EXERCISE 28 — AZIMUTH SUN (Numerical Solution)

  1. On 20th July 2008, AM at ship in DR 44˚ 31’ N 069˚ 42’ E, the azimuth of the sun was 100˚(C) when chron showed 04h 01m 52s. If the chron error was 04m 20s SLOW and variation was 8˚E, find the deviation for the ship’s head.

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NOTE:
  1. If LHA<180, P = LHA and If LHA>180, P = 360-LHA
  2. Naming of A, B, C: A is named opposite to the latitude when LHA is between 270 & 90 and same as latitude when LHA is between 90 & 270. B is named same as declination. For C, if A & B are of same names, add and retain names. If of contrary names  then subtract and retain name of larger one.

We know that:

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NOTES:

Naming of Azimuth:  The prefix N or S is the name same as C whereas suffix E or W depends on the value of LHA. If LHA is between 0 to 180, the body lies to the WEST and if it is between 180 and 360, the body lies to the EAST.

Azimuth = S 76.9˚E
T Az= 103.1˚ (T)
C Az = 100.0˚ (C)
Error = 3.1˚ E
Variation = 8.0˚ E
Deviation= 4.9˚ W

NOTE:

The calculation of deviation is elementary chartwork. It can be easily calculated by referring the following simple formulae: Error – Variation = Deviation

Where, if error & variation are of same names then subtract and retain name. If of contrary names, then add and retain the name of the larger one.

  1. On 22nd Sept 2008, PM at ship in DR 18˚ 20’ N 085˚ 40’ E, the azimuth of the sun was 265˚(C) when the GPS clock showed 10h 09m 38s. If variation was 2˚W, calculate the deviation of the compass.

d     h     m      s

GMT                   22    10    09    38
LIT (E)                 05    42    40
LMT                    22   15     52     18

GMT  22 Sept 10h 09m 38s

GHA (22d 10h)         331˚  51.4’                            Dec         N  00˚  05.6’
Incr. (09m 38s)         002˚  24.5’                       d(-1.0)                     00.2’
GHA                            334˚  15.9’                            Dec         N  00˚  05.4’
Long (E)                 (+)085˚  40.0’
LHA                             059˚  55.9’                             Lat            18˚  20’ N

P =  LHA
= 59˚  55.9’

We know that :

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Azimuth = S 79.7˚W

T Az          = 259.8˚ (T)
C Az          = 265.0˚ (C)
Error         = 5.2˚ W
Variation  = 2.0˚ W
Deviation = 3.2˚ W

  1. On Jan 19th 2008, in DR 40˚ 16’S 175˚ 31’ E, the azimuth of the sun was 267˚(C) at 1618 ship’s time. If the ships time difference was 11h 30m from GMT and variation was 2.3˚E, find the deviation for the ship’s head.

    d      h     m     s

Ship’s time           19     16   18   00
Diff. (1130hr)      (-)11   30   00
GMT                      19    04    48   00

GMT     19THJan 04h 48m 00s

GHA (19d 04h)         237˚  22.8’                                        Dec         S  20˚  28.4’
Incr. (48m 00s)        012˚  00.0’                                      d(-0.5)                 00.4’
GHA                          249˚  22.8’                                       Dec         N  20˚  28.0’
Long (E)                (+)175˚  31.0’
LHA                         064˚  53.8’                                          Lat            40˚  16’ S

We know that:
P = LHA
= 064˚  53.8’

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Azimuth = S 89.3˚W
T Az          = 289.3˚ (T)
C Az          = 287.0˚ (C)
Error         = 2.3˚ E
Variation  = 2.3˚ E
Deviation = NIL

  1. On 30th April 2008, in DR 00˚ 00’ 060˚ 12’W, the sun bore 080˚(C) when the GPS clock showed 11h 00m 52s. If the variation was 1˚W, find the deviation of the compass.

d      h       m      s

GMT                   30      11     00     52
LIT (W)                (-)      04     00    48
LMT                    30     07     00     04

GMT  30 April 11h 00m 52s

GHA (30d 11h)            345˚  42.5’                              Dec         N  14˚  57.8’
Incr. (00m 52s)         000˚  13.0’                              d(+0.7)              00.0’
GHA                           345˚  55.5’                                Dec         N  14˚  57.8’
Long (W)                (-)060˚  12.0’
LHA                           285˚  43.5’                                   Lat            00˚  00’

We know that :

P = (360˚ – LHA)
= (360˚ – 285˚  43.5’)
= 74˚  16.5’

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Azimuth = N 74.5˚E
T Az          = 074.5˚ (T)
C Az          = 080.0˚ (C)
Error         = 5.5˚ W
Variation  = 01˚ W
Deviation = 4.5˚ W

  1. On 31st August 2008, in DR 10˚ 11’ S 000˚ 00’, the azimuth of the sun was 282˚(C) when chron showed 05h 10m 25s. If the chron error was 00m 05s FAST and variation was 3˚E, find the deviation for the ship’s head.

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GMT  31 Aug 17h 10m20s

GHA (31d 17h)            074˚  58.4’                Dec         N  08˚  19.6’
Incr. (10m 20s)           002˚  35.0’            d(-0.9)                  00.2’
GHA                              077˚  33.4’               Dec           N  08˚  19.4’
Long (E)                        000˚  00.0’
LHA                               077˚  33.4 ’          Lat                 10˚  11’ S

P = LHA
= 77˚  33.4’

We know  that :

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Azimuth = N 79.4˚ W
T Az          = 280.6˚ (T)
C Az          = 282.0˚ (C)
Error         = 1.4˚W
Variation  =3.0˚ E
Deviation = 4.4˚ W

manas NIRENWAL

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