Great Circle Sailing - Ship...

CAPT. KHAN THE SHIP OFFICERS HANDBOOK

NAVIGATION

Great Circle Sailing

Great Circle Sailing is used for long ocean passages. For this purpose, the earth is considered a perfect spherical shape; therefore, the shortest distance between two points on its surface is the arc of the great circle containing two points. As the track is the circle, so the course is constantly changing, and the track must be broken down into a series of short rhumb lines at frequent intervals that can be used to sail on the Mercator chart. Doing this, the navigator would use the Gnomonic charts combined with the Mercator charts to draw the sailing track.

EQ Equator AB Great circle trackP Pole GreatcircleinitialcoursePA PolardistanceofA GreatcirclefinalcoursePB Polar distance of B

Procedure to use Gnomonic and

Mercator Charts for Great Circle Sailing

1. Plot departure and destination positions on the gnomonic chart; join two positions, since the great circle appears as a straight line on the gnomonic chart.

2. Choosethespecificintervalmeridianalongthetrackwherethecourse will be changed. Then plot the positions of intersection of the track and the meridian chosen on the Mercator chart.

3. Join all the plotted positions on the Mercator chart by a series of rhumb lines; the course and distance between each position can be solved by the plane sailing method.

As the great circle track line is plotted on the gnomonic chart, the vertex and the chosen intermediate positions can be read off directly from the chart. However, this is not as accurate as the calculation which will be shown later in this section.


NAVIGATION

Great circle distance

A B A B AB

cos AB cosPAcosPB sinPAsinPBcosPsinLat. sinLat. cosLat. cosLat. cosD.long.

= += +

( )1AB A B A B ABD cos sinLat. sinLat. cosLat. cosLat. cosD.Long.= +

Initial course

= = B A AB

A AB

sinLat sinLat cosDcosPB cosPAcos ABcossinPAsin AB cosLat sinD

1 B A AB

A AB

sinLat. sinLat. cosDcos

cosLat. sinD =

The initial course also can be calculated by following formulas:

1 B AB

AB

cosLat. sinD.Long.sin

sinD =

or1 AB

A B A AB

sinD. Long.tan

cosLat. tanLat. sinLat. cosD. Long. =

Final course

A B AB

B AB

sinLat sinLat cosDcosPA cosPBcos ABcossinPBsin AB cosLat sinD

= =

A B AB

B AB

sinLat sinLat cosDB arccos

cosLat sinD

=

Thefinalcoursealsocanbecalculatedbyfollowingformulas:

1 A AB

AB

cosLat. sinD.Long.sin

sinD =

or1 AB

B A B AB

sinD. Long.tan

cosLat. tanLat. sinLat. cosD. Long. =

The great circle calculations of initial and final courses result inquadrantal notation as cardinal compass. Corrected quadrant must be named in order to avoid mistakes when converting into three-figurenotation(0-360).


NAVIGATION

Rules to name Initial and Final

Courses of a Great Circle

The initial course always has same name as the initial latitude and eastorwestdirectionofthecourse.Thefinalcoursealwayshastheopposite name from final latitude unless initial position and finalpositionareindifferenthemispheres,whenthefinalwillhavesamenameasfinallatitudeandeastorwestdirectionofthecourse.

Summary Direction of course: Easterly

InitialLatitude

FinalLatitude

North South

NorthInitial course: NE

Final course: SE

Initial course: SE

Final course: NE

SouthInitial course: NE

Final course: SE

Initial course: SE

Final course: NE

Direction of course: Westerly

InitialLatitude

FinalLatitude

North South

NorthInitial course: NW

Final course: SW

Initial course: SW

Final course: NW

SouthInitial course: NW

Final course: SW

Initial course: SW

Final course: NW

Initial and final courses canalsobe foundbyusingABC tablesorABC computations, just like solving the azimuth of a celestial body by considering one position as the observers position and another as position of the celestial body. The azimuth would be the initial or finalcourse,dependingwhichisdesignated.Forexample,inordertofindinitialcourse,theinitialpositionisconsideredastheobserversposition,andthefinalpositionasthecelestialposition.Conversely,for finding the final course; the finalposition is consideredas theobservers position, and initial position as the celestial position. The course would be named as "C", and direction is the hour angle which is D. Long. between two positions.


