Gyrocompass

A gyrocompass is a type of non-magnetic

• they find
  magnetic north
  , and
• they have a greater degree of accuracy because they are unaffected by
  ferromagnetic materials, such as in a ship's steel hull, which distort the magnetic field.^[4]

Aircraft commonly use gyroscopic instruments (but not a gyrocompass) for navigation and attitude monitoring; for details, see flight instruments (specifically the heading indicator) and gyroscopic autopilot.

History

The first, not yet practical,

In the United States, Elmer Ambrose Sperry produced a workable gyrocompass system (1908: U.S. patent 1,242,065), and founded the Sperry Gyroscope Company. The unit was adopted by the U.S. Navy (1911^[3]), and played a major role in World War I. The Navy also began using Sperry's "Metal Mike": the first gyroscope-guided autopilot steering system. In the following decades, these and other Sperry devices were adopted by steamships such as the RMS Queen Mary, airplanes, and the warships of World War II. After his death in 1930, the Navy named the USS Sperry after him.

Meanwhile, in 1913, C. Plath (a Hamburg, Germany-based manufacturer of navigational equipment including sextants and magnetic compasses) developed the first gyrocompass to be installed on a commercial vessel. C. Plath sold many gyrocompasses to the Weems’ School for Navigation in Annapolis, MD, and soon the founders of each organization formed an alliance and became Weems & Plath.^[9]

Before the success of the gyrocompass, several attempts had been made in Europe to use a gyroscope instead. By 1880,

In 1923 Max Schuler published his paper containing his observation that if a gyrocompass possessed Schuler tuning such that it had an oscillation period of 84.4 minutes (which is the orbital period of a notional satellite orbiting around the Earth at sea level), then it could be rendered insensitive to lateral motion and maintain directional stability.^[10]

Operation

A

Another, more practical, method is to use weights to force the axis of the compass to remain horizontal (perpendicular to the direction of the center of the Earth), but otherwise allow it to rotate freely within the horizontal plane.[2]^[3] In this case, gravity will apply a torque forcing the compass's axis toward true north. Because the weights will confine the compass's axis to be horizontal with respect to the Earth's surface, the axis can never align with the Earth's axis (except on the Equator) and must realign itself as the Earth rotates. But with respect to the Earth's surface, the compass will appear to be stationary and pointing along the Earth's surface toward the true North Pole.

Since the gyrocompass's north-seeking function depends on the rotation around the axis of the Earth that causes

Mathematical model

We consider a gyrocompass as a gyroscope which is free to rotate about one of its symmetry axes, also the whole rotating gyroscope is free to rotate on the horizontal plane about the local vertical. Therefore there are two independent local rotations. In addition to these rotations we consider the rotation of the Earth about its north-south (NS) axis, and we model the planet as a perfect sphere. We neglect friction and also the rotation of the Earth about the Sun.

In this case a non-rotating observer located at the center of the Earth can be approximated as being an inertial frame. We establish cartesian coordinates ${\displaystyle (X_{1},Y_{1},Z_{1})}$ for such an observer (whom we name as 1-O), and the barycenter of the gyroscope is located at a distance ${\displaystyle R}$ from the center of the Earth.

First time-dependent rotation

Consider another (non-inertial) observer (the 2-O) located at the center of the Earth but rotating about the NS-axis by ${\displaystyle \Omega .}$ We establish coordinates attached to this observer as

$${\displaystyle {\begin{pmatrix}X_{2}\\Y_{2}\\Z_{2}\end{pmatrix}}={\begin{pmatrix}\cos \Omega t&\sin \Omega t&0\\-\sin \Omega t&\cos \Omega t&0\\0&0&1\end{pmatrix}}{\begin{pmatrix}X_{1}\\Y_{1}\\Z_{1}\end{pmatrix}}}$$

so that the unit ${\displaystyle {\hat {X}}_{1}}$ versor ${\displaystyle (X_{1}=1,Y_{1}=0,Z_{1}=0)^{T}}$ is mapped to the point ${\displaystyle (X_{2}=\cos \Omega t,Y_{2}=-\sin \Omega t,Z_{2}=0)^{T}}$. For the 2-O neither the Earth nor the barycenter of the gyroscope is moving. The rotation of 2-O relative to 1-O is performed with angular velocity ${\displaystyle {\vec {\Omega }}=(0,0,\Omega )^{T}}$. We suppose that the ${\displaystyle X_{2}}$ axis denotes points with zero longitude (the prime, or Greenwich, meridian).

