Calculate lens coverage?

-chrille- · Mar 14, 2024

For example. I got a 210mm lens from Fujinon. Coverage is 240mm at f/22 and covering angle is 59 deegres. How do I calculate bellows draw (or focus distance) this lens will cover a 8x10 sheet film?

How does aperture affect the lens coverage?

Also wonder what lens coverage actually means. For example my lens with 240mm coverage, when used on a 8x10 will it show a black vignetting circle or is it a soft degradation of the image toward the corners of the film?

koraks · Mar 14, 2024

Have you looked at this: https://en.wikipedia.org/wiki/Angle_of_view_(photography)
In particular, this diagram is usful:

Forget about 'object' and consider the angle between the red lines as the angle of view. You can use the formulas in the section I took that illustration from to determine coverage at different distances between the lens and the film plane (=bellows draw).

-chrille- said:
For example my lens with 240mm coverage, when used on a 8x10 will it show a black vignetting circle or is it a soft degradation of the image toward the corners of the film?

Depends on the lens design/topology. Most modern lenses in my experience give a hard transition between exposed area and total darkness.

ic-racer · Mar 14, 2024

Coverage is 240mm. Nothing to calculate.

Ian Grant · Mar 14, 2024

Coverage is not the same the circle of illumination, it is also dependent on aperture so it's best to look up the manufacturers figures.

The coverage in terms of the sharp image circle reduces as you open up. A lens I've been testing a 1902 Dallmeyer Stigmatic No 5 9" f6 is listed as covering 8"x5" at f6, and 12"x10" at f16. My 12" f6.8 Dagor cover 10"x8" at f6.8, 12"x10" at f16, and 17"x14" at f45.

Another factor is the design and the maximum aperture, so Plasmats will be sharper at wider apertures than a Tessar or type, followed my Triplets.
Carl Zeiss Jena recommended the f6.3 Tessars for critical work, f4.5 general use, f3.5 and later f2.9 for low light/high speed work.

Ian

Chan Tran · Mar 14, 2024

coverage is 240mm which doesn't cover the 8x10.

grain elevator · Mar 14, 2024

I think the op wants to know at what focus distance it will cover 8x10. At some distance, it inevitably will, the given figures are probably at infinity. I don't know how to calculate that distance, maybe someone else does.

Ian C · Mar 14, 2024

I agree with all the answers prior to this one.

Mr. Grant states in post #4 the essential fact that is often overlooked, misunderstood, or both. A lens projects a circle of image. But only the central part is photographically useable wide open.

This is generally given at infinity focus where the lens is at its closest usable position from the image plane. Nikon also expresses these as coverage angles—very useful in calculations. This gives the smallest diameter for the image circle at the given aperture.

Some makers of LF lenses, particularly Nikon, state both the wide-open and the infinity-focus image circle diameters. The usable wide-open-aperture image circle is much smaller than the circle of illumination. The region outside of the usable circle has substandard definition, and in the outermost regions, has significant light falloff, which is worse at the edges.

The photographically useable portion of the circle increases as the aperture is closed to a smaller opening. Many lenses reach a maximum-diameter infinity-focus image circle at f/22. Closing to a smaller opening than f/22 doesn’t meaningfully increase the size of the image circle.

Smaller-format wide-angle lenses, such as 75 mm and shorter on, say, the 4” x 5” format reach their maximum image circle at f/16. Examples are the Nikkor SW 75 mm and 65 mm SW wide-angle lenses.

Answer to Question: An 8” x 10” holder has a format diagonal of about 313.8 mm. For a lens that has a 59º coverage angle at f/22, the image distance (second nodal point to image plane), is 277.3 mm. The image circle will just cover the diagonals (the corners) at a subject distance of 865 mm (subject plane to first nodal point of the lens).

So, unless you’re doing some very close work, this lens is not useful on the 8” x 10” format.

koraks · Mar 14, 2024

Ian C said:
A lens projects a circle of image. But only the central part is photographically useable wide open.

Yes, but it has to be said that modern designs such as the Symmar-S have tend to be very usable close to the edge of the image circle. Older, simpler designs tend to go wonky long before that - which is explored artistically by many.

xkaes · Mar 14, 2024

I know there is a formula for this -- minimal extension to fill the format -- despite this going off on a tangent. I know this is not exactly it, but it's something like the focal length of the lens multiplied by the diagonal of the film format and then divided by the image circle at infinity. FL * FD / IC

reddesert · Mar 14, 2024

If the angle of view remains constant as the focus distance changes, then the diameter of coverage goes up as:

coverage = (1 + e/f) * coverage_at_infinity

where e = extension past infinity, f = focal length. (Edit to add: This isn't complicated optics, it's just similar triangles. The diameter of coverage is proportional to the lens distance from the film.)
The magnification m = e/f. Also, m = image_distance / subject_distance.