NAVIGATION

Example 1 Findthedistance,initialcourseandfinalcourse:

From A: 56 20 N 008 12 WTo B: 52 12 N 057 10 W

A BLat. 56 20 N Lat. 52 12 N = =

ABD.Long. 57 10 8 12 48 58 (W) = =

Distance

( )( )

1AB A B A B AB

1

D cos sinLat. sinLat. cosLat. cosLat. cosD. Long.

cos sin56 20 sin52 12 cos56 20 cos52 12 cos48 58

28 16.5

= +

= +

=

Distance 1696.5miles=

Initial course

1 B A AB

A AB

1


cosLat. sinD

sin52 12 sin56 20 cos28 16.5cos 77 25.4 77.4cos56 20 sin28 16.5

=

= = =

= = Initialcourse N77.4 W 282.6 T

Final course

1 A B AB

B AB

1

sinLat sinLat cosDcos

cosLat sinD


=

= = =

= = Finalcourse S62.0 W 242 T

Example 2 Find the distance, initial course and final course of great circlesailing:

From A : 33 22 S 113 08 ETo B: 10 51 S 049 16 E

A BLat. 33 22 S Lat. 10 51 S = =

ABD. Long. 113 08 49 16 63 52 (W) = =


NAVIGATION

Distance

( )( )

1AB A B A B AB

1



62 18.1

= +

= +

=


Initial course

1 B A AB

A AB

1


cosLat. sinD


=

= = =

= = Initialcourse S95.2 W 275.2 T

Final course

1 A B AB

B AB

1


cosLat sinD


=

= = =

= = Finalcourse N57.9 W 302.1 T

Example 3

Crossing 180 meridian

Find the distance and initial course of great circle sailing from Vancouver to Guam:

Vancouver(A): 49 12 N 122 50 WGuam(B): 13 30 N 145 15 E

A BLat. 49 12 N Lat. 13 30 N= =

( )ABD.Long. 360 122 50 145 1591 55 (W)

= +

=

Distance

( )( )

1AB A B A B AB

1



81 03.4

= +

= +

= Distance 4863.4miles=

Initial course

1 B A AB

A AB

1


cosLat. sinD


=

= = =

= = Initialcourse N79.7 W 280.3 T


NAVIGATION

Example4

Crossing 180 meridian

Find the distance and initial course of great circle sailing from Bluff Harbour to Easter Island:

BluffHarbour(A): 46 20 S 169 10 EEasterIsland(B): 26 25 S 105 15 W

A BLat. 46 20 S Lat. 26 25 S= =

( )ABD.Long. 360 169 10 105 1585 35 (E)

= +

=

Distance

( )( )

1AB A B A B AB

1



68 19.1

= +

= +

=


Initial course

1 B A AB

A AB

1


cosLat. sinD


=

= = =

= = Initialcourse S73.9 E 106.1 T

Example 5

Crossing the Equator and 180 meridian

Find the distance, initial course and final course of great circlesailing:

From A : 17 S 170 ETo B: 22 N 110 W

A BLat. 17 S Lat. 22 N= =

( )ABD.Long. 360 170 11080 (E)

= +

=

Since the name of the latitude of the destination is contrary to the latitude of departure, then the latitude of destination is treated as a negative quantity.

Distance

( )( )

1AB A B A B AB

1


cos sin17 sin( 22 ) cos17 cos( 22 )cos8087 27.2

= +

= +

=



NAVIGATION

Initial course

( )

1 B A AB

A AB

1


cosLat. sinD

sin 22 sin17 cos87 27.2cos

cos17 sin87 27.2

113 56.1 113.9

=

= = =

Initialcourse S113.9 E 066.1 T= =

Final course

1 A B AB

B AB

1


cosLat sinD

sin17 sin( 22 )cos87 27.2coscos( 22 )sin87 27.2

70 30.7 70.5

=

= = =

Finalcourse N70.5 E 070.5 T= =

Example6 Compare rhumb line distance with great circle distance:

Dunedin(A): 45 44 S 171 15 EPanama(B): 7 30 N 79 21 W

Rhumb line sailing

A A A

B B B

Lat. 45 44 S M.P. 3075.80 Long. 171 15 ELat. 7 30 N M.P. 448.24 Long. 79 21 WD. Lat. 3194 (N) D.M.P. 3524.04 D. Long. 6564 (E)

1 1D.Long. 6564C tan tan 61.8D.M.P. 3524.04

= = =

= = =

D. Lat. 3194Distance 6159.1milescosC cos61.8

Great circle sailing A BLat. 45 44 S Lat. 07 30 N= =

( )ABD.Long. 360 171 15 79 21 109 24= + =

( )( )

1AB A B A B AB

1



108 51.9

= +

= +

=


Difference 6531.9 6159.1 372.8miles= =


NAVIGATION

Vertex The arc of a great circle will always curve towards the nearest pole and away from the equator. The vertex is the point on a great circle that is closest to the pole; by knowing the latitude of the vertex, if it istoohigh(whichisusuallyassociateswithice,fog,coldandsevereweather),thenavigatormighthavetomodifythepassageplanforasafervoyage.Therearetwoverticesonagreatcircle,180apart;thenearer vertex is usually the chosen one for navigational calculation. The vertexs latitude is always numerically equal to or greater than the latitude of any other point on the great circle, including the latitude of departure and destination. At the vertex, the great circle isrunninginadirectionof090/270.Knowingthepositionofthevertex also helps in calculating the position of any intermediate position on the track of a great circle. In the spherical triangle APB, if anglesAandBarelessthan90;thevertexwilllieinsidethetrianglebetweenAandB,asshowninthe figure(1)below,andtheshipstrack passes through the vertex. IfeitherAorBisgreaterthan90,the vertex will lie outside the spherical triangle and on the side of the anglewhichisgreaterthan90,asshowninthefigure(2)below,andthe ships track does not pass through the vertex.

