Great-circle navigation

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Orthodromic course drawn on the Earth globe

Great-circle navigation or orthodromic navigation (related to orthodromic course; from Ancient Greek ορθός (orthós) 'right angle' and δρόμος (drómos) 'path') is the practice of navigating a vessel (a ship or aircraft) along a great circle. Such routes yield the shortest distance between two points on the globe.[1]

Course

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File:Sphere geodesic 4sigma.svg
Figure 1. The great circle path between (φ1, λ1) and (φ2, λ2).

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 = (φ11) and plans to travel the great circle to a point at point P2 = (φ22) (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

tanα1=cosϕ2sinλ12cosϕ1sinϕ2sinϕ1cosϕ2cosλ12,tanα2=cosϕ1sinλ12cosϕ2sinϕ1+sinϕ2cosϕ1cosλ12,

where λ12 = λ2 − λ1[note 1] and the quadrants of α12 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

tanσ12=(cosϕ1sinϕ2sinϕ1cosϕ2cosλ12)2+(cosϕ2sinλ12)2sinϕ1sinϕ2+cosϕ1cosϕ2cosλ12.[note 2][note 3]

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

Relation to geocentric coordinate system

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File:PositionAngle.svg
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 uE, uN 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 φsLua error: not enough memory. and geodetic longitude λsLua error: not enough memory., where φLua error: Internal error: The interpreter exited with status 1. is considered positive if north of the equator, and where λLua error: Internal error: The interpreter exited with status 1. is considered positive if east of Greenwich. In the geocentric coordinate system centered at the center of the sphere, the Cartesian components are

𝐬=R(cosφscosλscosφssinλssinφs)

and the target position is

𝐭=R(cosφtcosλtcosφtsinλtsinφt).

The North Pole is at

𝐍=R(001).

The minimum distance dLua error: Internal error: The interpreter exited with status 1. is the distance along a great circle that runs through sLua error: Internal error: The interpreter exited with status 1. and tLua error: Internal error: The interpreter exited with status 1.. It is calculated in a plane that contains the sphere center and the great circle,

ds,t=Rθs,t

where θLua error: Internal error: The interpreter exited with status 1. 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

𝐬𝐭=R2cosθs,t=R2(sinφssinφt+cosφscosφtcos(λtλs))

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

ds,N=Rθs,N=R(π/2φs)

If a ship starts at tLua error: Internal error: The interpreter exited with status 1. and sails straight to the North Pole, the travel distance is

dt,N=Rθt,n=R(π/2φt)

Derivation

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The cosine formula of spherical trigonometry[4] yields for the angle pLua error: Internal error: The interpreter exited with status 1. between the great circles through sLua error: Internal error: The interpreter exited with status 1. that point to the North on one hand and to tLua error: Internal error: The interpreter exited with status 1. on the other hand

cosθt,N=cosθs,tcosθs,N+sinθs,tsinθs,Ncosp.
sinφt=cosθs,tsinφs+sinθs,tcosφscosp.

The sine formula yields

sinpsinθt,N=sin(λtλs)sinθs,t.

Solving this for sin θs,tLua error: Internal error: The interpreter exited with status 1. and insertion in the previous formula gives an expression for the tangent of the position angle,

sinφt=cosθs,tsinφs+sin(λtλs)sinpcosφtcosφscosp;
tanp=sin(λtλs)cosφtcosφssinφtcosθs,tsinφs.

Further details

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Because the brief derivation gives an angle between 0 and πLua error: Internal error: The interpreter exited with status 1. 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 ≤ πLua error: Internal error: The interpreter exited with status 1..

The computation starts from a construction of the great circle between sLua error: Internal error: The interpreter exited with status 1. and tLua error: Internal error: The interpreter exited with status 1.. It lies in the plane that contains the sphere center, sLua error: Internal error: The interpreter exited with status 1. and tLua error: Internal error: The interpreter exited with status 1. and is constructed rotating sLua error: Internal error: The interpreter exited with status 1. by the angle θs,tLua error: Internal error: The interpreter exited with status 1. around an axis ωLua error: Internal error: The interpreter exited with status 1.. The axis is perpendicular to the plane of the great circle and computed by the normalized vector cross product of the two positions:

𝝎=1R2sinθs,t𝐬×𝐭=1sinθs,t(cosφssinλssinφtsinφscosφtsinλtsinφscosλtcosφtcosφssinφtcosλscosφscosφtsin(λtλs)).

