Switch to English Language Passer en langue française Omschakelen naar Nederlandse Taal Wechseln Sie zu deutschen Sprache Passa alla lingua italiana
Members: 71,867   Posts: 1,583,269   Online: 1120
      
Results 1 to 4 of 4
  1. #1
    gainer's Avatar
    Join Date
    Sep 2002
    Posts
    3,726
    Images
    2

    Hyperfocal distance

    If you thought my previous article was strange, wait til you read this. The hyperfocal distance of any lens is 2000 times the diameter of its aperture in millimeters, or 1 meter for each millimeter of lens aperture. This is based on an allowable blur circle of 1/2000 millimeter per millimeter of lens focal length. As you probably know, when the camera is focused at the hyperfocal distance, anything from half that distance to infinity is in acceptably sharp focus.

    We do not often calculate depth of field or hyperfocal distance, relying on tables computed by someone else. If we were to do so, we might be surprised to find that hyperfocal distance and depth of field depend only on the diameter of the aperture and the allowable blur, or circle of
    confusion.

    The 1941 edition of the War Department Technical Manual of Basic Photography says that the circle of confusion, for critical work, should not exceed 1/2000 of the focal length of the lens. I think they got that from Kodak. I think that is the number we use most of the time.
    This manual says that "the hyperfocal distance of any lens is found by multiplying the focal length squared by the reciprocal of the circle of confusion and dividing by the f/ value. This result divided by 12 will be the H. F. D. in feet." The manual assumes focal length and circle of confusion to be measured in inches. If we use millimeters, we divide the result by 1000 to get H. F. D. in meters.
    In equation form:
    hfd = [(F ^ 2 / c) / (f)]/ 1000
    where F is focal length, c is diameter of circle of confusion (or allowablable blur circle) and f is the f-stop number, defined as F/D, D being the diameter of the lens aperture.
    Substituting F/2000 for c and F/D for f, we have:
    hfd = [F * F * 2000/F)/ (F/D)]/1000

    F cancels out and the hyperfocal distance in meters for any camera is twice the diameter of the aperture in mm. Any camera lens with a 4 mm aperture will have an 8 meter hfd.
    Naturally, the exposure time for a 50 mm lens with a 4 mm aperture ( f/12.5 ) will be much less than for a 300 mm lens with a 4 mm aperture ( f/75 ), but the hyperfocal distances are the same.

    OK, so we don't know the aperture diameter. We do know the f-stop and the focal length. We can calculate hfd = 2D = 2F/f in a flash with our pocket calculator. No tables with columns for each lens we own, or involved calculations. I even know some old timers who remember how to do long
    division with pencil and paper.

  2. #2
    Sean's Avatar
    Join Date
    Aug 2002
    Location
    New Zealand
    Shooter
    Multi Format
    Posts
    8,593
    Blog Entries
    7
    Images
    15
    comments from the previous article system:

    By gainer - 03:02 AM, 09-14-2006 Rating: None
    My opening statement should have stated that every millimeter of aperture adds 2 meters of hyperfocal distance and one meter to the closest focus point.

    By Jordan - 03:57 AM, 09-14-2006 Rating: None
    Interesting little rule of thumb, and great for lenses without DOF scales. Thanks, Pat.

    By MichaelBriggs - 10:36 AM, 09-14-2006 Rating: None
    There are two approaches as to what diameter to use for circle of confusion in depth of field (and also depth of focus) equations. One is to use a diameter for the circle of confusion that is proprotional to the focal length of the lens, the other is to use a fixed diameter. In the second approach the value of the diameter depends on the format because of the enlargement factor to get a typical size print depends on the size of the format.
    The first approach, which you have used, is based on the assumption that the prints will be viewed at the correct perspective distance. This approach has the advantage of simplying the equations -- as you have noted, the focal length cancels out in some equations. It has the disadvantage that this assumption does not match how people view prints. People do not prints at varying distances as a function of the focal length taking lens!
    I think the other approach, of using a fixed diameter, by format, makes more sense. Typically, recent books seems to use this approach. There is an excellent derivation of depth of field in "Applied Photographic Optics" by Sidney Ray. He has a paragraph discussing the approach of using a circle of confusion that is a fraction of the focal length, describing it as old idea that causes confusion: "It gives values for depth of field which imply different enlargements for different focal lengths." (I.e., the issue of viewing at the center of persepctive). "The idea of C = f/1000 is now deprecated, and instead the value of C is taken as constant for a range of lenses for a given format."

