Great-circle navigation

The great circle path may be found using spherical trigonometry; this is the spherical version of the inverse geodetic problem.If a navigator begins at P₁ = (φ₁,λ₁) and plans to travel the great circle to a point at point P₂ = (φ₂,λ₂) (see Fig. 1, φ is the latitude, positive northward, and λ is the longitude, positive eastward), the initial and final courses α₁ and α₂ are given by formulas for solving a spherical triangle

${\displaystyle {\begin{aligned}\tan \alpha _{1}&={\frac {\cos \phi _{2}\sin \lambda _{12}}{\cos \phi _{1}\sin \phi _{2}-\sin \phi _{1}\cos \phi _{2}\cos \lambda _{12}}},\\\tan \alpha _{2}&={\frac {\cos \phi _{1}\sin \lambda _{12}}{-\cos \phi _{2}\sin \phi _{1}+\sin \phi _{2}\cos \phi _{1}\cos \lambda _{12}}},\\\end{aligned}}}$

where λ₁₂ = λ₂ − λ₁^{[note 1]}and the quadrants of α₁,α₂ are determined by the signs of the numerator and denominator in the tangent formulas (e.g., using the atan2 function).The central angle between the two points, σ₁₂, is given by

${\displaystyle \tan \sigma _{12}={\frac {\sqrt {(\cos \phi _{1}\sin \phi _{2}-\sin \phi _{1}\cos \phi _{2}\cos \lambda _{12})^{2}+(\cos \phi _{2}\sin \lambda _{12})^{2}}}{\sin \phi _{1}\sin \phi _{2}+\cos \phi _{1}\cos \phi _{2}\cos \lambda _{12}}}.}$ ^{[note 2][note 3]}

(The numerator of this formula contains the quantities that were used to determinetan α₁.)The distance along the great circle will then be s₁₂ = Rσ₁₂, where R is the assumed radiusof the Earth and σ₁₂ is expressed in radians.Using the mean Earth radius, R = R₁ ≈ 6,371 km (3,959 mi) yields results forthe distance s₁₂ which are within 1% of the geodesic length for the WGS84 ellipsoid; see Geodesics on an ellipsoid for details.

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 geocentric 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^[4] 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 thefollowing three axes: theaxis 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 shiftof 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_tand multiplication by sin θ_s,t,because these values are always positive and that operation does not change signs; then effectively

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

To find the way-points, that is the positions of selected points on the great circle betweenP₁ and P₂, we first extrapolate the great circle back to its node A, the pointat which the great circle crosses theequator in the northward direction: let the longitude of this point be λ₀ — see Fig 1. The azimuth at this point, α₀, is given by

${\displaystyle \tan \alpha _{0}={\frac {\sin \alpha _{1}\cos \phi _{1}}{\sqrt {\cos ^{2}\alpha _{1}+\sin ^{2}\alpha _{1}\sin ^{2}\phi _{1}}}}.}$ ^{[note 4]}

Let the angular distances along the great circle from A to P₁ and P₂ be σ₀₁ and σ₀₂ respectively. Then using Napier's rules we have

${\displaystyle \tan \sigma _{01}={\frac {\tan \phi _{1}}{\cos \alpha _{1}}}\qquad }$ (If φ₁ = 0 and α₁ = 1⁄2π, use σ₀₁ = 0).

This gives σ₀₁, whence σ₀₂ = σ₀₁ + σ₁₂.

The longitude at the node is found from

${\displaystyle {\begin{aligned}\tan \lambda _{01}&={\frac {\sin \alpha _{0}\sin \sigma _{01}}{\cos \sigma _{01}}},\\\lambda _{0}&=\lambda _{1}-\lambda _{01}.\end{aligned}}}$

Finally, calculate the position and azimuth at an arbitrary point, P (see Fig. 2), by the spherical version of the direct geodesic problem.^{[note 5]} Napier's rules give

${\displaystyle {\color {white}.\,\qquad )}\tan \phi ={\frac {\cos \alpha _{0}\sin \sigma }{\sqrt {\cos ^{2}\sigma +\sin ^{2}\alpha _{0}\sin ^{2}\sigma }}},}$ ^{[note 6]}

${\displaystyle {\begin{aligned}\tan(\lambda -\lambda _{0})&={\frac {\sin \alpha _{0}\sin \sigma }{\cos \sigma }},\\\tan \alpha &={\frac {\tan \alpha _{0}}{\cos \sigma }}.\end{aligned}}}$

The atan2 function should be used to determineσ₀₁,λ, and α.For example, to find themidpoint of the path, substitute σ = 1⁄2(σ₀₁ + σ₀₂); alternativelyto find the point a distance d from the starting point, take σ = σ₀₁ + d/R.Likewise, the vertex, the point on the greatcircle with greatest latitude, is found by substituting σ = +1⁄2π.It may be convenient to parameterize the route in terms of the longitude using

${\displaystyle \tan \phi =\cot \alpha _{0}\sin(\lambda -\lambda _{0}).}$ ^{[note 7]}

Latitudes at regular intervals of longitude can be found and the resulting positions transferred to the Mercator chartallowing the great circle to be approximated by a series of rhumb lines. The path determined in this waygives the great ellipse joining the end points, provided the coordinates ${\displaystyle (\phi ,\lambda )}$ are interpreted as geographic coordinates on the ellipsoid.

These formulas apply to a spherical model of the Earth. They are also used in solving for the great circle on the auxiliary sphere which is a device for finding the shortest path, or geodesic, on an ellipsoid of revolution; see the article on geodesics on an ellipsoid.

Compute the great circle route from Valparaíso,φ₁ = −33°,λ₁ = −71.6°, toShanghai,φ₂ = 31.4°,λ₂ = 121.8°.

The formulas for course and distance giveλ₁₂ = −166.6°,^{[note 8]}α₁ = −94.41°,α₂ = −78.42°, andσ₁₂ = 168.56°. Taking the earth radius to beR = 6371 km, the distance iss₁₂ = 18743 km.

To compute points along the route, first findα₀ = −56.74°,σ₀₁ = −96.76°,σ₀₂ = 71.8°,λ₀₁ = 98.07°, andλ₀ = −169.67°.Then to compute the midpoint of the route (for example), takeσ = 1⁄2(σ₀₁ + σ₀₂) = −12.48°, and solveforφ = −6.81°,λ = −159.18°, andα = −57.36°.

If the geodesic is computed accurately on the WGS84 ellipsoid,^[5] the resultsare α₁ = −94.82°, α₂ = −78.29°, ands₁₂ = 18752 km. The midpoint of the geodesic isφ = −7.07°, λ = −159.31°,α = −57.45°.

A straight line drawn on a gnomonic chart is a portion of a great circle. When this is transferred to a Mercator chart, it becomes a curve. The positions are transferred at a convenient interval of longitude and this track is plotted on the Mercator chart for navigation.

• Compass rose
• Great circle
• Great-circle distance
• Great ellipse
• Geodesics on an ellipsoid
• Geographical distance
• Isoazimuthal
• Loxodromic navigation
• Map
  • Portolan map
• Marine sandglass
• Rhumb line
• Spherical trigonometry
• Windrose network

• Great Circle – from MathWorld Great Circle description, figures, and equations. Mathworld, Wolfram Research, Inc. c1999
• Great Circle Map Interactive tool for plotting great circle routes on a sphere.
• Great Circle Mapper Interactive tool for plotting great circle routes.
• Great Circle Calculator deriving (initial) course and distance between two points.
• Great Circle Distance Graphical tool for drawing great circles over maps. Also shows distance and azimuth in a table.
• Google assistance program for orthodromic navigation