Position angle

The position angle of the point t at the point s is the angle at which the green and the dashed great circles intersect at s. The unit directions u_E, u_N and the rotation axis ω are marked by arrows.

Detailed evaluation of the optimum direction is possible if the sea surface is approximated by a sphere surface. The standard computation places the ship at a geodetic latitude φ_s and geodetic longitude λ_s, where φ is considered positive if north of the equator, and where λ is considered positive if east of Greenwich. In the global coordinate system centered at the center of the sphere, the Cartesian components are

${\displaystyle {\mathbf {s} }=R\left({\begin{array}{c}\cos \varphi _{s}\cos \lambda _{s}\\\cos \varphi _{s}\sin \lambda _{s}\\\sin \varphi _{s}\end{array}}\right)}$

and the target position is

${\displaystyle {\mathbf {t} }=R\left({\begin{array}{c}\cos \varphi _{t}\cos \lambda _{t}\\\cos \varphi _{t}\sin \lambda _{t}\\\sin \varphi _{t}\end{array}}\right).}$

The North Pole is at

${\displaystyle {\mathbf {N} }=R\left({\begin{array}{c}0\\0\\1\end{array}}\right).}$

The minimum distance d is the distance along a great circle that runs through s and t. It is calculated in a plane that contains the sphere center and the great circle,

${\displaystyle d_{s,t}=R\theta _{s,t}}$

where θ is the angular distance of two points viewed from the center of the sphere, measured in radians. The cosine of the angle is calculated by the dot product of the two vectors

${\displaystyle \mathbf {s} \cdot \mathbf {t} =R^{2}\cos \theta _{s,t}=R^{2}(\sin \varphi _{s}\sin \varphi _{t}+\cos \varphi _{s}\cos \varphi _{t}\cos(\lambda _{t}-\lambda _{s}))}$

If the ship steers straight to the North Pole, the travel distance is

${\displaystyle d_{s,N}=R\theta _{s,N}=R(\pi /2-\varphi _{s})}$

If a ship starts at t and swims straight to the North Pole, the travel distance is

${\displaystyle d_{t,N}=R\theta _{t,n}=R(\pi /2-\varphi _{t})}$

The cosine formula of spherical trigonometry [1] yields for the angle p between the great circles through s that point to the North on one hand and to t on the other hand

${\displaystyle \cos \theta _{t,N}=\cos \theta _{s,t}\cos \theta _{s,N}+\sin \theta _{s,t}\sin \theta _{s,N}\cos p.}$

${\displaystyle \sin \varphi _{t}=\cos \theta _{s,t}\sin \varphi _{s}+\sin \theta _{s,t}\cos \varphi _{s}\cos p.}$

The sine formula yields

${\displaystyle {\frac {\sin p}{\sin \theta _{t,N}}}={\frac {\sin(\lambda _{t}-\lambda _{s})}{\sin \theta _{s,t}}}.}$

Solving this for sin θ_s,t and insertion in the previous formula gives an expression for the tangent of the position angle,

${\displaystyle \sin \varphi _{t}=\cos \theta _{s,t}\sin \varphi _{s}+{\frac {\sin(\lambda _{t}-\lambda _{s})}{\sin p}}\cos \varphi _{t}\cos \varphi _{s}\cos p;}$

${\displaystyle \tan p={\frac {\sin(\lambda _{t}-\lambda _{s})\cos \varphi _{t}\cos \varphi _{s}}{\sin \varphi _{t}-\cos \theta _{s,t}\sin \varphi _{s}}}.}$

Because the brief derivation gives an angle between 0 and π which does not reveal the sign (west or east of north ?), a more explicit derivation is desirable which yields separately the sine and the cosine of p such that use of the correct branch of the inverse tangent allows to produce an angle in the full range -π≤p≤π.

The computation starts from a construction of the great circle between s and t. It lies in the plane that contains the sphere center, s and t and is constructed rotating s by the angle θ_s,t around an axis ω. The axis is perpendicular to the plane of the great circle and computed by the normalized vector cross product of the two positions:

${\displaystyle \mathbf {\omega } ={\frac {1}{R^{2}\sin \theta _{s,t}}}\mathbf {s} \times \mathbf {t} ={\frac {1}{\sin \theta _{s,t}}}\left({\begin{array}{c}\cos \varphi _{s}\sin \lambda _{s}\sin \varphi _{t}-\sin \varphi _{s}\cos \varphi _{t}\sin \lambda _{t}\\\sin \varphi _{s}\cos \lambda _{t}\cos \varphi _{t}-\cos \varphi _{s}\sin \varphi _{t}\cos \lambda _{s}\\\cos \varphi _{s}\cos \varphi _{t}\sin(\lambda _{t}-\lambda _{s})\end{array}}\right).}$

