Originally Posted by captaa
Q1
Assume a North polar stereographic chart whose grid is aligned with the Greenwich meridian.An aircraft flies from the geographic North pole for a
distance of 480 NM along the 110°E meridian, then follows a grid
track of 154° for a distance of 300 NM. Its position is now approximately:
A) 70°15'N 080°E B) 80°00'N 080°E C) 78°45'N 087°E D) 79°15'N 074°E
Flying for 480NM from North Pole(N) along 110°E, aircraft reaches A.
Departure along meridian=dLat° x 60
dLat°=Departure/60=480/60=8°
So, A's position is 82°N 110°E.
From A, aircraft flies a grid track of 154° which is is respect to the grid datum i.e. Greenwich meridian.
True track=360-angle NAR
RAY=26° (180°-154°)
NAY=70°
NAR=RAY+NAY=96°
So, True track from A = 264°
This track is predominantly westerly with a slight southward tilt.
Using departure formula, Departure=300NM. Find dLong?
dLong=Departure/cos (Mean Lat)
- A:Mean Lat=76°N, New point is at 89°E.
- B:Mean Lat=81°N, New point is at 78°E.
- C:Mean Lat=80°N, New point is at 81°E.
- D:Mean Lat=81°N, New point is at 78°E.
So, the closest I can get is to B) 80°00'N 080°E. There must be a better way of doing it, I hope.
The same question was also discussed in
another forum. And
they found out that it is not a calculation question, it is more of plotting question which requires accurate map plotting.