EXERCISE 36 – INTERCEPT STAR (Numerical Solution)

  1. On 31st Aug 2008, AM at ship in DR 40˚ 30’N 064˚ 56’E, the sextant altitude of the star DIPHDA was 21˚ 28.4’ when the chron (error 01m 06s FAST) showed 00h 21m 32s. If IE was 0.9’ off the arc & HE was 09m, find the direction of the LOP and the intercept.
q4p1
GMT     31 Aug 00h  20m  26s

GHA Ȣ(31d 00h)    339˚  32.7’                                           Dec         S  17˚  56.1’
Incr. (20m 26s)    005˚  07.3’                                            Lat            40˚  30’ N
GHA Ȣ                 344˚  40.0’
SHA *             (+) 348˚  59.1’
GHA *                  333˚  39.1’
Long (E)        (+) 064˚  56.0’
LHA  *                  038˚  35.1’

P  = LHA
= 38˚  35.1’

dai 4

We know that :
Cos CZD = ( Cos P × Cos Dec × Cos Lat) + ( Sin Lat × Sin Dec )
Cos CZD = ( Cos 038˚  35.1’ × Cos 17˚  56.1’ × Cos 40˚  30’ N ) + ( Sin 40˚  30’ N × Sin 17˚  56.1’ )
CZD = 68˚  33.6’

Sext Alt                         21˚ 28.4’
IE (OFF)                   (+) 00.9’
Observed Alt               21˚ 29.3’
Dip (HE 09m)          (-) 05.3’
App Alt                         21˚ 24.0’
T Corrn.                    (-)02.5’
T Alt                            21˚ 21.5’
TZD                               68˚ 38.5’

TZD             =   68˚ 38.5’
CZD             =   68˚ 33.6’
Intercept    =   4.9’ AWAY

q4p2

Azimuth = S 39.6˚ W

T Az          = 219.6˚ (T) 

LOP = 129.6˚ – 309.6˚

  1. On 22nd Sept 2008, PM at ship in DR 60˚ 10’N 092˚ 27’ E, the sextant altitude of the star ARCTURUS was 25˚ 01.0’ when the chron (error 05m 01s SLOW) showed 00h 46m 31s. If IE was 0.2’ on the arc & HE was 17m, find the direction of the LOP and the intercept.
q5p1
GMT     22 Sept 12h  51m  32s

GHA Ȣ(30d 12h)         181˚  43.3’                                           Dec         N  19˚  08.3’
Incr. (51m 32s)           012˚  55.1’                                           Lat            60˚  10’ N
GHA Ȣ                           194˚  38.4’
SHA *                     (+) 145˚  59.2’
GHA *                          340˚  37.6’
Long (E)              (+)092˚  27.0’
LHA  *                        073˚  04.6’

P  = LHA
= 73˚  04.6’

dai 5

We know that :
Cos CZD = ( Cos P × Cos Dec × Cos Lat) + ( Sin Lat × Sin Dec )
Cos CZD = ( Cos 73˚  04.6’ × Cos 19˚  08.3’ × Cos 60˚  10’ ) + ( Sin 60˚  10’ × Sin 19˚  08.3’ )
CZD = 65˚  05.3’

Sext Alt                         25˚ 01.0’
IE (ON)                     (-) 00.2’
Observed Alt               25˚ 00.8
Dip (HE 17m)          (-) 07.3’
App Alt                         24˚ 53.5’
T Corrn.                    (-)02.1’
T Alt                            24˚ 51.4’
TZD                               65˚ 08.6’

TZD             =   65˚ 08.6’
CZD             =   65˚ 05.3’
Intercept    =   03.3’ AWAY

We know that :

q5p2

Azimuth = S 85.2˚ W

T Az          = 265.2˚ (T)

LOP = 175.2˚ – 355.2˚

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