A right-handed tilted coordinate system with the center at the center of the sphere is given by the following three axes: the axis s, the axis

${\displaystyle \mathbf {s} _{\perp }=\omega \times {\frac {1}{R}}\mathbf {s} ={\frac {1}{\sin \theta _{s,t}}}\left({\begin{array}{c}\cos \varphi _{t}\cos \lambda _{t}(\sin ^{2}\varphi _{s}+\cos ^{2}\varphi _{s}\sin ^{2}\lambda _{s})-\cos \lambda _{s}(\sin \varphi _{s}\cos \varphi _{s}\sin \varphi _{t}+\cos ^{2}\varphi _{s}\sin \lambda _{s}\cos \varphi _{t}\sin \lambda _{t})\\\cos \varphi _{t}\sin \lambda _{t}(\sin ^{2}\varphi _{s}+\cos ^{2}\varphi _{s}\cos ^{2}\lambda _{s})-\sin \lambda _{s}(\sin \varphi _{s}\cos \varphi _{s}\sin \varphi _{t}+\cos ^{2}\varphi _{s}\cos \lambda _{s}\cos \varphi _{t}\cos \lambda _{t})\\\cos \varphi _{s}[\cos \varphi _{s}\sin \varphi _{t}-\sin \varphi _{s}\cos \varphi _{t}\cos(\lambda _{t}-\lambda _{s})]\end{array}}\right)}$

and the axis ω. A position along the great circle is

${\displaystyle \mathbf {s} (\theta )=\cos \theta \mathbf {s} +\sin \theta \mathbf {s} _{\perp },\quad 0\leq \theta \leq 2\pi .}$

The compass direction is given by inserting the two vectors s and s_⊥ and computing the gradient of the vector with respect to θ at θ=0.

${\displaystyle {\frac {\partial }{\partial \theta }}\mathbf {s} _{\mid \theta =0}=\mathbf {s} _{\perp }.}$

The angle p is given by splitting this direction along two orthogonal directions in the plane tangential to the sphere at the point s. The two directions are given by the partial derivatives of s with respect to φ and with respect to λ, normalized to unit length:

${\displaystyle \mathbf {u} _{N}=\left({\begin{array}{c}-\sin \varphi _{s}\cos \lambda _{s}\\-\sin \varphi _{s}\sin \lambda _{s}\\\cos \varphi _{s}\end{array}}\right);}$

${\displaystyle \mathbf {u} _{E}=\left({\begin{array}{c}-\sin \lambda _{s}\\\cos \lambda _{s}\\0\end{array}}\right);}$

${\displaystyle \mathbf {u} _{N}\cdot \mathbf {s} =\mathbf {u} _{E}\cdot \mathbf {u} _{N}=0}$

u_N points north and u_E points east at the position s. The position angle p projects s_⊥ into these two directions,

${\displaystyle \mathbf {s} _{\perp }=\cos p\,\mathbf {u} _{N}+\sin p\,\mathbf {u} _{E}}$,

where the positive sign means the positive position angles are defined to be north over east. The values of the cosine and sine of p are computed by multiplying this equation on both sides with the two unit vectors,

${\displaystyle \cos p=\mathbf {s} _{\perp }\cdot \mathbf {u} _{N}={\frac {1}{\sin \theta _{s,t}}}[\cos \varphi _{s}\sin \varphi _{t}-\sin \varphi _{s}\cos \varphi _{t}\cos(\lambda _{t}-\lambda _{s})];}$

${\displaystyle \sin p=\mathbf {s} _{\perp }\cdot \mathbf {u} _{E}={\frac {1}{\sin \theta _{s,t}}}[\cos \varphi _{t}\sin(\lambda _{t}-\lambda _{s})].}$

Instead of inserting the convoluted expression of s_⊥, the evaluation may employ that the triple product is invariant under a circular shift of the arguments:

${\displaystyle \cos p=(\mathbf {\omega } \times {\frac {1}{R}}\mathbf {s} )\cdot \mathbf {u} _{N}=\omega \cdot ({\frac {1}{R}}\mathbf {s} \times \mathbf {u} _{N}).}$

If atan2 is used to compute the value, one can reduce both expressions by division through cos φ_t and multiplication by sin θ_s,t, because these values are always positive and that operation does not change signs; then effectively

${\displaystyle \tan p={\frac {\sin(\lambda _{t}-\lambda _{s})}{\cos \varphi _{s}\tan \varphi _{t}-\sin \varphi _{s}\cos(\lambda _{t}-\lambda _{s})}}.}$