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Thread: 90mm for 6x17

  1. #21
    SMBooth's Avatar
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    Jbbooks, Your right the camera will not have movement, or at least I hope not....
    I have used a 90mm on a 6x17 and quite like the format. My inexperience with LF lens shows because the concept that a 120mm len can have 75deg (APO-S)or 105deg (SA) is quite confusing.

  2. #22
    jbbooks's Avatar
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    Quote Originally Posted by SMBooth View Post
    Jbbooks, Your right the camera will not have movement, or at least I hope not....
    I have used a 90mm on a 6x17 and quite like the format. My inexperience with LF lens shows because the concept that a 120mm len can have 75deg (APO-S)or 105deg (SA) is quite confusing.
    Well, it is confusing.

    From going back and looking at what you have posted, it appears that you have been looking for a lens with the largest possible image circle for a given focal length. For a camera that does not have movements, that is not necessary. Any lens in a given focal length that has adequate coverage, a large enough image circle to cover the diagonal dimension of the film, will be usable in your camera. The larger image circle of a lens with greater coverage is not usable beyond what can be captured on your film, if you do not have or need movements.

    Also, I think there is a tendency, when a lens is chosen for a panoramic format, to not be aware of how much the added width of the format, by itself, will effect the increase in subject width desired without the need to go to a shorter lens. In fact, how the use of a lens that, in a non-panoramic format, was needed to increase subject width results in overkill, so to speak, with the greater aspect ratio of a panoramic format. Isn’t that the advantage of the panoramic format, that you get the increased horizontal field of view at the same or larger image size without an increase in the vertical field of view that just using a shorter lens would give you? In other words, more of the width of the subject at the same level of detail and without the added sky and foreground in an image of a city skyline or a range of mountains in a landscape, for example? That was the reason for my comment about considering the use of a lens with a focal length longer than 90mm. However, if you have tried it and like the 90mm, then that is what you ought to use. In any event, with no movements, any image circle of about 180mm will suffice, no matter what focal length you choose.

    And, by the way, larger maximum aperture size aggravates the problem of exposure fall off with a given lens. In fact, using a smaller aperture will reduce or, even, eliminate most of the problem and the manufacturer will generally state that getting the maximum benefit from the use of a centerfilter will require stopping down to some smaller than maximum aperture for the lens for which it is intended.

  3. #23
    SMBooth's Avatar
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    Well the theory I was following was that a lens with larger IC will exhibit less fall off at the edge for the same f stop.
    Your concept of a longer len is quite sound, and perhaps I will experiment with a 135mm I already have.

  4. #24
    michael9793's Avatar
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    I'm using caltar 90mm for my da yi 6x17 camera and it works great. I use tri-x and develop it in acufine developer at 1600 ASA at 7 mins development. 74deg f.
    "Capturing an image is only one step of the long chain of events to create a beautiful Photograph” See my updated website: mandersenphotography.com

  5. #25
    Ole
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    Quote Originally Posted by SMBooth View Post
    ... because the concept that a 120mm len can have 75deg (APO-S)or 105deg (SA) is quite confusing.
    It is confusing. So I'll try to see if I can unmuddy the waters, at the risk of making it worse..


    Image circle is a result of the construction of the lens, and the focal length of the lens. The lens does not care what film format it is used on, so we put the film sizes aside for later.

    Lenses of the same construction have the same image angle, which when projected on the focal plane gives a circle with a diameter depending on the focal length of the lens.

    For the mathematically minded, tan(Theta/2)=D/(2xF) , where Theta is the image angle, D the diameter of the image circle, and F the focal length.

    Coverage also increases as the aperture is stopped down, since the unavoidable "weaknesses" towards the edge of the image circle become less important - using a smaller area of the glass means that a lot of aberrations are drastically reduced.


    And next illumination:

    The physical construction of the lens - barrel, filter mount, et cetera - mean that at full aperture, some of the light that would otherwise illuminate the edges of the image circle are blocked on the way. This is vignetting, not to be confused with light fall-off.

    Light fall-off is caused by geometry only, and even without a lens you can see it.
    Because when seen from the side the (round) aperture becomes a narrow oval, the area becomes smaller with the cosine of the angle. Because the light exiting the aperture travels farther and spreads over a larger area, it also is reduced by the square of the cosine of the angle (between the lens axis and the line the light travels from the aperture). And at the moment I can't remember where the fourth cosine drop-off is from, but I'm sure I will remember shortly after I hit "submit"...

