8x10 lens questions

[steve_williams3]

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_jarvis1]

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.

[wilhelm]

That's not quite right -- an 8x10 isn't actually 8"x10".

[nathan_congdon]

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.

 

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Nathan

[david_goldfarb]

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]

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_poulsen1]

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.

 

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Assume that "I" is the image circle, "V" is length of the small side,

and "H" is the length of the long side.

 

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

 

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

 

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"^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.

 

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Switch the formulas if you want maximum shift.

 

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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_poulsen1]

To clarify, "I" is the diameter of the image circle.

[pete_andrews]

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]

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.

[nathan_congdon]

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.

 

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Nathan