How To Calculate New Exposure Times for Different Size Enlargements in the Darkroom

August 05, 2016  •  3 Comments

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I received a great question from one of my Darkroom Newsletter Subscribers (thanks, Rey) and I thought I would create a blog post to share this information with others because I get this question from time to time.  

Rey wanted to know how to calculate the new exposure time when making a larger print in the darkroom.  As you can probably guess, as your light source in the enlarger gets farther away from the print, you will need more exposure time.  

There is no need to guess, or make test strips and waste a lot of time.  By using the inverse square law for light, the math will do the work for you.  

This is yet another reason why I love classic darkroom photography... I get to do cool math and figure things out!

I walk through an example below to illustrate the variables and math required to do this. You could very easily automate this with some spreadsheet software.

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Variables you need to use:
(A) Original exposure time
(B) Original dimension (long edge)
(C) New dimension (long edge)
(D) New exposure time

New Exposure Time (D) = ((C/B) x (C/B)) x A


New Exposure Time (D) = ((C / B)^2) X A (this version is faster because it calculates the square of C/B). 

Let’s say you made an 8x10 print at your exposure was 26 seconds
Now, you want to make the exact same print, but bigger, an 11x14 print.  

New Exposure Time = ((14/10) x (14/10)) x 26

If we break the math down it looks like this:
((1.4) x (1.4) x 26
(1.96) x 26 = 50.96 or 51 seconds 

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A possible point of confusion in your explanation(non-registered)
Tim, your formula works but ... (C/B)^2 will give the the square of C/B, which is correct. You might consider editing your text, getting rid of "this version is faster because it uses square root".
Using the formula for the square root would mean using the formula (C/B)^0.5 which would lead to an incorrect answer.
How about this "this version is faster because it calculates the square of C/B"??
Tim Layton Fine Art
Hi Rey, you are welcome. Thanks for the question and keep me posted on your progress.
Rey Castaneda(non-registered)
Thanks a lot Tim! I'll put this in practice in my next printing session. Great help!
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