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[
x
y
z
]
=
[
3
4
cos
t
−
1
4
sin
t
−
1
2
2
sin
2
t
3
4
sin
t
−
1
4
cos
t
−
1
2
sin
2
t
0
]
+
ρ
+
sin
(
t
−
ψ
)
−
1
2
sin
2
(
t
−
ψ
)
+
1
3
sin
3
(
t
−
ψ
)
−
1
4
sin
4
(
t
−
ψ
)
+
1
10
sin
5
(
t
−
ψ
)
10
(
(
sin
ϕ
)
[
0
0
1
]
+
(
−
1
)
⌊
t
−
2.356194
2
π
⌋
cos
ϕ
5
8
+
1
4
sin
2
2
t
+
3
8
sin
2
t
+
1
2
2
(
sin
2
t
)
(
sin
t
−
cos
t
)
[
−
3
4
cos
t
−
1
4
sin
t
+
1
2
2
sin
2
t
−
3
4
sin
t
−
1
4
cos
t
−
1
2
2
sin
2
t
0
]
)
{\displaystyle {\begin{bmatrix}x\\y\\z\end{bmatrix}}={\begin{bmatrix}{3 \over 4}\cos t-{1 \over 4}\sin t-{1 \over 2{\sqrt {2}}}\sin ^{2}t\\{3 \over 4}\sin t-{1 \over 4}\cos t-{1 \over 2}\sin ^{2}t\\0\end{bmatrix}}+{\rho +\sin(t-\psi )-{1 \over 2}\sin 2(t-\psi )+{1 \over 3}\sin 3(t-\psi )-{1 \over 4}\sin 4(t-\psi )+{1 \over 10}\sin 5(t-\psi ) \over 10}{\Biggl (}(\sin \phi ){\begin{bmatrix}0\\0\\1\end{bmatrix}}+{(-1)^{{\bigl \lfloor }{t-2.356194 \over 2\pi }{\bigr \rfloor }}\cos \phi \over {\sqrt {{5 \over 8}+{1 \over 4}\sin ^{2}2t+{3 \over 8}\sin 2t+{1 \over 2{\sqrt {2}}}(\sin 2t)(\sin t-\cos t)}}}{\begin{bmatrix}-{3 \over 4}\cos t-{1 \over 4}\sin t+{1 \over 2{\sqrt {2}}}\sin 2t\\-{3 \over 4}\sin t-{1 \over 4}\cos t-{1 \over 2{\sqrt {2}}}\sin 2t\\0\end{bmatrix}}{\Biggr )}}