Second and third fixed rotations

We now rotate about the ${\textstyle Z_{2}}$ axis, so that the ${\textstyle X_{3}}$-axis has the longitude of the barycenter. In this case we have

$${\displaystyle {\begin{pmatrix}X_{3}\\Y_{3}\\Z_{3}\end{pmatrix}}={\begin{pmatrix}\cos \Phi &\sin \Phi &0\\-\sin \Phi &\cos \Phi &0\\0&0&1\end{pmatrix}}{\begin{pmatrix}X_{2}\\Y_{2}\\Z_{2}\end{pmatrix}}.}$$

With the next rotation (about the axis ${\textstyle Y_{3}}$ of an angle ${\textstyle \delta }$, the co-latitude) we bring the ${\textstyle Z_{3}}$ axis along the local zenith (${\textstyle Z_{4}}$-axis) of the barycenter. This can be achieved by the following orthogonal matrix (with unit determinant)

$${\displaystyle {\begin{pmatrix}X_{4}\\Y_{4}\\Z_{4}\end{pmatrix}}={\begin{pmatrix}\cos \delta &0&-\sin \delta \\0&1&0\\\sin \delta &0&\cos \delta \end{pmatrix}}{\begin{pmatrix}X_{3}\\Y_{3}\\Z_{3}\end{pmatrix}},}$$

so that the ${\textstyle {\hat {Z}}_{3}}$ versor ${\textstyle (X_{3}=0,Y_{3}=0,Z_{3}=1)^{T}}$ is mapped to the point ${\textstyle (X_{4}=-\sin \delta ,Y_{4}=0,Z_{4}=\cos \delta )^{T}.}$

Constant translation

We now choose another coordinate basis whose origin is located at the barycenter of the gyroscope. This can be performed by the following translation along the zenith axis

$${\displaystyle {\begin{pmatrix}X_{5}\\Y_{5}\\Z_{5}\end{pmatrix}}={\begin{pmatrix}X_{4}\\Y_{4}\\Z_{4}\end{pmatrix}}-{\begin{pmatrix}0\\0\\R\end{pmatrix}},}$$

so that the origin of the new system, ${\displaystyle (X_{5}=0,Y_{5}=0,Z_{5}=0)^{T}}$ is located at the point ${\displaystyle (X_{4}=0,Y_{4}=0,Z_{4}=R)^{T},}$ and ${\displaystyle R}$ is the radius of the Earth. Now the ${\displaystyle X_{5}}$-axis points towards the south direction.

Fourth time-dependent rotation

Now we rotate about the zenith ${\displaystyle Z_{5}}$-axis so that the new coordinate system is attached to the structure of the gyroscope, so that for an observer at rest in this coordinate system, the gyrocompass is only rotating about its own axis of symmetry. In this case we find

$${\displaystyle {\begin{pmatrix}X_{6}\\Y_{6}\\Z_{6}\end{pmatrix}}={\begin{pmatrix}\cos \alpha &\sin \alpha &0\\-\sin \alpha &\cos \alpha &0\\0&0&1\end{pmatrix}}{\begin{pmatrix}X_{5}\\Y_{5}\\Z_{5}\end{pmatrix}}.}$$

The axis of symmetry of the gyrocompass is now along the ${\displaystyle X_{6}}$-axis.

Last time-dependent rotation

The last rotation is a rotation on the axis of symmetry of the gyroscope as in

$${\displaystyle {\begin{pmatrix}X_{7}\\Y_{7}\\Z_{7}\end{pmatrix}}={\begin{pmatrix}1&0&0\\0&\cos \psi &\sin \psi \\0&-\sin \psi &\cos \psi \end{pmatrix}}{\begin{pmatrix}X_{6}\\Y_{6}\\Z_{6}\end{pmatrix}}.}$$

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 potential energy is constant. Therefore its Lagrangian ${\displaystyle {\mathcal {L}}}$ corresponds to its kinetic energy ${\displaystyle K}$ only. We have

$${\displaystyle {\mathcal {L}}=K={\frac {1}{2}}{\vec {\omega }}^{T}I{\vec {\omega }}+{\frac {1}{2}}M{\vec {v}}_{\rm {CM}}^{2},}$$

