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)