Editing Gyrocompass (section)

== Dynamics of the system ==
Since the height of the gyroscope's barycenter does not change (and the origin of the coordinate system is located at this same point), its [[gravitational energy|gravitational potential energy]] is constant. Therefore its Lagrangian <math>\mathcal{L}</math> corresponds to its kinetic energy <math>K</math> only. We have
<math display="block">\mathcal{L}=K=\frac{1}{2} \vec{\omega}^{T}I\vec\omega+\frac{1}{2} M \vec{v}_{\rm CM}^{2},</math>
where <math>M</math> is the mass of the gyroscope, and
<math display="block">\vec{v}_{\rm CM}^{2}=\Omega^2 R^2 \sin^2\delta={\rm constant}</math>
is the squared inertial speed of the origin of the coordinates of the final coordinate system (i.e. the center of mass). This constant term does not affect the dynamics of the gyroscope and it can be neglected. On the other hand, the tensor of inertia is given by
<math display="block">I=\begin{pmatrix}
I_{1}&0&0\\
0 & I_{2}&0\\
0 &0  & I_{2}
\end{pmatrix}</math>
and
<math display="block">\begin{align}
\vec{\omega}&=\begin{pmatrix}
1 & 0 & 0\\
0 & \cos\psi & \sin\psi\\
0 & -\sin\psi & \cos\psi
\end{pmatrix} \begin{pmatrix} \dot{\psi}\\ 0\\ 0 \end{pmatrix}+\begin{pmatrix}
1 & 0 & 0\\
0 & \cos\psi & \sin\psi\\
0 & -\sin\psi & \cos\psi
\end{pmatrix} \begin{pmatrix}
\cos\alpha & \sin\alpha & 0\\
-\sin\alpha & \cos\alpha & 0\\
0 & 0 & 1
\end{pmatrix} \begin{pmatrix}
0\\
0\\
\dot{\alpha}
\end{pmatrix}\\
&\qquad + \begin{pmatrix}
1 & 0 & 0\\
0 & \cos\psi & \sin\psi\\
0 & -\sin\psi & \cos\psi
\end{pmatrix} \begin{pmatrix}
\cos\alpha & \sin\alpha & 0\\
-\sin\alpha & \cos\alpha & 0\\
0 & 0 & 1
\end{pmatrix} \begin{pmatrix}
\cos\delta & 0 & -\sin\delta\\
0 & 1 & 0\\
\sin\delta & 0 & \cos\delta
\end{pmatrix} \begin{pmatrix}
\cos\Phi & \sin\Phi & 0\\
-\sin\Phi & \cos\Phi & 0\\
0 & 0 & 1
\end{pmatrix} \begin{pmatrix}
\cos\Omega t & \sin\Omega t & 0\\
-\sin\Omega t & \cos\Omega t & 0\\
0 & 0 & 1
\end{pmatrix} \begin{pmatrix}
0\\
0\\
\Omega
\end{pmatrix}\\
&= \begin{pmatrix}
\dot{\psi}\\
0\\
0\\
\end{pmatrix}+ \begin{pmatrix}
0\\
\dot{\alpha}\sin\psi\\
\dot{\alpha}\cos\psi
\end{pmatrix}+ \begin{pmatrix}
-\Omega\sin\delta\cos\alpha\\
\Omega(\sin\delta\sin\alpha\cos\psi+\cos\delta\sin\psi)\\
\Omega(-\sin\delta\sin\alpha\sin\psi+\cos\delta\cos\psi)
\end{pmatrix}
\end{align}</math>

Therefore we find
<math display="block">\begin{align}
\mathcal{L} &= \frac{1}{2} \left [I_{1}\omega_{1}^{2}+I_{2} \left (\omega_{2}^{2}+\omega_{3}^{2} \right ) \right ]\\
            &= \frac{1}{2} I_{1} \left (\dot{\psi}-\Omega\sin\delta\cos\alpha \right )^{2} +\frac{1}{2} I_{2} \left \{ \left [\dot{\alpha}\sin\psi+\Omega(\sin\delta\sin\alpha\cos\psi+\cos\delta\sin\psi) \right ]^{2} + \left [\dot{\alpha}\cos\psi+\Omega(-\sin\delta\sin\alpha\sin\psi+\cos\delta\cos\psi) \right ]^{2} \right \} \\
            &= \frac{1}{2} I_{1} \left (\dot{\psi}-\Omega\sin\delta\cos\alpha \right )^{2}+\frac{1}{2} I_{2} \left \{ \dot{\alpha}^{2} +\Omega^{2} \left (\cos^{2}\delta+\sin^{2}\alpha\sin^{2}\delta \right ) +2\dot{\alpha}\Omega\cos\delta \right \}
\end{align}</math>