${\displaystyle M}$

$${\displaystyle {\vec {v}}_{\rm {CM}}^{2}=\Omega ^{2}R^{2}\sin ^{2}\delta ={\rm {constant}}}$$

$${\displaystyle I={\begin{pmatrix}I_{1}&0&0\\0&I_{2}&0\\0&0&I_{2}\end{pmatrix}}}$$

$${\displaystyle {\begin{aligned}{\vec {\omega }}&={\begin{pmatrix}1&0&0\\0&\cos \psi &\sin \psi \\0&-\sin \psi &\cos \psi \end{pmatrix}}{\begin{pmatrix}{\dot {\psi }}\\0\\0\end{pmatrix}}+{\begin{pmatrix}1&0&0\\0&\cos \psi &\sin \psi \\0&-\sin \psi &\cos \psi \end{pmatrix}}{\begin{pmatrix}\cos \alpha &\sin \alpha &0\\-\sin \alpha &\cos \alpha &0\\0&0&1\end{pmatrix}}{\begin{pmatrix}0\\0\\{\dot {\alpha }}\end{pmatrix}}\\&\qquad +{\begin{pmatrix}1&0&0\\0&\cos \psi &\sin \psi \\0&-\sin \psi &\cos \psi \end{pmatrix}}{\begin{pmatrix}\cos \alpha &\sin \alpha &0\\-\sin \alpha &\cos \alpha &0\\0&0&1\end{pmatrix}}{\begin{pmatrix}\cos \delta &0&-\sin \delta \\0&1&0\\\sin \delta &0&\cos \delta \end{pmatrix}}{\begin{pmatrix}\cos \Phi &\sin \Phi &0\\-\sin \Phi &\cos \Phi &0\\0&0&1\end{pmatrix}}{\begin{pmatrix}\cos \Omega t&\sin \Omega t&0\\-\sin \Omega t&\cos \Omega t&0\\0&0&1\end{pmatrix}}{\begin{pmatrix}0\\0\\\Omega \end{pmatrix}}\\&={\begin{pmatrix}{\dot {\psi }}\\0\\0\\\end{pmatrix}}+{\begin{pmatrix}0\\{\dot {\alpha }}\sin \psi \\{\dot {\alpha }}\cos \psi \end{pmatrix}}+{\begin{pmatrix}-\Omega \sin \delta \cos \alpha \\\Omega (\sin \delta \sin \alpha \cos \psi +\cos \delta \sin \psi )\\\Omega (-\sin \delta \sin \alpha \sin \psi +\cos \delta \cos \psi )\end{pmatrix}}\end{aligned}}}$$

Therefore we find

$${\displaystyle {\begin{aligned}{\mathcal {L}}&={\frac {1}{2}}\left[I_{1}\omega _{1}^{2}+I_{2}\left(\omega _{2}^{2}+\omega _{3}^{2}\right)\right]\\&={\frac {1}{2}}I_{1}\left({\dot {\psi }}-\Omega \sin \delta \cos \alpha \right)^{2}+{\frac {1}{2}}I_{2}\left\{\left[{\dot {\alpha }}\sin \psi +\Omega (\sin \delta \sin \alpha \cos \psi +\cos \delta \sin \psi )\right]^{2}+\left[{\dot {\alpha }}\cos \psi +\Omega (-\sin \delta \sin \alpha \sin \psi +\cos \delta \cos \psi )\right]^{2}\right\}\\&={\frac {1}{2}}I_{1}\left({\dot {\psi }}-\Omega \sin \delta \cos \alpha \right)^{2}+{\frac {1}{2}}I_{2}\left\{{\dot {\alpha }}^{2}+\Omega ^{2}\left(\cos ^{2}\delta +\sin ^{2}\alpha \sin ^{2}\delta \right)+2{\dot {\alpha }}\Omega \cos \delta \right\}\end{aligned}}}$$

The Lagrangian can be rewritten as

$${\displaystyle {\mathcal {L}}={\mathcal {L}}_{1}+{\frac {1}{2}}I_{2}\Omega ^{2}\cos ^{2}\delta +{\frac {d}{dt}}(I_{2}\alpha \Omega \cos \delta ),}$$

$${\displaystyle {\mathcal {L}}_{1}={\frac {1}{2}}I_{1}\left({\dot {\psi }}-\Omega \sin \delta \cos \alpha \right)^{2}+{\frac {1}{2}}I_{2}\left({\dot {\alpha }}^{2}+\Omega ^{2}\sin ^{2}\alpha \sin ^{2}\delta \right)}$$

