Image Processing Reference
In-Depth Information
a plane perpendicular to the red axis, which contains the points: P , K and Z .We
now define another plane, denoted Γ 2 , which contains the points: R , Z , W , X R ,
and which is perpendicular to the triangle R X R O . This plane contains all possible
values of rgb, and hence also P . Since P belongs to both planes, P must be part
of the line defined by the intersection of the two planes. This line contains Z and
is perpendicular to the triangle R X R O . That is, if you place your fingertip on Z
in Fig. D.3 (a) and lift it vertically, then you are moving along this line and will
eventually reach P . Having this in mind we now define the triangle R WX B , where
X B is the middle point of the line spanned by R and G , see Fig. D.3 (b). Note that
this figure is also part of Fig. D.1 (b).
From Fig. D.3 (b) follows that
= −− ZW
−− ZW
= −− WP
cos (H )
·
cos (H )
(D.16)
−− WP
· −− WP
−− WP
. Realizing that H =
From the definition of saturation we have
=
S
−− WP
60°
H we can express
as
−−−→
−− WP =
WX B
cos ( 60°
(D.17)
H)
Inserting Eqs. D.16 and D.17 into Eq. D.15 yields
· −−−→
1
−−−→
S
WX B ·
cos (H )
1
3
P r =
·
+
(D.18)
cos ( 60°
H)
R X R
−−−→
−−−→
WX B = −−−→
−−−→
R X R =
Since
WX R
and
WX R
/
1 / 3 we can reduce Eq. D.18 to
1
·
1
3
cos (H )
cos ( 60°
S
P r =
+
(D.19)
H)
After having calculated P b and P r we can find the last coordinate as P g =
P b
P r . The final step is to convert from rgb to RGB. We know that P r = P R /(P R +
P G + P B ) , P g = P G /(P R + P G + P B ) , P b = P B /(P R + P G + P B ) and that I =
(P R + P G + P B )/ 3. From this follows that P R =
1
3 IP b .
Substituting into Eqs. D.3 and D.19 yields the following expressions which are valid
when 0°
3 IP r , P G =
3 IP g , and P B =
H
120°:
P B = I I · S
(D.20)
1
cos (H )
cos ( 60°
S
·
P R =
I
·
+
(D.21)
H)
P G =
3 I P R P B
(D.22)
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