The Lagrangian can be rewritten as
<math display="block">\mathcal{L}=\mathcal{L}_{1}+\frac{1}{2} I_{2}\Omega^{2}\cos^{2}\delta+\frac{d}{dt}(I_{2}\alpha\Omega\cos\delta),</math>
where
<math display="block">\mathcal{L}_{1}=\frac{1}{2} I_{1} \left (\dot{\psi}-\Omega\sin\delta\cos\alpha \right )^{2}+\frac{1}{2} I_{2}\left (\dot{\alpha}^{2}+\Omega^{2}\sin^{2}\alpha\sin^{2}\delta \right )</math>
is the part of the Lagrangian responsible for the dynamics of the system. Then, since <math>\partial \mathcal{L}_1/\partial\psi = 0</math>, we find
<math display="block">L_{x}\equiv\frac{\partial \mathcal{L}_1}{\partial\dot{\psi}}=I_1 \left (\dot{\psi}-\Omega\sin\delta\cos\alpha \right )=\mathrm{constant}.</math>

Since the angular momentum <math>\vec L</math> of the gyrocompass is given by <math>\vec L=I\vec\omega,</math> we see that the constant <math>L_x</math> is the component of the angular momentum about the axis of symmetry. Furthermore, we find the equation of motion for the variable <math>\alpha</math> as
<math display="block">\frac{d}{dt} \left(\frac{\partial \mathcal{L}_{1}}{\partial\dot{\alpha}}\right)=\frac{\partial \mathcal{L}_{1}}{\partial\alpha},</math>
or
<math display="block">\begin{align}
I_{2}\ddot{\alpha} &=I_{1}\Omega \left (\dot{\psi}-\Omega\sin\delta\cos\alpha \right )\sin\delta\sin\alpha+\frac{1}{2} I_{2} \Omega^{2}\sin^{2}\delta\sin2\alpha\\
&=L_{x}\Omega\sin\delta\sin\alpha+\frac{1}{2} I_{2} \Omega^{2}\sin^{2}\delta\sin2\alpha
\end{align}</math>

=== Particular case: the poles ===
At the poles we find <math>\sin\delta=0,</math> and the equations of motion become
<math display="block">\begin{align}
L_{x} &=I_{1}\dot{\psi}=\mathrm{constant}\\
I_{2}\ddot{\alpha}&=0
\end{align}</math>

This simple solution implies that the gyroscope is uniformly rotating with constant angular velocity in both the vertical and symmetrical axis.

=== The general and physically relevant case ===
Let us suppose now that <math>\sin\delta\neq0</math> and that <math>\alpha\approx0</math>, that is the axis of the gyroscope is approximately along the north-south line, and let us find the parameter space (if it exists) for which the system admits stable small oscillations about this same line. If this situation occurs, the gyroscope will always be approximately aligned along the north-south line, giving direction. In this case we find
<math display="block">\begin{align}
L_{x}&\approx I_{1} \left (\dot{\psi}-\Omega\sin\delta \right )\\
I_{2}\ddot{\alpha}&\approx \left (L_{x}\Omega\sin\delta+I_{2} \Omega^{2}\sin^{2}\delta \right) \alpha
\end{align}</math>

Consider the case that
<math display="block">L_{x}<0,</math>
and, further, we allow for fast gyro-rotations, that is
<math display="block">\left |\dot{\psi} \right |\gg\Omega.</math>

Therefore, for fast spinning rotations, <math>L_x<0</math> implies <math>\dot\psi<0.</math> In this case, the equations of motion further simplify to
<math display="block">\begin{align}
L_{x}              &\approx -I_{1} \left |\dot{\psi} \right | \approx \mathrm{constant}\\
I_{2}\ddot{\alpha} &\approx -I_{1} \left |\dot{\psi} \right |\Omega \sin\delta\alpha
\end{align}</math>

Therefore we find small oscillations about the north-south line, as <math>\alpha\approx A\sin(\tilde\omega t+B)</math>, where the angular velocity of this harmonic motion of the axis of symmetry of the gyrocompass about the north-south line is given by
<math display="block">\tilde\omega=\sqrt{\frac{I_{1}\sin\delta}{I_{2}}}\sqrt{\left |\dot{\psi} \right |\Omega},</math>
which corresponds to a period for the oscillations given by
<math display="block">T=\frac{2\pi}{\sqrt{\left |\dot{\psi} \right |\Omega}}\sqrt{\frac{I_{2}}{I_{1}\sin\delta}}.</math>

Therefore <math>\tilde\omega</math> is proportional to the geometric mean of the Earth and spinning angular velocities. In order to have small oscillations we have required <math>\dot{\psi}<0</math>, so that the North is located along the right-hand-rule direction of the spinning axis, that is along the negative direction of the <math>X_7</math>-axis, the axis of symmetry. As a side result, on measuring <math>T</math> (and knowing <math>\dot{\psi}</math>), one can deduce the local co-latitude <math>\delta.</math>