Thevertex is90 from thepointwhere the trackof thegreatcircle cuts the equator.

The course where the great circle crosses the equator is equal to the co. latitude of the vertex.

AAVVVQVEQC

PNV

Initial positionGreat circle trackVertex(upperbranch)Vertex(lowerbranch)Latitude of vertexEquatorIntersection of GC track and equatorGC course at equator Co. Latitude of vertex V

= = NP V CV 90


NAVIGATION

The position of the vertex, and the distance from departure point to vertex,canbecalculatedbyusingNapiersRulesintherightangletriangle PVA:

For all formulas used for great circle vertex calculations, if the name of the latitude of any position, including the departure and destination, is contrary to the latitude of the vertex, then the latitude of those having a contrary name to the latitude of vertex is treated as a negative quantity.

Latitude of the vertex (Lat.V)

( )V A

sinPV cos coA cos(coPA) sinAsinPAcosLat. sin AcosLat.

= =

=

( )1V ALat. cos sin cosLat.= or ( )1V BLat. cos sin cosLat.=

Difference of longitude between

departure and vertex (D. Long.AV)

sin(coP) tanPV tan(coPA)= A

AVV

tanLcot PAcosP cosD.Long.cot PV tanL

= =

1 AAV

V

tanLat.D.Long. cos

tanLat. =

D. Long. can also be found by the formula:

AVV

sin(coA) cosPVcos(coP)cos A cosPVsinP

cos A cos AsinP sinD.Long.cosPV sinLat.

==

= =

1AV

V

cosD.Long. sinsinLat.

=

When using above formula, if the latitude of departure is contrary to the vertex, then the result has to be subtracted by 180 to get corrected D. Long.

Difference of longitude between

destination and vertex (D. Long.BV)

1 BBVV

tanLat.D.Long. cos

tanLat. =

or 1BVV

cosD.Long. sinsinLat.

=

Similarly, when using the above formula, if the latitude of destination is contrary to vertex, then the result has to be subtracted by 180 to get corrected D. Long.

Distance from departure position to

the vertex (DAV) AV A AV

sinAV cos(coPA)cos(coP) sinPAsinPsinD cosLat. sinD.Long.

= = =


NAVIGATION

( )1AV A AVD sin cosLat. sinD. Long.= or

sin(coP) cosAVcos(coA)cosP cos AVsin A

cosPcos AVsin A

==

=

1 AVAV

cosD. Long.D cos

sin =

Example7 Find the great circle distance, the initial course and the position of the vertex, and also the distance from departure position to the vertex:

From A : 34 55 S 56 10 WTo B: 33 55 S 18 25 E

A BLat. 34 55 S Lat. 33 55 S = =

ABD. Long. 056 10 18 25 74 35 (E) = + =

Distance

( )( )

1AB A B A B AB

1



59 58.9

= +

= +

=


Initial course

1 B A AB

A AB

1


cosLat. sinD


=

= = =

Initialcourse S67.5 E 112.5 T= =

Vertexs position

( ) ( )1 1V AV

Lat. cos sin coslat. cos sin67 30.4 cos34 55

Lat. 40 44.8 S

= =

=

1 1AV

V

V

tanLat tan34 55D. Long. cos cos 35 53.0 (E)tanLat. tan40 44.8

Long. 56 10 W 35 53.0 (E) 20 17 W

= = =

= =

V VVertex's Position: Lat. 44 44.8 S Long. 20 17.0 W = =


NAVIGATION


the vertex

( ) ( )1 1AV A AVD sin cosLat. sinD.Long. sin cos34 55 sin35 53.028 43.6

= =

=


Practical Method for Great Circle Sailing

It is not practical for a ship to sail along a great circle track, because she has to change course constantly in order to follow it. Therefore, the great circle is divided into equal segments by longitudes, and is then made up of a series of rhumb lines. The rhumb lines can be plotted on the Mercator chart and followed by the ship. The rule of thumb for selecting the equal interval D. Long. from the vertex is:

Short legs in lower latitudes,long legs in higher latitudes

ByusingNapiersrulesforthesphericalright-angletrianglePVX:

Latitude at the meridian cuts the great circle track

X VX V

sin(coP) tanPV tan(coPX)cosP tanPVcotPX

cosPcotPX tanPX cosPcotPVtanPV

tanLat. cosD.Long. tanLat.

