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In-Depth Information
˜
˜
To seek a fast solution to this maximisation problem, we can differentiate Q
φ t 1 )
in terms of the six transformation parameters, and set them to be zero individually.
Therefore, the optimal solutions for the six DOFs can be obtained by solving these
numerical equations:
(
φ t |
˜
˜
Q
(
φ t |
φ t 1 )
=
0
,
(21)
T d
which is available, and
˜
˜
˜
˜
Q
(
φ t |
φ t 1 )
∂θ
=
Q
(
φ t |
φ t 1 )
R
q
∂θ =
0
,
(22)
R
q
where the rotational vector
θ
has three Euler angles,
( θ x , θ y , θ z )
.Further,
˜
˜
Q
(
φ t |
φ t 1 )
˜
˜
˜
= i
j
p
(
α
˜
|
β
,
φ
)(
β
γ
)
f ˜
α
/
α
˜
,
(23)
t j
t i
t
1
t i
t j
xt
zt
R
The rotation matrix can be formulated using a unit quaternion that can be expressed
as q
=(
q 0 ,
q 1 ,
q 2 ,
q 3 )
:
q 0 2
q 1 2
q 2 2
q 3 2
+
2
(
q 1 q 2
q 0 q 3 )
2
(
q 1 q 3 +
q 0 q 2 )
.
q 0 2
q 1 2
q 2 2
q 3 2
R
=
2
(
q 1 q 2 +
q 0 q 3 )
+
2
(
q 2 q 3 +
q 0 q 1 )
(24)
q 0 2
q 1 2
q 2 2
q 3 2
2
(
q 1 q 3
q 0 q 2 )
2
(
q 2 q 3 +
q 0 q 1 )
+
The Levenberg-Marquardt (L-M) technique [23] is applied to search for the optimal
solutions, as this is a non-linear optimisation problem, solved by combining gradi-
ent descent and Gauss-Newton iteration. The variation of the rotation matrix from
frame to frame can be derived using a linear optimisation [3], where the incremental
rotation quaternion is computed at each frame with
Δ
1
q
=(
ζ , θ x /
2
, θ y /
2
, θ z /
2
) ,
(25)
ζ =( θ x 2
+ θ y 2
+ θ z 2
) /
4
.
Assuming a vector v
, then the derivatives of the rotation matrix with
respect to the quaternion can be obtained
=[
v 1 ,
v 2 ,
v 3 ]
c 1
c 4
c 3
c 2
R 1
,
q =
c 2
c 3
c 4
c 1
(26)
c 3
c 2
c 1
c 4
where
c 1
=
q 1 v 1
+
q 4 v 2
q 2 v 3
,
c 2
=
q 3 v 1
+
q 0 v 2
+
q 1 v 3
,
(27)
c 3 =
q 2 v 1
q 1 v 2 +
q 0 v 3 ,
c 4 =
q 1 v 1 +
q 2 v 2 +
q 3 v 3 .
 
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