Enlargement sizes and aspect ratios for the numerically impaired OK, so Im not a math Geek! I understand aspect ratios, and how to calculate them. I understand that for instance 35mm film would have an aspect ratio of 1:1.5 (35mm frame is 24mm in height x 36mm in width, which then equates to a ratio of 36/24 = 1.5). I understand that an 8x10 sheet of paper would have a aspect ratio of 1.25, (obviously 1.5 is larger than 1.25), and if you wanted to fill the entire 8" x 10" paper with an image, some cropping will necessarily occur at one end or the other, or both. Taking that same 35mm negative and this time using a 16x20 sheet of paper (aspect ratio of again 1.25) cropping would occur here too. Now, what I dont understand is how the following is calculated . 35mm film has a format ratio of 1:1.5, so a full-frame print on 16"x20" paper can measure no more than 13.33"x20", otherwise the image would be significantly cropped. And this If we take a full frame 35mm negative and enlarge it to make an 8x10" print, we enlarge the long dimension of the negative to match the long dimension of the paper and it results in a print measuring 6.6" wide How are those precise sizes calculated (13.33 and 6.6)? Is there some formula to calculate these sizes and to know what the actual width and height of the enlarged print would be? Thanks in advance

It's easier to think of 135 size film as 2:3 ratio (whole numbers)... or the short side is .6667x (or 2/3 times) that of the long side... or the long side is 1.5x (or 1 1/2 times) the short side. So... 6.6667x1.5=10... or 10x.6667=6.6667... or 13.3333x1.5= 20... or 20x.6667=13.3333.

It also helps if you convert both measurements into the same units. Either think of 35mm film as being 1" x 1.5" (it's close enough for most purposes) or think of the paper as being: 8x10 = 20cm x 25cm (200mm x 250mm), and 16x20 = 40cm x 50cm (400mm x 500mm). You calculate the magnification by dividing the length of long edge of the paper by the length of the long edge of the negative. You then determine how long the short dimension of the printed image is by multiplying the length of the short dimension of the negative by the magnification. Using the "inches" version: magnification = 10.0" /1.5" = 6.6666 length of short dimension of printed image = 1.0" x 6.6666= 6.666" Hope this helps. Matt

Algebra is your friend. Basic formula in this case would be: x=long side length of print/aspect ratio of negative. So in the 8x10 example x=10/1.5 or 6.66666. In the other example x=20/1.5 or 13.333333.

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You know my math professors told me the same thing way back in college but... instead of attending class I cut, and hung out at the track So where does that leave me now well Im still tearing up losing tickets, Im still wasting valuable paper due to poor cropping/composing and I still suck at math! Anyway, really, thanks one and all for all the information, it was much appreciated! (Now, anyone got a tip for the 3rd at Belmont? )

Or, you could just crank on your enlarger till it looks "about right". Crop a little, so what? No calculator required.

I do it the same as tim k. Whatever looks right for that particular neg and that paper size I go with.

Ira, Im not sure I really understood your question. However, if your problem is the difference in proportion between the 35mm frame and the standard paper sizes, than here are my comments: This difference, indeed turns out in either wasting paper (if you keep the proportion of the 35mm frame when enlarging), or in cropping (if you fill all the paper with image). Usually the cropping alternative is choused, because it also helps in correcting the composition. But, if you require a paper that matches (or gets closer) to the proportions of the 35mm frame, than cut the standard paper formats in two. The 8x10 becomes 2x(8x5), the 11x14 becomes 2x(11x7), the 12x16 becomes 2x(12x8), and so one. Now, check the proportions: 8/5=1.6; 11/7=1.57; 12/8=1.5; etc. Almost no paper waste. I hope I answered your question, phenix

............ and Exactly, I agree, that's what I've been doing for years. But somewhere I came across those statements which I included in my original post and was going nuts trying to figure out how those precise numbers were calculated. I figured (no pun intended) that if I actually knew in advance,what the actual dimensions would be, I could save time (and paper) by more careful composition and time and paper in the darkroom. Once again, thanks to all!

Here's what I think about "standard print sizes". They SUCK!! Print what feels right and don't worry about it!!

Yes, I agree And now I see that it gets worse with other formats. Now that I have had a Math Enlightenment, I figure that I am wasting about 20% of negative image area due to cropping when I am working with 6x6 and 8x10 and a little less than 10% when I shoot 645 and enlarge on 8x10. Yup paper sizes suck! (Wow... now that I no longer feel numerically impaired, maybe I can attempt to fix the economy... maybe I can screw around with those numbers... yeah right )

In the interest of saving the planet, going green and all, you just don't want to be going around wasting your negatives. You need a 4x5. The math is so simple even I can do it. 4x5=8x10=no waste.

Plenty of people do NOT crop and just don't fill the entire paper. Nothing wrong with that at all and a healthy amount of print border almost always looks good.

I stopped fretting over cropping issues years ago. I print what pleases me, I crop in my viewfinder. BTW- I get plenty of test strips from printing full frame 6x6 by trimming my paper before printing. Rick