nicholas_f._jones Posted December 13, 2002 Share Posted December 13, 2002 While I'm awaiting the arrival of parts for my new enlarger I'd like to calculate the maximum print size I can get given the following: 300mm enlarging lens, 8x10 negative, and maximum distance between film carrier and easel sitting on baseboard of ca. 57". Does anyone have an equation that will yield the maximum print size under these conditions? Link to comment Share on other sites More sharing options...
dave_schneider Posted December 13, 2002 Share Posted December 13, 2002 You will be able to get an enlargement of 18x23 or a magnification of about 2.3. The lens will be extended 430mm from the film plane and the easel will be 990mm from the nodal point of the lens. This will put the negative 56 inches above the easel. You can use Bob Wheeler's Vade Mecum programs to calculate all these numbers. I have written a program for my Palm Pilot which I keep with me in the darkroom to calculate changes in magnification based on height and the resulting exposure changes when changing magnifications. Read Bob Wheeler's notes on the programs, it will give you all the information you need even if you don't use his software. Link to comment Share on other sites More sharing options...
leonard_evens Posted December 14, 2002 Share Posted December 14, 2002 Usually this would be expressed in terms of the distance from the lens to the easel, not the film to the easel. If u is the distance from the lens to the easel, f is the focal length of the lens, and E is the enlargement factor, then u/f = 1 + E or E = u/f - 1 Of course you have to measure u and f in the same units. If u is 57 inches, then to get it in mm, multiply by 25.4 to get 1448 mm. So E = 1448/300 - 1 = 3.83. Your 8 x 10 negative image area is probably slightly smaller than 8 x 10 inches, so you will have to measure it and multiply by 3.83 to find the print dimensions. Now if you really mean to use the film to carrier distance, it is a little more complicated. It requires solving a quadratic equation. I would have to check, but I got the following from some quick calculation. Let S be the distance from the film plane to the easel. Then E = (S - 2f + sqrt(s^2 - 4Sf))/(2f) Putting 1448 mm in for S, yields an enlargement factor of 2.41. Perhaps someone will check my algebra. Link to comment Share on other sites More sharing options...
nicholas_f._jones Posted December 17, 2002 Author Share Posted December 17, 2002 Dave and Leonard, Thanks so much for your timely and informed responses. The printout of this thread is going into my technique file this morning. Leonard, I did mean film carrier-to-easel distance because under present circumstances that's the easier measurement to take when the chassis is racked up to its maximum height. Now that I know for sure that I can get 20x24" I can make an informed decision about what size easel to purchase and can plan for future printing and mounting. Link to comment Share on other sites More sharing options...
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