K.Haroon Site Admin
Joined: 11 Dec 2005 Posts: 1268

Posted: Thu Jun 14, 2012 4:56 pm Post subject: Questions on Lat/Long Calculations 


Given:The coordinates of the heliport at Issy les Moulineaux are:N48o50 E002o16.5 The coordinates of the antipodes are:
a) S41o10 W177o43.5
b) S48o50 E177o43.5
c) S48o50 W177o43.5 < Correct
d) S41o10 E177o43.5
In geography, the antipodes of any place on Earth is the point on the Earth's surface which is diametrically opposite to it. Two points that are antipodal to one another are connected by a straight line running through the centre of the Earth.
Opposite of N48°50' E002°16.5' is:
Long E2.275°+180 = 182.275° showing that 2.275° has crossed the 180E/W meridian and has gone into the other hemisphere, so it will be:
1802.275 = 177.725°W or 177°43.5'W
N48°50' will simply be S48°50' so option (c) is correct. To visualize you can draw a diagram viewing the globe from above (for change of long).
An aircraft at latitude 02o20N tracks 180o(T) for 685 km. On completion of the flight the latitude will be:
a) 03°50S < Correct
b) 04°10S
c) 04°30S
d) 09°05S
685 km / 1.85 = 370NM
370NM / 60 = 6.17° or 6°10'
From a position of 2°20'N travelling towards south (180°T) by 6°10' will take the aircraft through the equator.
2°20' Out of 6°10' will be in northern hemisphere and the remaining will be in the southern hemisphere:
6°10'
2°20'

or
5°70
2°20'

3°50'S
An aircraft departing A(N40°00' E080°00') flies a constant true track of 270° at a ground speed of 120 kt. What are the coordinates of the position reached in 6 HR?
b) N40°00' E064°20' < Correct
Departure (NM) = Change of Longitude (in minutes) x Cosine of Latitude
Distance 120 x 6 = 720
Cos 40 = 0.766
Change of Long = 720/0.766 = 940' or
940/60 = 15.66° or 15°40'
Flying West (270°) from E080°00' means the new long will be:
80°00'
15°40'

or
79°60
15°40'

64°20'E
An aircraft flies the following rhumb line tracks and distances from position 04o00N 030o00W: 600 NM South, then 600 NM East, then 600 NM North, then 600 NM West. The final position of the aircraft is:
a) 04°00N 029°58W < Correct
b) 04°00N 030°02W
c) 04°00N 030°00W
d) 03°58N 030°02W
From 4N 30W going south by 600nm (10 deg) will end up at 6S 30W
From 6S 30W going east by 600nm (long change = 600/cos6 = 10.06 deg) will end up at 6S 19.94W
From 6S 19.94W going north by 600nm (10 deg) will end up at 4N 19.94W
From 4N 19.94W going west by 600nm (long change = 600/cos4 = 10.03 deg) will end up at 6S 29.97W or 6°S 29°58'W
An aircraft starts at position 0411.0S 17812.2W and heads True North for 2950nm, then turns 90o left maintaining a rhumb line track for 314 km. The aircraft's final position is:
a) 5500.0N 17412.2W
b) 4500.0N 17412.2W
c) 5500.0N 17713.8E
d) 4500.0N 17713.8E < Correct
Starting Position = 04°11.0'S 178°12.2'W
2950nm North = 2950/60 = 49.16° (takes the aircraft in the northern hemisphere)
49.16  4.18 = 44.98 = 177°48'
From 45N a 90deg left turn for 314km makes the aircraft cross 180EW anti meridian.
314km/1.85 = 169.7nm
Change of long = 169.7/cos45 = 240' or 240/60 = 4°
178.20W + 4° = 182.2
360182.2 = 177.8 or 177°48'
Final Position = 45N 177°48'E
5 hours 20 minutes and 20 seconds hours time difference is equivalent to which change of longitude:
a) 81o 30
b) 78o 15
c) 79o 10
d) 80o 05 < Correct
5 hours 20 minutes and 20 seconds = 5.338 Hours
1 hour = 15 deg
5.338 hrs = 15 x 5.338 = 80.08 deg or 80°05'
Given:
Position A is N00o E100o
Position B is 240o(T), 200 NM from A
What is the position of B?
a) S01o40 E101o40
b) N01o40 E097o07
c) S01o40 E097o07 < Correct
d) N01o40 E101o40
Angles:
240180 = 60
270240 = 30
New Position:
Sin 30 x 200 = 100/60 = 1.66 deg South
Cos 30 x 200 = 173.2/60 = 2.88 deg West
An aircraft departs a point 0400N 17000W and flies 600 nm South, followed by 600 nm East, then 600 nm North, then 600 nm West. What is its final position?
a) 0400N 17000W
b) 0600S 17000W
c) 0400N 16958.1W < Correct
d) 0400N 17001.8W
Starting at 0400N 17000W
Flying 600nm South (600/60 = 10 deg) the position becomes 0600S 17000W
Flying 600nm East (600/cos6 = 10.055 deg. 17010.055 = 159.945) the position becomes 0600S 159.945W
Flying 600nm North (600/60 = 10 deg) the position becomes 0400N 159.945W
Flying 600nm West (600/cos4 = 10.024 deg. 159.945+10.024 = 169.969) the position becomes 04°00'N 169°58.1'W
An aircraft at position 2700N 17000W travels 3000 km on a track of 180T, then 3000 km on a track of 090T, then 3000 km on a track of 000T, then 3000 km on a track of 270T. What is its final position?
a) 2700N 17000W
b) 0000N 17000W
c) 2700N 17318W < Correct
d) 2700N 14300W
Same as the previous question but this time without calculations.
3000 km South means 27 deg (after doing the required conversions). Now the aircraft flies east on the equator. Meridians dont converge on the equator. Then it flies north for 3000 km i.e. back to Lat 27N. Then it flies west for 3000 km. Meridians will converge here. For the same distance on equator there will be more change of longitudes here. So position will be to the west of the original starting position. There is only one option (c) suggesting that.
As the INS position of the departure aerodrome, coordinates 35o32.7N 139o46.3W are input instead of 35o32.7N 139o46.3E. When the aircraft subsequently passes point 52o N 180oW, the longitude value show on the INS will be:
a) 080o27.4W
b) 099o32.6W < Correct
c) 099o32.6 E
d) 080o27.4 E
Change of longitude when aircraft reaches 180W = 180139.77 = 40.23
Longitude shown on INS = 139.7740.23 = 99.54 or 99°32.4'W 
