Great Circle Sailing Notes

DECK 122 (NAVIGATION-II) Great circle sailing

Lindbergh Chart of the Great circle sailing chart of the North Atlantic Ocean 1926

GREAT CIRCLE SAILING

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A great circle is a circle which cuts the a sphere into two equal halves and its centre is coincident with the centre of the sphere.

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Plane passing through centre of the sphere

Great circle

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http://upload.wikimedia.org/wikipedia/en/9/91/Great_circle.png


The equator is a great circle.

A Great circles cross the

equator at two points 180° apart.

All longitutes are great circle.

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P Show the great circles

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PA, PB, AB is an arc of Great

circle

PAB is an spherical triangle

O is the centre of the sphere

The lenght of side AB is angle

AOB

Angle O is not equal to angle P


P

A

B

O

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P

A

B

Vs

Vn

Equator

Prime meridian

P=Elevated Pole (i.e. pole chosen for the triangle) Angle P=D.Long from A to B (E or W) Side PA=Angular distance of A from the Elevated Pole 'P'. For example if elevated pole is North Pole and A is in north latitude then PA = 90°-LAT A. If elevated pole is North Pole and A is in south latitude then PA = 90° +LAT A. Side PB = Angular distance of B from the Elevated Pole 'B'

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North elevated pole

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North elevated pole

The elevated pole chosen can be in either hemisphere.

P

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South elevated pole

Equator

PB=90-Lat B

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South elevated pole

A

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South elevated pole

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To find distance AB: Cos AB = Cos P x Sin PB x Sin PA + Cos PB x Cos PA

P is Elevated Pole (i.e. pole chosen for the triangle) Angle P = D.Long from A to B (E or W) PA = Co Lat A PB = Co Lat B Co Lat in the same hemisphere (90-Lat) Co lat in the opposite = hemisphere(90+Lat)

P

A

B

Vs

Vn

Equator

Prime meridian

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You may prefer to use the adjusted Marc St Hilaire Formula

Cos AB = Cos P x Cos Lat A x Cos Lat B ± Sin Lat A x Sin Lat B

P

A

B

Vs

Vn

Equator

Prime meridian

(+ if A and B have same name) (- if A and B have different names)

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To find initial course A; Cos A = (Cos PB - Cos PA x Cos AB) / (Sin PA x Sin AB)

To find final course – reciprocal of B; Cos B = (Cos PA - Cos PB x Cos AB )/ (Sin PB x Sin AB)

Vessel is sailing from A to B AB = distance PAB or angle A = initial course PBA or angle B = reciprocal of final course

P

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The principal advantage of calculating great circles this way is that once PA and PB have been calculated, the rest can be left to the calculator and no ambiguity concerning sides or angles bigger or less than 90° will occur.

When calculating spherical triangles it is best to convert all sides and angles into decimal angles.

This can be done using the ° '" button on your calculator, or by dividing the minutes by 60. Always work to 3 decimal places of a degree when using decimal angles.

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DECK 122 (NAVIGATION-II) Great circle sailing-Example

Lat A = 34° 27’ N

Lat B = 41° 23’ S

D.Long = 105° 44’

North elevated pole

PA = 90° – 34° 27’

PA = 55° 33’

Calculator

Press 90 Press °’’’Press – Press 34 Press°’’’Press 27°’’’

A

B

P

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Lat A = 34° 27’ N

Lat B = 41° 23’ S

D.Long = 105° 44’ E

North elevated pole

PA = 90° – 34° 27’

PA = 55° 33’ = 55.55

PB = 90° + 41° 23’

PB = 131° 23’= 131.383

A

B

P

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Lat A = 34° 27’ N

Lat B = 41° 23’ S

D.Long = 105° 44’ E=105.733

PA = 55° 33’=55.55

PB = 131° 23’ = 131.383 Cos AB= Cos 105° 44’ x Sin 131° 23’ x Sin 55° 33’ + Cos 131° 23’ x Cos 55° 33’

Cos AB = - 0.541743104

Press shift Press cos Press Answer Press enter

AB = 122.802 To convert degress Press shift Press °’’’

or Press °’’’ Press enter 122° 48’ 07”

Distance AB = 122.802 x 60 = 7368.1 mile.

A

B


P

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PA = 55° 33’

PB = 131° 23’

AB = 122° 48’ 07” Becarefull when transferring the formula to the calculator!

Use ( and ) or divide sin PA and Sin AB !

