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ON 1st Dec 2008, in DR 06˚ 35.0’ N 064˚ 18.0’ W, owing to a hazy horizon to the South, a back angle observation of the sun’s LL on the meridian was made and the sextant altitude was found to be 118˚ 11.8’. If HE was 14m, and IE was 2.4’ on the arc, required the latitude and the direction of the PL.
Points to pounder prior to solving problems of this chapter.
- True Altitude is named same as Azimuth
- MZD is named opposite to T Alt.
- If MZD and Dec are of same name so add and retain same name and If of opposite names then subtract the smaller one from the larger one and retain the name of the larger one.
- Whether ‘d’ correction is to be added or subtracted is known by inspecting the Dec for the required hour and the next hour. If Dec is increasing, d correction is to be added and vice-versa.
- In order to know the naming of T Alt and MZD, one should draw a diagram and easily know and understand the naming system. Drawing the diagram in exams may leave a positive impression, that the candidate has understood the fundamentals correctly irrespective of clerical errors.
I wanna know, in question no. 1, the Dr longitude given is West, but why is the LIT in the calculation East
That’s a typing mistake otherwise the calculation is correct
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.
Where can I find a free 2008 Nautical Almanac (NP314-08). Thanks.
u can google it
Please explain how to decide true altitude position N or S when DR lat is not given
True alt N or S is same as azimuth at mer pass
How can identify that true alt “s” or N”
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.
In the 5th solutions how the sextant altitude is 61°27.5′. I couldn’t understand
same as bearing
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