. Although when all the variables are plugged into the K equation, it comes out to be 12.478 so A2*K = 3194, so there's some rounding going on.

 

[Diapositivo]

You might recall the preferred tone reproduction curves in the Holm example. A projected transparency needs to be more contrasty than one on a light table. I believe this is surround. The determination of Munsell's 18% and CIE's 19.77% were determined under very strict conditions. Ones that normally won't be encountered in real life. Viewing conditions is an important, yet frequently overlooked, aspect of photography.

This is from Holm.
View attachment 153468

Back to meter calibration and image placement. This is from Holm which I've posted before, but it's a key piece of information.

"That the luminance range of a statistically average scene is 160:1 (log range 2.2), and the resulting exposure range on the image capture medium is 80:1 (log range 1.9), corresponding to a camera flare factor of 2.

That the mean log luminance of a statistically average scene is approximately 0.95 log units below the highlight log luminance (edge of detail in white) and 1.25 log units above the shadow log luminance (edge of detail in black), and that this mean luminance is assumed to be the luminance metered, directly or indirectly, for exposure determination. These values result in the mean luminance correlating with a Lambertian scene reflectance of 12% for 100% highlight reflectance."

The calibration Luminance for the exposure meter is at a value that would equal a 12% reflectance on a sunny day. This value will equal Hg as 8/ISO. For different luminance values, the exposure calculator will adjust the f/stop and shutter speed to equal the same exposure at the film plane as with the calibration luminance. You have your slide film, you know where Hg will fall and the density on the curve corresponding to Hg exposure is known. I saw one of your posts where you believe K = 12.5 corresponds to 16% reflectance. Can you say where that is from?

First, the answer to your question: the correlation between K = 12.5 and target grey 16% comes from this source:
http://dpanswers.com/content/tech_kfactor.php
which I quated a couple times in this thread.

So, as far as my doubts are concerned, I think that we have now firmly that Hg is the "target grey" given by the light meter, and that it is linked to the film by the speed equation. Pluggin in the ISO speed to the ISO equation we derive, for slides, HR and subtracting 1/3 EV of exposure we have where Hg falls on the exposure axis. Then the plot will give us density Dg.
For negative material the speed equation gives us directly the Hg value once the ISO speed is plugged in, and then the plot gives us Dg.

The only doubt which remains to clarify is where exactly is the target grey for a light meter in terms of reflectance. This is loosely defined in standards, so each manufacturer is free to choose, within a certain range, a lighter or darker shade of grey.
I do believe that sums up with a higher, or lower, K factor (as in the exposure equation) because that's the overall adjustment factor that aligns together film speed, light, photographic system, and - shall I add - target grey or "correct exposure" for a reflected light meter.

I still think there is a bit too much difference between 12% and 18%.

It is possible that the average reflection of the average scene is 12% but I don't know where is the source for this estimation. It is then possible that light meter manufacturers set for 16.5% or 18% grey for practical reasons, i.e. to "help" the photographer not to blow up highlights with slides, but I doubt it. (With negatives, the difference is so small that it doesn't matter anyway). I doubt it because a reflected spot meter would be calibrated without "help" and would be calibrated, in that case, to 12%. But Minolta, and the entire world, consider 18% to be the reference and calibration grey.

[Many resources exist on the internet saying that light meters is not really calibrated on 18% reflectance, but they usually make confusion between K and k, one example being this by Tom Hogan:
http://www.bythom.com/graycards.htm ]

What surprise me is that if you take any 12% reflective grey and I ask you what tone of grey that is, you would answer without hesitation: "a dark grey"!
Other things don't match. The human eye sets the average grey at 18% and I don't think the human eye is "off-set" in respect to the average grey of the average scene. Our eyes are calibrated to the world around us. To our eyes, 12% is a dark shade.

If average grey was 12%, a 4% black would be less than 1.5 EV below middle grey and a 90% white (such as a white shirt) would be basically 3 EV above middle grey. It's a middle that it is not very much in the middle from a photographic exposure point of view.

The reason why a light table requires "less contrast" than a slide projection MIGHT be on the fact that a light projection never obtains a blackest "black point" because the light of the projector lightens, to a certain extent, the room.
On the light table there is no such effect. Dmin exactly sets the blackpoint and Dmax exactly sets the whitepoint. That helps perceived contrast. Basically you have a larger perceived distance between black point and white point for the same image.

