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  1. #11

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    Magnification is

    m = [d + squareroot(d^2 – 4df)]/2f – 1

    d = negative-to-print distance

    f = focal length

  2. #12
    MattKing's Avatar
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    If you need larger, one of the special purpose wide angle enlarger lenses will give that to you.
    Matt

    “Photography is a complex and fluid medium, and its many factors are not applied in simple sequence. Rather, the process may be likened to the art of the juggler in keeping many balls in the air at one time!”

    Ansel Adams, from the introduction to The Negative - The New Ansel Adams Photography Series / Book 2

  3. #13
    Jeff Bannow's Avatar
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    Thanks all.
    - Jeff (& sometimes Eva, too) - http://www.jeffbannow.com

  4. #14
    Martin Aislabie's Avatar
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    You may find that the limiting factor to maximum print size quickly becomes the Enlarger Head to Column/Wall distance

    Just something to look out for

    Martin

  5. #15
    Jeff Bannow's Avatar
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    Definitely something I'll need to watch for. I assume for a 40" print the head needs to be 20" from the wall?
    - Jeff (& sometimes Eva, too) - http://www.jeffbannow.com

  6. #16

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    Quote Originally Posted by Ian C View Post
    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”)
    So I am trying to figure out how to hit the long dimension of paper size 30"x40" and the short of 42" & 56" rolls in terms of neg to wall distance, enlarger will be horizontal, landscape orientation. For example, my mural lenses will be the 50mm Rodagon G for 35mm, 105mm Rodagon G for 6x6 and 150mm Apo N for 4x5. So what distance from the neg stage would I need to attain a monster 56"x70" enlargement from 4x5 via the 150mm lens or 42x42" from 6x6 with the 105mm lens, etc....

    Is there a different formula?

    I am just trying to figure out the size of room needed to do max enlargements on Ilford 30"x40", 42" & 56"roll paper, not counting the enormous room to soup the prints in. Seems like a good mural room should be at least 10'x10 feet, soup room *much* larger.

  7. #17

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    Without doing the math transformation to calculate distance, just put the formula into Excel.
    You can calculate the magnification you need to get to the print size. Print size / negative frame size = magnification.
    You have the focal length of the lens.
    Just start putting in different d (distance) values till the magnification value comes out to the number you need.
    It is a trial and error process, but you can narrow in on the distance pretty fast.

  8. #18

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    I did a google search and found this

    easel_to_carrier_distance = focal_length * (2 + 1/M + M)

    where M is the negative to print magnification.

    For a 7.2 X enlargement, the distance with a 300 mm lens would be 300 * 9.34 = 2801 mm = 110.3 inches.
    For a 240, the distance would be 240 * 9.34 = 2241 mm = 88.25 inches.

    Validate the formula by testing it on your enlarger as you have it setup.

  9. #19

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    So what distance from the neg stage would I need to attain a monster 56"x70" enlargement from 4x5 via the 150mm lens
    The 4” x 5” film image is about 95mm x 120mm as limited by the window in the film holder when the photo was made in the camera. You’d need 15X to do this. Realistically, you’ll likely have to project an image slightly larger than the intended print for proper coverage if you intend to make a borderless print. I’d plan on making the small dimension of the projection at least 57”. That requires 15.24X.

    The negative-to-print distance in this situation with a 150mm lens is 2596mm = 102.2”. The formula of post #18 gives the same value and is easy to use.
    Last edited by Ian C; 02-22-2014 at 10:11 AM. Click to view previous post history.

  10. #20

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    Quote Originally Posted by Ian C View Post
    Magnification is

    m = [d + squareroot(d^2 – 4df)]/2f – 1

    d = negative-to-print distance

    f = focal length
    The formula works assuming that front and rear nodal planes of the lens are the same. But I think for enlarging lenses they are close enough for the purpose. Camera lenses especially those for 35mm would not be close.

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