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How to determine max enlarging size for my setup?
I'm going to build a drop table for my enlarger, and would like to figure out how big I can go to determine how big to make the table.
I'm using a Durst L1200, and will be enlarging mostly 6x6 and 4x5. If I know the maximum distance from the lens to the floor, and the focal length of the lens, can I calculate the biggest print size?
Lens:
6x6 - 80mm or 90mm
4x5 - 135mm or 150mm
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Raise the head to maximum height and give us the negative-to-easel distance, focal length, and format.
With this data calculating the maximum print size to within a millimeter or two is simple.
Lens-to-easel distance is imprecise. Negative-to-easel distance is easy to measure accurately.
If you give us the numbers it will take a few seconds to calculate the answer.
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I can do that. I'll measure this afternoon when I get home.
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Even easier. Don't enlarge more than 3x the negative. 
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 Originally Posted by Robert Hall
Even easier. Don't enlarge more than 3x the negative.
That would be no fun! I suppose I could cut up my roll of 42" paper into 8x10s or something ....
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No, no, no. You've got it all wrong. What you need is a 42" negative!
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Originally Posted by Robert Hall
No, no, no. You've got it all wrong. What you need is a 42" negative!
Now that would be impressive!
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Originally Posted by Ian C
Raise the head to maximum height and give us the negative-to-easel distance, focal length, and format.
With this data calculating the maximum print size to within a millimeter or two is simple.
Lens-to-easel distance is imprecise. Negative-to-easel distance is easy to measure accurately.
If you give us the numbers it will take a few seconds to calculate the answer.
Maximum from negative carrier to floor is about 65". Thanks.
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For any format you’ll always get the greatest magnification with the shortest focal length lens that’s specifically designed for the format.
In order to calculate the largest projection possible you must know the projection distance (negative-to-easel), the dimensions of the window opening in the negative carrier, and calculate the magnification m that the negative-to-easel distance gives for the focal length of the enlarging lens.
The largest projection is m times the dimensions of the carrier’s widow size.
I’ll use the dimensions in my Omega negative carriers as typical. Omega 6 x 6cm carriers have 55.5mm x 55.5mm opening and the 4” x 5” carrier window is 92.5mm x 120.1mm.
If you placed a 1” tall easel on the floor you have a 64” = 1625.6mm negative-to-print distance.
For the stated projecting distance you’ll get the following focal length, magnification, and projection size combinations:
6 x 6cm Negative
80mm, 18.3X, 1013.7mm x 1013.7mm (39.9” x 39.9”)
90mm, 16X, 888mm x 888mm (35” x 35”)
4” x 5” Negative
135mm, 9.9X, 919.5mm x 1193.9mm (36.2” x 47”)
150mm, 8.7X, 806.8mm x 1047.6mm (31.8” x 41.2”)
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Thanks Ian. What's the formula you used here?
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