==

= =

=

( )1X VX VLat. tan cosD.Long. tanLat.=

Course at the meridian cuts the great circle track V VX

sin(coX) cosPVcos(coP)cosX cosPVsinP sinLat. sinD.Long.

== =

( )1 V VXX cos sinLat. sinD.Long.=

The longitude can also be selected as the equal interval distance on the great circle from the vertex, and the position can be calculated by usingNapiersrules:

X V VX

sin(coPX) cosPVcosVXcosPX cosPVcosVX sinLat. sinLat. cosD

== =

( )1X V VXLat. sin sinLat. cosD=

XVX

V

sin(coP) tan(coPX)tanPVtanLat.

cosP cotPXtanPV cosD.Long.tanLat.

=

= =

1 XVX

V

tanLatD. Long. cos

tanLat =


NAVIGATION

Example8 Find:thegreatcircledistance;initialcourse;finalcourse;positionofthevertex;andthelatitudesthatcutintermediatemeridiansat5intervals, starting from the departure meridian:

From A : 51 25 N 9 30 WTo B: 46 00 N 49 00 W

A BLat. 51 25 N Lat. 46 00 N= =

ABD.Long. 49 00 W 9 30 W39 30 (W)

=

=

Distance

( )( )

1AB A B A B AB

1



26 17.1

= +

= +

=


Initial course

1 B A AB

A AB

1


cosLat. sinD

sin46 00 sin51 25 cos39 30cos 86 09.9 86.2cos51 25 sin39 30

=

= = =

Initialcourse N86.2 W 273.8 T= =

Final course

1 A B AB

B AB

1


cosLat sinD

sin51 25 sin46 00'cos39 30cos 63 36.5 63.6cos46 00 sin39 30

=

= = =

Finalcourse S63.6 W 243.6 T= =

Vertexs Position

( )( )

1V A

1

Lat. cos sin coslat.

cos sin86 09.9 cos51 25 51 31.1 N

=

= =

1 AAV

V

tanLat tan51 25D.Long. cos 4 54.0 (W)tanLat tan51 31.1

= = =

VLong. 9 30.0 W 4 54.0 (W) 14 24.0 W= + =

V VVertex's Position: Lat. 51 31.1 N Long. 14 24.0 W= =

Waypoints ForintervalsofD.Long.of5fromdepartureposition(A)

ALong. 9 30.0 W= VLong. 14 24.0 W=


NAVIGATION

Latitude of any x position can be calculated by formula:


VX

X XX V

1

2

3

4

5

6

D. Long.Longitude LatitudePosition

(Long. ) (Lat. )(Long. Long. )X 14 30 W 0 06 51 31.1 NX 19 30 W 5 06 51 24.5 NX 24 30 W 10 06 51 04.9 NX 29 30 W 15 06 50 32.1 NX 34 30 W 20 06 49 45.2 NX 39 30 W 25 06 48

7

43.4 NX 44 30 W 30 06 47 25.4 N

Example9 Find: the great circle distance; initial course; position of the vertex; distance from departure position to the vertex; and the positions wherethemeridiansof140 W,160 W,180and160 E cut the track on the great circle:

From A : 48 24 N 124 44 WTo B: 34 50 N 139 50 E

A BLat. 48 24 N Lat. 34 50 N = =

( )ABD.Long. 360 124 44 139 5095 26 (W)

= +

=

Distance

( )( )

1AB A B A B AB

1



57 56.6

= +

= +

=


Initial course

1 B A AB

A AB

1


cosLat. sinD

sin34 50 sin48 24 cos67 56.6coscos48 24 sin67 56.6

61 50.6 61.8

=

= = =



NAVIGATION

Vertexs Position

( )( )

1V A

1

Lat. cos sin cosLat.

cos sin61 50.6 cos48 24 54 10.3 N

=

= =

1 AAV

V

tanLat tan48 24D.Long. cos 35 35.6 (W)tanLat tan54 10.3

= = =

VLong. 124 44 W 35 35.6 160 19.6 W = + =

V VVertex's Position: Lat. 54 10.3 N Long. 160 19.6 W = =


the vertex

( )( )

1AV A AV

1

D sin cosLat. sinD.Long.

sin cos48 24 sin35 35.6

22 43.9

=

=

= Distance 1363.9miles=

Position where the meridian cut the great

circle track ( )

1X VX VLat. tan cosD. Long. tanLat.

