English: A Klein bottle split lengthwise. The split does not separate it into two disconnected parts; instead it creates an orientable strip of twice the length of the original Klein bottle.

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

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

Graphed using Grapher.