Image Circle Coverage and Angle of View - how to calculate?

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Originally Posted by Emmanuel BIGLER

Is there a minimum practical image circle one should look for with a lens for 4x5 if one doesn't want to run out of movements for landcapes? Looking at the 120mm Schneider Apo Symmar L again – is a 189mm image circle considered sufficient or restrictive?

In fact, my understanding is that for decades photographers were happy with tessar-type lenses which cover in diameter their focal length plus 15%, not more.

For example I have offered to my brother-in-law a vintage Voigtländer 9x12 cm (84x114, diagonal = 141 mm) Avus plate camera, the lens is a 135mm skopar, a tessar-type, the camera has a vertical shift, the amount of shift actually permitted was quite small, with respect to modern standards, may be 10mm only !

However imagine that you want to shoot a conventional landscape with 1/3 of ground and 2/3 of skies or vice versa from a LF camera with a perfectly horizontal setup.
From this starting point where the horizon is just located in the middle of the image, since you are a proud owner of a brand-new LF camera with all possible movements, you do not want to tilt the whole camera up or down for framing 1/3 - 2/3 (this is what your photographically ignorant brother-in-law does routinely with his point 'n shoot camera, shame on him ;-) )

Hence you have to shift the standards of the camera, the amount of required shift is 1/6th of the vertical side of the image. And for a 150mm tessar in 4x5", in landscape orientation, 1/6th of the small side = about 16 mm this is already off-limits with respect to what the tessar lens can achieve without running out of image circle : a tessar-type covering 60°, with a f.l. of 150mm, the lens can be shifted by + or - 12.6 mm on the long side of the 4x5" film and 15.3 mm along the short side.

The consequence is that modern lens manufacturers consider "standard' view camera lenses with a minimum of 70°-75° of angle, covering their focal length + 40% to 50%. With such a standard 150 mm lens covering 70°, no problem for a perfect 1/3-2/3 composition by shifting only, an image including of course some vertical buildings you would not like to "see falling on you" ;-)

Regarding a lens with 189mm of image circle, this is 150 plus 26%, so you are ready for making a perfectly "vertically-correct" 1/3-2/3 landscape composition in 4x5" !
precisely, an image circle of 189 mm allows you, in theory on 4x5", to shift by + or -22 mm on the long side and + or - 26 on the short side, i.e. about one inch.

Now if you find that your excellent 120mm lens does not cover enough, you'll just have to switch to the smaller 9x12cm European film format : film holders are totally compatible with 4x5 holders !

Have fun ! Once you've started to use the vertical shift, you'll wonder how you could have lived so long with a camera offering no movements at all ;-)

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I'm learning!

And looking at the Schneider 120mm F/5.6 Symmar Apo (earlier version) this is I think 179mm movement, so even less leeway for movement on 4x5.

There are a couple of quick approximations I use to give myself an idea of how much tilting and shifting I can do with a particular lens.

For tilting, I subtract the field of view (which is related to focal length and the diameter of the format), from the coverage and divide by two to find the limits of movement on the diagonal, which is where those limits will be most critical. For example, if the field of view of a 120mm lens on 4x5 is about 64 degrees. Subtracted from the 74 degree coverage of a 120mm lens that has a 180mm image circle, I get 10 degrees to spare. If I tilted along the diagonal of the frame I could tilt 5 degrees one way or the other before losing coverage. Along horizontal or vertical axes, however, I can tilt a bit more--maybe 7 degrees each way. Again, that's an approximation, but it gives you an idea. Providing a horizontal focus plane with a camera three feet away from that plane (such as focusing on the ground with a camera height of 3 feet), and assuming a vertical film plane, a 120mm lens will need to be tilted 7.5 degrees--a lens with a 180mm image circle will be at its limits in this scenario.

Shifting is easier. I just subtract the diameter of the film from the image circle. 180-150= 30mm, which means I can shift the film on the diagonal 15mm towards one corner (from center) before running out of image. I can shift somewhat more for horizontal and vertical shifts.

Since the coverage and image circle are provided in most lens data, it's easy enough to compare those with the field of view for that focal length and the format diameter, respectively, just to get an idea.

In my first days of photography, I had my "aha!" moment when trying to enlarge a negative made with a Yashica TLR. The enlarger (in someone else's darkroom) had a 50mm lens for enlarging 35mm negatives, even though it was a 6x6 enlarger. I couldn't do it--the image appearing on the easel had no corners. That's when I first understood the difference between focal length, field of view, and coverage. And that's when I first understood why a 50mm lens for 35mm film was cheap while a 50mm for 6x6 was expensive.

Rick "who loves extreme wide-angle views and uses lenses at their limits" Denney

Hello there!

I hope it is okay to bump such an old thread but this seems to be the best opportunity to get an answer to a question I got : Is there a way to calculate the total possible image circle of a lens? Which parameters would you need?

I understood that this formula gives the image circle diameter / radius

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Originally Posted by Emmanuel BIGLER

image circle diameter = 2 * f * tan((total angle of view) /2)
image circle radius = f * tan((total angle of view) /2)

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So if I get the total possible angle of view, I could calculate the total possible image circle. To my understanding the total possible angle of view gets limited by either the Lens Body/Dimensions or the Entry/Exit Pupil Dimensions or both?.

I'd be very thankful if anybody got some hints. :)

Use algebra to solve the equation of Emmanuel BIGLER for angle of view.

http://www.mathsisfun.com/algebra/tr...n-cos-tan.html

I know how to use algebra but what I don't know is how to calculate the total possible angle of view :)

You cannot calculate the "total possible" angle of view (= the useful image circle) of a lens. Unless you are the optical designer of the lens. The lens image circle depends entirely on its optical design.

I guess you are referring to a lens for which you have no published data on image circle? In that case you cannot calculate the image circle, you have to measure it on an optical bench.

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Originally Posted by ic-racer

I guess you are referring to a lens for which you have no published data on image circle? In that case you cannot calculate the image circle, you have to measure it on an optical bench.

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Yes that is what I tried to do. Too bad. Even if I know about the principal planes and the entry and exit pupils? Okay then :) Thanks for your answer.

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Originally Posted by Tedman

I know about the principal planes and the entry and exit pupils

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Yes, in that case just feed the info into your optical design software and look at the Vignetting Plot.

Attachment 112025 Image from OpTaliX

Nope, the vignetting plot shows how illumination drops off with angle off axis. This has nothing to do with coverage as usually understood.

To find coverage one needs MTF plots or to measure resolution at a range of distance off axis. At a range of apertures.