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

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• 05-27-2013, 02:33 PM
Stephen Benskin
The first example didn't use the reciprocal of the anitlog. You only used the anitlog.
• 05-27-2013, 02:38 PM
Michael R 1974
Whoa!!! :D

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

• 05-27-2013, 03:04 PM
Stephen Benskin
Quote:

Originally Posted by Michael R 1974
Whoa!!! :D

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
• 05-27-2013, 07:28 PM
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.
• 05-27-2013, 07:38 PM
Stephen Benskin
Quote:

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
• 05-27-2013, 08:03 PM
Michael R 1974
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?
• 05-27-2013, 08:19 PM
Stephen Benskin
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.

Attachment 69372
• 05-27-2013, 08:23 PM
Michael R 1974
Apologies I was thinking of I as intensity, not illuminance. Amazing :(
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