Thus, if you have a 210 mm lens, with coverage at infinity = 240mm, and you want to cover 8x10, which is a film diagonal of about 315mm, you need extension e:

315 mm = (1 + e / 210) * 240

so extension e = 66 mm. Your magnification is m = 66 / 210 = 0.31. The image distance is 210+66 = 276mm, the object distance = image_distance / m = 276 / 0.31 = 890 mm, so the distance from film to subject is approx 1170 mm.

Note that since the magnification is about 1/3 and you have a piece of 8x10 film, the subject area will be about 24x30". This is almost a head and torso portrait, pretty large for such a small camera to subject distance. A 210mm lens is fairly wide for 8x10 and because the film is so large, even working at magnification 1/3 is a large physical area (Compared to say 35mm format, where to get magnification 1/3 you'd probably be using a macro or close-up lens).

Bronson Dugnutt · Mar 15, 2024

Here's a question that's in the same ballpark: what is the effective f-stop of the lens below when used at infinity?

Pentax 67 SMC 300mm front group used as a barrel lens in a speed graphic board. The complete lens had an aperture stop behind what is now the rear element and an additional negative element behind that.

Is it as simple as (170mm / 76mm) = 2.2? Or does the odd rear node and protrusion of the lens into the camera necessitate some additional considerations? Being a telephoto design, my naïve intuition is that any additional light afforded to the film plane by the protruding rear section is being 'eaten' by the optics thus rendering the simple calculation valid.

koraks · Mar 15, 2024

Bronson Dugnutt said:
what is the effective f-stop of the lens below when used at infinity?

AFAIK for calculating the numerical aperture you work with the diameter of the aperture as projected through the front of the lens. You can take a ruler or caliper, face the lens head-on and guesstimate the aperture diameter that way. This is generally a little less than the diameter of the front lens element/group.

Focus setting/distance doesn't matter; this is only useful to determine focal length by focusing at infinity. Neither does the length of the lens as such matter, etc.

At least, according to my (limited) understanding of the matter.

Bronson Dugnutt said:
Is it as simple as (170mm / 76mm) = 2.2?

That sounds about right; since you know what lens you've got, it's a little easier. You know it used to be a 300/4 lens, originally. This means the projected aperture diameter was 300/4 = 75mm - indeed quite close to the 76mm you found, but note it's a smidgeon less. If the front group alone gives a focal length of 170mm, the wide-open numerical aperture will be 170/75 = f/2.27
A sanity check proves this should be about right since you're using the same front element but the focal length is now roughly half that of the original lens, and indeed your new combination is roughly a stop faster.

Alan Edward Klein · Mar 15, 2024

ic-racer said:
Coverage is 240mm. Nothing to calculate.

What does that mean?

koraks · Mar 15, 2024

Alan Edward Klein said:
What does that mean?

Well, OP asked for the coverage of the lens. He also stated the image circle is 240mm. That leaves very little to calculate.

However, I interpreted the question like this:

grain elevator said:
I think the op wants to know at what focus distance it will cover 8x10.

Ian C · Mar 15, 2024

The lens projects a fixed coverage angle at a given aperture. At infinity focus, the lens is as close as it can be to the image plane and still produce a well-defined image. The diameter of the circle of good definition is the infinity-focus image-circle diameter.

When we move the lens farther from the image plane to focus on a subject at a finite distance, the image circle increases in diameter. This is what is asked in post #1. Essentially, “At what subject distance will the lens fully cover the diagonals of the film?”

Of course, he asked in terms of “bellows extension. “That sort of muddies the water insofar as “bellows extension” can be interpreted in more than one way. Is it the total amount the bellows are opened from the fully closed position? Or is it the distance the bellows are opened beyond the infinity focus position for the given lens? I’ve read or heard other interpretations as well. We need to state our questions and answers in a way that eliminates ambiguity. As it’s used in conversations on Photrio, “bellows extension” is not generally defined in a mutually-agreed-upon manner.

That’s why I add defining phrases, such as “second nodal point to image plane” and “subject plane to first nodal point of the lens” which are measurable and prevent confusion.

My definition of "bellows extension" is the distance from the second nodal point to image plane at any subject distance. In other words, The Image Distance. At infinity focus, the extension, which is the image distance = focal length of the lens.

Mark J · Mar 15, 2024

OK, so the lens will continue to cover 59° when focused down, within a reasonable range of mag.
You can estimate the lens coverage by using the lens formula, 1/f = 1/u + 1/v where 'f' is the focal length.
This is strictly accurate for a 'thin' lens but is close enough for this problem.
8x10" diagonal is 312.5 mm according to Schneider.

We can just lob some numbers into the equation to show how this works -
for 3000mm object distance ( we can call this 'u' ) we can solve for 'v' in the equation above and get 225.8mm
The image circle will increase in proportion to 225.8/210 if the coverage angle stays the same.
Hence in this case you get about 258mm circle.
This isn't enough...

