I get the concept behind CoC, I'm just trying to wrap my head around the formula. I've seen the simple approach ("d/1500"), and I've read that "1500" is based on a standard print size of 8x10" and a viewing distance of 10"? But I want to actually plug in my own values for print size and distance. How do I calculate the "1500"?


The 1500 is a bit arbitrary. The idea is that the bigger the print the farther away it is. In other words to look at a whole image comfortably you hold it/stay at a distance such that the viewing angle of the diagonal is always around 60°.

The human eye can, in ideal condition, and at the center of the field of view, distinguish about half a minute of arc, so that would be 1/7200th of a 60° angle of view, but in practice a visual acuity of 2' of arc is used which leads to 1/(60*30)=1/1800th of the diagonal.


The circle of confusion standard is all about the magnification of the blur point, and how visibly apparent that makes any misfocus in the final image to a viewer.

In that sense viewing distance relative to the print size is also magnification... making the blur point (circle of least confusion) more/less visible. It also depends on the visual acuity of the viewer and the viewing/lighting conditions. If any of those things are going to be non-standard then you need to use a different CoC.

The easy way to determine your new CoC requirement is to determine the magnification variable you are introducing... i.e. if you are going to enlarge the image by 2x you would divide the standard CoC by 2x; and the same if you were going to view an image from half standard distance (1/2 the diagonal)... those are both actually the same thing.

This is essentially how the CoC for different image formats is determined. E.g. .015mm for M4/3 compared to .03mm for FF; because the 4/3 sensor has to be enlarged 2x more for any given output size.

There's also an app for that...

Also note that the CoC standard requires less than 2MP of recorded resolution... which is why no calculator even considers sensor resolution.


The d/1500 is related to sensor size enlarged to viewing size, in the result as seen in an 8x10 inch print at 10 inches by the human eye. Many do not realize that enlargement size of sensor size, or viewing distance, is a factor, but they are.
d is the sensor diagonal.

Greater viewing distance allows greater CoC size (harder to see smaller blur).
Greater print size requires smaller CoC size (easier to see enlarged blur).

Everything is computed on the diagonals (which is the lens diameter on the sensor). 8x10 inch diagonal is 12.8 inches, or 325.3 mm.

If the print is different size than 8x10, CoC is multiplied by new diagonal / (325 mm or 12.8 inches).

If the viewing distance is different then 10 inches (254 mm), CoC is multiplied by (254 mm or 10 inches) / new distance.


I never heard of D/1500 rule-of-thumb, being curious, I will attempt to reverse engineer. We typically examining prints at standard reading distance, about 500mm.At this distance a circle size of ½ millimeter is appropriate i.e., 0.5mm.

The diagonal measure of a full frame is 43.27mm. 43.27 ÷ 1500 = 0.0288mm, in other words, this formula yields a circle of confusion size of 0.0288mm at the focal plane of the camera.

Math reveals that we can enlarge the resulting image 17.3 X. Such an enlargement will yield an image 16 inches by 24 inches. Give the above parameter delivers an image will have a circle size of ½ mm.


All of these answers were great and helped me arrive at the formula, which I basically boiled down to this:

The classic “d/1500” assumes the ratio of print diagonal to viewing distance is 1:1. If instead the ratio is different (e.g. a 5’ print viewed at 7.5’; a 1.5 ratio), then you multiply d/1500 by that ratio. For the example, that’s (43.3mm/1500) * 1.5 = 0.0432 CoC, where 43.3 is of course the the Full Frame sensor diagonal.

You can also go with D/1750 or D/1000 or whatever tolerance you prefer.


It assumes a negative/sensor size of 36x24mm.

It assumes a viewing distance of 10" (25-30cm).

It assumes an enlargement size of 8x10" to 8x12", depending on if the 3:2 aspect ratio negative is cropped to 5:4 or not. In either case, the full 24mm height of the negative is enlarged to 8", for an enlargement ratio of approximately 8.5:1.

Lastly, it assumes a visual acuity of 20/20 vision by the viewer.

Depending on the amount of light, which affects pupil diameter, which in turn determines diffraction limiting, as well as depending on the exact wavelength (our eyes can resolve red light better than blue) a person with 20/20 vision is assumed to be able to resolve black/white line pairs somewhere between 20-60 arcseconds, or 0.33-1.0 arcminutes, or 0.0556-0.01667 degrees.

But the world is not usually made up of black/white line pairs. It's made up of irregularly shaped objects that do not have 100% contrast alternating between pure white and pure black. So the acceptable circle of confusion is based on the assumption that we can only resolve "real world" things that are about 3.5 times wider than those 1 arcminute line pairs.

At a viewing distance of 25cm (10"), 210 arcseconds (3.5 arcminutes or 0.0583 degrees) works out to about.255mm. Factor in an 8.5X enlargement ratio, and that means 0.03mm on a negative 24mm high.

Since a 24x36mm film frame has a diagonal of 43.26mm. d/1500 works out to 0.029mm. However, if we crop the part of the negative printed to 30x24mm, the diagonal is reduced to 38.42mm, which works out to 0.026mm. But in both cases, we're using the same enlargement ratio, which is roughly 8.5:1. So it can be a bit arbitrary depending on the aspect ratio of the image, even though the ability of our eyes to perceive angular differences doesn't change with the aspect ratio of the photograph we're viewing.

Some lens manufacturers recommend d/1750 instead of d/1500. They assume the viewer has better than 20/20 vision.

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