Metering

[avishek_aiyar]

Hello,

 

I have possibly what is a dumb question. I have been trying to get a handle on this, but so far, I am not convinced.

 

I am told that every camera meter is calibrated to render an image as 18% gray (or 12% luminance, but I don't want to get hung up

on details that may not be critical to my question).

 

So, for example...an 8 bit black and white image would have 256 gradations between black and white with the 18% gray or the mid

tone occupying the center of the histogram.

 

So, I point the camera meter (in spot metering mode to keep things simple) at an object, it takes a reflected light reading and

compares that to 12/18% gray that it is calibrated to, and then overexposes or underexposes accordingly to render it 18% gray or

place it in zone v as ansel Adams would say.

 

1. That would mean every black and white image metered using say my nikon d300 would have a perfectly centered histogram?

Assuming of course the camera meter is trustworthy. Or if we are picky about details, then slightly off center to the left since it is

actually calibrated to 12% gray?

 

2. If again, I leave exposure to the cameras meter, why do I end up wth overexposed and underexposed shots (histogram to the far

right and far left)? All exposures should be perfect unless the camera gets a wrong reflected light reading in the first place? Or it is off

calibration.

 

This is is clearly not the case, since my meter screws up more often than not. But I can't logically understand the metering system of

the camera. Bear in mind that I am not taking about complex systems like matrix metering.

 

The other question would be...if I took my d300 and pointed it at a kodak 18% gray card, what kind of a histogram can I expect?

 

I am sorry of this sounds a bit muddled, but I am still trying to grasp this concept. I am trying to understand the zone system and

hence these experiments/questions.

 

Thanks a lot.

 

Avi

[SCL]

<p>In examples 1) and 2) you missed the point of your previous paragraph. The object being metered upon will be exposed as if it is reflecting approximately 18% of the light shed upon it....that doesn't mean it will be perfectly exposed. For instance if the object is solid black it will appear as 18% grey. If the object is solid white it will appear as if it is 18% grey. Neither exposure represents an accurate representation of the object.<br>

The camera's metering is trustworthy...you just need to think through what is actually happening and how you want to compensate for it. Since you have a digital camera, and it will cost you nothing extra, try experimenting with exposing an egg and then do the same for a solid black object.., then try the same shot with both in the same picture, it should become clear within a few shots. Nothing like real life to clarify theory.</p>

[gdw]

<p>Answer to number 1, absolutely not. Answer to number 2, it's called pilot error.</p>

<p>You point a reflected light meter toward white and use that exposure the white will be middle gray in your photograph, the middle gray will be dark gray and the shadows will be blocked.</p>

<p>You point a reflected light meter towrd black and use that exposure the black will be middle gray in your photograph, the middle gray will be light gray and the highlights will be blown.</p>

<p>You point the reflected light meter toward middle gray and use that exposure the middle gray will be middle gray in your photograph, the white will be white and the black will be black.</p>

<p>You are highly unlikely to ever have a perfectly centered histogram. There is no "correct" shape for a histogram.</p>

<p>The ONLY way a light meter will give you correct exposure is to: 1 point it at middle gray; 2 point it at an area that reflects the same amount of light as middle gray; 3 point it at an area the contains equal amounts of white and black that averages out to be middle gray; 4 apply a manual exposure compensation to make up for the lightness or darkness of the area that you are reading--the pilot part. The light meter doesn't know any of this. That is the reason that if you want correct exposure the photographer has to know all of this. That's what the histogram and the blinking blown highlight tool is for.</p>

<p>Until you learn how to use it keep the histogram as far to the right as possible without it begining to climb up the right side of the box. In post processing you can darken an image without losing quality. You almost always pick up noise and lose quailty if you have to lighten an image.</p>

<p>And when you don't have anything else to do waste all the time you want on 12% gray nonsense--it is absolutely meaningless in the real world. It's primary purpose is for technogeeks to band about to convience themselves that they are so terribly smart. Really has nothing to do with photography.</p>

