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Thread: Approximate ISO-EI speed test with in-camera (LF) contacting

1. The first example didn't use the reciprocal of the anitlog. You only used the anitlog.

2. Whoa!!!

I think I must have read Stephen's post (#4) incorrectly and put the brackets in the wrong place or something? Is this formula correct or not:

H= 1/10^D * I

H = I / 10^D

or is it

H = 10^D * I

3. Originally Posted by Michael R 1974
Whoa!!!

I think I must have read Stephen's post (#4) incorrectly and put the brackets in the wrong place or something? Is this formula correct or not:

H= 1/10^D * I

H = I / 10^D

or is it

H = 10^D * I

H= 1/10^D * I

4. What the heck is going on here?

H = 0.08

1/10^0.3 = 0.5

0.08 = 0.5 * I

I = 0.08 / 0.5

I = 0.16

This can't be. I'm losing my mind.

5. Originally Posted by Michael R 1974
What the heck is going on here?

H = 0.08

1/10^0.3 = 0.5

0.08 = 0.5 * I

I = 0.08 / 0.5

I = 0.16

This can't be. I'm losing my mind.
H = 0.08 * 0.5 = 0.04

6. So my first substitution was wrong (H=0.08). I wasn't using the equation properly.

Need to rethink this.

I'm not understanding what this formula does. H is exposure in lux-s. Then on the right side of the equation we have I in lux, and then the other term which is the reciprocal anti-log of D. Somehow multiplying I (lux) by the reciprocal anti-log of D gets us to lux-s? How?

7. Both are in lux sec. Don't over think it. Light falling on the step tablet, the density of the step tablet, and light that passed through.

Use this for a reference.

8. Apologies I was thinking of I as intensity, not illuminance. Amazing

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