General Nav problem
Thread Starter
Joined: Feb 2003
Posts: 178
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From: Near EGCC
General Nav problem
Just having some problems with G Nav:
A lambert conformal conic chart has a constant of the cone on 0.75. the initial course of a straight line track drawn on this chart from A 40N 50W to B is 043T at A. Course at B is 055T. What is the longitude of B
a) 34w <--answer
b) 36w
c) 38W
d) 41W
Help in working this question would be much appreciated.
Many thanks,
Wing Bar
A lambert conformal conic chart has a constant of the cone on 0.75. the initial course of a straight line track drawn on this chart from A 40N 50W to B is 043T at A. Course at B is 055T. What is the longitude of B
a) 34w <--answer
b) 36w
c) 38W
d) 41W
Help in working this question would be much appreciated.
Many thanks,
Wing Bar
Jet Blast Rat
Joined: Jan 2001
Posts: 2,081
Likes: 0
From: Sarfend-on-Sea
Change of straight line course is chart convergence, since on a straight line direction doesn't change, so the only reason course can change is if the north reference, the meridian, changes. Angle between meridians is chart convergence. Therefore convergence is 12 degrees.
Lambert chart convergence = ch. long x constant of the cone
ch. long = convergence / constant of the cone
=12 / 0.75
=16 degrees
Track is Easterly, so the westerly longitude must decrease, 50 W - 16 = 34 W
This question has been asked before, as are many G Nav questions, and Alex, Paul and I usually answer. Are other schools' G Nav instructors not answering their students' questions?
Send Clowns
Gen Nav Instructor
BCFT
Lambert chart convergence = ch. long x constant of the cone
ch. long = convergence / constant of the cone
=12 / 0.75
=16 degrees
Track is Easterly, so the westerly longitude must decrease, 50 W - 16 = 34 W
This question has been asked before, as are many G Nav questions, and Alex, Paul and I usually answer. Are other schools' G Nav instructors not answering their students' questions?
Send Clowns
Gen Nav Instructor
BCFT
