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Euler angles describe the orientation of a rigid body through three successive angular rotations. Their convention specifies the axes and order of the rotations. For a symmetric top, a common convention takes $\theta$ as the inclination, $\varphi$ as precession and $\psi$ as spin about the body axis.

The two cyclic Euler angles give conserved canonical momentaHere $L_3$ is the body-axis angular momentum and $L_z$ the vertical angular momentum. Thus ${\omega }_3=L_3/I_3$ is constant. Since the Lagrangian has no explicit time dependence, the energy is also conserved:These are the three independent conserved quantities furnished by the continuous symmetries of the heavy symmetric top.

For nonvertical uniform precession of a heavy symmetric top, write $\dot{\varphi }=\Omega$ and fix $\theta$. The $\theta$ Euler-Lagrange equation reduces toFor $\cos \theta \ne 0$ its solutions areFor $0<\theta <\pi /2$, uniform precession therefore requires $|{\omega }_3|\ge 2\sqrt{I_{1}Mgl\cos \theta }/I_3$. For $\pi /2<\theta <\pi$, the discriminant is positive for every spin, including zero. At $\theta =\pi /2$, the condition is $\Omega =Mgl/(I_3{\omega }_3)$, requiring nonzero spin. Once this condition holds, choose $L_z=I_1\Omega {\sin }^2\theta +L_3\cos \theta$ and $\dot{\psi }={\omega }_3-\Omega \cos \theta$; both cyclic equations are then satisfied. The exactly vertical configurations are also solutions for arbitrary spin, but the azimuthal member of the Euler angles is degenerate there, so their existence is separate from this nonvertical discriminant condition.

In the Euler angles for a symmetric top convention, $\theta$ is the inclination of the body symmetry axis ${\mathbf{e}}_3$ from the fixed vertical, $\varphi$ is the azimuth of the line of nodes in the fixed horizontal plane, and $\psi$ is rotation about ${\mathbf{e}}_3$ measured from that line: