Circle of confusion formula:
Viewing to naked eye from distance D
c = 2,5 a D
with visual acuity a = 0,0004
c = D / 1000
Circle of diffusion of digital camera:
c = 1,25 r = 2,5 i  with i the size of pixel.
Conventional viewing distance:
25cm to naked eye, or 250mm
The circle of confusion is then c = 0.25mm
Exposure time:
It is expressed in seconds: : t = 1 / T for a shutter speed T. For example:
t = 1/ T = 1/250 = 0,004 seconds
Move by rotation:
Let's consider an horizontal rotation of the camera on its axis.
The rotation speed is W in turns per minute [O/mn].
It takes w = 60/W seconds for one turn, so w is the rotation speed expressed in seconds per turn [s/O].
 The picture will be clear if the slip of the image of a dot is less than the circle of confusion c of the imager. c is linked to the focal length F, to the shutter time t, and to the rotation speed by:
2 p t / w < c / F with p = pi = 3,14116
Therefore comes the relations:
w > 2 p t F / c
 and  W < 60 c / (2 p t F)
NB: c and F are expressed in the same unit.
t is in seconds


Another way to calculate is to say that the exposure time shall not exceed the time to offset a circle of confusion distance. With a digital image the number of pixels to run a complete turn is N and as c=2,5i the relation is
t <  w  2,5i / Ni  or w > t N / 2,5
Also, N = 360 I / A with A angle of field of the lens, and the relation becomes
w > t 360 I / 2,5 A
To view the picture on a computer screen, the sharpness criteria is less stringent and allows to decrease the size of the file. It means the rotation can be faster.
With a 24x36mm film cameral, 35mm lens
 w > 2 p (1/250) 35 / 1,25 x 0.015
So w > 47 seconds
With a digital camera, sensor1.8"
lens 7.3mm, 2272x1704 pixels ( 4 Mpx)
a pixel is 0,00031mm and c=2,5 i
 w > 2 p (1/250) 7.3 / (2,5 . 0,0032)
So w > 23 seconds


On a computer screen
Let's have a 360° panoramic view with a file 6000 pixel length which is shot at 1/250' of second,
t N / 2,5 = (1/250) 5000 /2,5
It needs a rotation w>8 so more than 8 seconds only for a turn.
The calculations here opposite corresponds to the maximum sharpness criteria whilst the one of the screen to the sharpness just necessary.
Transversal move of the subject:
Let's consider the move transversal and perpendicular to the optical axis of a subject at the distance H of the camera.
The subject is moving at the speed S in m/s
NB: c and F are expressed in the same unit.
H is given in meters.



The picture will be clear if the slip of the image of the subject is less than the circle of confusion c of the imager. c is linked to the focal length F, to the shutter time t, to the distance H and to the speed S by:
S t /H < c / F
Therefore,  t < c H / (F S)




Tables of depth of field:
  The Depth of Field tables (DoF) calculate the increase of the converging spot outside the focal plan, and compare this value to a preset of the circle of confusion. When closing the aperture, the converging spot decreases. It is only a geometry concern.
The digital cameras have the reputation to have important depth of field. Do you think this is correct?
Nevertheless, it remains true that the depth of field is always smaller on large enlargement, than on a postcard size print.
The depth of field depends of the enlargement ratio of the print.
The values of the circle of confusion of the lens makers are, as it is said:
0.03mm for 24x36 mm frames
0.05mm for medium size
In the books of authors, all reliable and competent, I have noted the following values for 24x36mm:
   "0,020" "0,030" "0,033" "0,036" These values haven't followed the improvements in quality of films and lenses.
The following values of circle of confusion are those traditionally used for calculations of depth of field and hyperfocal.
In 24x36mm (35mm film)
c = 0,036 mm
In medium format from 4,5x6 to 6x9cm, (films 120 and 220)
c= 0,056 mm
In large format from 9x12 to20x25cm,
c= 0,120 mm




Circle of confusion of an enlargement:

 The enlargement must be clear when viewed  close to 25cm, whatever is its size. The circle of confusion is then 0,25mm. What is the circle of confusion necessary on the film or on the sensor?
(Note that the examples are here only by chance)
The necessary circle of confusion of the film, or of the sensor is:
c = 0,25/ P = 0,25 U / X
in which P is the enlargement ratio, get by:
P = X/U = Y/V = Z/W
X, Y, Z and U, V, W are respectively the dimensions and the diagonal of the enlargement and of the imager.
This relation is also: X / 0,25 = U / c
Resolution necessary to a negative 24x36mm
Enlargement 30x40cm
c = 0,25 x 36 / 400 =  0,022mm
All films having resolutions below 0,20 this enlargement can be done in best conditions.


From a digital camera, 1/1,8" sensor
4 Mpx, width of sensor 7,2mm
Enlargement 30x40cm,
c = 0,25 x 7,2 / 400 = 0,0045mm
How many pixels I are necessary ?
I = U/i = (2,5 i X/0,25) / i
I = 10 X = 4000 pixels
A 10Mpx sensor pixels is necessary.

Maximum enlargement of a negative:

We can calculate the maximum print size of a negative of known circle of confusion c starting with the geometrical relation:
 X /0,25 = U / c  
D print size
U negative size
0,25 = visual circle of confusion at 25cm
or Xmax = 0,25 U / c
From a negative of 35mm film, with a camera capable of 0,015mm resolution and imposing a conventional viewing distance of 25cm:
Xmax = 0,25x36 /1.25 x 0.015 = 480mm


Maximum enlargement of a digital image:

We can calculate the maximum print size X of an image got with a digital camera which circle of confusion c is known, and of I x J pixels. The size of pixels being i, and 0,25mm the visual circle of confusion at 25cm, the relation is:
X / 0,25 =  I.i / c
with c = 2,5 i from which:
Xmax = I / 10  in mm
From a digital camera with sensor 1/1.8" of 2272x1704 pixels ( 4 Mpx) and imposing a conventional viewing distance of 25cm:
Xmax = 2272 / 10 =  227 mm




Definition of print:
The best print will be got for a circle of confusion equal to the visual circle of confusion of eye at 25cm, which is 0,25mm.
Therefore, c = 0,25 = 2,5 (25,4/N)
which gives: N = 254 ppi
It is unnecessary to take higher printing definition.
Note that most soft indicate dpi for ppi. See printers
For very large prints it is unlikely to view it at such 25cm short distance, so the circle of confusion has to fit the viewing distance. Here is an example:
The enlargement width W is 1m. The normal viewing distance is 125 cm but someone would come closer to check a detail. That could be 60cm only. So the circle of confusion at that distance will be c = 0.0001 x 600 = 0,6 mm.
The pixels needed will be: N = 2.5 x W / c = 4166 pixels
The picture shall be 4166 x 3125 which is 13Mpx at least.
Optimized print size:
The size to print must be X x Y mm. Which file size is necessary for a quality print viewed at 25 cm?
  Definition of printer N = 254
  Size of file: I x J pixels
Knowing that there is need of N/25,4 pixels per mm
The sizes of I and J with X and Y in mm are:
I = 10 X  pixels and J = 10 Y  pixels
For large prints displayed at home on the walls, the minimum viewing distance will usually be 50cm. Then the circle of confusion necessary on the print is 0,5 mm. Now the number of pixels needed is N = 2.5 x width / c = 2.5 x 25.4 / 0.5
This is 127 ppi
With a 10Mpx image (3648 x 2736 px), the maximum size of the enlargment will be (3648 / 127 x 25.4)  ~ 70 x 50 cm
Width of enlargement  X = 2,5 ( I / c )