Non-spiralling overdamped central oscillator trajectory ($\Delta {\varphi }_{\infty }=\arctan [\omega /(\gamma +\sqrt{{\gamma }^2-{\omega }^2})]<\pi /4$)

For the initially circular damped planar oscillator with overdamping $\gamma >\omega$, put $\alpha =\sqrt{{\gamma }^2-{\omega }^2}$. The exact solution replaces $\cos \nu t,\sin \nu t/\nu$ by $\cosh \alpha t,\sinh \alpha t/\alpha$. Resolving along the starting radius and its anticlockwise tangent yieldsIts limiting angle is the title value. Both coordinates remain positive when $c>0$, so the trajectory returns to the origin without repeated revolutions. It initially moves tangentially and bends toward its slow decaying eigenmode, rather than following a radial straight line from the start.