=

VX

X XX V

1

2

3

4


(Long. ) (Lat. )(Long. Long. )X 140 W 20 19.6 52 24.4 NX 160 W 0 19.6 54 10.3 NX 180 W 19 40.4 52 31.3 NX 160 E 39 40.4 46 50.0 N

Example10

Crossing Equator

Findthegreatcircledistance;initialcourse;finalcourse;longitudewhere the great circle crosses the equator; nearest vertex; and the waypointsat10intervalsfrom130Wto170W:

From A : 30 N 120 WTo B: 20 S 173 W

A BLat. 30 N Lat. 20 S= =

ABD.Long. 173 120 53 (W)= =



NAVIGATION

Distance

( )( ) ( )

1AB A B A B AB

1


cos sin30 sin 20 cos30 cos 20 cos53

71 24.8

= +

= + =


Initial course

( )

1 B A AB

A AB

1


cosLat. sinD


cos30 sin71 24.8

127 38.9 127.6

=

=

= =


Final course

( )( )

1 A B AB

B AB

1


cosLat sinD

sin30 sin 20 cos71 24.8cos

cos 20 sin71 24.8

46 51.7 46.9

=

=

= =


Vertexs position Inthiscase,theanglePABisgreaterthan90;therefore,thevertexdoes not lie between A and B, but outside of the A side. From the right-angled triangle PVA:

( )( )

1V A

1

Lat. cos sin cosLat.

cos sin127 38.9 cos30 46 42.6 N

=

= =

1 AAV

V

tanLat tan30D.Long. cos 57 03.1 (E)tanLat tan46 42.6

= = =

VLong. 120 W 57 03.1 062 56.9 W= + =


Longitude where great circle crosses

the Equator

At the pole, the meridian of the position where the great circle crossestheequatoris90withthemeridianofthevertex.Wecanprove this as follows:

X VX VtanLat. cosD.Long. tanLat.=

At equator, Lat.X equals zero so XtanLat. 0=

VX V V V

VX VX

cosD.Long. tanLat. 0 Lat. 0 tanLat. 0cosD.Long. 0 D.Long. 90

=

= =


NAVIGATION

E EX V VXLongitude Long. D. Long.

62 56.9 W 90 152 56.9 W

= +

= + =

Meridian where great circle crosses the equatoris152 56.9 W

Positions where the meridian cut the great

circle track ( )

1X VX VLat. tan cosD.Long. tanLat.

=

VX

X XX V

1

2

3

4

5


(Long. ) (Lat. )(Long. Long. )X 130 W 67 03.0 22 29.1 NX 140 W 77 03.0 13 22.8 NX 150 W 87 03.0 03 07.5 NX 160 W 97 03.0 07 25.5 SX 170 W 107 03.0 17 17.4 S

Example 11

Crossing Equator and 180 Meridian

From A : 45 N 100 WTo B: 30 S 130 E

Find:1. Great circle distance2. Initial course3. Final course4. Vertexsposition5. Course at equator6. Longitudewhencrossingequator7. Latitudewhencrossing180meridian8. Courseandlatitudeforevery10longitude9. Mercatorsailingcourseanddistance10. Comparedistancebetweensailingmethods.


NAVIGATION

A BLat. 45 N Lat. 30 S= =

( )ABD.Long. 360 100 130 130 (W)= + =


Distance

( ) ( )

1AB A B A B AB

1


cos sin45 sin 30 cos45 cos 30 cos130

138 20.8

= + = +

=


Initial course

( )

( )

B A AB1

A AB

1

sin Lat. sinLat. cosDcos

cosLat. sinD


cos45 sin138 20.8

86 32.6 86.5

=

=

= =

Initialcourse N86.5 W 273.5 T = =

Final course

( )( )

( )( )

A B AB1

B AB

1

sinLat. sin Lat. cosDcos

cos Lat. sinD

sin45 sin 30 cos138 20.8cos

cos 30 sin138 20.8

54 35.3 54.6

=

=

= =


Vertexs position

( )( )

1V A

1

Lat. cos sin AcosLat.

cos sin86 32.6 cos45 45 06.2 N

=

= =

1 AAV

V

tanLat tan45D.Long. cos 4 52.9 (W)tanLat tan45 06.2

= = =

VLong. 100 00 W 4 52.9 (W) 104 52.9 W= + =



NAVIGATION

Great circle course when crossing the

Equator

At the equator, the angle between a great circle track and theequator equals the latitude of the vertex.

Latitude of vertex =

( )V

Course 180180 coLat.180 90 45 06.2

224 53.8224.9 T

= + = += +

= =

Longitude of position at Equator

The meridian of the position at the equator and the meridian of the vertexwouldmakeanangleof90atthepole.

( )Longitude 360 104 52.9 90 165 07.1 E= + =

Latitude when crossing 180

Meridian

( )

( )

1X VX V

VX1

X

Lat. tan cosD.Long. tanLat.

