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• 01-18-2012, 04:05 AM
RalphLambrecht
Quote:

Originally Posted by johnnywalker
I use the area of the print: double the area, double the light (twice as long an exposure, or 1 f-stop); quadruple the area = 4 X the exposure or two f-stops.

CHECK PAGE 511 FOR A GRAPHICAL METHOD!
• 01-20-2012, 05:11 AM
RalphLambrecht
Exposure Height and Exposure Correction Whenever the enlarger head is raised or lowered, and the negative magnification is changed, print exposure must be corrected. In ‘Tables and Templates’, you will find a chart to determine the magnification of your enlargement and another to estimate the exposure compensation required to accommodate a change in enlarger height. Strictly speaking, projected print exposures fail to follow the inverse-square law, but they follow the inverse square of the lens-to-paper distance if the paper reciprocity failure is ignored, in which case, a new theoretical exposure time (t2) is given by:
• 01-22-2012, 03:27 AM
RalphLambrecht
Quote:

Originally Posted by George Collier
Following the lens to paper measurement ratio method (for those of us without metering devices), wouldn't it be more accurate to measure the difference in the lamp to paper difference, which is really the material issue, and would be a different ratio, given the re-focusing necessary?

EnlargerHeightExpTimeEqn.eps
• 01-22-2012, 03:40 AM
RalphLambrecht
Quote:

Originally Posted by ic-racer

The equation I use is:

new_time = old_time x (new_M +1)^2 / (old_M+1)^2

where M = new magnification (print/neg) and m = old magnification (print/neg)

The exposure time factor would be:

Factor = (M + 1)^2 / (m + 1)^2

I wonder if Nicholas could supply the magnification in the above examples. I would like to see how this equation (converted from 'factor' to stops) compares to the metered values measured by Nicholas.

it's allhere
• 01-22-2012, 07:02 AM
kapro
Quote:

Originally Posted by Bruce Osgood
You're dealing with the Inverse Square Law. Moving a light source farther from the subject it diminishes proportionately to the distance.

I've worked out a simple Excel program that I simply plug in three variables and come up with New Elevated Exposure

New elevated exposure equals:

((New elevation / Old Elevation) ^2) X Original Exposure

Supposing your 8x10 elevation from lens to paper was 24 inches @ 16 seconds and your New elevation is 30
inches, then: ((24/30)^2) = 2.44
2.44 X Original Exposure (16 seconds) = 39.4 New Elevated Exposure.

Either my calculator doesn't work or I should return university diploma:

((New elevation / Old Elevation) ^2) X Original Exposure

Supposing your 8x10 elevation from lens to paper was 24 inches @ 16 seconds and your New elevation is 30
inches, then: ((30/24)^2) = (1,25)^2)=1,5625

1,56 X Original Exposure (16 seconds) = 25s New Elevated Exposure.
• 04-18-2014, 04:46 PM
RalphLambrecht
I cannot attach the file with the info you desire because it's too big for this forum.send me a private email to rlambrec@ymail.com and you shall receive a copy.:)
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