User:AugPi/sandbox

${\begin{aligned}{\begin{bmatrix}x\\y\\z\end{bmatrix}}=&{\begin{bmatrix}{3 \over 4}\cos t-{1 \over 4}\sin t-{1 \over 2{\sqrt {2}}}\sin ^{2}t\\{3 \over 4}\sin t-{1 \over 4}\cos t-{1 \over 2}\sin ^{2}t\\0\end{bmatrix}}\\&+{\rho +\sin(t-\psi )-{1 \over 2}\sin 2(t-\psi )+{1 \over 3}\sin 3(t-\psi )-{1 \over 4}\sin 4(t-\psi )+{1 \over 10}\sin 5(t-\psi ) \over 10}\\&\qquad \qquad \cdot {\Biggl (}(\sin \phi ){\begin{bmatrix}0\\0\\1\end{bmatrix}}+{(-1)^{{\bigl \lfloor }{t-2.356194 \over 2\pi }{\bigr \rfloor }}\cos \phi \over {\sqrt {{5 \over 8}+{1 \over 4}\sin ^{2}2t+{3 \over 8}\sin 2t+{1 \over 2{\sqrt {2}}}(\sin 2t)(\sin t-\cos t)}}}{\begin{bmatrix}-{3 \over 4}\cos t-{1 \over 4}\sin t+{1 \over 2{\sqrt {2}}}\sin 2t\\-{3 \over 4}\sin t-{1 \over 4}\cos t-{1 \over 2{\sqrt {2}}}\sin 2t\\0\end{bmatrix}}{\Biggr )}\end{aligned}}$

where $0\leq t<2\pi$, $0\leq \phi <2\pi$, $\psi =5.497786=\pi +2.356194$ is a constant, and $\rho =2.4$ is slightly variable (e.g., it could be set to 2.3 instead). The constant 2.356194... is the solution to the equations $-{\tfrac {3}{4}}\cos t-{\tfrac {1}{4}}\sin t+{\tfrac {1}{2{\sqrt {2}}}}\sin 2t=0=-{\tfrac {3}{4}}\sin t-{\tfrac {1}{4}}\cos t-{\tfrac {1}{2{\sqrt {2}}}}\sin 2t$.