D.Long. 180 104 52.9 75 07.1

Lat. tan cos75 07.1 tan45 06.2 14 27.2 N

=

= =

= =

Latitude at every 10 of Longitude


VX

X XX V

1

2

3

4

5

6


(Long. ) (Lat. )(Long. Long. )X 110 W 5 07.1 44 59.4 NX 120 W 15 07.1 44 05.7 NX 130 W 25 07.1 42 15.7 NX 140 W 35 07.1 39 23.0 NX 150 W 45 07.1 35 18.4 NX 160 W 55 07.1 29

7

8

9

10

11

12

51.3 NX 170 W 65 07.1 22 53.6 NX 180 75 07.1 14 27.2 NX 170 E 85 07.1 4 52.9 NX 160 E 95 07.1 5 07.0 SX 150 E 105 07.1 14 40.1 SX 140 E 115 07.1 23 04.5 S


NAVIGATION

Rhumb line course and distance at every

10 of Longitude

0

1

2

3

Long. Lat. Mean Lat. D. Lat. Dep. Co. Dist.X 100 W 45 00.0 N

44 59.7 N 0.6 424.3 269.9 424.3X 110 W 44 59.4 N

44 32.6 N 53.7 427.6 262.8 431.0X 120 W 44 05.7 N

43 10.7 N 110.0 437.5 255.9 451.1X 130 W 42 15.7 N

40 49.4 N 172.

4

5

6

7

7 454.0 249.2 485.7X 140 W 39 23.0 N

37 20.7 N 244.6 477.0 242.9 536.1X 150 W 35 18.4 N

32 34.9 N 327.1 505.6 237.1 602.2X 160 W 29 51.3 N

26 22.5 N 417.7 537.5 232.1 680.7X 170 W 22 53.6 N

18 40.4 N 506.4 568.4 228.3 761.

8

9

10

11

12

3X 180 14 27.2 N

9 40.1 N 574.3 591.5 225.9 824.4X 170 E 4 52.9 N

0 07.1 S 599.9 600.0 225.0 848.5X 160 E 5 07.0 S

9 53.6 S 573.1 591.1 225.9 823.3X 150 E 14 40.1 S

18 52.3 S 504.4 567.8 228.4 759.5X 140 E 23 04.5 S

26

13

32.3 S 415.5 536.8 232.3 678.8X 130 E 30 00.0 S

Totaldistance:8306.9miles

Mercator sailing course and distance

From A : 45 N 100 WTo B: 30 S 130 E

A A A

B B B

Lat. 45 N M.P. 3013.38 Long. 100 WLat. S M.P. 1876.67 Long. 130 ED. Lat. 4500 (S) D.M.P. 4890.05 D. Long. 7800 (W)

30

1 1D.Long. 7800C tan tan 57.9D.M.P. 4890.05

= = =


NAVIGATION

Course S57.9 W 180 237.9 T= + =

D. Lat 4500Distance 8471.8milescosC cos57.9

= = =

Compare distances between methods of

sailings

Greatcircledistance:8300.8milesSeries rhumblinecourse:8306.9milesOnerhumblinecourse:8471.8miles

Wecansee there isnotmuchdifferencewhenbreakingdownthegreat circle track by a series of rhumb lines for convenience (6.1miles difference), but it still saves compared to Mercator sailing164.9miles.


NAVIGATION

Composite Great Circle Sailing

The great circle track is always curved toward the nearest pole, where its vertex is the point nearest to the pole. In very high latitudes, the track of a great circle cannot go beyond a certain latitude due to navigational restrictions, e.g., ice, fog, severe weather, etc. In such cases,thesailingtrackofagreatcirclehastobemodified.Thetrackthen consists of the combined parts of great circles and the parallel of limiting latitude, which is called Composite Great Circle Sailing. So, composite great circle sailing is a combination of great circle sailing and parallel sailing.

Initial course

( ) ( )1V1

A

sinPV cos coA cos coPA sin AsinPAcosLat.sinPV

sin AsinPA cosLat.

= =

= =

1 V

A

cosLat.sin

cosLat. =

Final course

( ) ( )2V2

B

sinPV cos coPB cos coB sinPBsinBcosLat.sinPV

sinBsinPB cosLat.

= =

= =

1 V

B

cosLat.sin

cosLat. =


NAVIGATION

D. Long. between departure,

destination positions and

vertices

( ) ( )1 1A

1 1V

sin coP tanPV tan coPAtanLat.

cosP tanPV cot PAtanLat.

=

= =

1

1 AAV

V

tanLat.D.Long. cos

tanLat. =

( ) ( )2 2B

2 2V

sin coP tanPV tan coPBtanLat.

cosP tanPV cot PBtanLat.

=

= =

2

1 BBV

V

tanLat.D.Long. cos

tanLat. =

Distance from departure position and destination to

limiting latitude

( ) 1 11 1

A1

1 V

sin coPA cos AV cosPVcosPA cos AV cosPV

sinLat.cosPAcos AVcosPV sinLat.

=

=

= =

( ) 2 22 2

B2

2 V

sin coPB cosBV cosPVcosPB cosBV cosPV

sinLat.cosPBcosBVcosPV sinLat.