[avishek_aiyar]

<p>Stephen and Gary,<br>

Thanks a lot for your replies. I am beginning to get clarity now.<br>

Gary....agree with your comment about 12% gray......not important in the real world. <br>

Thanks.<br>

Avi</p>

 

[rodeo_joe1]

<p>"So, I point the camera meter (in spot metering mode to keep things simple) at an object" - Ahh! but it has to be the right object with exactly 18% reflectance for that theory to work.</p>

<p>Somewhere along the way, 18% average reflectivity and a grey mid-tone got totally confused. They're actually completely separate things that even Ansel Adams apparently couldn't unpick.</p>

<p>The 18% reflectivity figure was originally derived by Kodak from a statistical study of the huge amount of (amateur) film passing through their processing facilities. I don't know the exact date of the study offhand, but it was a long time ago. Anyway, the result of this study showed that the <em>average snapshot</em> contained brightness values that averaged to a total reflectance of 18% of the incident light. This in due course became the standard reflectance value to which most lightmeters, including early TTL camera metering systems, were calibrated. It's <em>not</em> the same thing as a mid grey tone as I'll explain, but it<em> is</em> the reflectance value of a standard metering greycard.</p>

<p>So why is 18% not mid grey? To explain: The mid grey value of Adams's Zone system is a Zone V tone, which if you work it out is actually 3 zones below a near-pure white Zone VIII, or more scientifically 3 stops below a 100% Lambertian reflectance value. Now 3 stops down from 100% reflectance is actually 12.5 % reflectance, and this is half a stop different again from the 18% meter calibration reflectance. Sorry you Zone workers, but you can't have total equivalence between zones and stops and still insist that Zone V is the same as 18% reflection! In fact I'm being generous in giving Zone VIII a scientific value of 100% reflectance, by Adams's visualisation definition it's more like 90%, and that takes Zone V even lower.</p>

<p>This is all water under the bridge now and things have moved on. Our TTL meters no longer simply average the entire scene, and haven't done for many years. We also don't aspire to take the average snapshot that Kodak evaluated so long ago. Having said that, one of the most reliable metering methods is to ignore TTL readings and use a handheld incident meter. Always applying a little bit of thought and pre-visualisation of course.</p>

[johnw63]

<p>The only way to have a nice centered and balanced historgram is to shoot a scene that is nice and balanced with light, medium , and dark stuff AND that the medium stuff is just about middle grey. Otherwise the graph will be shifted one way or the other with spikes to show masses of one tone or another.</p>

<p> </p>

[orcama60]

<p>Avishek, any light meter including the one into the camera, is to tell the proper exposure for medium gray ( 18 % ) or zone 5. According to Bruce Barnbaum, it should not be called "light meter" but gray meter instead and I agree with him. If you want to be precise on your exposure, I do recommend to use the zone system, and the book, The Art of Photography by Bruce Barnbaum is the best book you can read about photography and it will explain you in details everything you need to know about your question or everything else. A quick statement, if you have black and white in a scene, and you expose for the black, the black will be in zone 5 - which is not black - ( practically medium gray - 18% ) but the white will be overexposed ( zone 10 aprox ). If you do the inverse, the white will be in zone 5 and the black in zone 1 or 2 ( completely black with no definition and / or texture - underexposed and the white will be too dark ). So you need to make two exposures and then calculate the average so you can have the perfect one. But for this, you need to get familiar with the zone system which I am trying to apply in every shot from now on ( after I read that book ) and I am getting excellent results than before. Remember, your camera only sees in one aperture and your eye in several apertures at the same time. What you see is not what the camera sees, so you need to learn how the camera sees and go from there using the zone system which is very simple and very useful and powerful. Buy the book, study it and you will understand a lot of stuff that would be impossible to explain with a few words. Your best friend in photography is the zone system and the histogram by the way. Buy the book and you will not regret. Happy learning and shooting !! </p>