EXERCISE 37- LONGITUDE BY CHRONOMETER STAR (Numerical Solution) |

# EXERCISE 37- LONGITUDE BY CHRONOMETER STAR (Numerical Solution)

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 longitude where it crosses the DR latitude.

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’

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’

###### P = 55˚ 00.8’

Since, sight is AFTER mer. Pass. ,so LHA = P

LHA          = 055˚ 00.8’
GHA         = 162˚ 18.5’
Long. W  = 107˚ 17.7’

###### Obs. Long. = 107˚ 17.7’ W

Azimuth = N 81.5˚ W

T Az          = 278.5˚ (T)

Hence ,LOP  = ( TAz  ±  90 )
=  008.5˚ – 188.5˚

###### GMT     22 Sept 12h  51m  32s

GHA Ȣ(30d 12h)      180˚  44.2’                                           Dec         N  19˚  08.3’
Incr. (51m 32s)       012˚  54.6’                                             Lat            60˚  10’ N
GHA Ȣ                     193˚  38.8’
SHA *                (+) 145˚  59.2’
GHA *                     339˚  38.0’

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’

We know that :

###### P = 73˚ 11.2’

Since, sight is AFTER mer. Pass. ,so LHA = P

LHA          = 073˚ 11.2’
GHA         = 339˚ 38.0’
Long. E    = 093˚ 33.2’

Obs. Long. = 093˚ 33.2’ E

###### Azimuth = N 81.5˚ W

T Az          = 278.5˚ (T)

Hence ,LOP  = ( TAz  ±  90 )
=  008.5˚ – 188.5˚

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 longitude where it cuts the DR latitude.

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’

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’

###### P     = 46˚ 30.0’

Since, sight is before mer. Pass. ,

So LHA = (360 – P)

LHA          = 313˚ 30.0’
GHA         = 142˚ 36.5’
Long. E    = 172˚ 53.5’

###### Obs. Long. = 172˚ 53.5’ E

Azimuth = N 79.8˚ E

T Az          = 079.8˚ (T)

LOP = 169.8˚ – 349.8˚

###### 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’

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’

We know that:

P = 30˚ 18.0’

Since, sight is AFTER mer. Pass. ,
So , LHA = P

LHA          = 030˚ 18.0’
GHA         = 070˚ 38.0’
Long. W   = 040˚ 20.0’

###### Obs. Long. = 040 ˚ 20.0’ W

Azimuth = N 64.7˚ W

T Az          = 295.3˚ (T)

Hence , LOP  = ( TAz  ±  90 )
LOP = 025.3˚ – 205.3˚

###### 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’

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’

We know that :

P = 38˚ 45.2’

Since, sight is AFTER mer. Pass. ,
So LHA = P

LHA          = 038˚ 45.2’
GHA         = 333˚ 39.1’
Long. E    = 065˚ 06.1’

###### Obs. Long. = 065˚ 06.1’ E

We know that :

Azimuth = S 39.8˚ W

T Az          = 219.8˚ (T)

Hence , LOP  = ( TAz  ±  90 )

LOP = 129.8˚ – 309.8˚