• Members 487 posts
    May 9, 2023, 2:06 p.m.

    I am posting this for the benefit of anyone who may be interested.

    Over the past few years, I have found that I much prefer using the hyperfocal field size instead of the hyperfocal distance. The hyperfocal field size is the diameter of the field of view at the hyperfocal distance.

    The hyperfocal field size depends only on the entrance pupil of the lens and the circle of confusion (expressed as a fraction of the sensor size). The focal length of the lens, the f-number and the sensor size are all irrelevant.
    Screenshot 2023-05-09 at 14.26.24.png

    Most depth-of-field calculators today use a circle of confusion of 0.030mm for full-frame sensors. This corresponds to R=1440.

    I prefer to use R=2000 as it makes the arithmetic simpler and is also more appropriate for modern digital cameras with pixel counts of 12Mp or more. It corresponds to a circle of confusion of 0.022mm for full-frame. Dedicated pixel peepers may like to try R=4000 or even more.

    For example, using R=2000, a 10mm lens at f/8 has an entrance pupil of 1.25mm and a hyperfocal field size of 2.5 metres; while a 200mm lens at f/4 (or a 50mm lens at f/1) has an entrance pupil of 50mm and a hyperfocal field size of 100 metres.

    Using the hyperfocal field size makes it easy to work out how blurred a distant background will be. If the actual field size in the plane of focus is, say, one tenth of the hyperfocal field size, then a background at infinity will be blurred to ten times the size of the circle of confusion. That blurring will be clearly visible, but not large.

    On the other hand, if the field size is one hundredth of the hyperfocal field size, then the blurring of a background at infinity will be 100 times the size of the circle of confusion. That is a much larger amount of blurring.

    Examples in next post.

    Screenshot 2023-05-09 at 14.26.24.png

    PNG, 56.3 KB, uploaded by TomAxford on May 9, 2023.

  • Members 487 posts
    May 9, 2023, 2:07 p.m.

    Examples

    75mm f/1.8, EP=42mm, HFD=84m, AFD=21cm, so HFD = 400 x AFD (approximately)
    (EP = entrance pupil, HFD = hyperfocal field diameter, AFD = actual field diameter in the plane of focus)
    _5090437x.JPG

    75mm f/7.1, EP=11mm, HFD=22m, AFD=21cm, so HFD = 100 x AFD (approximately)
    _5090438x.JPG

    75mm f/14, EP=5.4mm, HFD=11m, AFD=21cm, so HFD = 50 x AFD (approximately)
    _5090439x.JPG

    Same lens focussed on background from same position. The image is very low contrast because the lens is looking through the water drops on the window. The image is slightly smaller because of focus breathing in this lens.
    _5090440x.JPG

    _5090437x.JPG

    JPG, 580.2 KB, uploaded by TomAxford on May 9, 2023.

    _5090438x.JPG

    JPG, 556.6 KB, uploaded by TomAxford on May 9, 2023.

    _5090439x.JPG

    JPG, 553.3 KB, uploaded by TomAxford on May 9, 2023.

    _5090440x.JPG

    JPG, 920.4 KB, uploaded by TomAxford on May 9, 2023.

  • Removed user
    May 12, 2023, 4:05 p.m.

    Good to see someone demonstrating the importance of aperture size in basic photography!

    I was into blur some time ago and made a spreadsheet that indeed shows the dependency of blur on the effective aperture diameter:

    blur.jpg

    My main reference for that was Merklinger's excellent paper on focusing:

    www.trenholm.org/hmmerk/TIAOOFe.pdf

    When the aperture size got dumbed-down into the ubiquitous "f-number", it's effect on basic imaging faded away from general knowledge - with many thinking that the same f-number gives one the same image, duh.

    blur.jpg

    JPG, 399.7 KB, uploaded by xpatUSA on May 12, 2023.

  • Members 535 posts
    May 12, 2023, 4:10 p.m.