A right-handed tilted coordinate system with the center at the center of the sphere is given by the following three axes: the axis sLua error: Internal error: The interpreter exited with status 1., the axis

𝐬=ω×1R𝐬=1sinθs,t(cosφtcosλt(sin2φs+cos2φssin2λs)cosλs(sinφscosφssinφt+cos2φssinλscosφtsinλt)cosφtsinλt(sin2φs+cos2φscos2λs)sinλs(sinφscosφssinφt+cos2φscosλscosφtcosλt)cosφs[cosφssinφtsinφscosφtcos(λtλs)])

and the axis ωLua error: Internal error: The interpreter exited with status 1.. A position along the great circle is

𝐬(θ)=cosθ𝐬+sinθ𝐬,0θ2π.

The compass direction is given by inserting the two vectors sLua error: Internal error: The interpreter exited with status 1. and sLua error: Internal error: The interpreter exited with status 1. and computing the gradient of the vector with respect to θLua error: Internal error: The interpreter exited with status 1. at θ=0Lua error: Internal error: The interpreter exited with status 1..

θ𝐬θ=0=𝐬.

The angle pLua error: Internal error: The interpreter exited with status 1. is given by splitting this direction along two orthogonal directions in the plane tangential to the sphere at the point sLua error: Internal error: The interpreter exited with status 1.. The two directions are given by the partial derivatives of sLua error: Internal error: The interpreter exited with status 1. with respect to φLua error: Internal error: The interpreter exited with status 1. and with respect to λLua error: Internal error: The interpreter exited with status 1., normalized to unit length:

𝐮N=(sinφscosλssinφssinλscosφs);
𝐮E=(sinλscosλs0);
𝐮N𝐬=𝐮E𝐮N=0

uNLua error: Internal error: The interpreter exited with status 1. points north and uELua error: Internal error: The interpreter exited with status 1. points east at the position sLua error: Internal error: The interpreter exited with status 1.. The position angle pLua error: Internal error: The interpreter exited with status 1. projects sLua error: Internal error: The interpreter exited with status 1. into these two directions,

𝐬=cosp𝐮N+sinp𝐮E,

where the positive sign means the positive position angles are defined to be north over east. The values of the cosine and sine of pLua error: Internal error: The interpreter exited with status 1. are computed by multiplying this equation on both sides with the two unit vectors,

cosp=𝐬𝐮N=1sinθs,t[cosφssinφtsinφscosφtcos(λtλs)];
sinp=𝐬𝐮E=1sinθs,t[cosφtsin(λtλs)].

Instead of inserting the convoluted expression of sLua error: Internal error: The interpreter exited with status 1., the evaluation may employ that the triple product is invariant under a circular shift of the arguments:

cosp=(𝝎×1R𝐬)𝐮N=ω(1R𝐬×𝐮N).

If atan2 is used to compute the value, one can reduce both expressions by division through cos φtLua error: Internal error: The interpreter exited with status 1. and multiplication by sin θs,tLua error: Internal error: The interpreter exited with status 1., because these values are always positive and that operation does not change signs; then effectively

tanp=sin(λtλs)cosφstanφtsinφscos(λtλs).

Finding way-points

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To find the way-points, that is the positions of selected points on the great circle between P1 and P2, we first extrapolate the great circle back to its node A, the point at which the great circle crosses the equator in the northward direction: let the longitude of this point be λ0 — see Fig 1. The azimuth at this point, α0, is given by

tanα0=sinα1cosϕ1cos2α1+sin2α1sin2ϕ1.[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

tanσ01=tanϕ1cosα1(If φ1 = 0 and α1 = 12π, use σ01 = 0).

This gives σ01, whence σ02 = σ01 + σ12.

The longitude at the node is found from

tanλ01=sinα0sinσ01cosσ01,λ0=λ1λ01.
File:Sphere geodesic 2sigma.svg
Figure 2. The great circle path between a node (an equator crossing) and an arbitrary point (φ,λ).

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

.)tanϕ=cosα0sinσcos2σ+sin2α0sin2σ,[note 6]
tan(λλ0)=sinα0sinσcosσ,tanα=tanα0cosσ.