    By gainer - 03:51 PM, 09-14-2006 Rating: None
    The only mathematical complication introduced by making the circle of confusion variable is to change the 2000 to whatever factor you think proper for the lens you have at hand. If you think 1345 appropriate, then the 2 would be replaced by 1.345. The hyperfocal distance would be less and the depth of field would be greater, but the picture would be fuzzier.
    The idea that you can know ahead of time how any person will view a picture, let alone assign a viewing scheme to all viewers for any one picture, is at best beside the point. If you have some knowledge about the best circle of confusion that can be produced by a given lens, by all means use it, but even that will not force the viewer to use it. Nothing in all the literature I have read, including Hardy and Perrin, "The Principles of Optics" includes viewpoint in the discussion of depth of field. Hyperfocal distance is a property of the lens. The equation for hyperfocal distance according to Hardy and Perrin is:
    hfd = DF / c
    c being diameter of circle of confusion.
    If you have knowledge of c for the lens at hand and its focal length you can calculate F/c easily enough and multiply it by D.
    The whole business of depth of field is rather arbitrary. We know there is only one point of sharpest focus, and we know that the image of a point, if we could obtain a true point source, would not be a point. By moving the plane of focus and plotting the resulting diameter of the image we can get a plot from which we can by mathematical or graphical means find the minimum diameter and the variation of diameter with movement of the point source. This has nothing to do with the eyesight of the experimenter, who is equipped with a microscope and whatever other tools are needed. Where does the eyesight or viewpoint of a potential viewer of a print enter?
    The way to use subjective judgement of in or out of focus is to show a projected image, let the observer change the focus until the image is judged just out of focus in both directions, recording the decision points. You will be hard pressed to find a way to use those statistics to do what you want to do.
    At the same time, we know there will be some lenses that cannot meet the F/2000 requirement. Would we use them for anything but soft focus portraits? In fact, the depth of field of such lenses will seem greater than for fine lenses.

    By jstraw - 09:25 PM, 09-14-2006 Rating: None
    Any chance for a non-math-major's guide for figuring out HFD for a given lens?
    Is it reasonable to gather that for any given lens, there could be a mark on the focusing rails that indicates that lens, focused at it's HFD?

    By gainer - 10:12 PM, 09-14-2006 Rating: None
    I can't think of anything easier than dividing the focal length (in millimeters) by the f# you want to use and multiplying the result by 2. Focus the camera on something that many meters away and everything from half that distance on out will be in acceptably sharp focus.
    I will work on a formula for marking the focusing rail. It's not hard to do, I just haven't had enough coffee yet today.

    By jstraw - 01:49 AM, 09-15-2006 Rating: None
    I guess the mark on the rail foa given lens would move with each change of aperture, eh?
    I came across this: http://www.nikonians.org/html/resou...yperfocal2.html

    By MichaelBriggs - 04:57 AM, 09-15-2006 Rating: None
    "The only mathematical complication introduced by making the circle of confusion variable is to change the 2000 to whatever factor you think proper for the lens you have at hand." If you do this (as I suggest, since I think a constant diameter is more appropriate), then the derivation of your article falls apart. The focal length of the lens no longer cancels, so the hyperfocal distance is not simply a function of the aperture diameter.

