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 = 47˚ 52.9’
Since, sight is before mer. Pass. ,so LHA = 360 – P
LHA = 312˚ 08.1’
GHA = 161˚ 01.9’
Long. E = 151˚ 06.2’
Obs. Long. = 151˚ 06.2’ E
Azimuth = N 65.9˚ E
T Az = 065.9˚ (T)
LOP = 155.9˚ – 335.9˚
-
On 30th April 2008, in DR 00˚ 20’N 060˚ 12’W, the sextant altitude of the sun’s UL East of the Meridian was 44˚ 13.4’ when the GPS clock showed 01h 00m 52s. If IE was 3.1’ off the arc & HE was 20m, find the direction of the LOP and the longitude where it cuts the DR latitude.
d h m s
GMT 30 13 00 52
LIT (W) (-) 04 00 48
LMT 30 09 00 04
can you tell how you have calculated gmt in 1st question
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
Its wrong
Can u tell how did you got the GMT time to be 13h 00m 52s in the last question?
No
No its wrong
thank you very much Lovepreet for notifying us.. we will look into it & correct as required.best wishes for your exams.
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 …
Thank.you lovepreet..can you please specify the numerical where it is wrong.