EXERCISE 32 — LONG BY CHRONOMETER SUN (Numerical Solution)

GMT     30 April 13h  00m  52s

GHA (30d 13h)                015˚  42.7’                                           Dec         N  14˚  59.3’
Incr. (00m 52s)               000˚  13.0’                                        d(+0.7)                00.0’
GHA                                 015˚  55.7’                                         Dec         N  14˚  59.3’
Lat                                       00˚  20’ N 

Sext Alt                                44˚ 13.4’
IE (off)                             (+)       03.1’
Observed Alt                     44˚ 16.5’
Dip (HE 20m)               (-)        07.9’
App Alt                              44˚ 08.6’
T Corrn. UL                  (-)         16.8’
T Alt                                  43˚ 51.8’

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Vikrant_sharma

12 Comments

  • @vikarant_sharma diagram in first question is wrong obs long value is smaller than DR LONG so it should be drawn in left side of DR long

  • 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

  • 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

    • Calculate latitude course and speed error for following ship
      A)ship steering 230°T at 22.0 knots in latitude 41°24S
      B)ship steering 140°T at 15.0 knots in latitude 56°00N
      Please solve this problem argent machinical project hy

  • 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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