Image Processing Reference
In-Depth Information
However, pulling
λ
from Eq. (13.48) and substituting it in Eqs. (13.46) and (13.47)
reduces the number of equations to 2.
XM
11
+
YM
12
+
ZM
13
+
M
14
−
x
(
XM
31
+
YM
32
+
ZM
33
+
M
34
)=0(13.49)
XM
21
+
YM
22
+
ZM
23
+
M
24
−
y
(
XM
31
+
YM
32
+
ZM
33
+
M
34
)=0(13.50)
Now let
c
(
p
,
p
)=(
X,Y,Z,
1
,
0
,
0
,
0
,
0
,
c
)
T
−
cX,
−
cY,
−
cZ,
−
(13.51)
r
(
p
,
p
)=(0
,
0
,
0
,
0
,X,Y,Z,
1
,
r
)
T
−
rX,
−
rY,
−
rZ,
−
(13.52)
D vectors produced by the pair of vectors
p
,
p
representing a
point and its image according to Eq. (13.42). Furthermore, let
where
c
,
r
are 12
−
m
=(
M
11
,M
12
,M
13
,M
14
,M
21
,M
22
,M
23
,M
24
,M
31
,M
32
,M
33
,M
34
)
T
,
(13.53)
which is a vector version of the unknown matrix
M
. Consequently, we can express
Eqs. (13.49) and (13.50) as the equation pair
c
T
m
=0
(13.54)
r
T
m
=0
(13.55)
where we have omitted displaying the dependency of
c
,
r
on
p
,
p
for convenience.
Geometrically, the homogeneous equation
6
s
T
m
=0
(13.56)
represents the equation of a (hyper)plane in
E
12
, where the constant vector
m
is the
normal of the plane. For a corresponding pair
p
,
p
, we can obtain two points in this
12D space, i.e.,
s
1
=
c
and
s
2
=
r
given by Eqs. (13.54) and (13.55), that satisfy
Eq. (13.56).
Let
be a set of correspondence vectors.
7
S
=
{
c
,
r
}
Because of measurement
noise, members of
will only approximately satisfy Eq. (13.56). Then, a TLS solu-
tion is the preferable procedure to find
m
. In that, one attempts to minimize
e
(
m
)=
S
|
s
T
m
|
2
h
(
s
)
d
s
=
m
T
ss
T
h
(
s
)
d
s
m
,
with
m
=1
,
S
S
(13.57)
which is the linear symmetry problem in
E
12
, see theorem 12.1. Here
h
is a func-
tion that represents the certainty on the correspondence data
s
, which is equivalent
6
An equation is homogeneous if it is of the form
Ax
=
0
, with
x
being the unknown vector.
Only if the null space of
A
is nonnil, does another solution than
x
=
0
exist.
7
The set can be a dense set, e.g., a jointly observed surface in the world frame, and in
the digital image frame. Although the term “digital” implies sampling, strictly speaking,
sampling is not necessary for the conclusions of this chapter. This is because even the
transformation from (
x, y,
1)
T
to (
c, r,
1)
T
is continuous!
