EXERCISE 34 – AZIMUTH STAR (Numerical Solution) |

# EXERCISE 34 – AZIMUTH STAR (Numerical Solution)

###### GMT     05 March 22h  35m  16s

GHA Ȣ(05d 22h)            134˚  01.0’                                           Dec         N  08˚  53.2’
Incr. (35m 16s)             008˚  50.4’
GHAȢ                              142˚  51.4’
SHA  *                          (+)062˚  12.4’
GHA *                            205˚  03.8’
Long (E)                     (+)083˚  46.0’
LHA  *                            288˚  49.8’

###### NOTE:

Working is very similar to that of azimuth- Sun, except that GHA Ȣ and SHA * have been used, which can be taken from the nautical almanac.

P = (360˚ – LHA)
= (360˚ – 288˚ 49.8’)
= 71˚  10.2’

We know that :

###### Azimuth = N 73.8˚ E

T Az          = 073.8˚ (T)
C Az          = 078.0˚ (C)
Error         = 4.2˚ W
Variation  = 3.0˚ W
Deviation = 1.2˚ W

1. ###### On 30th Nov 2008, PM at ship in DR 48˚ 57’ N 173˚ 18’ W, the azimuth of the star VEGA was 296˚(C) when the GPS clock showed 07h 39m 22s. If variation was 1˚E, calculate the deviation of the compass.

d       h      m      s
GMT                        01    07      39    22
LIT (W)                    (-)     11      33    12
LMT                         30    20    06     10

###### GMT  01 Dec 07h 39m 22s

GHA Ȣ(01d 07h)             175˚  30.7’                                           Dec *        N  38˚  47.6’
Incr. (39m 22s)            009˚  52.1’7                                             Lat            48˚  57’ N
GHA  Ȣ                            185˚  22.8’
Long (W)                   (-) 173˚  18.0’
LHA  Ȣ                             012   04.8’
SHA  *                      (+)080˚  41.7’
LHA  *                           092˚  46.5’

P  = LHA
= 92˚  46.5’

We know that :

###### Azimuth = N 60.53˚W

T Az          = 299.47˚ (T)
C Az          = 296.00˚ (C)
Error         = 3.5˚ E
Variation  = 1.0˚ E
Deviation = 2.5˚E

1. ###### On 01ST May 2008, AM at ship in DR 62˚ 11’S 179˚ 58’ E, the azimuth of the star SPICA was 312˚(C) when the GPS clock showed 13h 00m 08s. If variation was 10˚E, calculate the deviation of the ship’s head.

d     h     m     s
GMT                       30   13    00  08
LIT (E)                            11   59   52
LMT                        01   01  00   00

###### GMT                      30d 13h 00m 08s

GHA Ȣ(30d 13h)       053˚  50.6’                                           Dec *         S  11˚  12.5’
Incr. (00m 08s)        000˚  02.0’
GHA Ȣ                       053˚  52.6’
SHA  *(+)                  158˚  35.1’
GHA *                       212˚  27.7’
Long (E)             (+)  179˚  58.0’
LHA  *                      032˚  25.7’

P = LHA
= 032˚  25.7’

###### Azimuth =N 39.4˚W

T Az          = 320.6˚ (T)
C Az          = 312.0˚ (C)
Error         = 8.6˚ E
Variation  = 10.0˚ E
Deviation = 1.4˚ W

1. ###### On 13th Sept 2008 at ship in DR 30˚ 46’ N 090˚ 36’ W, the star RASALHAGUE bore 275˚(C) at 04h 36m GMT. If variation was 05˚W, calculate the deviation of the compass.

d      h       m      s
GMT                      14     04     36    00
LIT (W)                  (-)     06     02    24
LMT                       13     22     33   36

###### GMT                  14d 04h 36m 00s

GHA Ȣ(14d 04h)            053˚  30.5’                                           Dec *         N  12˚  33.3’
Incr. (36m 00s)             009˚  01.5’
GHA Ȣ                            062˚  32.0’
SHA  *                       (+)096˚  09.7’
GHA *                           158˚  41.7’
Long (W)                  (-)  090˚  36.0’
LHA  *                            068˚  05.7’

P = LHA
= 68˚  05.7’

We know that :

###### Azimuth = N 89.9˚W

T Az          = 270.1˚ (T)
C Az          = 275.0˚ (C)
Error         = 5.1˚ W
Variation  = 5.0˚ W
Deviation = 0.1˚ W

1. ###### On 21st Jan 2008 at 0320 ship’s time, in DR 64˚ 12’ N 112˚ 18’ E, the azimuth of the star DENEB bore 034.5˚(C). If variation was 4˚E, and the ship’s time was 7h ahead of GMT, find the deviation.

d       h       m      s
Ship’s time                21      03     20   00
Diff. (7 hr)(-)                       07     00   00
GMT                           20     20     20   00

###### GMT                       20d 20h 20m 00s

GHA Ȣ(20d 20h)            059˚  34.8’                                           Dec *         N  45˚  18.5’
Incr. (20m 00s)             005˚  00.8’
GHA Ȣ                              064˚  35.6’
SHA  *                          (+)049˚  34.9’
GHA *                               114˚  10.5’
Long (E)                      (+)112˚  18.0’
LHA  *                             226˚  28.5’

P = (360 – LHA)
= 133˚  31.5’

We know that :

###### Azimuth = N 34.4˚E

T Az          = 034.4˚ (T)
C Az          = 034.5˚ (C)
Error         = 0.1˚ W
Variation  = 4.0˚ E
Deviation = 4.1˚ W

### 1 Comment

• Pallav says:

I found a very usefull.
Keep it up
Sir!!