Great Circle - Chief Mate

8/11/2019 Great Circle - Chief Mate

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1

GREAT

CIRCLE

SAILINGS


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Great circle v Mercator


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Newcastle to San Francisco enters

the Arctic Circle 6633 North


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G.C. routes in REDNote the Icebergs


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Great circle v Rhumb line

1. ADVANTAGEGreat Circle shorter distance than Rhumbline, therefore saving Fuel and Time.

2. DISADVANTAGE

Great Circle may take you into highLatitudes and therefore encounter bad

weather with risk of damage to the vesseland cargo and a slower steaming speed.


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Basic Great Circle Drawing

DLong

Equator

North Pole


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Naming Courses

Initial course = Elevated pole + direction of

travel

Final course = Opposite elevated pole +

direction of travel


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Cosine formulae for Great Circle

Cos AB = (Cos P x Sin PA x Sin PB) + (Cos PA x Cos PB)

Cos AB =

C A B

(Cos P x Sin PA x Sin PB)

A B

+ (Cos PA x Cos PB)


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How Do We Find ?

Angle PA?

Angle PA = (90- Lat A)

Angle PB?

Angle PB = (90- Lat B)

Angle P?

Angle P = (DLong A to B)


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Basic Great Circle Drawing

DLong

Equator

North Pole


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Example 1

Find the great circle distance, initial and final

tracks between the following places

A22 10N 074 56Wto

B49 27N 010 46W

Draw a sketch and find angles PA, PB & P (DLong)


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Example 1

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P

A

B

2210N

074 56W

4927N

01046W

Equator


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P 90 00N P 90 00N 074 56W

A 22 10N B 49 27N 010 46W

PA 67 50N PB 40 33N P 064 10E

C A B A B


(Cos 06410 x Sin 6750 x Sin 4033) + (Cos 6750 x Cos 4033)

AB = 56 41 55.73 = 56 41.9 x 60 = 3401.9287 = 3401.9N.M.

Distance = 3401.9 / 60 = 5641.9 SAVE in D


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Example 1

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P

A

B

2210N

074 56W

4927N

01046W

Equator

DLong

6410E


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Initial Course

NOTE: NO Formulae Given

You will have to learn this!!!!!!

B A D

Cos A = (Cos PBCos PA x Cos AB)(Sin PA x Sin AB)

A D

Remember (Cos B-AD)

(Sin AD)

Can use ABC tables


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Initial Course

Initial course = N44 26 06.64E = N4426.1E = 044.4(T)


Cos A = (Cos 4033.0 Cos 6750.0 x Cos 56 41.9)

(Sin 6750.0 PA x Sin 56 41.9)

SAVE in X


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Final Course

Final course = S9412 34.56E = S9412.6E = 85.8(T)

Cos B = (Cos PACos PB x Cos AB)(Sin PB x Sin AB)

Cos B = (Cos 6750.0 Cos 4033.0 x Cos 56 41.9)

(Sin 4033.0 PA x Sin 56 41.9)

SAVE in Y if required


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Comparison

Initial course = 044.4(T)

Final course = 85.8(T)

Comparing distances by Mercator

Great Circle distance 3401.9

Mercator distance 3483.1Difference 81.2


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Example 2



A39 00S 020 00W

to

B40 00S 142 00E



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Example 2

20

P

A B3900S020 00W

4000S

14200E

Equator


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Initial Course

Initial course = S13 52 54E = S1352.9E = 166.1(T)


Cos A = (Cos 5000.0 Cos 5100.0 x Cos 99 18.2)

(Sin 5100.0 PA x Sin 99 18.2)

SAVE in X


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Final Course

Final course = N14 05 13.06E = N1405.2E = 14.1(T)


Cos B = (Cos 5100.0 Cos 5000.0 x Cos 99 18.2)

(Sin 5000.0 PA x Sin 99 18.2)



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Example 3



A04 07N 098 55E

to

B41 00S 033 58E


A


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Example 3

25P

A

B

4100S

03358E

0407N

09855EEquator


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P 90 00N P 90 00S 098 55E

A 04 07N B 41 00S 033 58E

PA 94 07S PB 49 00S P 064 57W

C A B A B



AB = 74 14 19.49 = 7414.3 x 60 = 4454.325 = 4454.3N.M.