Initial course N 120.8 E so Course = 120.8 T

Final course S 70.8 E = 109.2 T

A

B

To find initial course A; Cos A = (Cos PB - Cos PA x Cos AB) / (Sin PA x Sin AB)

To find final course – reciprocal of B; Cos B = (Cos PA - Cos PB x Cos AB )/ (Sin PB x Sin AB)

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Find the initial and final course and total distance from;

(A) California 35° 10’ N – 120° 45’ W to

(B) Aucland 36° 51’ S – 174° 49’ E.

To find Dlong:

120° 45’+ 174° 49’ = Ans

360°- Ans = 64° 26’ E

Dlong= 64° 26’ E

= 64°.433 E

A

B

P

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(A) California 35 10 N – 120 45 W to

(B) Aucland 36 51 S – 174 49 E.

P = 64° 26’

PA = 90° - 35°10’ = 54° 50’

PB = 90°+ 36° 51’=126° 51’

Distance = 93° 37’.1 x 60

AB = 5617.1 mile.

A

B

P


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(A)California 35 10 N – 120 45 W to

(B) Aucland 36 51 S – 174 49 E.

P = 64° 26’

PA = 54° 50’

PB = 126° 51’

AB = 93° 37’.1

a = N 133.67 W

Initial Course C = 226°.3 T

A

B

P

To find initial course A; Cos A = (Cos PB - Cos PA x Cos AB )/ Sin PA x Sin AB

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Find the initial and final course and total distance from; (A)

California 35 10 N – 120 45 W

to (B) Aucland 36 51 S – 174 49 E.

P = 64° 26’

PA = 54° 50’

PB = 126° 51’

AB = 93° 37’.1

b = N 47.63 E

Recip Co or Final Co = S 47.53 W

C = 227.6° T

A

B

P

To find final course – reciprocal of B; Cos B = (Cos PA - Cos PB x Cos AB )/ Sin PB x Sin AB

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Vertex

Maximum Latitude that the

great circle reaches is known as

the vertex.

Vertex north and Vertex south

DECK 122 (NAVIGATION-II) Great circle sailing- Napier's Rules

P

A

B

Vs

Vn

The latitude of the vertex equals the angle between the great circle and the equator at the intersection of the great circle and the equator.

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Vertex of a Great Circle The vertex of a great circle is the maximum latitude point

of the great circle. The vertex has the following properties: There is a maximum latitude point in both the northern

and southern hemispheres; these points have the same value of latitude (eg if northern vertex = 40°N then southern vertex = 40°S).

The longitudes of the vertices are 180° apart (e.g. if one is in 20°W, the other is in 160°E).

At the vertex the course on the great circle is exactly 090°T or 270°T, depending on whether you are proceeding towards the east or the west. This means that the angle between the great circle and the meridian at the vertex is always 90°.

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Vertex before the start position

Sailing A to B

<90

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Vertex between the start and end position

Sailing A to B

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Vertex After the final Position

Sailing A to B

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Rule:

A and B less than 90° vertex between A and

B

A bigger than 90° , vertex before the A.

B bigger than 90° , vertex after the B.

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Position of the Vertex and use of Napier's Rules

The basic form of Napier's Rules is used to resolve the following:

Finding the position of the vertex of a great circle

Solving the great circle legs of a composite great circle

Resolving any other right angled spherical triangle, be it terrestrial or celestial

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PA = Polar distance of A = (90° - Lat of A) PV = Polar distance of V = (90° - Lat of V) VA = Arc of great circle.

P

V

A

Vertex

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PA = Polar distance of A = (90° - Lat of A) PV = Polar distance of V = (90° - Lat of V) VA = Arc of great circle.

P

V

A

Vertex

1

1

2

2

3 3

4

4

5

5

6

6

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To find Lat. of vertex We know; A = Initial course and PA = Polar distance of A = (90° - Lat of A)

P

V

A

Vertex

1

1

2

2

3 3

4

4

5

5

6

6

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Sine of middle part = Product of Tan of Adjacent Parts

Sine of middle part = Product of Cos of Opposite Parts

Sin PV = Cos Co A x Cos Co PA Cosine of a complementary angle is its sine

e.g. Cos Co 30° = Sin 30°

Sin PV = Sin A x Sin PA

Lat of vertex

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Sin mid part = Product of tan of adjacents.

Sin Co PA = Tan Co A x Tan Co P

Cos PA = 1 / Tan A x 1/ Tan P

Multiple by Tan P

Tan p x Cos PA = 1/Tan A

Tan P = 1 / (Tan A x Cos PA)

This gives us P, the D.Long between A and V, and hence the longitude of V.