The atan2 function should be used to determine σ01, λ, and α. For example, to find the midpoint of the path, substitute σ = 1201 + σ02); alternatively to find the point a distance d from the starting point, take σ = σ01 + d/R. Likewise, the vertex, the point on the great circle with greatest latitude, is found by substituting σ = +12π. It may be convenient to parameterize the route in terms of the longitude using

tanϕ=cotα0sin(λλ0).[note 7]

Latitudes at regular intervals of longitude can be found and the resulting positions transferred to the Mercator chart allowing the great circle to be approximated by a series of rhumb lines. The path determined in this way gives the great ellipse joining the end points, provided the coordinates (ϕ,λ) 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.

Example

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

The formulas for course and distance give λ12 = −166.6°,[note 8] α1 = −94.41°, α2 = −78.42°, and σ12 = 168.56°. Taking the earth radius to be R = 6371 km, the distance is s12 = 18743 km.

To compute points along the route, first find α0 = −56.74°, σ01 = −96.76°, σ02 = 71.8°, λ01 = 98.07°, and λ0 = −169.67°. Then to compute the midpoint of the route (for example), take σ = 1201 + σ02) = −12.48°, and solve for φ = −6.81°, λ = −159.18°, and α = −57.36°.

If the geodesic is computed accurately on the WGS84 ellipsoid,[5] the results are α1 = −94.82°, α2 = −78.29°, and s12 = 18752 km. The midpoint of the geodesic is φ = −7.07°, λ = −159.31°, α = −57.45°.

Gnomonic chart

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File:Admiralty Chart No 132 Gnomonic Chart of Indian and Southern Oceans, Published 1914.jpg
Admiralty Gnomonic Chart of the Indian and Southern Oceans, for use in plotting great circle tracks

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.

See also

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Notes

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  1. ^ In the article on great-circle distances, the notation Δλ = λ12 and Δσ = σ12 is used. The notation in this article is needed to deal with differences between other points, e.g., λ01.
  2. ^ A simpler formula is
    cosσ12=sinϕ1sinϕ2+cosϕ1cosϕ2cosλ12;
    however, this is numerically less accurate if σ12 small.
  3. ^ These equations for α1212 are suitable for implementation on modern calculators and computers. For hand computations with logarithms, Delambre's analogies[2] were usually used:
    cos12(α2+α1)sin12σ12=sin12(ϕ2ϕ1)cos12λ12,sin12(α2+α1)sin12σ12=cos12(ϕ2+ϕ1)sin12λ12,cos12(α2α1)cos12σ12=cos12(ϕ2ϕ1)cos12λ12,sin12(α2α1)cos12σ12=sin12(ϕ2+ϕ1)sin12λ12.
    McCaw[3] refers to these equations as being in "logarithmic form", by which he means that all the trigonometric terms appear as products; this minimizes the number of table lookups required. Furthermore, the redundancy in these formulas serves as a check in hand calculations. If using these equations to determine the shorter path on the great circle, it is necessary to ensure that |λ12| ≤ π (otherwise the longer path is found).
  4. ^ A simpler formula is
    sinα0=sinα1cosϕ1;
    however, this is less accurate α0 ≈ ±12π.
  5. ^ The direct geodesic problem, finding the position of P2 given P1, α1, and s12, can also be solved by formulas for solving a spherical triangle, as follows,
    σ12=s12/R,sinϕ2=sinϕ1cosσ12+cosϕ1sinσ12cosα1,ortanϕ2=sinϕ1cosσ12+cosϕ1sinσ12cosα1(cosϕ1cosσ12sinϕ1sinσ12cosα1)2+(sinσ12sinα1)2,tanλ12=sinσ12sinα1cosϕ1cosσ12sinϕ1sinσ12cosα1,λ2=λ1+λ12,tanα2=sinα1cosσ12cosα1tanϕ1sinσ12.
    The solution for way-points given in the main text is more general than this solution in that it allows way-points at specified longitudes to be found. In addition, the solution for σ (i.e., the position of the node) is needed when finding geodesics on an ellipsoid by means of the auxiliary sphere.
  6. ^ A simpler formula is
    sinϕ=cosα0sinσ;
    however, this is less accurate when φ ≈ ±12π
  7. ^ The following is used: cosσ=cosϕcos(λλ0)
  8. ^ λ12 is reduced to the range [−180°, 180°] by adding or subtracting 360° as necessary

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References

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  5. ^ Lua error: Internal error: The interpreter exited with status 1.

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