View Full Version : 8x10 lens questions

Steve Williams_812
8-Apr-2000, 08:40
I am considering a Kodak 190mm F6.3 Wide Field Ektar for use on 8x10. I know that at f 22 it has an image circle of about 320mm. Can anyone tell me how much movement this would allow me?

Chad Jarvis
8-Apr-2000, 09:33
Pythagoras would be disappointed. The diagonal of an 8x10 is roughly 325mm (that's the square root of the sum of 8 squared and 10 squared). Working with a 320mm image circle, you are already in a deficit.

8-Apr-2000, 11:32
That's not quite right -- an 8x10 isn't actually 8"x10".

Natha Congdon
8-Apr-2000, 16:00
I have always understood that 8 X 10 film and holders really measured 8 X 10, and that the diagonal was indeed 325 mm as cited above. (The math is definitely right to three digits, as I just checked it). All web sites I've seen which give info on available rise/fall/shift for different 8 X 10 lenses cite 325 mm as the minimum image circle. Maybe you're thinking about the fact that there is a very small incursion onto the film from the edges of the holder, but I believe the 8 X 10 film I buy, anyway, measures exactly 8 X 10". Certainly it prints perectly proportionally on 16 X 20 paper. I'd measure if I had a negative here in front of me.


David A. Goldfarb
8-Apr-2000, 17:40
I think Bill is referring to that small incursion of the edges of the film holder. That said, I think most users of the 190mm WF Ektar consider it suitable for 8x10 with no movement. I don't have one myself, but I have the 10" (254mm) WF Ektar, and it's a great lens.

chuck k
8-Apr-2000, 18:44
I've been using the Kodak Ektar f7.7 203mm lens on 8x10 and it works very well. I was told that it would not cover 8x10, but it does, and makes a nice longer lens for the 4x5.

neil poulsen
9-Apr-2000, 13:46
I derived the following formulae to calculate the maximum rise with a VxH (e.g. 4x5, 5x7, 8x10, etc.) negative in the landscape or portrait orientation. If interested, I can forward a derivation, which would include a diagram, etc.

Assume that "I" is the image circle, "V" is length of the small side, and "H" is the length of the long side.

MaxRiseLand = (SQRT(I^2-H^2) - V)/2

MaxRisePort = (SQRT(I^2-V^2) - H)/2

"^2" denotes the square and SQRT denotes the square-root. I believe these formulae will plug directly into Excel. If you calculate these by hand, take the squares before you subtract.

Switch the formulas if you want maximum shift.

If you find that the argument of the SQRT is negative, or if the resulting calculation is negative, then the format is too large for the negative size that you've specified.

neil poulsen
9-Apr-2000, 14:00
To clarify, "I" is the diameter of the image circle.

Pete Andrews
10-Apr-2000, 06:13
I was prompted to measure a sheet of 5"x4" film the other day, and guess what? It wasn't 5" by 4". It was actually 99.5mm by 125mm, well short of 5 by 4. So you 10" by 8" guys might be getting short-changed as well. Take a ruler to that film!

Ron Shaw
10-Apr-2000, 12:19
The image circle is 320mm when focused at infinity. How often do you do that? At 1:1, it is 640mm. At a typical hyperfocal setting, it may be a good deal larger than 320mm.

Natha Congdon
10-Apr-2000, 16:23
For all you camera nerds who just had to know (like me!): I measured both Ilford and Fuji 8 X 10 films yesterday, and they were very slightly different, but basically 7 7/8 X 9 7/8. This is counting the whole sheet, not allowing for incursion by the holder. The measured diagonal to cover the entire sheet was exactly 320 mm for both. Obviously, the point made by another poster above,that the image circle is far wider for closer targets, will be the most relevant fact in governing what you can get away with in lens coverage under normal circumstances.