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1. Originally Posted by Michael R 1974
Is it not reading the illuminance (intensity) of reflected light and then using Exposure = Illuminance x Time?

Edit: I feel like I'm missing something here...
As Ben said, some of us can get unnecessary too technical but if you feel like you're missing something then yes you do. While it's simple Exposure=Illumninance x Time but the meter reads the luminance( and not Illuminance) of reflected light and thus there is a relatively complex process to arrive at what will be the Illuminance at the film plane.

2. That isn't what I meant by missing something. I know the variables involved. I meant it might be a Benskin-esque "plain sight" trap.

3. Originally Posted by Stephen Benskin
Take a look at his flare graph. He starts it at 0.10 over Fb+f and has the amount of flare based from that point, when it is actually based off a stop below. A flare factor of 2 will not double the exposure at 0.10. His exposure testing also uses Zones and stopping down 4 stops.

What about the technique of shooting a step tablet with a camera, whether it is inside like with Shaffer's method or outside like WBM? The camera is the exposing device and the exposure meter determines the exposure. Wouldn't it be beneficial to understand what the film plane exposure should be (Hg)?

Let's say we are testing a 125 speed film. What should the exposure be at the speed point? What should the exposure be at the metered exposure point? What is the difference between the speed point constant and the exposure constant?
Stephen, for the speed point we know the exposure should be Hm in S=0.8/Hm when the ISO conditions are satisfied. Determing Hm is what I find tricky. Simplifying without ISO requirements, even when just targetting an arbitrary fixed density speed point with a step tablet test (in camera, out of camera, whatever), I find it somewhat less than straight forward.

4. Originally Posted by Michael R 1974
Stephen, for the speed point we know the exposure should be Hm in S=0.8/Hm when the ISO conditions are satisfied. Determing Hm is what I find tricky. Simplifying without ISO requirements, even when just targetting an arbitrary fixed density speed point with a step tablet test (in camera, out of camera, whatever), I find it somewhat less than straight forward.
I'm speaking more theoretically. Under the ISO conditions, what should Hm be if the film speed was 125? What would the metered exposure point be? And what is the difference between the two?

This isn't as hard or overly technical as some are suggesting, and by understanding a few basic rules of exposure, it's possible to evaluate the validity of a test method like Schaffer's or WBM. To start with, what are the exposure instructions for the WBM method? It's a simple job of comparing the expected results with the two known exposure points in the above question.

5. Books don't seem to cover this. The more technical books assume their readers are familiar with the values of the variables and don't bother to show examples. More general photography books usually don't attempt to cover it, so the opportunity of working with actual numbers associated with exposure falls through the cracks. I think this deprives people of a very useful tool for analysis or simply for a better understanding of the process. How can someone think to properly analyze something like the ISO speed standard when they don't have the necessary tools.

6. Isn't it just 10 times? So instead of .8 it's 8...

7. Originally Posted by Bill Burk
Isn't it just 10 times? So instead of .8 it's 8...
Yup. So what would Hm and Hg be for a 125 speed then?

8. So, the speed point would be..

125 = 0.8 / Hm
multiply both sides of equation by Hm... 125 Hm = 0.8
divide both sides of equation by 125... Hm = 0.8 / 125
Hm = 0.0064

Likewise, the metered point would be...
Hg = 0.064

Ten times... Seems extremely arbitrary or lucky to be such a round easy to remember number.

9. Originally Posted by Bill Burk
So, the speed point would be..

125 = 0.8 / Hm
multiply both sides of equation by Hm... 125 Hm = 0.8
divide both sides of equation by 125... Hm = 0.8 / 125
Hm = 0.0064

Likewise, the metered point would be...
Hg = 0.064

Ten times... Seems extremely arbitrary or lucky to be such a round easy to remember number.
So it takes 0.064 Lux.Sec to produce a density of .10 plus fog on the film?

10. When the rest of the conditions are met, but yes, that is the idea.

I should have labeled the exposure units...

metre-candela-seconds

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