EXERCISE 30 — LATITUDE BY MERIDIAN ALTITUDE SUN (Numerical Solution)

  1. On21st Jan 2008 AM, in DR 24˚ 36.0’ S 110˚ 20.0’ W,the sextant altitude of the Suns’sLL was 85˚ 03.5’. If IE was 1.6’ off the arc and HE was 10m, find the latitude and state the direction of the PL (LOP).

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  1. On 1st Sept 2008, in DR equator 50˚ 27.0’ E, the sextant meridian altitude of the sun’s UL was 82˚ 10.4’. If IE was 2.4’ on the arc and HE was 17m, required the latitude and the PL (LOP).

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  1. ON 1st May 2008, in DR longitude 179˚ 58.0’ E, the observed altitude of the sun’s LL on the meridian was 64˚ 35.9’ South of the observer. If HE was 15m, find the latitude and the PL.

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  1. ON 14th Sept 2008, in DR longitude 116˚ 27.0’ W, the sextant meridian altitude of the sun’s UL North of the observer was 70˚ 29.8’. If IE was 3.2’ off the arc and HE was 12m, find the latitude and state the direction of the PL (LOP).

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13 Comments

  • In 3rd problem how to find true altitude position if lat is not given

    How to find dec is greater than lat or dec smaller than lat

    • To find true altitude you don’t need latitude and to identify dec and lat which greater is look at mzd and dec poles. If mzd n and dec n then plus if differences poles minus. Answer will depend on either dec and mzd who have highers value.

    • it is named North or South, with respect to observer and the declination, if observer is north and declination is south then Cel body is south, so you can use this.

      • On the 21st jan.2008, the dip(HE10m) 05.6′ is subtracted from the observed Alt. where is the 05.6′ to arrived at the App Alt and 16.1′ Tcorrn.LL added to App Alt, to arrive at TAlt, MZD 04 44.4 S, not in the solution and Dec 19 54.4 s also not in the solution and finally you arrived at an answer
        latitude 24 38. 8′ where are all these figures from

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