Find the level on the right where visual noise to the eye "is comparable in strength" with the noise on the left. This shows the relative importance of chrominance noise w.r.t. the luminance noise for your eye. For myself, I do not think that the number is much above 10%.
More precise, the noise in the left half is normally distributed in the
plane LUMA=const
; if you take orthogonal coordinate system
in this plane, two components of the noise are independent and normally distributed with
the given number as the deviation. The noise in the right half is normally distributed
on the line R'=G'=B'
; the deviation of each
coordinate is the given number.
Do not even think of trying to do this if your graphic system is below 24-bit. I do not expect that this may survive printing with less than perfect printers... I won't trust the data for noise=40 very much, since due to non-linearity of gamma-curves, high visible noise on the left may be due to difference of luma vs. luminosity; while luma is constant, luminosity has a significant noise there. This non-linearity may also explain visible red tint on the left hand side of the last graph.
Noise=4
Noise=10
Noise=20
Noise=40
random-noise.pl 4 >rand4.png
random-noise.pl 10 >rand10.png
random-noise.pl 20 >rand20.png
random-noise.pl 40 >rand40.png
Details of implementation: one call to rand()
corresponds to
one point on the left, and two points on the right. This system's rand()
is BSD random()
; it has 128 bits of internal state, and randbits=31
.
Ilya Zakharevich; HOME