ENLARGEMENT |
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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)
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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
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- Resolution necessary to a negative 24x36mm
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- Enlargement 30x40cm
- c = 0,25 x 36 / 400 = 0,022mm
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- All films having resolutions below 0,20 this enlargement
can be done in best conditions.
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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.
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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
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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
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