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as constraints on the intrinsic camera parameters. In (
1.41
), the expression
A
−
T
is
an abbreviation for
(A
T
)
−
1
.
Zhang (
1999a
) derives a closed-form solution for the extrinsic and intrinsic cam-
era parameters by defining the symmetric matrix
B
A
−
T
A
−
1
,
=
(1.42)
which can alternatively be defined by a six-dimensional vector
b
=
(B
11
,B
12
,
(h
i
1
,h
i
2
,h
i
3
)
T
B
22
,B
13
,B
23
,B
33
)
. With the notation
h
i
=
for the column vec-
tors
h
i
of the homography
H
, we obtain
h
i
B
h
j
=
v
ij
b
,
(1.43)
where the six-dimensional vector
v
ij
corresponds to
v
ij
=
h
i
2
h
j
3
,h
i
3
h
j
3
)
T
.
(1.44)
(h
i
1
h
j
1
,h
i
1
h
i
2
+
h
j
1
,h
i
2
h
j
2
,h
i
3
h
j
1
+
h
i
1
h
j
3
,h
i
3
h
j
2
+
Equation (
1.41
) is now rewritten in the following form:
v
12
(
v
11
−
v
22
)
T
b
=
0
.
(1.45)
Acquiring
n
images of the planar calibration rig yields
n
equations of the form
(
1.45
), leading to the homogeneous linear equation
V
b
=
0
(1.46)
for
b
, where
V
isamatrixofsize2
n
3, (
1.46
) yields a solution
for
b
which is unique up to a scale factor. Zhang (
1999a
) shows that for
n
×
6. As long as
n
≤
2im-
ages and an image sensor without skew, corresponding to the matrix element
A
12
being zero, adding the appropriate constraint
(
0
,
1
,
0
,
0
,
0
,
0
)
b
=
0 also yields a so-
lution for
b
in this special case. If only a single calibration image is available, Zhang
(
1999a
) proposes to assume a pixel sensor without skew (
A
12
=
=
0), set the principal
point given by
u
0
and
v
0
equal to the image centre, and estimate only the two ma-
trix elements
A
11
and
A
22
from the calibration image. It is well known from linear
algebra that the solution to a homogeneous linear equation of the form (
1.46
) cor-
responds to the eigenvector of the 6
6matrix
V
T
V
which belongs to the smallest
×
eigenvalue.
Using the obtained value of
b
, Zhang (
1999a
) determines the intrinsic camera
parameters based on the relation
B
=
νA
−
T
A
, where
ν
is a scale factor, as follows:
B
11
B
23
)/
B
11
B
22
−
B
12
v
0
=
A
23
=
(B
12
B
13
−
B
33
−
B
13
+
B
11
B
23
)
/B
11
ν
=
v
0
(B
12
B
13
−
A
11
=
√
ν/B
11
α
u
=
νB
11
/
B
11
B
22
−
(1.47)
B
12
α
v
=
A
22
=
α
u
cot
θ
=
A
12
=−
B
12
α
u
α
v
/ν
u
0
=
B
13
α
u
/ν
A
13
=
A
12
v
0
/α
v
−
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