PPRuNe Forums - View Single Post - india DGCA ATPL exam
View Single Post
Old 28th Apr 2012, 07:19
  #327 (permalink)  
3greens 1inthemirror
 
Join Date: Sep 2011
Location: ZZZZ
Age: 24
Posts: 69
Likes: 0
Received 0 Likes on 0 Posts
1) Hdg = 120 (M)
Island is 15 degrees to left.
Hdg to Island = 105 deg

T V M D C
88 17W 105

Thus, True brg to Island is 88 deg

2) conv = 12 deHdg tg (55-43)
conv = dlong x sin lat
since constant of cone is 0.75 , sin lat = 0.75

Thus, dlong = 12/0.75 = 16 deg
( since course is 043 which means the a/c is moving east)
50 W - 16 = 34 W

3) Scale formula - 6 cm /[ 1 (deg) x 60 (nm) x 185200 cm]
= 1/1 852 000 ~ 1/1 850 000

{1 deg = 60 nm; 1 nm = 1.852 km = 1852 m = 185200 cm}

4) Same as 1) Hdg to ground feature = 355 - 20 = 325 M
Apply CDMVT, True hdg to ground feature = 340 deg
Hdg OF a/c from ground feature = 340 - 180(reciprocal)
= 160 deg

5) contant of cone = sin (lat)
sin (lat) = 0.3955
Thus, lat = 23 deg 18 minutes
3greens 1inthemirror is offline