= Gyroscope vector geometry
{title2=$\boldsymbol\tau_O=d\mathbf L_O/dt$}
Choose a fixed support point $O$ and nonrotating axes in an <inertial frame>. All <angular momenta>, <torques>, and <moments of inertia> below refer to $O$. Let $\widehat{\mathbf e}$ point along the axle from $O$ toward the <centre of mass>, and let $\widehat{\mathbf z}$ point vertically upward. Write $\theta$ for their <angle>. For a rapidly spinning <symmetric top>, the dominant <angular momentum> is $\mathbf L_O\simeq L_s\widehat{\mathbf e}$, where $L_s=I_3\omega_3$ is signed and $I_3$ is the axial <principal moment of inertia>. The sign of the axial <angular velocity> $\omega_3$ follows the <right-hand rule>.
\Image[/gyroscope-wheel-torque-vector-geometry.png]
{title=A fixed wheel support separates the downward gravitational force from the transverse torque and sideways precession}
With <mass> $M$, <centre of mass> distance $d$, and constant <gravitational acceleration> $g$, the gravitational <force> and its <torque> are
$$
\mathbf r=d\widehat{\mathbf e},\qquad \mathbf F_g=-Mg\widehat{\mathbf z},\qquad
\boldsymbol\tau_O=\mathbf r\times\mathbf F_g=Mgd\,\widehat{\mathbf z}\times\widehat{\mathbf e}.
$$
An ideal support applies a <force> at $O$, so its <lever arm> and <torque> about $O$ vanish. At the instant the axle lies in the $xz$ plane with a positive $x$ component, the gravitational <torque> points along $+y$. It adds $d\mathbf L_O=\boldsymbol\tau_O\,dt$ toward $+y$, rather than along the downward <force>. The <dot product> identity $d|\mathbf L_O|^2/dt=2\mathbf L_O\cdot\boldsymbol\tau_O$ shows why perpendicular <torque> changes direction without instantly changing magnitude. This vector balance and the slow-turning relation $\boldsymbol\tau_O=\boldsymbol\Omega\times\mathbf L_O$ are developed in <The Feynman Lectures on Physics, Volume I by Richard Feynman and others/Chapter 20: Rotation in space> (sections 20-2 and 20-3, equations 20.13–20.15).
For steady <precession> about the vertical, $\boldsymbol\Omega=\Omega\widehat{\mathbf z}$ and $d\widehat{\mathbf e}/dt=\Omega\widehat{\mathbf z}\times\widehat{\mathbf e}$. Equating the two <cross products> gives
$$
\Omega\simeq\frac{Mgd}{L_s}=\frac{Mgd}{I_3\omega_3},\qquad
|\boldsymbol\tau_O|=Mgd\sin\theta.
$$
The <angular momentum> tip traces a <circle> of <radius> $|L_s|\sin\theta$: its sideways <speed> is $|\Omega L_s|\sin\theta$. Both that <radius> and the gravitational <lever arm> contain $\sin\theta$, explaining its cancellation for $0<\theta<\pi$. Exactly vertical alignment has zero gravitational <torque> and no distinguishable azimuthal <precession> to infer by this cancellation. Positive $L_s$ gives positive <precession>, counterclockwise viewed from above; reversing the spin reverses the slow <precession>. <University Physics Volume 1 by William Moebs and others/11.4 Precession of a Gyroscope>[University Physics Volume 1, section 11.4, equation 11.12 and the torque and precession-circle derivation]{ref}
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