Prints have the less perceived contrast;
Slide projections have a much better perceived contrast;
Light table has a perfect perceived contrast.

In all three cases, the eye sets the middle in the middle between white and black, at the same point. It's the sensation of "snappiness" which is different.

 

[Diapositivo]

I believe Ralph Lambrecht is working on finding the statistics that represent the average reflectance of HIS photography.

I don't know. But the average reflectance of his photography, for a large enough number of pictures, must coincide with the average reflectance of our photography, unless he's specialized on winter season photography.

 

[Diapositivo]

What I want to know is if anyone's taken any pictures lately. I just pointed my meter and read the dial. It worked great!

That's because you used an incident light meter, or an averaging reflected light meter presumably calibrated on 18% grey.

If you used a spot reflected meter, and considered the calibration grey to be 8%, and use slides, it wouldn't work at all IMHO and experience.

 

[Diapositivo]

RobC, I found a text of yours, I infer, in largeformatphotography.info, may I reproduce it here: the thread was about the calibration of light meters for different colour temperatures (tungsten, daylight) but I think this post by Rob is interesting:

<<
Mark,

I enquired about calibration of my Minolta Spotmeter F sometime ago and this is the response I received. It may be of some use to you but makes no mention of colour temp.

<--

The meter is designed on the assumption that the midpoint of the film
characteristic curve is receiving an exposure level of 0.1 lux.seconds.
The formula used to calculate exposure is :

2^EV= (B*S)/K where

B= luminance in cd/m2
S=ISO sensitivity of film
K=calibration constant (calculated to be 14)

-->
>>

K is 14 and we know that because Minolta states that in the instruction manual.

What I notice is that the "midpoint" (that's obviously not the midpoint, but the "exposure point" given by the lightmeter) is receiving an exposure level of 0.1 lux.seconds.

That would equate to an Hg of -1.0 expressed as logarithm. We expect for 100 ISO this to be -1.1.There's 1/3 EV of difference.

EDIT: That shouldn't surprise us as, in fact, that might confirm that when K=14 the "middle grey of reference" or target grey, is a bit lighter, so the corresponding exposure is a bit less than what we calculated with the speed equation given by Stephan, which is if I get it right calculated for a K of 12.5 or so. But the difference between K = 14 and K = 12.5 is around 1/6 not 1/3 of EV (0.1 log).

It might be that the speed equation is calculated using K = 10.76, and so the difference in exposure between K = 10.76 and K = 14 is exactly 1/3 EV -> 0.1 log exposure.

That, for negative film, would make Hg fall for 100 ISO at -1.1 when using K = 10.76 and fall at -1.0 when using K = 14 which is what Minolta does.

That would make all things match very very well but for the exposure for slide film.

I calculated Hg for slides at -1.2 (1/3 of EV less than speed point) which can mean I am again misunderstanding the way I should use the speed equation given by Stephan (or, the correction for slide film to be exposed at 1/3 less than its speed point).

And now that I think about it, this 1/3 EV exposure less than speed point for slides (HR different from Hg) does not convince me much.
A light meter gives me a value which is good for negatives and slides. You don't have a "slide setting" in the lightmeter.
A film speed is described with an ISO number which has the same meaning, for the light meter, for slides and negative film.
So the speed equation must give me Hg for both slides and film.

The speed point for slides, as for negatives, is used only to determine ISO speed. Then we forget about it. The K value will give us a Hg which is the same for slide and negatives (and digital).

Hg will be, if the assumption underlined above is right, -1.1 when K = 10.76 (Weston, "old equation"), -1.0 when K = 14 (Minolta, Pentax), and somewhere in between for K = 12.5 (Sekonic, Canon, Nikon).

That would make all the box tick very, very, very nicely in my poor old fart's brain.

 

[RobC]

What I want to know is if anyone's taken any pictures lately. I just pointed my meter and read the dial. It worked great!

unlikely

 

[RobC]

you keep claiming you don't have a densitometer but you don't need one, you have a psot meter which I hope measures down to 1/10 stop.

Your problem is yourself because you refuse to do a practical evaluation. well I'll tell you one last time with a bit of extra added to make it really simple for you.