    "Nothing in all the literature I have read, including Hardy and Perrin, "The Principles of Optics" includes viewpoint in the discussion of depth of field." A derivation of depth of field MUST include a discussion of print viewing distance, for this is what converts blur circle diameter into visual angular resolution. Either you have read incomplete derivations, or missed the implications. To quote Sidney Ray: "Depth of field is therefore determined by the geometry of the taking, enlarging and viewing conditions related to the circle of confusion standard adopted (Figure 22.3)" (page 217 of the 3rd ed. of Applied Photographic Optics).
    "Where does the eyesight or viewpoint of a potential viewer of a print enter?" In converting from linear diameter of the blur circle to angular units. to compare to the resolution of human vision. If you don't understand this, you don't understand the deriviation of depth of field.
    "The idea that you can know ahead of time how any person will view a picture, let alone assign a viewing scheme to all viewers for any one picture, is at best beside the point." It's true that no single assumption will always be correctly describe how a viewer will view a print. But your equation will be more accurate if your assumption matches how people typically view prints. I think people virtually never adjust their viewing distance based on the focal length of the taking distance, which is what the center of perspective idea assumes, which is the idea that leads to the diameter of the circle of confusion being a fraction of the focal length. I think non-photographers typically view small prints at reading distance, and large prints at about their diagonal. Photographers do this, and also view all prints closely.

    By gainer - 05:09 AM, 09-15-2006 Rating: None
    Yes. These marks would get pretty crowded very soon. You could contrive separate focusing and hfd scales to be chenged when you change lenses, but I think after a while you would say "Oh the heck with it" and resort to the calculator. Once you calculate the hfd for the lens and aperture you want to use, you could place a stick that far from the lensboard and focus on the stick with the lens wide open. Then close down to the planned aperture.
    If you the hfd in feet, it's 166.7* F/f-stop and use F in inches. For a 12 inch lens,
    hfd = 2000 / f-stop.
    At f-64, hfd = 31.25 feet
    If you are happy with F/1000 for the blur circle,
    hfd = 1000 / f-stop and the hfd is 15.6 feet.

    By gainer - 05:14 PM, 09-15-2006 Rating: None
    Michael,
    To paraphrase, or maybe quote, Aristotle, the argument from authority is the weakest argument.
    I understand the concept of circle of confusion. It will exist in any optical system . If you want authority, Hardy and Perrin are as high as you need go when it comes to optics, especially photographic optics. There is a circle of confusion that is a property of the camera's lens that is not determined by that of the viewer of any print eventually made from a negative from that camera. That circle limits the maximum definition one can get at the point of best focus. Ultimately, it is limited by diffraction. The best any camera can do does not depend on its user or the viewer of any photos from it. In the best of all worlds, the circle of confusion depends only on the diameter of the lens aperture.
    The aperture that is used to estimate mathematically the depth of field is the aperture of the camera lens. The object of the calculation is to make a photo that shows in a flat plane all that the photographer wants to show to a hypothetical viewer of a scene which the photographer saw with a scanning, autofocusing, recording system with a field of view of about one degree. The viewer has the same organic system, but it is not controlled by the photographer. The photo cannot show all that the photographer saw. The only control the photographer has over what the viewer will see is-- well, you know the rest.
    The question we seem to be facing is similar to what audio enthusiasts face in the beginning times of high fidelity. We knew the bandwith of the average human hearing, and for some time we argued that it was only necessary to present that bandwidth for the listener to experience concert hall presence. The problem is that filtering the sound, either intentionally or by necessity, to fit human hearing doubled the filtration because the human listener was still in the loop.
    If we start with the idea that the human can only see so much and provide only that much in our photos, we are cheating the viewer. You have seen, I'm sure, equations that represent the effects of optical systems in series. That is why I contend that the circle of confusion to use in estimating depth of field and hyperfocal distance is that of the camera, not of any potential viewer. Furthermore, the camera will do what it will do. No equation I have rattling around in my ignorant brain can make it do otherwise. It is best that the equations I use do closely describe what the camera wants to do, not what some hypothetical viewer desires. When you come right down to it, no matter what you calculate or how you calculate it, what you see on the ground glass using a magnifier and what you do in the darkroom determine what you choose to put on the film. The guardian angels of photographers do the rest.

