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Measure $\theta$ from the upward vertical, let ${\mathbf{e}}_3$ point from the support along the symmetry axis, and use the support as the reference point for all moments of inertia. Choose ${\mathbf{e}}_2=(\hat{\mathbf{z}}-\cos \theta {\mathbf{e}}_3)/\sin \theta$ and ${\mathbf{e}}_1={\mathbf{e}}_2\times {\mathbf{e}}_3$. These form an instantaneous principal-axis frame; its transverse axes need not spin with marks painted on the body. The three infinitesimal Euler angles rotations add as vectors:Resolving the vertical direction givesA convention matters here: in genuinely material-fixed transverse axes, the first two components are rotated by the spin angle $\chi$. For one conventional choice they are $\dot{\theta }\cos \chi +\dot{\varphi }\sin \theta \sin \chi$ and $-\dot{\theta }\sin \chi +\dot{\varphi }\sin \theta \cos \chi$. Since $I_1=I_2$, that rotation changes neither kinetic energy nor the axial and vertical conserved momenta. The simplified displayed components use the instantaneous transverse frame just defined.^[ref]

For a frictionless fixed support, the support does no work and gravity is conservative, so the total energy is constant. The heavy symmetric top Lagrangian isThe two cyclic coordinates $\chi ,\varphi$ giveEquivalently, gravitational torque has no component along the vertical or symmetry axis. Eliminate $\dot{\varphi }=(J_z-J_3\cos \theta )/(I_1{\sin }^2\theta )$ and the axial spin from $E=T+V$:This is the heavy symmetric top reduction at fixed $J_3,J_z$. A constant inclination must satisfy the reduced equation of motion $I_1\ddot{\theta }=-U_{\mathrm{eff}}^{\prime }$, not merely have zero instantaneous velocity. Differentiation at fixed conserved momenta gives, for $0<\theta <\pi$ and $\Omega =\dot{\varphi }$,Therefore uniform precession of a heavy symmetric top requiresFor a prescribed upward inclination $0<\theta <\pi /2$, real precession rates exist only if the quadratic discriminant is nonnegative:For positive spin the absolute-value sign can be dropped. The two roots are the slow and fast branches and coalesce at the threshold. The familiar upright-limit critical magnitude is $2\sqrt{I_{1}mgh}$; strict inequality gives gyroscopic stabilization of an inverted symmetric top. This limiting stability condition should not be confused with a positive minimum over every possible inclination. As $\theta$ approaches $\pi /2$ from below, the fixed-inclination threshold approaches zero. At the horizontal angle $\Omega =mgh/J_3$ for $J_3\ne 0$; below the horizontal the quadratic admits real solutions even at $J_3=0$. The stated minimum therefore needs the inclination convention made explicit.