1
\$\begingroup\$

I'm shooting images on my DSLR and wondering what the circle of confusion would be if it were to be viewed only on an iPhone.

As I understand it, the CoC formula can be expressed as: (Acuity * Viewing Distance) / (Display Diagonal / Camera Sensor Diagonal)

So for the classic example of an 8x10 print at 10" using a full frame sensor: (0.001 * 254mm) / (322mm / 43.3mm) = 0.034 mm CoC

I found a study that claimed typical phone viewing distance is around 269mm. The diagonal of my iPhone is 126mm when cropped to 3:2. So for a full frame sensor: (0.001 * 269) / (126 / 43.3) = 0.087 mm CoC.

Does that seem right?

\$\endgroup\$
3
  • \$\begingroup\$ I don't care what generation of iPhone (or any other model) you have, but it does not have a full frame sensor... \$\endgroup\$
    – twalberg
    Nov 12, 2021 at 20:21
  • \$\begingroup\$ 😁 That would be the image sensor. Assuming a full frame DSLR in this scenario. \$\endgroup\$
    – zakray
    Nov 12, 2021 at 20:57
  • \$\begingroup\$ Which iPhone? Different models have different pixel densities. \$\endgroup\$
    – Michael C
    Nov 15, 2021 at 1:04

3 Answers 3

1
\$\begingroup\$

Because the viewing size of the image is ~ 1/3 standard, and the viewing distance is ~ standard; the CoC requirement for the recording medium would be ~ 1/3 as demanding (3x larger)... i.e. .09mm instead of .03mm for a FF camera.

So yes, your result makes sense.


I edited the question to clarify what you were actually asking

\$\endgroup\$
14
  • \$\begingroup\$ I see what you're saying... but I'm confused as to why 10" is the standard distance in the full frame formula. \$\endgroup\$
    – zakray
    Nov 13, 2021 at 17:14
  • \$\begingroup\$ And for that matter, why an 8x10 print? That's not even the same aspect ratio as a 3:2 sensor, wouldn't that throw off the diagonal calculations? \$\endgroup\$
    – zakray
    Nov 13, 2021 at 17:57
  • \$\begingroup\$ I realize you've read it's based on 10" distance for an 8x10, and you've probably read that it is based on 20/20 vision; both are not true... \$\endgroup\$ Nov 13, 2021 at 19:02
  • \$\begingroup\$ The standard is based on viewing any image of any size and aspect ratio from a distance approx equal to its' diagonal measurement (12.8" for an 8x10). It is the image diagonal because that relates to the diameter of the image circle required to create it, and when viewed at that distance the image occupies the approx 60˚ circular primary field of view of a human. And furthermore, a "normal lens" records an approx 55-60˚ diagonal field of view regardless of format... so it looks "normal" when viewed occupying that 60˚ human FoV. All of this existed before 35mm was even invented... \$\endgroup\$ Nov 13, 2021 at 19:12
  • \$\begingroup\$ Thank you Steven, this is very helpful. Let me come clean-- I work in film where the concept of a "standard distance" based on 60º isn't as useful. More commonly, people are in a theater where the screen occupies a larger field of view than a photo in a gallery would. I'd like to be able to input a screen diagonal, a distance from the screen (usually the middle of the theater), and a sensor diagonal (often Super35mm film which is 27.54mm). What formula can I use to do this? \$\endgroup\$
    – zakray
    Nov 13, 2021 at 21:06
2
\$\begingroup\$

The physics that encompasses circle of confusion size is based on the resolving power of the human eye. As you know, objects appear smaller with distance. If a friend holds a coin 1 inch in diameter and walks away, the apparent size of the coin shrinks. At a distance of 3000 diameters away, the coin now appears only as a point of reflected light, not as a disk. Additionally, a wagon wheel 3 feet in diameter becomes a point at 9000 feet (1 ¾ miles). The 3000 X diameter rule-of-thumb is based on human eyes with 20/20 vision in bright sun. Change any of these parameters and the rules change . Because of the contrast of photos and their typical viewing conditions, the industry uses 1000 X the diameter. That works out to 3.4 minutes of arc, equivalent to 1/100 of an inch viewed from 10 inches (0.25mm or 1/50 of an inch viewed from 20 inches (0.5mm viewed from 500mm).

