# EXERCISE 32 — LONG BY CHRONOMETER SUN (Numerical Solution)

1. ###### On 19th Jan 2008, PM at ship in DR 40˚ 16’S 175˚ 31’E, the sextant altitude of the sun’s LL was 43˚ 27.4’ when the GPS clock showed 03h 48m 00s. If IE was 1.5’ on the arc & HE was 22m, find the direction of the LOP and the longitude where it cuts the DR latitude.

d     h       m       s
GMT                       19   03     48    00
LIT (E)                    (+)    11     58   04
LMT                       19     15    46   04

###### GMT     19 Jan 03h  48m  00s

GHA (19d 03h)            222˚  23.0’                                           Dec          S  20˚  28.9’
Incr. (48m 00s)          012˚  00.0’                                           d(-0.5)                 00.4’
GHA                             234˚  23.0’                                           Dec          S  20˚  28.5’
Lat                              40˚  16’ S

Sext Alt                               43˚ 27.4’
IE (on)                              (-)       01.5’
Observed Alt                      43˚ 25.9’
Dip (HE 22m)                 (-)        08.3’
App Alt                               43˚ 17.6’
T Corrn. LL                     (+)       15.2’
T Alt                                   43˚ 32.8’

NOTE:

In the following formula, if the LAT and DEC are of same name then sign is (-), If of contrary names then sign is (+).

P = 49˚ 38.8’

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

NOTE:

• Before meridian passage (Mer. Pass.) , LHA wil be between 180 and 360. After meridian passage, LHA will be between 000 and 180.So, to calculate LHA,
• If it given that the sight is taken AM at ship or meridian of east, then P = LHA.
• If it is given that the sight is taken at PM at ship, then P = 360 – LHA
• When P>90, the minus sign obtained for the value of A is to be ignored and is taken care by changing the name of A.

LHA          = 049˚ 38.8’
GHA         = 234˚ 23.0’
Long. W   = 184˚ 44.2’

Obs. Long. = 175˚ 15.8’ E

We know that:

Azimuth = N 80.3˚ W
T Az= 279.7˚ (T)
LOP = 009.7˚ – 189.7˚

#### Vikrant_sharma

• Karanvir says:

The value of P is wrong. It should be 52°0.1′. Hence LHA would be 52°0.1′ and longitude = 004°29.6’E

• Retkiousk says:

Its wrong

• Gyanendra says:

Can u tell how did you got the GMT time to be 13h 00m 52s in the last question?

• thank you very much Lovepreet for notifying us.. we will look into it & correct as required.best wishes for your exams.

• Lovepreet Singh says:

solutions of formula cosP is wrongg … it matches the ans in the textbook bt when we try to solve it ans is slight different.. so it is humble request to u to pls correct the answers because students are facing difficulties when trying to get help from youe site …

• Manas says:

Thank.you lovepreet..can you please specify the numerical where it is wrong.

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