EXERCISE 28 — AZIMUTH SUN Numerical Solutions
4. 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’
![](https://indianmerchantnavy.org/wp-content/uploads/2023/01/image-10.png)
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