View Full Version : Film area

David Nash
24-Jun-2001, 09:50

Just a quick question. How many times greater is the surface area of a 5x4" or 10x8" negative compared with a 35mm neg? I know I've read the answer many times before, but I can't find the answer.


David Nash

Eric Boutilier-Brown
24-Jun-2001, 10:34
A 35mm neg is 1"x1.5" is size, leading to an area of 1.5 square inches. A 4x5 is (approximatly) 4"x5" which has an area of 20 square inches - 13 times larger. An 8x10 shee is 80 square inches, or 53 times larger in area.

David Haardt
24-Jun-2001, 15:54
To be even more exact, a 35mm negative is 24x36mm large (~ 0.94x1.42") thus has a size of 864 square millimetres. A 4x5" negative is 4x5" (102x127mm) large, thus has 12954 square millimeters. A 8x10" negative is 8x10" (203x254mm) large, thus has 51562 square millimeters. So a 4x5" negative is approximately 14.99x as large as a 35mm negative, a 8x10" approximately 59.68x. So 15x and 60x as good figures to have in mind ;-)


Doug Paramore
25-Jun-2001, 09:49
If you enlarge the 35mm neg, the usable image area is even less, due to the odd shape of the film. You lose a bit at each end when you go to the 8x10 shape, less so with 5x7. I ain't smart enough to figure out the specifics.


Stewart Ethier
25-Jun-2001, 18:21
The 4x5 image area is 3.75x4.75 (ANSI standard), or 17.8125 square inches, not 20 square inches. Multiply by 25.4^2 to get 11,491.9125 square millimeters. 35mm is, as mentioned above, 24x36 = 864 square millimeters. Therefore, the ratio is 13.30082465, or about 13.3.

For 8x10, the image area is 7.7x9.7 (ANSI standard), or 74.69 square inches, not 80 square inches. The corresponding ratio is 55.77199120, or about 55.8.

Pete Andrews
27-Jun-2001, 08:31
This business of film area is a complete red-herring. 5"x4" is NOT 13.75 times better than 35mm just because it has that much greater area. It's only 4 times better. The linear enlargement needed for a given print size is the only factor that should be taken into consideration when comparing formats.Knowing the relative areas of different film sizes is only of any possible use in calculating how much processing solution you need.