  3. #3
    Sean's Avatar
    Join Date
    Aug 2002
    Location
    New Zealand
    Shooter
    Multi Format
    Posts
    8,593
    Blog Entries
    7
    Images
    15
    comments from the previous article system:

    By jstraw - 06:01 PM, 09-15-2006 Rating: None
    HFD for a 210mm lens at f32:
    210/32*2*3.28 (for feet) = 43.5'
    Do I understand this correctly?

    By gainer - 08:57 PM, 09-15-2006 Rating: None
    Good enough for government work, Jay. You could stop down to 45 or so and still be more than twice as good as the average human eye. At 45 your hfd would be about 18 feet, so your estimated depth of field would be from about 9 ft to infinity.

    By jstraw - 09:32 PM, 09-15-2006 Rating: None
    Yes, and at f-45 my aperture would be 4.66mm too (as close to 4mm as I can get), right?
    I still need to know if I got the math right before I apply it to every lens and aperture!

    By Helen B - 11:54 PM, 09-15-2006 Rating: None
    Jay,
    The criterion that Patrick is using to determine hyperfocal distance is one of constant subject definition, not one of constant apparent print sharpness. Both criteria are valid in their own way, but I think that it is important to understand the difference.
    If you are using 4x5, then a c-o-c of 210/2000 mm (ie 0.105 mm) would be regarded as about right. Applying the f/2000 rule to a 90 mm lens at f/32 would give you a hyperfocal distance of 18.5 feet, with a c-o-c of 0.045 mm. If a c-o-c of 0.1 mm is acceptable (which it was for 210 mm lens), then the hyperfocal distance drops to 8.3 ft.
    The opposite happens with longer lenses - the f/2000 criterion will give you negatives of increasing blurriness, and a 2x enlargement of a picture shot with a 300 mm lens using that criterion would look unsharp at normal viewing distances.
    So before you start marking your rail for all your lenses at all usable apertures, you might like to consider which of these two simple ways of determining hyperfocal distance suit you best:
    c-o-c proportional to lens focal length (eg f/2000), or
    c-o-c fixed for one format (eg 0.1 mm for 4x5).
    Best,
    Helen

    By jstraw - 02:23 AM, 09-16-2006 Rating: None
    I'm getting a little lost in my limited math skills. Could someone show me the math for an example a 210mm lens at f32 in both methods? I'm not able to extrapolate this from the algabraic representation, it seems.

    By Helen B - 03:16 AM, 09-16-2006 Rating: None
    Jay,
    It just so happens that for a 210 mm lens on 4x5 the two methods give the same result - if one method is to use f/2000 and the other is to use 0.1 mm for all focal lengths on 4x5. At any focal length outside about 180 to 210 mm approximately the two methods will give dissimilar results, give or take a foot or two.
    Your example
    210/32*2*3.28 (for feet) = 43.5'
    is the correct way of using the f/2000 method.
    If you use 0.1 mm for 4x5 (I assume that you are using 4x5) then
    HFD = f^2/(N.c) / 1000 / 0.3048
    HFD in feet
    f = 210 mm
    N = 32
    c = 0.1 mm
    HFD = 45.2 ft.
    Best,
    Helen

    By jstraw - 03:46 AM, 09-16-2006 Rating: None
    Thanks Helen!

    By MichaelBriggs - 07:03 AM, 09-16-2006 Rating: None
    Patrick, I am not arguing from authority -- that is an unfounded insult. I have been explaining the physical basis of the depth of field equations (i.e., depth of field in the print does depend on how the print is viewed, which relates to what circle of confusion should be used). I have also provide a pointer to what I feel is an excellent deriviation, since I am not about to write several page, along with diagrams. Plus I have quoted several sentences to show that I am not the only person with these ideas. But first I have explained the physical princples, which you don't seem to understand.

