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Differentiating the spherical-geodesic first integral givesThe same equation follows from the affine geodesic equation after eliminating the arc-length derivatives. The squared first-derivative equation also fixes the normalization constant; a second-order solution must retain that first integral.

Use arc length $s$ as the affine parameter, and put a dot over derivatives with respect to $s$. The geodesic Lagrangian on the sphere isThe cyclic coordinate $\varphi$ gives $\partial L/\partial \dot{\varphi }=a^2{\sin }^2\theta \dot{\varphi }=\text{constant}$. With the constant normalized as in the paper,Here $h$ has units of inverse length, because $s$ is physical arc length. Unit-speed normalization givesFor $h\ne 0$, division by ${\dot{\varphi }}^2$ yields the spherical-geodesic first integralThe turning latitudes satisfy $\sin \theta =|ah|$. In the usual polar-coordinate range $0\le \theta \le \pi$,The maximum departure from the equator, often the geometric quantity intended by an angular amplitude, is insteadThus polar angle and signed latitude must not be interchanged. At $|ah|=1$ the geodesic is the equator. At $h=0$ it is a meridian through the poles; $\varphi$ cannot serve as a global parameter and the divided equation is inapplicable. The limiting polar range is then $[0,\pi ]$.

The original PDF prints an ambiguous derivative with the square on $\theta$ rather than a conventional derivative notation. If the intended equation was second order, differentiating the first integral, or eliminating $\dot{\theta }$ from the Euler-Lagrange equation, gives the equally useful formIt has the same normalized geodesics and the same turning angles. Giving both forms resolves the typographical ambiguity without pretending that a second derivative is the square of a first derivative. Geometrically these trajectories are great circles, with inclination $\arccos |ah|$ to the equator.

With polar angle $0\le \theta \le \pi$ and $0<|ah|\le 1$, the spherical-geodesic first integral gives ${\theta }_{\min }=\arcsin |ah|$ and ${\theta }_{\max }=\pi -\arcsin |ah|$. The maximum absolute latitude is the different angle $\arccos |ah|$. At $|ah|=1$ the great circle is the equator; at $h=0$ it is a meridian reaching both poles.