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To find the position of the vertex you will first have to find the great circle initial course angle A. This will be found by the cosine rule

We will then know two parts of the triangle and can find any other part. The parts we know are Angle A and the Co-Latitude of A (PA).

We need to find PV (when taken from 90°, PV will give the latitude of the vertex), and angle VPA (the D.long between A and V) which is applied to the known longitude of A to give the longitude of the vertex.

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A vessel sails on a great circle from A 40° 00'N 50° 00'W to B 43° 00'N 015° 00'W. Find the initial course and the position of the vertex.

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A vessel sails on a great circle from A 40° 00'N 50° 00'W to B 43° 00'N 015° 00'W. Find the initial course and the position of the vertex.

First find AB and initial course

D.Long = 35° E = P

PA = 50°

PB = 47°

Cos AB= Cos 35xSin47xSin50+Cos47xCos50

26° 11’ 36’’

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Initial course

PA = 50°

PB = 47°

AB= 26° 11’ 36’’

Cos A = (Cos PB-Cos PAxCos AB)/(SinPAxSinAB)

Course = N 71,87 E

Course = 71,87° T

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PA = 50°

PB = 47°

AB= 26° 11’ 36’’

Initial Course = 71,87 T


PV = 46,72= 46 43’ 12”

Lat of vertex = 90- 46 43’ 12”

Lat of vertex = 43° 16’ 48” N

P

V

A

50

71,87

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Initial course

PA = 50°

PB = 47°

AB= 26° 11’ 36’’

Course = 71,87 T

Lat of vertex = 43° 16’ 48” N

Tan P = 1 / Tan A x Cos PA

P = D.Long=26° 59’ 38” E

Long of vertex = 50W- 26° 59’ 38” E=23° 00’ 22” W

P

V

A

50

71,87

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Solution of right-angled spherical triangles to find latitudes of intermediate points along great circle tracks.

In practice, a GC route is approximated by following a succession of rhumb lines between points on the GC. We can use Napier's Rules to find these intermediate points.

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P

V

A

L

LV

PA = Co Lat A PV = Co Lat V

We know PV and P (the D.long from V to longitude of L). We need to find PL, and hence Lat L. Sin Mid Part = Tan Adjacents Sin Co P = Tan PV x Tan Co PL Cos P = Tan PV x Tan Lat L Cos P / Tan PV = Tan Lat L

Tan Lat L = Cos P / Tan PV

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Find the great circle distance and the initial and final courses from Wellington (A) 41° 38' S 175° 28' E to Panama (B) 07°24'N 079° 55 'W

Find also the position of the vertex and the latitude of a point on the great circle in longitude 140°W

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Find the great circle distance and the initial and final courses from Wellington (A) 41° 38' S 175° 28' E to Panama (B) 07°24'N 079° 55 'W

Draw the sketch

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Draw the sketch B 7° 24'N

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A =41° 38' S 175° 28' E

B =07°24'N 079° 5 5 'W

PA =

PB =

P =

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A =41° 38' S 175° 28' E

B =07°24'N 079° 55 'W

South elevated pole

PA = 48° 22’

PB = 97° 24’

P = 104° 37’ = D.long= 104° 37’ E

AB = ?


Long A : 175° 28' E Long B : 079° 55 'W Dlong : 255 23 W 360 D.Long : 104° 37’ E

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A =41° 38' S 175° 28' E

B =07°24'N 079° 5 5 'W

PA = 48° 22’

PB = 97° 24’

P = 104° 37’

Cos AB = Cos 104 37x Sin 48 22 x Sin 97 24+Cos 48 22x Cos 97 24

AB=105.819*60=6349.2 mile=105° 49’ 10”

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A =41° 38' S 175° 28' E; B =07°24'N 079° 5 5 'W

PA = 48° 22’

PB = 97° 24’

P = 104° 37’

AB=105° 49’ 10”=105.819°=6349.2

Initial course:


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A =41° 38' S 175° 28' E ; B=07°24'N 079° 5 5 'W

PA = 48° 22’

PB = 97° 24’

P = 104° 37’

AB=105° 49’ 10”=105.819°

Course: Cos A= (Cos PB-CosPAxCosAB) / Sin PA x Sin AB

A = 85.828 (angle)

Intial course = S 85.8 E = 180-85.8= 094.2 T

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A =41° 38' S 175° 28' E ; B=07°24'N 079° 5 5 'W

