EXERCISE 31 — INTERCEPT SUN (Numerical Solution)

1. On 5^th March 2008, AM at ship in DR 38˚ 11’S 151˚ 10’E, the sextant altitude of the sun’s LL was 35˚ 59.1’ when the chron (error 00m 46s SLOW) showed 10h 54m 54s. If IE was 1.3’ off the arc & HE was 30m, find the intercept & the direction of the LOP.

GMT   – 04 March 22h 55m 40s

GHA (04d 22h)         147˚  06.9’                                           Dec         S  06˚  01.3’
Incr. (55m 40s)        013˚  55.0’                                            d(-1.0)                00.9’
GHA                          161˚  01.9’                                             Dec        S  06˚  00.4’
Long (E)              (+) 151˚  10.0’
LHA                         312˚  11.9’                                              Lat               38˚  11’ S

P = (360˚ – LHA)
=  ( 360˚ – 312˚  11.9’)
= 47˚  48.1’

NOTE:

• While using below formula, one must be careful about   +   signs.
• If Lat and Dec are of same name then sign is (+).  
• If of contrary names then it has to be (-).

We know that :
Cos CZD = ( Cos P × Cos Dec × Cos Lat) + ( Sin Lat × Sin Dec )
Cos CZD = ( Cos 47˚ 48.1’ × Cos 06˚ 00.4’ × Cos 38˚ 11’ ) + ( Sin 38˚ 11’ × Sin 06˚ 0.4’ )
CZD = 53˚  51.7’

Sext Alt                            35˚ 59.1’
IE (off)                              (+) 01.3’
Observed Alt                  36˚ 00.4’
Dip (HE 30m)                  (-)  09.6’
App Alt                           35˚ 50.8’
T Corrn. LL                       (+)14.9’
T Alt                                 36˚ 05.7’
TZD                                 53˚ 54.3’

TZD              =   53˚ 54.3’
CZD             =   53˚ 51.7’
Intercept    =   02.6’ AWAY