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.


  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:



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


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

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