Calculate lens coverage?

[Alan Edward Klein]

A projected image with a diameter of 240mm.



That this is the diameter of the projected image circle at f/22 with focus at infinity.

So a lens that has a circle of 240mm cannot be used on 8x10 cameras because 240mm equals 9.45 inches? That's under 10 inches.

 

[koraks]

So a lens that has a circle of 240mm cannot be used on 8x10 cameras because 240mm equals 9.45 inches? That's under 10 inches.

Very simply put, yes - sort of.

Suitability of the format depends on the image diagonal. The diagonal of 8x10" = 12.8" or 325mm. So you need a lens that projects a (usable) image circle of at least 325 in order to obtain a corner-to-corner usable image on 8x10".

There are some caveats. Of course, wat constitutes usable will vary between people and situations. Some may feel it's OK if the corners go black since they'll crop it out or exhibit it for artistic effect. This makes the requirements a little more lax.

Furthermore, the image circle is the smallest at infinity focus (it's actually even smaller beyond infinity, but since nothing will be sharp, it's a moot point). As you focus closer, the lens extends further away from the image plane and hence, the image circle grows. For this reason, a lens with an image circle of 240mm at infinity can be perfectly usable for close-up works on 8x10". In fact, it's a very effective way to make (extreme) closeups on large format without running out of bellows draw, and somewhat limiting the effects of light falloff as a result of the extension factor. But this utility is limited to the focus range where the image circle will be adequate.

So it depends on a couple of things.

 

[Ian Grant]

A projected image with a diameter of 240mm.



That this is the diameter of the projected image circle at f/22 with focus at infinity.

The Image Circle figures published by manufacturers are not the same as the projected image diameter, this us the circle of illumination. It is important to differentiate between the two.

Perhaps an example. A 105mm lens like a Ross Xpres has a Circle of Illumination of over 160mm diameter at Infinity, so fully illuminates the screen of a 5x4 camera. however you can see the sharpness is dropping off at the corners at full aperture. a 5x4 film test shows it's still not sharp at the edges and corners at f16. But this lens is designed for a 6x9 Ensign Selfix 820.

Emmet Gowin made use of the Circle of Illumination in a series of images he made in the early 1970s.

So the Image Circle figures are the diameter of acceptable sharpness at a chosen aperture determined by the manufacturer.

Ian

 

[koraks]

@Ian Grant yes, see the first page of this thread. Thanks for adding another example; I think we're also all aware of how esp. wet plate photographers exploit the extremes of the image circles of 'vintage' lenses, which, arguably, have very little usable image circle if you apply modern criteria to it. However, as I stated on page 1, in my experience, modern LF lenses tend to be optically quite good close to the edges of their image circles. I'm referring to lenses like Symmar-S, Nikkor-W, Super Angulons etc.

 

[Ian C]

Here’s a simple way to calculate the extension (image distance) that just covers the diagonal of the format opening in the film holder.

At a given aperture and knowing:

I = manufacturer's published infinity coverage diameter at the specified aperture

f = focal length of lens

D = diagonal to be covered

We can calculate the required image distance (which I call the extension E) as

E = f*D/I

This is the same formula given in post #9 using different symbols.

Example using the values given in post #10, I = 240 mm, f = 210 mm, D = 315 mm,

E = 210 mm*315 mm/240 mm = 275.65 mm

Which agrees with the value calculated in post #10.

In post 9, the author commented, “I know this is not exactly it, . . .”

In fact, that is “exactly it.”

This relies on nothing more complicated than two similar triangles.

 

[xkaes]

In post 9, the author commented, “I know this is not exactly it, . . .”

In fact, that is “exactly it.”

THANKS, it's nice to know that my memory isn't as bad as some people want me to believe.

 

[reddesert]

I wouldn't measure its focal delta at 1:1, a telephoto like this will have very poor correction, try something like 2m object.
This lens will not be very well corrected without the rear element, it must have a fair bit of field curvature and some distortion ?

It's a good point that this lens especially is likely to do poorly at 1:1, and its aberrations could make determining the best focus and applying simple lens equations difficult. Measuring the extension past infinity at magnification say 1:3 or 1:4 and applying e=f*m to infer the focal length would be better.

I did run some test numbers before posting on the inferring-focal-length-from-extension method. The issue is that when you are focused further away, like 1:10 for a ~200mm focal length and 2000mm subject distance, you have to measure the extension and/or the magnification pretty accurately to get the focal length without a significant error - not hard on an optical bench with a precision stage, harder with typical photographic gear.

 

[xkaes]

I wouldn't measure its focal delta at 1:1, a telephoto like this will have very poor correction, try something like 2m object.
This lens will not be very well corrected without the rear element, it must have a fair bit of field curvature and some distortion ?

None of the Fujinon 210mm lenses were telephoto lenses. They had flange focal lengths from 189mm to 209mm.

http://www.subclub.org/fujinon/byfl.htm

 

[Mark J]

Yes, something like 1:4 or 1:5 would be a good compromise.
In industry, you'd feed in a known angle off-axis , and measure the image height ; then find 'f' from h = f.tan(theta)
(Usually not going beyond about 25% of the full field, to avoid distortion coming in.)
However this takes a bit more set-up and/or equipment.

 

[Mark J]

None of the Fujinon 210mm lenses were telephoto lenses. They had flange focal lengths from 189mm to 209mm.

http://www.subclub.org/fujinon/byfl.htm

I know.
This was a reply to Bronson's query re. the Pentax 300 Tele.

 

[Bronson Dugnutt]

I wouldn't measure its focal delta at 1:1, a telephoto like this will have very poor correction, try something like 2m object.
This lens will not be very well corrected without the rear element, it must have a fair bit of field curvature and some distortion ?

Yep, loads of distortion. I posted some example images taken on 4x5, 8x10 film & paper, and 120 roll film 6x8 back in this thread. It is more useful in a lens board than in its original housing despite being a mess optically. I have a series of negative glass elements that will change the effective focal length (and coverage) when applied to the rear of the lens. It doesn't do much to correct flaws, but is fun to play with.

It is unreasonably fast at f/~2.2 even with 1/1000s on the Speed Graphic but allows for neat stuff like hand holding paper negative exposures. I don't think I'll bother with an ND filter