    By jstraw - 02:38 PM, 09-16-2006 Rating: None
    Helen, what are your thoughts on how one decides which method one employs? Is there a break point for longer lenses prior to which you would use f/2000 and after which you would use 0.1?

    By gainer - 09:24 PM, 09-16-2006 Rating: None
    Michael, talk about unfounded insults!
    If there is anything I do not understand about depth of field, it is your attempt to explain it. The only control the photographer has over it in the field is the iris and the focusing knob. It will not make any difference in the resultant print whether the viewer of the print views it one way or another. The depth of field seen in the print is fixed. In the field, we hope to b able to predict how our camera controls will affect depth of field and the resultant artistic use or not of selective focus.
    The controls we have are inherent in the camera and its lenses and shutters. The interactions of these controls are describable by the geometry of optics. Our equations are dependent on the knowledge of optical principles and the limitations of optics. Any print I make shows what depth of field my camera captured, no matter who views it or how, if we can define the boundaries of that field by the line where blur is greater by some factor than that at the point of sharpest focus. If we must use a magnifying glass to find that point, then for practical purposes it does not exist and the print is uniformly sharp or blurred.
    If the argument is about the definition of blur circle, so be it. We may never agree. In my derivation, I started with long recognized principles that had culminated in an equation that appeared in a text by two well known and respected experts in photographic optics. Even they recognized that the size of the blur circle is somewhat arbitrary. The definition I used was from a War Department technical manul, which was primarilly intended for users of 4x5 and larger cameras for aerial mapping and such. There are two problems with such a definition. First, the blur circle in any case is variable with aperture. Second, an arbitrary choice of the constant may put the desired blur circle beyond the limit imposed by diffraction for a given aperture.
    I agree that the equation uses too many USA's for critical work, but the depth of field scales inscribed on the focusing rings of most cameras before autofocusing may suffer from the same ailment.
    What would you suggest we use as the mathematical definition of the circle of confusion? We could use the angular resolution defined by the Airy circle and calculate the distance that angle subtends in the image plane viewed from the aperture. That would be pretty accurate for a diffraction-limited lens. It would be conservative in the sense that depth of field might be smaller than actual.
    You called me stupid before I suggested you were arguing from authority. Nobody who has known me has ever called me stupid or accused me of not being able to understand anything I undertook to understand. That includes 30 years as an Aeronautical Space Engineer for NACA-NASA during which I did many kinds of research including making star chrts for Mercury astronauts and general simulation and human factors research.
    Enough, already.

  4. #4
    Sean's Avatar
    Join Date
    Aug 2002
    Location
    New Zealand
    Shooter
    Multi Format
    Posts
    8,593
    Blog Entries
    7
    Images
    15
    By Helen B - 11:56 PM, 09-16-2006 Rating: None
    Jay,
    In reply to your question, I pretty much follow the current Kodak suggestion (also used by many others), which is to use a constant c-o-c for a format as a baseline, then adjust it for the particular situation by using the focus setting for a wider aperture if necessary. Kodak revised their c-o-c calculation for general photography from being based on lens focal length to being based on the same formula, but using the normal focal length for the format and fixing the c-o-c at that - so Kodak use 0.15 mm for 4x5, which I find a little too large. That means that in the field I use the Kodak calculators, but with at least one stop wider aperture than is set on the lens.
    Then in practice you use that as the baseline to balance up the requirements for deep focus, proposed degree of enlargement, comparative effect of diffraction and bias to near or far focus.
    Best,
    Helen

    By jstraw - 02:34 AM, 09-17-2006 Rating: None
    Thank you.



 

APUG PARTNERS EQUALLY FUNDING OUR COMMUNITY:



Contact Us  |  Support Us!  |  Advertise  |  Site Terms  |  Archive  —   Search  |  Mobile Device Access  |  RSS  |  Facebook  |  Linkedin