I've already told you the offset is 3 stops, probably from the speed point. But you don't know where the speed point is. Well it really doesn't matter because whatever you point your meter at will result in a middlish grey. Most likely not 18%. doesn't matter. You also claim 3 stops is too much. I have idea why, you've already looked at the Minolta manual and there is a shift from center towards highlights positioning of middle.
If it is three stops and the scale from whit to black was 6 stop it would be in the middle of useable range from WHITE.
Here's what you do. YOU TEST. Without a test there is absolutely no way for meter to get 18% since it doesn't know how far from white the result will be.
The test is as follows. Yu set up camera pointed at an even middleish toned subject. in that subject you place a an 18% grey card perpendiclular to the camera lens and a white card.
You meter the middleish toned subject and expose for it without any adjustment. All you need to do is make sure the grey card and white card fill a large percentage of the frame.
Now you have one slide. Now you take your slide and put it on a light box. You then take your spot meter and from directly above you meter the slide.
You meter the subject that you metered, you meter the grey card and you meter the white in the slide. So you have three readings. Now you can compare the offsets between each of the readings which tell you a lot more than you already know. i.e. How far the subject is from white and how far the grey card is from white. It should be 2 1/2 stops. Then all yo need to do is tweak exposure until Grey Card IS 2 1/2 STOPS from WHITE card.

 

[RobC]

you keep claiming you don't have a densitometer but you don't need one, you have a psot meter which I hope measures down to 1/10 stop.

Your problem is yourself because you refuse to do a practical evaluation. well I'll tell you one last time with a bit of extra added to make it really simple for you.

I've already told you the offset is 3 stops, probably from the speed point. But you don't know where the speed point is. Well it really doesn't matter because whatever you point your meter at will result in a middlish grey. Most likely not 18%. doesn't matter. You also claim 3 stops is too much. I have idea why, you've already looked at the Minolta manual and there is a shift from center towards highlights positioning of middle.
If it is three stops and the scale from whit to black was 6 stop it would be in the middle of useable range from WHITE.
Here's what you do. YOU TEST. Without a test there is absolutely no way for meter to get 18% since it doesn't know how far from white the result will be.
The test is as follows. Yu set up camera pointed at an even middleish toned subject. in that subject you place a an 18% grey card perpendiclular to the camera lens and a white card.
You meter the middleish toned subject and expose for it without any adjustment. All you need to do is make sure the grey card and white card fill a large percentage of the frame.
Now you have one slide. Now you take your slide and put it on a light box. You then take your spot meter and from directly above you meter the slide.
You meter the subject that you metered, you meter the grey card and you meter the white in the slide. So you have three readings. Now you can compare the offsets between each of the readings which tell you a lot more than you already know. i.e. How far the subject is from white and how far the grey card is from white. It should be 2 1/2 stops. Then all yo need to do is tweak exposure until Grey Card IS 2 1/2 STOPS from WHITE card.

 

[Diapositivo]

Maybe this will help. It uses the exposure meter calibration Illuminance value and reflectance values down to 8%. Where does the exposure meter calibration Luminance value fall?

View attachment 153469

I don't understand the first and third column, and the general meaning of this table. My fault, I know.
But I understand that the middle column is the reflectivity of a target.
So the calibration is on the line with 0.178 I would say.

But, again, I'm not an engineer designing light meters. What matters to me is that Mr Minolta tells me that his light meters give me an exposure = LogH -1.0 (he means with ISO 100) and he tells me that in the e-mail to RobC.
For ISO 100 Hg is -1.0.
For ISO 200 Hg is -1.3.
For ISO 400 Hg is -1.6.
...

That is, if we follow the indication of the instrument produced by Mr. Minolta.

Rob, you naughty boy, you should have known the answer to my question much earlier in the thread!

When you asked me what I wanted to know, and I told you I wanted to know the density of the target grey, your answer should have been: that's the density corresponding to LogH -1.0 because that's what Minolta told me.

If we take e.g. a film curve like Fujifilm Provia (not that any other curve will give different numbers), we see that for an ISO 100 slide film the density corresponding to LogH -1.0 is exactly 1. That means an opacity of 10, and that means a transparency of 1/10 = 10%.

What remains to work through is why the transparency on slide film is 10% and not higher to render an 18% grey. That might be due optimization for viewing conditions, human vision behaviours etc.

 

[markbarendt]

I wanted to know the density of the target grey, your answer should have been: that's the density corresponding to LogH -1.0 because that's what Minolta told me.

That can only true when you know exactly how the photo will be displayed.

This is from Holm.
View attachment 153468

 

[Diapositivo]

That can only true when you know exactly how the photo will be displayed.

Given an exposure value (Log H -1.0) and a film characteristic curve, from the film curve you know which density corresponds to that exposure value.

