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Please can someone explain how the Angle of View (in degrees) relates to the Image Circle Coverage (in mm) as quoted in View Camera Magazine lens tables etc?
I realise that the term 'angle of view' has different meaning when used by view camera users compared with 35mm users, and understand the 35mm users version (2*arctan (format diameter/focal length).
However I am struggling with the relationship for view cameras where I see 72 degrees coverage etc quoted.
Thanks for any help on this.
Use focal length f as your adjacent leg, and half the coverage degrees as the adjacent angle phi, then image circle c = 2 f tan(phi).
They hardly relate at all. The coverage circle is determined by the design of the lens itself. Angle of view is a completely different thing; it's a function of focal length and format size.Quote:
Originally Posted by newmoon2night
BTW, that's not "format diameter" in your equation, it's "d" which is the length of film in the direction you want to calculate angle of view. This is required because most film formats are rectangular and therefore have a different angle of view for the long side vs. the short side. The correct equation is alpha = 2arctan (d/2f), and it works for all formats. Don't forget to use consistent units of measure for "d" and "f."
Hello !
In large format (LF) only a part of the total image circle is used for a given setting of shift+tilts.
When all movements are set to zero there is not difference with the situation encountered with smaller formats with no movements.
However the angle of coverage listed in LF lens specs is the total possible angle of view either with the biggest film surface with no movements that can be illuminated by the lens, or when exploring the total image circle with a smaller format by shifting/tilting the front or the rear standard for perpective and depth of field control.
the formulae are the same, namely
image circle diameter = 2 * f * tan((total angle of view) /2)
image circle radius = f * tan((total angle of view) /2)
total angle of view = 2 * atan( (image circle diameter) / (2 * f))
total angle of view = 2 * atan( (image circle radius) / f)
The actual angle of view like for 35 mm and MF cameras with no movements is limited by the film format diameter and not by the total image circle delivered by the lens. In fact even in 35mm and MF, the actual image circle is slightly bigger than the format diameter. For example when Hasseblad introduced the FlexBody, they had to disclose the total image circle of their Zeiss lens products, something that had nevered been presented before ;)
I do not know if there is a precise terminology for total angle of view and the actual (film format limited) angle of view, I hope this is clear.
Example : for 90° of total angle of view, plus or minus 45° with respect to the optical axis, the image circle diameter is equal to twice the focal length.
There was a 75mm Biogon lens covering actually 165mm, i.e. slightly more than 152mm, the diameter pf the 4x5" image format (94x120 mm actually). Hence the total image circle of the 75mm biogon was not 90° but 95°
A lens with 72° of image circle will cover in diameter its focal length plus 45%, a 70° covers the focal length plus 40%, a 75° covers the focal length plus 50%.
Summary
total angle | example | image diameter in f units
60° tessar@f/22 1.15 f
70 old symmar 1.40 f
72° apo sironar N 1.45 f
apo-symmar
75° apo-symmar-L 1.50 f
apo-sironar-S
80° old-angulon 1.68 f
90° Biogon 2.0 f
95° Biogon (actual) 2.18 f
100° super-angulon F/8 2.38 f
102° grandagon-N F/6.8 2.47 f
105° super symmar XL 2.6 f
super angulon F/5.6
grandagon-N F/4.5
110° super angulon XL 2.85 f
apo-grandagon 45/55
120° apo grandagon 35 3.46 f
super angulon xl 38
135° old Goerz hypergon 4.8 f !!!!!
One additional comment.
The above formulas apply when focused at infinity. When focused close-up, the angle of coverage or angle of view doesn't change, but the diameters of the corresponding circles are larger. You must multiply by the ratio of the bellows extension to the focal length.
Thanks for your advice and putting me right on this formula. I had been using 76.5 as D in my calculation, which is half the 4x5 diagonal (I think). If I use half of 127mm I will get the angle horizontally then?Quote:
Originally Posted by Bruce Watson
If I'm not mistaken this angle of view, as quoted by most 35mm format lens manufacturers, is calculated on the diagonal measurement, but most illustrations are on the basis that one is looking down from above (i.e. at the long side) which isn't the same ° measurement.
Having seen the explanations I realise now that image circle coverage and angle of coverage don't relate, and this is where I was getting completely confused ... I just couldn't see the connection!
Thanks again.
Thanks Jim for the formula, Emanuel for the extensive workings, and for Leonard for explaining why lens manufacturers generally give two measurements, of which one is at ∞
However my calculation skills are lacking, and when I enter the formula in Google to use their calculator I am getting meaningless results. I was using the following data as an example:
It’s the Schneider 120mm F/5.6 Apo Symmar L, which is quoted as having 75° angle of coverage at ∞, and 189mm image circle at F/22.
So presumably it should be 2*120*tan(75/2) should give an answer of 183mm? The answer I am getting is -48.12. Where am I going wrong anyone?!
