The first example didn't use the reciprocal of the anitlog. You only used the anitlog.
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The first example didn't use the reciprocal of the anitlog. You only used the anitlog.
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
Sorry about this, guys.
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.
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?
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
Apologies I was thinking of I as intensity, not illuminance. Amazing :(