EXERCISE 36 – INTERCEPT STAR (Numerical Solution)

  1. On 30thApril 2008, PM at ship in DR 34˚ 18’S 040˚ 20’W, the sextant altitude of the star SIRIUS was 57˚ 51.9’ at 08h 53m 03s chron time. If CE was 01m 40s FAST, IE was 1.2’ on the arc & HE was 21m, find the direction of the LOP and the intercept.
q1p1
GMT     30 April 20h  51m  23s

GHA Ȣ(30d 20h)             147˚  06.9’                                   Dec         S  16˚  43.7’
Incr. (51m 23s)               012˚  52.9’                                    Lat            34˚  18’ S
GHA Ȣ                            172˚  00.7’
SHA *                       (+) 258˚  37.3’
GHA *                            070˚  38.0’
Long (W)                 (-) 040˚  20.0’
LHA  *                         030˚  18.0’

P  = LHA
= 030˚  18.0’

 dai 1

We know that :
Cos CZD = ( Cos P × Cos Dec × Cos Lat) + ( Sin Lat × Sin Dec )
Cos CZD = ( Cos 030˚  18.0’ × Cos 16˚  43.7’ × Cos 34˚  18’ ) + ( Sin 34˚  18’ × Sin 16˚  43.7’ )
CZD = 32˚  18.0’

Sext Alt                             57˚ 51.9’
IE (ON)                                (-) 01.2’
Observed Alt                    57˚ 50.7’
Dip (HE 21m)                     (-) 08.1’
App Alt                          57˚ 42.6’
T Corrn.                             (-)00.6’
T Alt                                57˚ 42.0’
TZD                               32˚ 18.0’

TZD             =   32˚ 18.0’
CZD             =   32˚ 18.0’
Intercept    =   NIL

q1p2

Azimuth = N 64.7˚ W

T Az          = 295.3˚ (T)

LOP = 025.3˚ – 205.3˚

  1. On 19th Jan 2008, at about 1900 at ship in DR 00˚ 02’N 170˚ 50’E, the sextant altitude of the star BETELGUESE was 43˚ 11.1’ at 07h 33m 44s by GPS clock. If IE was 1.3’ off the arc & HE was 18m, find the direction of the LOP and the intercept.

                                      d       h    m   s
GMT                             19    07  33  44
LIT (E)                          (+)    11  23  20
LMT                             19    18  57  04

GMT     19 Jan 07h  33m  44s

GHA Ȣ(19d 07h)             223˚  03.7’                                           Dec         N  07˚  24.6’
Incr. (33m 44s)             008˚  27.4’                                             Lat            00˚  02’ N
GHA Ȣ                                 231˚  31.1’
SHA *                         (+) 271˚  05.4’
GHA *                              142˚  36.5’
Long (E)                 (+) 170˚  50.0’
LHA  *                             313˚  26.5’

P  = (360˚ – LHA)
= (46˚ 33.5’)

dai 2

We know that :
Cos CZD = ( Cos P × Cos Dec × Cos Lat) + ( Sin Lat × Sin Dec )
Cos CZD = ( Cos 46˚ 33.5’ × Cos 07˚  24.6’ × Cos 00˚  02’ ) + ( Sin 00˚  02’ × Sin 07˚  24.6’ )
CZD = 47˚  00.2’

Sext Alt                            43˚ 11.1’
IE (OFF)                         (+) 01.3’
Observed Alt                43˚ 12.4’
Dip (HE 18m)               (-) 07.5’
App Alt                         43˚ 04.9’
T Corrn.                          (-)01.0’
T Alt                              43˚ 03.9’
TZD                               46˚ 56.1’

TZD             =   46˚ 56.1’
CZD             =   47˚ 00.2’
Intercept    =   04.1’ TOWARDS

q2p1

Azimuth = N 79.9˚ E

T Az = 079.9˚ (T)

LOP = 169.9˚ – 349.9˚

  1. On 29th Nov 2008, AM at ship in DR 25˚ 30’S 107˚ 20’W, the sextant altitude of the star RIGEL was 35˚ 10.3’ when the GPS clock showed 11h 29m 20s. If IE was 2.8’ on the arc & HE was 12m, find the direction of the LOP and the intercept.

                                      d       h    m   s
GMT                            29    11  29  20
LIT (W)                        (-)    07  09  20
LMT                             29    04  20  00

GMT     29  Nov  11h  29m  20s

GHA Ȣ(29d 11h)              233˚  42.3’                                           Dec         S  08˚  11.4’
Incr. (29m 20s)              007˚  21.2’                                           Lat            25˚  30’ S
GHA Ȣ                                241˚  03.5’
SHA *                          (+) 281˚  15.0’
GHA *                                162˚  18.5’
Long (W)                  (-) 107˚  20.0’
LHA  *                             054˚  58.5’

P  = LHA
= 54˚  58.5’

dai 3

Cos CZD = ( Cos P × Cos Dec × Cos Lat) + ( Sin Lat × Sin Dec )
Cos CZD = ( Cos 054˚  58.5’ × Cos 08˚  11.4’ × Cos 25˚  30’  ) + ( Sin 25˚  30’  × Sin 08˚  11.4’ )
CZD = 54˚  57.9’

Sext Alt                             35˚ 10.3’
IE (ON)                             (-) 02.8’
Observed Alt                35˚ 07.5’
Dip (HE 12m)                (-) 06.1’
App Alt                          35˚ 01.4’
T Corrn.                          (-)01.4’
T Alt                              35˚ 00.0’
TZD                              55˚ 00.0’

TZD             =   55˚ 00.0’
CZD             =   54˚ 57.9’
Intercept    =   2.1’ AWAY

q3p1

Azimuth = N 81.9˚ W

T Az          = 278.1˚ (T)

LOP = 008.1˚ – 188.1˚

  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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