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’

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NOTE:

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

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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:

e32q1p2

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

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Vikrant_sharma

11 Comments

  • 2nd and 3rd point of the note is also wrong. Correct one is when the sun (and star) is east of the meridian it’s AM, at that time LHA is between 180-360 so P= 360°-LHA. when the sun and (the star) is west of the meridian at that time LHA is between 0°-180° , and LHA=P . Correct ans of the question is P= 49°38.8°… long=175°15.8°…. az=N80.1W or 279.9° true

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  • 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 …

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