I am curious about the depth of field of the famous saying "f/8 and be there".

"f/8 and be there" is often associated with hyperfocal distance on 35mm, which means infinite depth of field, but Weegee's original configuration is different:

  1. Speed Graphic (4x5 inch)
  2. 127mm Kodak Ektar Lens (equivalent angle of view to 42mm in 35mm)
  3. f/16
  4. a distance of 10 feet
  5. print size (8x10 inch)
  6. printed on Newspaper (the print quality of Newspaper affects Circle of Confusion)
  7. viewing distance when reading Newspaper (personally my viewing distance when reading Newspaper and looking at a photo is different)
  8. B&W Kodak Super Pancro Press Film
  • 2
    \$\begingroup\$ Seems to me you've defined your circle of confusion adequately, so you could use any of a number of online DOF calculators. The DOF would certainly be wider than the likely error of an experienced press photographer estimating distance... \$\endgroup\$
    – Zeiss Ikon
    May 4 at 17:54
  • 1
    \$\begingroup\$ It was always my understanding that "f/8 and be there" was a kind of nonsense saying, not actual advice. I took from it that the most important thing was to actually be ready, at the decisive moment. The right place at the right time, so to speak, being more important than specific settings. (f/8 being a "safe" aperture to recommend at the same time.) \$\endgroup\$
    – osullic
    May 4 at 21:04
  • \$\begingroup\$ @ZeissIkon Hmm, the equivalent question is what is the estimated circle of confusion for 4x5 inch film printed as 8x10 inch photo on Newspaper of Weegee's time? \$\endgroup\$
    – weakish
    May 5 at 14:31
  • 1
    \$\begingroup\$ I think so. Which mostly boils down to the halftone screen size they used (72 dpi, IIRC). \$\endgroup\$
    – Zeiss Ikon
    May 5 at 14:57

1 Answer 1


Depth of field calculations are based on assumptions that may or may not be accurate for a given situation.

The size of the circle of confusion is a circle too small to be the resolved at a specified viewing distance. For photographic work, OK to use ½ mm viewed from 20 inches.

We expose film making a negative and then enlarge this negative to make an 8x10 inch print. The 4x5 inch film you are asking about requires 2x enlargement to make the 8x10 display print. If this is the criteria, then the circle size on the film is ¼ mm.

If a 127mm lens is mounted, and set to f/8, the hyperfocal distance is 26.9 feet for a circle size of ¼ mm. We divide the hyperfocal distance by 2 to find the near distance with acceptable focus, infinity is the far distance. Thus, the span of acceptable focus is 13.45 feet to infinity.

If this setup, f/8 is focused at 10 feet, then the span of the depth of field is 7.34 feet to 15.7 feet.

If this camera is set to f/16: The hyperfocal distance is 13.6 feet. The span of acceptable focus is 6.8 feet to infinity.

  • \$\begingroup\$ "For photographic work, OK to use ½ mm viewed from 20 inches." I'd like to confirm whether this takes the print quality in to account. Nowadays Newspaper is usually 85 lpi and I do not know the printing resolution of Newspaper from Weegee's time (but it would be lower than 85 lpi I guess). \$\endgroup\$
    – weakish
    May 5 at 14:27
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    \$\begingroup\$ @weakish Not much has changed with regard to the resolution of newsprint over the past century. It's never pretended to be a high resolution medium. Those who published on newsprint never felt the need to increase it because it was sufficient to print text in a size that was readable by subscribers. \$\endgroup\$
    – Michael C
    May 5 at 14:39
  • \$\begingroup\$ Thanks to @ZeissIkon, the halftone screen they used is 72 lpi. That is about 2 lp/mm. And "½ mm viewed from 20 inches" is also about 2 lp/mm. So the printing resolution is sufficient for a view distance of 20 inches. \$\endgroup\$
    – weakish
    May 6 at 7:21

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