${\displaystyle \partial {\mathcal {L}}_{1}/\partial \psi =0}$

$${\displaystyle L_{x}\equiv {\frac {\partial {\mathcal {L}}_{1}}{\partial {\dot {\psi }}}}=I_{1}\left({\dot {\psi }}-\Omega \sin \delta \cos \alpha \right)=\mathrm {constant} .}$$

Since the angular momentum ${\displaystyle {\vec {L}}}$ of the gyrocompass is given by ${\displaystyle {\vec {L}}=I{\vec {\omega }},}$ we see that the constant ${\displaystyle L_{x}}$ is the component of the angular momentum about the axis of symmetry. Furthermore, we find the equation of motion for the variable ${\displaystyle \alpha }$ as

$${\displaystyle {\frac {d}{dt}}\left({\frac {\partial {\mathcal {L}}_{1}}{\partial {\dot {\alpha }}}}\right)={\frac {\partial {\mathcal {L}}_{1}}{\partial \alpha }},}$$

$${\displaystyle {\begin{aligned}I_{2}{\ddot {\alpha }}&=I_{1}\Omega \left({\dot {\psi }}-\Omega \sin \delta \cos \alpha \right)\sin \delta \sin \alpha +{\frac {1}{2}}I_{2}\Omega ^{2}\sin ^{2}\delta \sin 2\alpha \\&=L_{x}\Omega \sin \delta \sin \alpha +{\frac {1}{2}}I_{2}\Omega ^{2}\sin ^{2}\delta \sin 2\alpha \end{aligned}}}$$

Particular case: the poles

At the poles we find ${\displaystyle \sin \delta =0,}$ and the equations of motion become

$${\displaystyle {\begin{aligned}L_{x}&=I_{1}{\dot {\psi }}=\mathrm {constant} \\I_{2}{\ddot {\alpha }}&=0\end{aligned}}}$$

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 ${\displaystyle \sin \delta \neq 0}$ and that ${\displaystyle \alpha \approx 0}$, 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

$${\displaystyle {\begin{aligned}L_{x}&\approx I_{1}\left({\dot {\psi }}-\Omega \sin \delta \right)\\I_{2}{\ddot {\alpha }}&\approx \left(L_{x}\Omega \sin \delta +I_{2}\Omega ^{2}\sin ^{2}\delta \right)\alpha \end{aligned}}}$$

Consider the case that

$${\displaystyle L_{x}<0,}$$

$${\displaystyle \left|{\dot {\psi }}\right|\gg \Omega .}$$

Therefore, for fast spinning rotations, ${\displaystyle L_{x}<0}$ implies ${\displaystyle {\dot {\psi }}<0.}$ In this case, the equations of motion further simplify to

$${\displaystyle {\begin{aligned}L_{x}&\approx -I_{1}\left|{\dot {\psi }}\right|\approx \mathrm {constant} \\I_{2}{\ddot {\alpha }}&\approx -I_{1}\left|{\dot {\psi }}\right|\Omega \sin \delta \alpha \end{aligned}}}$$

Therefore we find small oscillations about the north-south line, as ${\displaystyle \alpha \approx A\sin({\tilde {\omega }}t+B)}$, where the angular velocity of this harmonic motion of the axis of symmetry of the gyrocompass about the north-south line is given by

$${\displaystyle {\tilde {\omega }}={\sqrt {\frac {I_{1}\sin \delta }{I_{2}}}}{\sqrt {\left|{\dot {\psi }}\right|\Omega }},}$$

$${\displaystyle T={\frac {2\pi }{\sqrt {\left|{\dot {\psi }}\right|\Omega }}}{\sqrt {\frac {I_{2}}{I_{1}\sin \delta }}}.}$$

Therefore ${\displaystyle {\tilde {\omega }}}$ is proportional to the geometric mean of the Earth and spinning angular velocities. In order to have small oscillations we have required ${\displaystyle {\dot {\psi }}<0}$, so that the North is located along the right-hand-rule direction of the spinning axis, that is along the negative direction of the ${\displaystyle X_{7}}$-axis, the axis of symmetry. As a side result, on measuring ${\displaystyle T}$ (and knowing ${\displaystyle {\dot {\psi }}}$), one can deduce the local co-latitude ${\displaystyle \delta .}$

See also

• Acronyms and abbreviations in avionics
• Heading indicator, also known as direction indicator, a lightweight gyroscope (not a gyrocompass) used on aircraft
• HRG gyrocompass
• Fluxgate compass
• Fibre optic gyrocompass
• Inertial navigation system, a more complex system that also incorporates accelerometers
• Schuler tuning
• Binnacle

Notes

References