PA = 48° 22’

PB = 97° 24’

P = 104° 37’

AB=105° 49’ 10”=105.819°


Cos B= (Cos PA-CosPBxCosAB) / Sin PB x Sin AB

A = 48.738 (angle)

Final course= N 48.7 E= 048.7 T

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A =41° 38' S 175° 28' E ; B=07°24'N 079° 5 5 'W

PA = 48° 22’ PB = 97° 24’ P = 104° 37’

AB=105° 49’ 10”=105.819°

A= S 85.8 E , B= N 48.7 E



Latitute of vertex: Sin Mid Part= Cos opposite parts

Sin PV = Cos Co A x Cos Co PA


Sin PV = Sin 85.828 x Sin 48° 22’

PV = 48.196°

Lat V = 90-48.196 = 41.804 = 41 48.2 S

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A =41° 38' S 175° 28' E ; B=07°24'N 079° 5 5 'W

PA = 48° 22’ PB = 97° 24’ P = 104° 37’

AB=105° 49’ 10”=105.819°



Longitute of vertex: Sin Mid Part = Tan Adjacent Parts

Sin Co PA = Tan Co A x Tan Co P

Cos PA = 1 / Tan A x 1 / Tan P

Tan P = 1 / (Tan A x Cos PA)

P = 6.266° or 6° 15’ 57”

Longitude of vertex = 175° 28'E + 6°15'.9E = 181°43'.9E or 178° 16'.1W

Longitute : 178° 16’.1 W

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To calculate the latitude of a point on the great circle in 140°W. In triangle VPL:

The polar angle P is 178° 16.1 W - 140°W =38°16.6’ or 38.268

PV = 48.196°

Sin Mid Part = Tan Adjacent Parts

Sin Co P = Tan PV x Tan PL

Cos P = Tan PV x (1/Tan Lat L)

Tan Lat L = Cos P / Tan PV

Tan L = 0.70208

Lat of point is 35.072° or 35° 04'.3S in longitude 140° West

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To calculate the latitude of a point on the great circle in 85°W

In triangle VPL: The polar angle P is 178° 16MW-85°W = 93° 16'.1 or 93.268° PV = 48.196°

Tan Lat L = Cos P / Tan PV = 0.05098

Lat of point is-2.918° S =2.918°N =2° 55’.1 N in longitude 85°W

(- sign means go to the opposite latitude from the pole used in the calculation.)

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DECK 122 (NAVIGATION-II) Great circle sailing- Self assessment test

Find the great circle distance and the initial and final courses from Dondra Head, South of Sri Lanka 05° 48' N 80° 36' E to Cape Leeuwin in Western Australia 34° 26' S 115° 04' E. Find the position of the vertex and the latitude of a point on the track in longitude 100°E

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A=05° 48' N 80° 36' E

B= 34° 26' S 115° 04' E

PA=90- 05° 48‘N = 84° 12’

PB=90+34° 26'S = 124° 26’

D.Long= 115° 04' E- 80° 36' E =34° 28’ E or 34.467

AB=51.729°=x60=3103.7 mile


P

A

B


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A=05° 48' N 80° 36' E ; B= 34° 26' S 115° 04' E

PA=90- 05° 48‘N = 84° 12’

PB=90+34° 26'S = 124° 26’

D.Long= 115° 04' E- 80° 36' E =34° 28’ E or 34.467

AB=51.729°=x60=3103.7 mile

A=N143.5E

Course= 143.5 T


P

A

B


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A=05° 48' N 80° 36' E ; B= 34° 26' S 115° 04' E

PA=84° 12’ PB= 124° 26’

D.Long=34° 28’ E or 34.467

AB=51.729°=x60=3103.7 mile

A=N143.5E Course= 143.5 T

B=45.817°

Final course=S45.817°E or 134.2° T


P

A

B

To find final course – reciprocal of B; Cos B = (Cos PA - Cos PB x Cos AB )/ Sin PB x Sin AB

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A=05° 48' N 80° 36' E ; B= 34° 26' S 115° 04' E

PA=84° 12’ PB= 124° 26’

D.Long=34° 28’ E or 34.467

AB=51.729°=x60=3103.7 mile

A=N143.5E Inital Course= 143.5 T

B=45.817° Final course=S45.817°E or 134.2° T

Where is vertex?


