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solving a minimum acceleration problem . Let us recall the essential of this method.
We assume that the current position of the camera with respect to its desired position
is given by the rotation matrix
R
(
t
) and the translation vector
b
(
t
). In this case, the
collineation matrix is given by
K
+
R
(
t
)+
b
d
f
(
t
)
n
fT
K
G
(
t
) ∝
.
1 vector [
v
T
T
]
T
,where
v
denotes the time derivative of
b
We denote
U
the 6
×
ω
and
is defined by (7.4). In [23], the problems, denoted
PC1
and
PC2
, of finding
a path of the collineation matrix corresponding to the minimum energy and mini-
mum acceleration problem respectively have been solved. These problems can be
formulated as
ω
•
(PC1)
find
G
(
t
) minimizing:
J
1
=
1
0
U
T
U
dt
,
subject to (7.4),
v
=
b
and with boundary conditions:
⎧
⎨
⎩
G
(0)
∝
G
0
,
G
(1)
∝
I
3
×
3
;
•
(PC2)
find
G
(
t
) minimizing:
J
2
=
1
0
U
T
U
dt
,
subject to (7.4),
v
=
b
and with boundary conditions:
⎧
⎨
G
(0)
∝
G
0
,
G
(1)
∝
I
3
×
3
,
⎩
,
U
(0)=
0
6
×
1
.
U
(1)=
0
6
×
1
In this case, the camera velocity is constrained to be
0
at the beginning and the
end of the task. The boundary conditions are verified if
R
(0)=
R
0
,
b
(0)=
b
0
,
R
(1)=
I
3
×
3
and
b
(1)=
0
. The solutions of
PC1
and
PC2
are given by the following
proposition [23].
Proposition 7.1.
The optimal path of the collineation matrix in the sense of
PC1
and
PC2
is given by
G
(
t
)
∝
(1
−
q
(
t
))
Φ
0
+(
G
0
−
Φ
0
)
Γ
(
θ
0
,
t
)
(7.9)
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