=

=

= =

1 A1

V

sinLat.AV cos

sinLat. =

1 B2

V

sinLat.BV cos

sinLat. =

Distance along limiting latitude

1 2 V V V1 2V V D.Long. cosLat.=

Where ( )1 2 1 2V V AB AV BVD.Long. D.Long. D.Long. D.Long.= +


NAVIGATION

Example 12 Findtheinitialcourse,finalcourse,meridiansoftheverticesandthetotal distance of the following great circle positions if the limiting latitudeis38S:

From A: 34 55 S 056 10 WTo B: 33 55 S 018 25 E

Initial course

1 1V

A

cosLat. cos38sin sin 73 56.8cosLat. cos34 55

Initial course S73.9 E 106.1 T

= = = = =

Final course

1 1V

B

cosLat. cos38sin sin 71 43.7cosLat. cos33 55

Final course N71.7 E 071.7 T

= = = = =

Meridians of the vertices

1

1 1AAV

V

tanLat. tan34 55D.Long. cos cos 26 41.4 (E)tanLat. tan38

= = =

1VLong. 56 10 W 26 41.4 (E) 29 28.6 W = =

2

1 1BBV

V

tanLat. tan33 55D.Long. cos cos 30 36.8 (W)tanLat. tan38

= = =

2VLong. 30 36.8 (W) 18 25 E 12 11.8 W = =

Distance AV1 and BV2

1 1A1

V

sinLat. sin34 55AV cos cos 21 31.7sinLat. sin38

= = =

1DistanceAV 1296.7miles=

1 1B2

V

sinLat. sin33 55BV cos cos 24 59.9sinLat. sin38

= = =

2DistanceBV 1499.9miles=


NAVIGATION

Distance V1V2

( )( ) ( )

1 2 1 2V V AB AV BVD.Long. D.Long. D.Long. D.Long.

56 10 18 25 26 41.4 30 36.8

17 16.8 1036.8

= +

= + +

= =

1 21 2 V V V

1 2

V V D.Long. cosLat. 1036.8 cos38 817

DistanceV V 817miles

= = =

=

Total Distance 1296.7 1499.9 817 3613.6 miles= + + =

Example 13 Acompositegreatcircleroutefrom3540 N14000 Eto3730N12000W. Limited latitude is 45. Find initial and final course,longitudes of the vertices and total distance:

From A: 35 40 N 140 ETo B: 37 30 N 120 W

Initial course

1 1V

A

cosLat. cos45sin sin 60.5cosLat. cos35 40

= = = Initial Course N60.5 E 060.5 T= =

Final course

1 1V

B

cosLat. cos45sin sin 63cosLat. cos37 30

= = = Final Course S63 E 180 63 117 T= = =

Meridians (longitudes) of the

vertices 1

1 1AAV

V

tanLat. tan35 40D.Long. cos cos 44 08.2 (E)tanLat. tan45

= = =

( )1V

Long. 360 140 E 44 08.2 175 51.8 W= + =


NAVIGATION

2

1 1BBV

V

tanLat. tan37 30D.Long. cos cos 39 53.1 (W)tanLat. tan45

= = =

2VLong. 120 W 39 53.1 159 53.1 W= + =

Distance AV1 and BV2

1 1A1

V

sinLat. sin35 40AV cos cos 34 27.2sinLat. sin45

= = =

1DistanceAV 2067.2miles=

1 1B2

V

sinLat. sin37 30BV cos cos 30 34.8sinLat. sin45

= = =

2DistanceBV 1834.8miles=

Distance V1V2

1 2 1 2V V V VD. Long. Long. Long.

175 51.8 159 53.1 15 58.7 958.7

=

= = =

1 21 2 V V VV V D.Long. cosLat. 958.7 cos45 677.9= = =

1 2V VDistance 677.9miles=

Total Distance 2067.2 1834.8 677.9 4579.9 miles= + + =

Using ABC Tables for Great Circle

Sailing

The ABC tables can be used to find the initial and final coursessimilarly to finding theazimuth.For finding the initial course, thedeparture latitude is used as DR latitude; destination latitude isused as declination, and D. Long. is used as hour angle with direction eastorwest.Similarlyforfindingfinalcourse,destinationlatitudebecomesDR latitude, departure latitudebecomesdestination andsame hour angle except direction is opposite in initial course case. Thenthecourseisnamedaccordingtothedirection(EastorWest),instead of the size of the hour angle.