    Thanks!!

    May you upload it, Please?

    ( you know, I'm lazy )

  • Removed user
    May 12, 2023, 4:32 p.m.

    I'll try ... ... No, can't be done. If you are seriously interested, I can put it up on my website.

  • Removed user
    May 12, 2023, 5:18 p.m.

    Tom, is your 'R' the same factor as is commonly used for figuring a CoC based on sensor size, e.g.

    (my emphasis)

    www.dicklyon.com/tech/Photography/DepthOfField-Lyon.pdf

  • Members 83 posts
    May 12, 2023, 5:32 p.m.

    Thanks for posting. That is interesting stuff. It is useful to know what size object will be visible or blurred at infinity.

    While you write

    it seems to me you are calculating the diameter of the entrance pupil based upon (focal length)/(F Number), or at least those are the 2 numbers I usually see in lens specifications and not entrance pupil diameter. Please correct me if I got this wrong.

    Circle of confusion has been assigned a value in more than one way. Traditionally, based upon standard print size and standard viewing distance and standard eyesight. More recently sometimes it is sometimes based upon the spacing of pixels on the sensor chip (spacing important and sensor size irrelevant) instead of print size and viewing distance.

  • Members 487 posts
    May 12, 2023, 5:43 p.m.

    Yes, exactly the same. I called it R for convenience. As far as I am aware, there is no commonly used symbol for it, and R seemed as good as any.

  • Members 487 posts
    May 12, 2023, 5:56 p.m.

    You are quite correct. I meant than once you know the entrance pupil, the focal length and f-number can be forgotten as they are no longer required.

    Also, if you are given an unknown lens, if you measure the entrance pupil diameter then it is not necessary to also measure the focal length (if you just want to calculate the hyperfocal field diameter).

    Yes, if you base the CoC on the pixel spacing, then that is an absolute size independent of the sensor size.

    However, if, instead of the pixel spacing, you are given the number of pixels along each side of the image, then that determines a CoC that is not absolute, but relative to the image size.

    I think whatever way we do things, there will always be some particular cases in which a different approach is more convenient.

  • Removed user
    May 12, 2023, 6:26 p.m.

    Thanks, I had just now done a calc for full-frame and 0.030mm and sure enough 'R' equaled 1442.

  • Removed user
    June 25, 2023, 11:42 p.m.

    Once popular was the so-called "Zeiss formula" which had Tom's 'R' equal to 1730:

    en.wikipedia.org/wiki/Zeiss_formula

  • Members 487 posts
    June 26, 2023, 6:46 a.m.

    Thanks for that link. I was not aware of the "Zeiss formula" although I was aware that a CoC of 0.025mm had sometimes been used instead of 0.030mm for FF cameras.

    It is interesting that the formula c = d/1730 has been given a name (the Zeiss formula), yet that formula does not seem to be widely used in its own right. Also, no name or symbol is given to the denominator in the formula.

    The more general form of this formula c = d/R seems much more useful to me, particularly with modern digital images that can be viewed at almost any size desired.

  • Members 54 posts
    July 3, 2023, 7:27 p.m.

    Sounds interesting! Please do mount it on your website!

  • Members 300 posts
    July 4, 2023, 11:07 a.m.

    R seems not to be the right letter to your formula. It usually means radius in geometry. en.wikipedia.org/wiki/Radius

    en.wikipedia.org/wiki/Zeiss_formula says:
    "The formula is c = d / 1730, where d is the diagonal measure of a camera format, film, sensor, or print, and c the maximum acceptable diameter of the circle of confusion." bolded by me.

    That was interesting and seemed too coarse, so I counted the CoC for a print on my wall. The print size is 77x57 cm and the diagonal is ~ 96cm.
    960 mm/1730 = 0.55 mm. In my print there's smaller details visible than 0.5 mm. I think the Z(eiss) factor should be 2*1730 or more.

    Do I misunderstand the case?