The diagonal measure comes into play, we tend to view photographs from a distance about equal to their diagonal measure. Consider a print 8x10 inch made from 35mm full frame. We must enlarge the 35mm image 8x to obtain this enlargement. The diagonal measure of an 8x10 is 13 inches. We tend to view this size print from about 13 inches. Given these considerations, depth-of-field tables are likely based on a circle size of ½ millimeter. To accomplish, in the camera, the maximum permissible circle size becomes 0.5mm ÷ 8 = 0.1mm.

The answer to your question is – the subpixels (red – green – blue) glowing dots on the phone should be no larger than 0.5mm, however, smaller is better. Viewed from 10 inches they need to be 1/000 of an inch = 0.25mm.

Let me add – a neat way the industry uses for depth-of-field tables is to express the circle of confusion size as a fraction of the focal length. This method roughly takes into account how much magnification will likely be applied to produce the viewed image. Often the circle size chosen is 1/1000 of the focal length. For critical work Leica uses 1/1500 of the focal length and Kodak used 1/1750 from very critical work.

\$\endgroup\$
4
  • \$\begingroup\$ "The physics that encompasses circle of confusion size is based on the resolving power of the human eye" No. Sorry, but that's a logical-sounding fallacy. You're talking anatomy and we're talking photography. CoC is the result of a calculation based on actual measurements. It is completely objective. When you bring the human visual system into the picture you're making it subjective and it's no longer the same discussion. \$\endgroup\$
    – DaveInAZ
    Nov 16, 2021 at 18:31
  • \$\begingroup\$ DaveinAZ You should read Phototropic Lenses by C.B. Neblettle -- Photographic Principles by Lewis Larmore - Neblette's Handbook of Photography - The basis is distance to resolve a disk versus seeing it as a dimensionless point. Accepted is 3000 diameters distance 20/20 vision bright sunlight - too stringent for photography we generally use 1/1000 of the viewing distance 3.4 minutes of arc, equivalent to 1/100 of an inch viewed from 10 inches. What criteria would you use? \$\endgroup\$ Nov 16, 2021 at 19:40
  • \$\begingroup\$ "...distance to resolve a disk versus seeing it...". No, you're still talking about perception. "Seeing" is a subjective activity and CoC is math. It exists outside us, whether anyone perceives it, or not. The math factors in a certain 'visual acuity' value, but that's just a framework to allow comparison, as is the 20/20 'standard'. And CoC is for capturing images. \$\endgroup\$
    – DaveInAZ
    Nov 16, 2021 at 21:54
  • \$\begingroup\$ Lets make sure to keep the discussion polite please. \$\endgroup\$
    – AJ Henderson
    Nov 17, 2021 at 0:20
0
\$\begingroup\$

Circle of Confusion is a term that applies only to capturing images through optical lenses. It doesn't apply to displaying images, so the question is a non sequitur.

If you're trying to determine the optimal viewing distance from your phone's display, that depends on your eyes.

\$\endgroup\$
3
  • \$\begingroup\$ Doesn't the circle necessarily depend on the enlargement size of the print (to use the classic terminology?). To be clear I'm talking about capturing an image with my full-frame DSLR and then displaying that on an iPhone. \$\endgroup\$
    – zakray
    Nov 12, 2021 at 22:58
  • \$\begingroup\$ @zakray That might have been an important fact to include in your original question, as the natural assumption the way it's written now would be that the photo in question was taken on the phone by the phones sensor. Hence my comment regarding your phone not having a full-frame sensor.... \$\endgroup\$
    – twalberg
    Nov 12, 2021 at 23:48
  • \$\begingroup\$ Thanks, I've updated the question \$\endgroup\$
    – zakray
    Nov 13, 2021 at 0:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.