    A perfect pinhole would show fall-off proportional to cos^4(Phi), where Phi is the angle mentioned above - and Theta/2 is the limit of coverage and thus Phi.

    Wide-angle lenses of "modern" design - the ones with huge front elements - use a thick strong negative outer element to "tilt" the image of the aperture so that the effect of that particular light loss becomes less.
    In short, all reasonably well constructed lenses have fall-off somewhere between cos^3(Phi) and cos^4(Phi) - that's just basic optics, really.
    Note that the angle is very prominent in all the formulas here: That's why a "normal" lens doesn't show prominent light loss, whereas wide angle lenses do. It's simply that Phi becomes smaller with a longer focal length on the same film format!

    Now let's look at otherwise similar lenses - let's say a 90mm lens for 617 format. The diagonal of the format is 179mm, so I'll round it off to 180mm since I like to be able to do calculations like these in my head.

    The necessary image diameter is 180mm, which gives a radius of 90mm. D/2F equals 1 in this case, which gives us a Phi (or Theta/2 ) of 45 degrees, so we need a lens with at least 90 degrees angle of coverage.

    Cos^4(45) = 0.25
    Cos^3(45) = 0.35

    A simple lens, like indeed an old Angulon 90mm, will lose two full stops in the corners of a 617 frame. A lens of more modern construction (like a Super Angulon), will approach 1.5 stops loss, but no real lenses are quite as good as that. The real light loss will be somewhere between 1.5 and 2 stops for all 90mm WA lenses used on 617 format!

    Lens manufacturers give curves showing the light fall-off, usually with the cos^4(Phi) curve drawn in to show that their lenses do better than this. But you have to remember that the curves show 0-100 diameter, where 100 is the manufacturer's intended diameter of coverage (or the angle of coverage, depending on who made the diagram). Reading and comparing curves from different manufacturers is not always straightforward - you may end up hand-drawing the curves to your own scale to make sense of it.

    To summarise - all lenses have light fall-off, and depending on the type of lens it will always be between cos^3 and cos^4. The only reason this only becomes apparent on wide angle lenses is that only there does the angle get large enough to see the difference. At 30 degrees off axis, cos^4 gives 0.56 and cos^3 gives 0.65 - a relatively much smaller difference. About a stop in either case. Even closer to the axis, say at 10 degrees, we get 0.94 and 0.955 - a difference you wouldn't notice.
    -- Ole Tjugen, Luddite Elitist
    Norway

  6. #26
    michael9793's Avatar
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    Again Ole, you amaze me with your knowledge.
    "Capturing an image is only one step of the long chain of events to create a beautiful Photograph” See my updated website: mandersenphotography.com

  7. #27
    SMBooth's Avatar
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    Yes, I meant to thank Ole for his clear explanation.

  8. #28
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    What about strongly retrofocus lenses? For example, I have a 20/2.8 designed for a 35mm SLR and it exhibits very little falloff despite covering about 95 degrees. Assuming they have the strongly-negative front element (one less cos) and a rear node that's about as far from the film as a film-diagonal, would I be right in guessing that these have closer to cos^2(phi)*cos(phi/k) falloff, where k is the pupil ratio?

    If so, is there any particular reason other than cost that we don't see such lenses being manufactured and sold for large format? There are tele lenses sold for LF that mean you can use a shorter bellows, why not retrofocus lenses that allow you to go stupidly-wide without recessed lens plates and centre filters?

    Is it that smaller formats tend to have much larger exit pupils compared to the film size than larger formats, which further reduces the falloff? And that we wouldn't get that benefit with LF?

  9. #29
    David A. Goldfarb's Avatar
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    Retrofocus lenses do exhibit less falloff of illumination, and modern LF wide lenses tandem to be slightly retrofocus mainly to provide more working room, without incurring too much of a penalty in distortion or field curvature.
    flickr--http://www.flickr.com/photos/davidagoldfarb/
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  10. #30
    Ole
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    To make an LF lens retrofocus enough to make a significant difference to the evenness of illumination, it would be impractically big and heavy. Compare a retrofocus and a non-retrofocus 20mm lens for 35mm format, and remember that the mounting/barrel is a much greater part of the total weight in that format than in LF lenses.

    A good LF retrofocus WA lens could easily be made, but how many would like a 75m f:8 lens weighing 4 kg?
    -- Ole Tjugen, Luddite Elitist
    Norway

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