Reading Emanuel’s post, am I right in understanding that a 120mm lens needs a bigger angle of coverage to provide movements than say a 210mm lens, in order to provide reasonable movements for 4x5?
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?
Finally I want to make a composing frame for a 180mm lens (and possibly a second one for a 120mm lens), out of 2" x 1" wood - basically a simple small piece of wood with a hole cut in it in proportion to 4x5 and with the hole cut at an angle fanning out (the angle needs to be 46° I think!), which I can hold up about an inch away (a little more than a thumb's width!) from my eye. Do I need to adjust this angle because of the one inch distance from my eye, or will the difference be minimal?
newmoon, you forgot to convert degrees to radians.
Thank youQuote:
Originally Posted by Dan Fromm
Answer 184.158! That's what I was expecting!
I hadn't picked up on anyone mentioning radians in previous posts, but presumably the post referring to Phi was the clue?
Unfortunately I'm not an engineer, my school days are long gone, and my work does not involve radians or degrees!
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 ;-)
I'm learning!Quote:
Originally Posted by Emmanuel BIGLER
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
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?.Quote:
Originally Posted by Emmanuel BIGLER
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.
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.Quote:
Originally Posted by ic-racer
Yes, in that case just feed the info into your optical design software and look at the Vignetting Plot.Quote:
Originally Posted by Tedman
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.
The diagram posted in #19 is a Vignetting Plot for Mechanical Vignetting, not Natural (cos) Vignetting. The curves look different.
Here is a clearer example:
Attachment 112063
Hey thanks for your answers so far...
Am I right that in your example the effective pupil area is the diameter of the Circle of Coverage?
Would I really need to use an Optics Design software to find out about that?
My plan was to draw the lens with it's principal planes, pupil positions, image plane for infinity focus etc and draw some rays. The first part(drawing the lens) is no problem but I don't know what rays define the (maximum) Image Circle. I'll try to upload an example later today.
Tedman, given an unknown lens with unknown design all you can do to find the diameter of the circle of acceptable (to you) sharpness is to shoot film and (a) examine the negatives at the same magnification as the final print or (b) print to the size you're aiming for and examine the print. You can't calculate what you want with no data but you can always measure. There's no magic involved, just ask the lens what it can do. And ask it with film, don't look at ground glass.
I'm late coming to this thread but I had to do some digging around to figure out what the hell you gentlemen were getting at. For those of us who are mathematically challenged, I made a simplified explanation that I could reference before purchasing new lenses. There will be more information like this coming to my new website in the future, just in case someone finds it useful and could use help in other areas of large format photography.
HOW TO DETERMINE THE ANGLE OF VIEW FOR VINTAGE LENSES
The Angle of View changes based on the size of the film. So the first step is find the diagonal of the film you're using.
Diagonal = √(width²+height²)
Next you might want to convert this to millimeters (perhaps because it is more accurate?). Then you use the Angle of View Formula below.
2arctan(diagonal of film/(2*focal length))
The last step is converting radians to degrees, then round to the nearest whole number.
RESOURCES:
1. Diagonal calculator: http://www.mathopenref.com/rectanglediagonals.html
Attachment 152542
The computational procedure you laid out gives the smallest angle covered that will cover a format given focal length. What does that have to do with anything?
Oh, and by the way, I don't know where you got your table of film sizes by format but it is wrong. You may have found a table of plate sizes. The usable area of a piece of sheet film is smaller than the usable area of a plate of the same nominal size.
Dan, I found the chart on a blog located at http://jbhphoto.com/blog/2011/01/29/film-diagonal/.
As for the maths involved, I'm not completely confident, other than I used it to calculate known angles of view for lenses that I had specs on and the numbers matched. It has nothing to do with image circle or coverage, only the angle of view. I use this to determine how "wide", "normal" or "telephoto" a lens is compared to other lenses I own. Personally, I like a certain angle for portraiture and like to stick tightly to that range to maintain a certain look no matter what camera system I'm using.
If you have any additional insight regarding image coverage, or if my formula is incorrect, I am very interested to learn more. Also, if I need to edit my post above as to not send folks astray I would be obliged to do so.
Christoph, that something is posted on a site doesn't make it true.
As was pointed out earlier in this discussion, there are many definitions of coverage. Lens makers' coverage claims are often more narrower than users report.
Movements are very important to photographers who use view cameras. Your formulas, which start from the film's diagonal, give no information about a lens/format combination will allow movements.
The above formula is an easier to read, more clearly written version of what Bruce Watson was saying on page one back in May of 2009. The formula is not meant to give any information regarding movements and image circle coverage. It provides a way to compare lenses and their focal length rendering across all film formats, which is why I found this thread to begin with.
Quote:
Originally Posted by Bruce Watson
Christoph, the following Chart...
Is always an excellent Reference (for 'Angle of View'):
Large Format Angle of View -- Format Comparison Chart.
http://lensn2shutter.com/angleofviewchart.html
--
Thank-you!
Best regards, -Tim.