The problem you are referring to regards the way you use the light meter. You might optimize your exposure for a certain display method and therefore change your exposure. That might suggest you a different exposure than the one suggested by the light meter in order to arrive to the different density suggested by Holm.

For instance, to optimize your exposure for light table according to Holm, you must end up with density 1.0 for zone V which is exactly where Mr. Minolta places target grey.

Holm's table is, I think, something very empiric and "tentative", a rough guideline.

 

[Stephen Benskin]

I don't understand the first and third column, and the general meaning of this table. My fault, I know.
But I understand that the middle column is the reflectivity of a target.
So the calibration is on the line with 0.178 I would say.

First column is reflection density and the 3rd column is Luminance. The calibration Luminance for the exposure meter standard is where the red arrow is.

 

[Stephen Benskin]

But, again, I'm not an engineer designing light meters. What matters to me is that Mr Minolta tells me that his light meters give me an exposure = LogH -1.0 (he means with ISO 100) and he tells me that in the e-mail to RobC.
For ISO 100 Hg is -1.0.

That's the exposure for 100 speed reversal color film - 10 / 100 = 0.10 lxs

 

[Diapositivo]

First column is reflection density and the 3rd column is Luminance. The calibration Luminance for the exposure meter standard is where the red arrow is.

I suppose you mean:
Exposure meter standards want the device to be calibrated with a Luminance of 3.183 (cd/m2?).
That, in that table, corresponds to a reflectivity of 12.1%.
Ergo: Light meters calibrated according to this standard are calibrated for an average reflectivity of 12.1%.

Which arises the questions:
Why Mr. Minolta tells us (I think he tells us) that its lightmeters are calibrated for 18% reflectivity? (OK that should be asked to him );
What Hg would result @ 100 ISO from a calibration to 12.1%? (Mr. Minolta tells us squarely his instruments give LogH = -1.0);
Is it possible that Minolta lightmeters simply are calibrated to a different standard?
Is it possible that the standards for calibration of lightmers to which you refer are not used in photography? (Those might be standard for other uses of light meters, you know, all the other applications).

Besides, there is something I don't understand.
How can a calibration Luminance be directly correlated to a certain reflectivity, as the table does.
There certainly are a lot of other variables that must be populated before you arrive, in a calibration process, from a certain Luminance used in calibration, to a certain density. A relation must be defined somewhere!

 

[Diapositivo]

That's the exposure for 100 speed reversal color film - 10 / 100 = 0.10 lxs

Well, yes. But he doesn't tell us that, Mr. Minolta. Mr. Minolta gives us that exposure for slides and negatives. And it's 1/3 EV less exposure than what Hg is with the speed equation.

Mr. Minolta explicitly tells us that light meter gives an exposure as HMinolta = -1.0.
Mr. Minolta explicitly tells us that he uses a K = 14. He doesn't tell us he does it because it matches recommended exposure for slide film, but I guess he does it for that reason. After all, K = 14 is outside of the ISO recommended range for values of K!
The source I referred to says that a calibration for K = 14 yields an exposure that is 1/3 EV less than for K = 10.64.

What I construct from all that is that:
Standards require K = 10.64 but then, they also require a 1/3 less EV than nominal speed for slide film!
Minolta uses K = 14 and that results in an exposure which is the desired one for slides.
That also corresponds to a calibration to a 18% grey.
Sekonic meters are calibrated for K = 12.5 which is 1/6 EV above Minolta exposure, and half way between Minolta and standard ISO. A compromise between negative and slide film.
So Minolta light meters are thought for slide film, Sekonic are a compromise between slide and negative (or slide and ISO) and that is why there is the little discrepancy.

The final answer is: Light meters with K = 14 are calibrated to a 18% grey and are spot on for slide film. Light meters with K = 12.5 are calibrated to a 15.7% grey, expose 1/6 EV more than Minolta/Pentax and are halfway between calibration for slide and calibration for negative as recommended by ISO standards.
ISO standard would require a K = 10.64 which would correspond to a 13.5% grey and would result in 1/3 EV more exposure than Minolta/Pentax, if we trust http://dpanswers.com/content/tech_kfactor.php

This is my understanding so far.

PS In fact, Mr. Minolta, in the manual of the Spotmeter, makes explicit reference to the dynamic range of slide film. He doesn't say it explicitly, but it is a device which is thought with the slide user in mind.