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  1. #11
    arigram's Avatar
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    Ok, I am a bit confused, but I'll give it a shot.
    The problem with trying it right now is that my current setup doesn't allow me such a magnification. The roof of the darkroom is pretty low, so if I wanted such enlargements I would need to construct a lower or a variable height desk.
    aristotelis grammatikakis
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  2. #12

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    A competent scientist or engineer knows when they understand a problem well enough to use theory & equations rather than experiment. Do you think that Boeing and Airbus design airplanes by building full scale models and testing to see whether they break, beefing up parts if they break, or reducing weight if they don't break?

    Nick, 127 and GMA have all given correct equations. Nick appears to have divided by 8 when his equation called by multiplying by 8.

    Here is another equation addressing the problem (derivable from the others): The total distance from negative to paper is d_tot = f * (m+1)^2 / m. Here "^2" means squared. This equation neglects the separation of the principal planes of the lens, which is likely to be only a couple of mm and so is reasonable to neglect.

    If your 6x6 cm negative is actually 56 mm square, then m is 480 mm / 56 mm = 8.6 (rounding up), giving d_tot = 80 mm * (8.6 + 1)^2 / 8.6 = 80 mm * 92.2 / 8.6 = 860 mm.

    As GMA pointed out, even if you aim to print full frame, you someday will probably wish to crop a negative. If we guess a 20% crop (45x45 mm portion of the negative), then m becomes 10.7. As Nick pointed out, the focal length of the lens might be slightly different from the nominal value. A likely difference is about 1 mm, so we can allow for this with margin by using f = 82 mm. (The focal length is constant and does not vary as the bellows is racked out.)

    Using these refined values for m and f, d_tot = 82 mm * (10.7 + 1) ^2 / 10.7 = 1050 mm. This is the distance from the negative to the baseboard.

    The largest uncertainty in these calculations is the largest crop that might be wanted, which determines what value should be used for m. The uncertainty is also present in an experimental determination of the distance -- if you pick some magnifcation for the experiment, you might later find that it doesn't allow for a crop that you want to do.

    An alternative solution to the problem of getting larger size prints would be to use a wide-angle enlarging lens. These type of lenses cover a given format with a shorter than normal focal length. They cost more and don't show up frequently on the used market. Light falloff in the corners will be worse. For example, the 60 mm f4 Rodagon-WA is rated by Rodenstock to cover 6x6 cm. This lens (for m=10.7) woudl reduce d_tot to 800 mm.

  3. #13

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    Okay Nick, I've been trying to figure the "wrong math" of .14cm to get 9cm. Is the formula you listed originally wrong or am I misreading sumthin'?

  4. #14

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    Nevermind, I see. Divide first, then multiply by f. 9/8 then x80 = 9cm

  5. #15
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    Right on! Joe.

  6. #16
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    arugran -
    I suggest that you cut the top of the shelf where the enlarger stands so that it can be lifted off. Then, place rails in a "breadboard" type arrangement so that a board can be slid in on top of the rails to lower the enlarger. You can make a series of these "rails". The ceiling of my darkroom is too low to make 16x20 prints from a 4x5 negative using my 150mm enlargiing lens; I find this arrangement works well. Then, if you have to "crop" the image but still want a 16x20 print, just lower the board to the next lower pair of rails (and so on).

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