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Initial Course

Initial course = S45 16 11.54W = S4516.2W = 225.3(T)


Cos A = (Cos 4900.0 Cos 9407.0 x Cos 7414.3)

(Sin 9407.0 PA x Sin 7414.3)

SAVE in X


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Final Course

Final course = N110 07 56.71W = N11007.9W = 249.9(T)


Cos B = (Cos 9407.0 Cos 4900.0 x Cos 7414.3)

(Sin 4900.0 PA x Sin 7414.3)



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Example 4



A33 52S 151 16E

to

B12 04S 077 14W



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Example 4

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P

A

B

3352S

151 16E

1204S

07714W

Equator


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P 90 00S P 90 00S 151 16E

A 33 52S B 12 04S 077 14W

PA 56 08S PB 77 56S P 228 30W

P 131 30E

C A B A B



AB = 114 55 55.95 = 11455.9 x 60 = 6895.932 = 6895.9N.M.



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Initial Course

Initial course = S53 52 12.88E = S5352.2E = 126.1(T)


Cos A = (Cos 7756.0 Cos 5608.0 x Cos 11455.9)

(Sin 5608.0 PA x Sin 11455.9)

SAVE in X


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Final Course

Final course = N43 17 55.43W = N4317.9E = 043.3(T)


Cos B = (Cos 5608.0 Cos 7756.0 x Cos 11455.9)

(Sin 7756.0 PA x Sin 11455.9)



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Example 5 (SQA July 2011)



A54 21S 064 18W

to

B04 06N 008 15E


B


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Example 5

35P

A

B

5421S

06418W

0406N

00815EEquator

PrimeMeridian


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P 90 00N P 90 00N 064 18W

A 54 21S B 04 06N 008 15E

PA 35 39S PB 94 06S P 072 33E

C A B A B



AB = 83 19 31.46 = 8319.5 x 60 = 4999.524 = 4999.5N.M.



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Initial Course

Initial course = S106 39 32.46E = S10639.5E = 073.3(T)


Cos A = (Cos 9406.0 Cos 3539.0 x Cos 8319.5)

(Sin 3539.0 PA x Sin 8319.5)

SAVE in X


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Final Course

Final course = N34 02 31.63E = N3402.5E = 34.0(T)

Cos B = (Cos PACos PB x Cos AB)

(Sin PB x Sin AB)

Cos B = (Cos 3539.0 Cos 9406.0 x Cos 8319.5)

(Sin 9406.0 PA x Sin 8319.5)



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SQA Dec 2012 Q3

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3 i) Fi d th Rh b li di t b t


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3ai) Find the Rhumb line distance between

49 47N 006 27W to 18 20N 067 50W

Start Lat 49 47.0N MP 3436.41 Start Long 006 27WFinal Lat 18 20.0N MP 1111.91 Final Long 067 50W

DLat 31 27.0S DMP 2324.50 DLong 61 23W

DLat 1887S DLong 3683W

Tan course = Dlong / DMP

Tan course = 3683 / 2324.50 = S5744.5W = 237.7T

Dist = DLat / Cos CourseDist = 1887 / Cos 57.74 = 3535.5

Track 237.7T and Distance 3535.5

Use ANSwer buttonLeave 57.74221104on your displayType 1887 / Cos ANS


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3aiiDraw a sketch

41

P

B

A

1820N

067 50W

4947N

00627W

Equator


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P 90 00N P 90 00N 006 27W

A 49 47N B 18 20N 067 50W

PA 40 13N PB 71 40N P 061 23W

C A B A B



AB = 57 44 29.64 x 60 = 3464.5N.M.



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P

B

A

1820N

067 50W

4947N

00627W

Equator

DLong

6123W


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3aiii - Initial Course

Initial course = N99 48.5W = 260.2(T)

Cos A = (Cos PBCos PA x Cos AB)

(Sin PA x Sin AB)

Cos A = (Cos 7140.0 Cos 4013.0 x Cos 5744.5)

(Sin 4013.0 PA x Sin 5744.5)


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3bSteaming times

Rhumb line 3535.3 +1810 = 5345.3

Steaming time 5345.3 / 17kts = 13d 02h 26m

GC 3464.5 + 1810 = 5274.5

Adverse current 17kts3 kts = 14kts x 72hrs = 1008Remainder of passage 5274.5 1008 = 4266.5

Steaming time 4266.5 / 17kts = 10d 10h 58m

Total GC steam time = 10d 10h 58m + 72hrs = 13d 10h 58m

R/L 13d 02h 26m

GC 13d 10h 58m

R/L 08h 32m quicker 45


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For more examples seePosition of the Vertex


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Great Circle - Chief Mate