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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.
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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).
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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.
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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
sinLat. sinLat. cosDcos
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
sin56 20 sin52 12 cos28 16.5cos 61 58.7 62.0cos52 12 sin28 16.5
=
= = =
= = 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) = =
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
NAVIGATION
Distance
( )( )
1AB A B A B AB
1
D cos sinLat. sinLat. cosLat. cosLat. cosD. Long.
cos sin33 22 sin10 51 cos33 22 cos10 51 cos63 52
62 18.1
= +
= +
=
Distance 3738.1miles=
Initial course
1 B A AB
A AB
1
sinLat. sinLat. cosDcos
cosLat. sinD
sin10 51 sin33 22 cos62 18.1cos 95 13.8 95.2cos33 22 sin62 18.1
=
= = =
= = Initialcourse S95.2 W 275.2 T
Final course
1 A B AB
B AB
1
sinLat sinLat cosDcos
cosLat sinD
sin33 22 sin10 51 cos62 18.1cos 57 52.1 57.9cos10 51 sin62 18.1
=
= = =
= = 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
D cos sinLat. sinLat. cosLat. cosLat. cosD. Long.
cos sin49 12 sin13 30 cos49 12 cos13 30 cos91 55
81 03.4
= +
= +
= Distance 4863.4miles=
Initial course
1 B A AB
A AB
1
sinLat. sinLat. cosDcos
cosLat. sinD
sin13 30 sin49 12 cos81 03.4cos 79 40.1 79.7cos49 12 sin81 03.4
=
= = =
= = Initialcourse N79.7 W 280.3 T
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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
D cos sinLat. sinLat. cosLat. cosLat. cosD. Long.
cos sin46 20 sin26 25 cos46 20 cos26 25 cos85 35
68 19.1
= +
= +
=
Distance 4099.1miles=
Initial course
1 B A AB
A AB
1
sinLat. sinLat. cosDcos
cosLat. sinD
sin26 25 sin46 20 cos68 19.1cos 73 55.5 73.9cos46 20 sin68 19.1
=
= = =
= = 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
D cos sinLat. sinLat. cosLat. cosLat. cosD. Long.
cos sin17 sin( 22 ) cos17 cos( 22 )cos8087 27.2
= +
= +
=
Distance 5247.2miles=
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
NAVIGATION
Initial course
( )
1 B A AB
A AB
1
sinLat. sinLat. cosDcos
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
sinLat sinLat cosDcos
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
D cos sinLat. sinLat. cosLat. cosLat. cosD. Long.
cos sin45 44 sin7 30 cos45 44 cos7 30 cos109 24
108 51.9
= +
= +
=
Distance 6531.9miles=
Difference 6531.9 6159.1 372.8miles= =
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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.
= = =
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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
D cos sinLat. sinLat. cosLat. cosLat. cosD. Long.
cos sin34 55 sin33 55 cos34 55 cos33 55 cos74 35
59 58.9
= +
= +
=
Distance 3598.9miles=
Initial course
1 B A AB
A AB
1
sinLat. sinLat. cosDcos
cosLat. sinD
sin33 55 sin34 55 cos59 58.9cos 67 30.4 67.5cos34 55 sin59 58.9
=
= = =
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 = =
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
NAVIGATION
Distance from departure position to
the vertex
( ) ( )1 1AV A AVD sin cosLat. sinD.Long. sin cos34 55 sin35 53.028 43.6
= =
=
Distance 1723.6miles=
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 =
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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
D cos sinLat. sinLat. cosLat. cosLat. cosD. Long.
cos sin51 25 sin46 00 cos51 25 cos46 00 cos39 30
26 17.1
= +
= +
=
Distance 1577.1miles=
Initial course
1 B A AB
A AB
1
sinLat. sinLat. cosDcos
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
sinLat sinLat cosDcos
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=
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
NAVIGATION
Latitude of any x position can be calculated by formula:
( )1X VX VLat. tan cosD.Long. tanLat.=
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
D cos sinLat. sinLat. cosLat. cosLat. cosD. Long.
cos sin48 24 sin34 50 cos48 24 cos34 50 cos95 26
57 56.6
= +
= +
=
Distance 4076.6miles=
Initial course
1 B A AB
A AB
1
sinLat. sinLat. cosDcos
cosLat. sinD
sin34 50 sin48 24 cos67 56.6coscos48 24 sin67 56.6
61 50.6 61.8
=
= = =
Initialcourse N61.8 W 298.2 T= =
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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 = =
Distance from departure position to
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
D. Long.Longitude LatitudePosition
(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)= =
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.
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
NAVIGATION
Distance
( )( ) ( )
1AB A B A B AB
1
D cos sinLat. sinLat. cosLat. cosLat. cosD. Long.
cos sin30 sin 20 cos30 cos 20 cos53
71 24.8
= +
= + =
Distance 4284.8miles=
Initial course
( )
1 B A AB
A AB
1
sinLat. sinLat. cosDcos
cosLat. sinD
sin 20 sin30 cos71 24.8cos
cos30 sin71 24.8
127 38.9 127.6
=
=
= =
Initialcourse N127.6 W 232.4 T= =
Final course
( )( )
1 A B AB
B AB
1
sinLat sinLat cosDcos
cosLat sinD
sin30 sin 20 cos71 24.8cos
cos 20 sin71 24.8
46 51.7 46.9
=
=
= =
Finalcourse S46.9 W 226.9 T= =
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= + =
V VVertex's Position: Lat. 46 42.6 N Long. 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
=
= =
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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
D. Long.Longitude LatitudePosition
(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.
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
NAVIGATION
A BLat. 45 N Lat. 30 S= =
( )ABD.Long. 360 100 130 130 (W)= + =
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
D cos sinLat. sinLat. cosLat. cosLat. cosD. Long.
cos sin45 sin 30 cos45 cos 30 cos130
138 20.8
= + = +
=
Distance 8300.8miles=
Initial course
( )
( )
B A AB1
A AB
1
sin Lat. sinLat. cosDcos
cosLat. sinD
sin 30 sin45 cos138 20.8cos
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
=
=
= =
Finalcourse S54.6 W 234.6 T= =
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= + =
V VVertex's Position: Lat. 45 06.2 N Long. 104 52.9 W= =
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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
( )1X VX VLat. tan cosD.Long. tanLat.=
VX
X XX V
1
2
3
4
5
6
D. Long.Longitude LatitudePosition
(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
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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
= = =
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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.
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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. =
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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.= +
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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=
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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= + =
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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.
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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
= = =
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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)= = =
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
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
1 1Azimuth( ) tan tanC cosLat. 0.639498 cos10 51
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
1 1Azimuth( ) tan tanC cosLat. 0.402442 cos56 20
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
tanLat. 56 20B 1.990286N C 0.868291NsinP sin48 58
= = = = = =
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CAPT. KHAN THE SHIP OFFICERS HANDBOOK
NAVIGATION
1 1
B
1 1Azimuth( ) tan tanC cosLat. 0.868291 cos52 12
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
tanLat. 21 19B 0.970820N C 1.716873NsinP sin23 42
= = = = = =
1 1
A
1 1Azimuth( ) tan tanC cosLat. 1.716873 cos18 08
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
1 1Azimuth( ) tan tanC cosLat. 1.703712 cos21 19
32 12.8
= = =
= = = =
Azimuth S32 12.8 W S32.2 WFinalcourse N32.2 E 032.2 T