View Full Version : 8x10 to 11x14 Lens Equivalencies

14-Oct-2011, 10:52
Hi all,

I know this info must be out there but I'm having a hard time finding it. If someone could post an answer or a link to where it might already be answered I'd appreciate it! I'm looking for a chart (or formula is it exists) to figure out rough equivalent focal lengths when going from 8x10 to 11x14. Thank you for the help!!

Also, can someone please give me an idea of the focal lenght ranges that equate to Wide, Normal, and Long lenses on 11x14? Thanks again!

E. von Hoegh
14-Oct-2011, 11:05
My favorite cheatsheet for AOV.

When I'm near a computer, that is.


Normal is the format diagonal. Say 420mm.

Oren Grad
14-Oct-2011, 11:12
11x14 is pretty easy, because its diagonal is very close to 10x the diagonal of a 35mm frame. So if you have well-developed intuitions as to what constitutes wide, normal and long for 35mm, you're all set.

FWIW, here's roughly how I think of 11x14:

Ultrawide: 210mm or shorter
Wide: 240-300mm
Semiwide: 350-360mm
Normal: 420-480mm
Long: Anything longer than 480mm

PS: The above is thinking in strictly mathematical terms. YMMV in terms of actual usage. Some users find that they perceive larger formats differently, so that the lens they'd choose to make a particular type of picture on different formats isn't necessarily the literal equivalent. An obvious example is that very long lenses are both harder to find and harder to use on 11x14 than on smaller formats. But many users find themselves shifting toward somewhat shorter focal lengths more generally.

14-Oct-2011, 11:14
You guys are awesome! Thank you!

Kerry L. Thalmann
14-Oct-2011, 12:26
All you need is the good, old Pythagorean Theorem (a^2 +b^2 = c^2, that's supposed to be a squared plus b squared equals c sqared) and you can do it yourself. Since you are only after relative comparsions, you can use the nominal film dimensions (for actual image area, I usually subtract 1/4" from the nominal dimensions to account for a 1/8" border around the edges of the film).

For 8x10, the format diagonal:

c = sqrt(8^2 + 10^2)
c = sqrt(164)
c = 12.8"

For 11x14, the format diagonal:

c = sqrt(11^2 + 14^2)
c = sqrt(317)
c = 17.8"

So, the ratio of the format diagonals is:

17.8/12.8 = 1.39

So, simply multiply any 8x10 focal length by 1.39 (or you can use 1.4 to simplify things of you don't have a calculator handy) to get the equivalent 11x14 focal length.

For example, a 300mm lens on 8x10 is equivelant to 300mm x 1.39 = 417mm on 11x14. So something like a 420mm Fujinon L would be equivalent to a 300mm "normal" lens on 8x10.


16-Oct-2011, 11:39
or...if you go by the square root of area ratios...that's another one that works out pretty much same as diagonal---liek for 4x5 which is FOUR times the area of 8x10, you get sqrt(4) = 2...so DOUBLE your 4x5 focal length for 8x10

from 8x10 to 11x14 you get ratio 11x14/8x10 = 154/80 = 1.925...take sqrt(1.925) and get 1.39...same as above!!!! hey hey!!!

but you can SEE that 4x5 is DOUBLE every dimension of 8x10..just like 16x20 equivalent will be DOUBLE the 8x10...11x14 is in the middle...not a factor of two, but a factor of sqrt(2)...just like it's one "stop" bigger...