P

A

B

V

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A=05° 48' N 80° 36' E ; B= 34° 26' S 115° 04' E

PA=84° 12’ PB= 124° 26’

D.Long=34° 28’ E or 34.467

AB=51.729°=x60=3103.7 mile

A=N143.52E Inital Course= 143.5 T

B=45.817° Final course=S45.817°E or 134.2° T

pAv=180-143.52=36.48


P

A

B

V

PV

pAv vPa

PA

AV

☺

☺

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2. Find the great circle distance and the initial and final courses from Fastnet Island 51°16'N 9° 3 6 'W to Mona Passage 18°28'N 67° 3 2 ' W .Find the position of the vertex and the latitude of a point on the track in longitude 20°W

3. Find the great circle distance and the initial and final courses from Strait of Magellan 52° 23' S 68° 18' W to Cape Town 33° 53'S 18° 2 0 ' E. Find the position of the vertex and the latitude of a point on the track in longitude 0°E

4. Find the great circle distance and the initial and final courses from Durban 29° 53'S 31° 0 4 ' E to Fremantle 32° 04'S 115° 2 6 ' E Find the position of the vertex and the latitude of a point on the track in longitude 100°E

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DECK 122 (NAVIGATION-II) Composite GC sailing

It is not always possible or desirable to travel along a great circle for some of the following reasons.

The great circle track may pass through high latitudes where weather is likely to be rough and the ship may encounter large waves and swell.

The great circle track may pass over land.

The saving of distance is small in low latitudes, or if the course is nearly north/south.

A great circle track may take the ship into head winds and adverse currents.

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Composite great circle sailing means travelling between two places by the shortest route with the restriction of not going north or south of a limiting latitude.

V1 V2

P

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Sin Mid Part = Tan Adjacent Parts or Sin Mid Part = Cos Opposite Parts


A

P

V1

To find the longitude of V, by finding D.long 'P' Sin Mid Part = Tan Adjacent Parts Sin Co P = Tan PV1 x Tan Co PA Cos P = Tan PV1 / Tan PA 71 SAK


Similarly, in the second triangle:

Cos P = Tan PV2 / Tan PB

to find the longitude of V2

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To find initial course A

Sin Mid Part = Cos Opposites

Sin PV1 = Cos Co A x Cos Co PA

Sin PV1 = Sin A x Sin PA

Sin A = Sin PV1 / Sin PA

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To find distance AV

Sin Mid Part = Cos Opposites

Sin Co PA = Cos PV1 x Cos AV1

Cos PA = Cos PV1 x Cos AV1

Cos AV1 = Cos PA / Cos PV1

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DECK 122 (NAVIGATION-II) Composite GC sailing-Example

Find the distance from Durban to Fremantle by composite great circle course using 35° S as the, limiting latitude. Find also the initial and final courses.

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Find the distance from Durban to Fremantle by composite great circle course using 35° S as the, limiting latitude. Find also the initial and final courses.

A Lat: 29 53 S Long: 31 04 E PA=60 07 PA=60.117

B Lat: 32 04 S Lonf: 115 26 E PB=57 56 PA=57.933

D.Long 84 22 E PV=55 PV=55.000

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A Lat: 29 53 S Long: 31 04 E PA=60 07 PA=60.117

B Lat: 32 04 S Lonf: 115 26 E PB=57 56 PA=57.933

D.Long 84 22 E PV=55 PV=55.000

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A Lat: 29 53 S Long: 31 04 E PA=60 07 PA=60.117

B Lat: 32 04 S Lonf: 115 26 E PB=57 56 PA=57.933

D.Long 84 22 E PV=55 PV=55.000

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A Lat: 29 53 S Long: 31 04 E PA=60 07 PA=60.117

B Lat: 32 04 S Lonf: 115 26 E PB=57 56 PA=57.933

D.Long 84 22 E PV=55 PV=55.000

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A Lat: 29 53 S Long: 31 04 E PA=60 07 PA=60.117

B Lat: 32 04 S Lonf: 115 26 E PB=57 56 PA=57.933

D.Long 84 22 E PV=55 PV=55.000

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A Lat: 29 53 S Long: 31 04 E PA=60 07 PA=60.117

B Lat: 32 04 S Lonf: 115 26 E PB=57 56 PA=57.933

D.Long 84 22 E PV=55 PV=55.000

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A Lat: 29 53 S Long: 31 04 E PA=60 07 PA=60.117

B Lat: 32 04 S Lonf: 115 26 E PB=57 56 PA=57.933

D.Long 84 22 E PV=55 PV=55.000

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Documents

Great Circle Sailing Notes