I have done a bit more re-arrangement of the equation, working back from the required increase in image circle, and can estimate that you need an object to lens distance of about 885mm.
This will give you a lens to image distance of 275.3mm and an image circle of 312mm
The mag is 1:3.2
The bellows extension is 275.3-210 = 65.3mm

I'm not sure this will be within a 'reasonable range of mag' for a Tessar, but it's a rough guide.

xkaes · Mar 15, 2024

66mm VS 65.3mm. Should I lose sleep?

Doremus Scudder · Mar 15, 2024

Alan Edward Klein said:
What does that mean?

It means simply that the poster didn't fully understand the question. Indeed, at infinity focus, the coverage of the lens in question is a circle with a diameter of 240mm, as per the manufacturer's specifications.

What needs calculation is how that coverage grows when the lens is extended further from the film and focus moves from infinity to closer objects. At some distance, since the lens projects a cone of light, the resulting circle will cover 8x10. The question is, at what distance. That has been more than adequately answered by reddesert and others.

Don't let this confuse you.

Doremus

Mark J · Mar 15, 2024

Yes, I missed the one post ( from Reddesert ) where he'd calculated it !
So... I just gave a different explanation.... there is no meaningful difference between 66 and 65.3, agreed.

Alan Edward Klein · Mar 15, 2024

koraks said:
Well, OP asked for the coverage of the lens. He also stated the image circle is 240mm. That leaves very little to calculate.

However, I interpreted the question like this:

Doremus Scudder said:
It means simply that the poster didn't fully understand the question. Indeed, at infinity focus, the coverage of the lens in question is a circle with a diameter of 240mm, as per the manufacturer's specifications.

What needs calculation is how that coverage grows when the lens is extended further from the film and focus moves from infinity to closer objects. At some distance, since the lens projects a cone of light, the resulting circle will cover 8x10. The question is, at what distance. That has been more than adequately answered by reddesert and others.

Don't let this confuse you.

Doremus

What does an image circle of 240mm mean and when? That's the equivalent of 9.45 inches.

Also, what is meant on the charts when it says image circle is let's say 212mm @ f22?

Large Format lenses specs

koraks · Mar 15, 2024

Alan Edward Klein said:
What does an image circle of 240mm mean and when?

A projected image with a diameter of 240mm.

Alan Edward Klein said:
Also, what is meant on the charts when it says image circle is let's say 212mm @ f22?

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

xkaes · Mar 15, 2024

Most lens tests are based with the lens at infinity at f22, but there are some exceptions, and most are the circle of of good definition, but that's the manufacturer's definition.

reddesert · Mar 15, 2024

Bronson Dugnutt said:
Here's a question that's in the same ballpark: what is the effective f-stop of the lens below when used at infinity?

Pentax 67 SMC 300mm front group used as a barrel lens in a speed graphic board. The complete lens had an aperture stop behind what is now the rear element and an additional negative element behind that.

View attachment 365463

Is it as simple as (170mm / 76mm) = 2.2? Or does the odd rear node and protrusion of the lens into the camera necessitate some additional considerations? Being a telephoto design, my naïve intuition is that any additional light afforded to the film plane by the protruding rear section is being 'eaten' by the optics thus rendering the simple calculation valid.

The f-number is the focal length divided by the apparent diameter of the entrance aperture. Your numbers for the bellows draw and diameter of the front element are probably pretty close for that lens, although there are lens designs where there would be differences. For ex, for most shorter-than-telephoto lenses, the entrance aperture is smaller than the entire front element. For highly asymmetric lenses, sometimes the flange-focal distance is fairly different from the focal length.

You can actually measure these numbers more accurately. For the entrance aperture, just lay a ruler across the front of the lens and measure the apparent diameter of the clear aperture as seen from the front.

For the focal length, you can get pretty close by measuring how far the lens has to extend past infinity to focus on an object of known distance or magnification ratio, and using simple lens equations. This can require a little care (and math) to get an accurate result. For an example, one way to do this is to measure the bellows draw at infinity, and to measure it at 1:1 magnification. The extension past infinity e = f*m, so measuring the extension needed to get to 1:1 gives you the accurate focal length.

The reason that focal length can be different from the bellows draw at infinity, is that the bellows draw is just the mechanical flange to focal distance, while what you want is the distance of the rear principal plane to the film, and the principal plane can be offset from the mechanical center of the lens. This is true for lenses that have a strong meniscus shape, for example.

The mechanical configuration of the lens - how far the elements protrude forward/back, and so on - doesn't affect the f-number.

Mark J · Mar 15, 2024

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 ?

-chrille- · Mar 16, 2024

Thank you very much for the intresting and comprehensive answers. Some even done the calculation which makes me feel a little spoiled

Calculate lens coverage?

-chrille-

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