EXERCISE 28 — AZIMUTH SUN (Numerical Solution) |

# EXERCISE 28 — AZIMUTH SUN (Numerical Solution)

###### 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:

###### 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 :

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’

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’

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

###### 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 :

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