GREAT CIRCLE SAILING - RSW...

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 PressPress 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 - 3510 = 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 = 40N then southern vertex = 40S).

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

At the vertex the course on the great circle is exactly 090T or 270T, 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


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

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AB=1571.6 mile 42 SAK


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) 0724'N 079 55 'W

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

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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) 0724'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 =0724'N 079 5 5 'W

PA =

PB =

P =

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A =41 38' S 175 28' E

B =0724'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 =0724'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 =0724'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=0724'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=0724'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=0724'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=0724'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 + 615'.9E = 18143'.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 140W. In triangle VPL:

The polar angle P is 178 16.1 W - 140W =3816.6 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 85W

In triangle VPL: The polar angle P is 178 16MW-85W = 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 =2.918N =2 55.1 N in longitude 85W

(- 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 100E

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A=05 48' N 80 36' E

B= 34 26' S 115 04' E

PA=90- 05 48N = 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 48N = 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=S45.817E 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=S45.817E 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=S45.817E 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 5116'N 9 3 6 'W to Mona Passage 1828'N 67 3 2 ' W .Find the position of the vertex and the latitude of a point on the track in longitude 20W

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 0E

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 100E

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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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GREAT CIRCLE SAILING - RSW...

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