NAVIGATION

Example14 UsingtheABCtable,findtheinitialandfinalcoursesofgreatcirclesailing:

From A: 48 24 N 124 44 WTo B: 34 50 N 139 50 E

( )ABD.Long. 360 124 44 139 50 95 26 (W) = + =

Initial course

B

A

AB

Lat. 48 24 N A 0.11NLat. 34 50 N B 0.70N

D.Long. 95 26 (W) C 0.81N

Azimuth N61.7 WInitial Course N61.7 W 298.3 T

= = =

Final course

B

A

AB

Lat. 34 50 N A 0.07NLat. 48 24 N B 1.13N

D.Long. 95 26 (E) C 1.20N

Azimuth N45.4 EFinal Course S45.4 W 225.4 T

= = =

Example 15 Findinitialcourseandfinalcourseofgreatcirclesailing:

From A: 33 22 S 113 08 ETo B: 10 51 S 049 16 E

.ABD.Long 113 08 49 16 63 52 (W) = =

Initial course

B

A

AB

Lat. 33 22 S A 0.32NLat. 10 51 S B 0.21 S

D.Long. 63 52 (W) C 0.11N

Azimuth N84.8 WInitial Course N84.8 W 275.2 T

= = =

Final course

B

A

AB

Lat. 10 51 S A 0.09NLat. 33 22 S B 0.73 S

D.Long. 63 52 (E) C 0.64 S

Azimuth S57.9 EFinal Course N57.9 W 302.1 T

= = =


NAVIGATION

The values A, B, and azimuth can also be computed by formulas:

AtanLat.AtanP

=

BtanLat.BsinP

=

1 1

A

Azimuth tanC cosLat.

=

WherePisdifference of longitude Between A and B positions, then the above example can be solved as follows:

Initial course

A

B

Lat. : latitude of observerLat. : declination of celestial bodyD. Long.:hourangle(W)Initial course: azimuth

A BLat. 33 22 S Lat. 10 51 S= =

ABD.Long. P LHA 63 52 (W)= = =

A

B

tanLat. 33 22 A 0.323093NA 0.323093NtanP tan63 52 B 0.213489 S

tanLat. 10 51B 0.213489S C 0.109603NsinP sin63 52

= = = = = =

1 1

A

1 1Azimuth( ) tan tanC cosLat. 0.109603 cos33 22

84 46.2

= = =

= = = =

Azimuth N84 46.2 W N84.8 WInitialcourse N84.8 W 275.2 T

Final course

B

A

Lat. : latitude of observerLat. : declination of celestial bodyD. Long.:hourangle(oppositedirectionofHAininitialcourse)Final course: azimuth

At final position, direction of movement of the ship is away from initial position, not toward, so the azimuth found must be added to or subtracted from 180 to find correct heading of the ship.

= = B ALat. 10 51 N Lat. 33 22 N

ABD.Long. P LHA 48 58 (E)= = =


NAVIGATION

B

A

tanLat. 10 51 A 0.094034NA 0.094034NtanP tan63 52 B 0.733532 S

tanLat. 33 22B 0.733532S C 0.639498 SsinP sin63 52

= = = = = =

1 1

B


57 52.1

= = =

= = = =

Azimuth S57 52.1 E S57.9 EFinalcourse N57.9 W 302.1 T

Example16 Find initialcourseandfinalcoursebyusingtheABCcomputationformula:

From A: 56 20 N 008 12 WTo B: 52 12 N 057 10 W

Initial course A BLat. 56 20 N Lat. 52 12 N = =

ABD.Long. P LHA 48 58 (W)= = =

tanLat. 56 20 A 1.306619 SA 1.306619StanP tan48 58 B 1.709061N

tanLat. 52 12B 1.709061N C 0.402442NsinP sin48 58

= = = = = =

1 1

A


77 25.4

= = =

= = = =

Azimuth N77 25.4 W N77.4 WInitialcourse N77.4 W 282.6 T

Final course = = B ALat. 52 12 N Lat. 56 20 N

ABD.Long. P LHA 48 58 (E)= = =

B

A

tanLat. 52 12 A 1.121995 SA 1.121995StanP tan48 58 B 1.990286N


= = = = = =


NAVIGATION

1 1

B


61 58.7

= = =

= = = =

Azimuth N61 58.7 E N62 EFinalcourse S62 W 242 T

Example17 FindinitialcourseandfinalcourseofgreatcirclesailingfromSuvato Honolulu:

Suva(A): 18 08 S 178 26 EHonolulu(B): 21 19 N 157 52 W

Initial course = = A BLat. 18 08 S Lat. 21 19 N

= = = ()ABD.Long. P LHA 23 42

A

B

tanLat. 18 08 A 0.746053NA 0.746053NtanP tan23 42 B 0.970820N


= = = = = =

1 1

A


31 30.2

= = =

= = = =

Azimuth N31 30.2 E N31.5 EInitialcourse N31.5 E 031.5 T

Final course = = B ALat. 21 19 N Lat. 18 08 S

= = = ( )ABD.Long. P LHA 23 42 W

tanLat. 21 19 A 0.888943SA 0.888943StanP tan23 42 B 0.814769S

tanLat. 18 08B 0.814769S C 1.703712SsinP sin23 42

= = = = = =

1 1

B


32 12.8

= = =

= = = =

Azimuth S32 12.8 W S32.2 WFinalcourse N32.2 E 032.2 T

Date post:	06-Feb-2018
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