EXERCISE 28 — AZIMUTH SUN (Numerical Solution)
- 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.
NOTE:
- If LHA<180, P = LHA and If LHA>180, P = 360-LHA
- 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.
- 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
- 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
- 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
- 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.
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
