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► The rhumb-line distance between points A (60° 00 minutes N 002° 30 minutes E) and B (60° 00 minutes N 007° 30 minutes W) is:
Answer: 300 NM
Departure = Ch Long (10° x 60) x Cos(60°) = 300 NM
► The great circle distance between position A (59° 34,1 minutes N 008° 08.4 minutes E) and B (30° 25.9 minutes N 171° 51.6 minutes W) is:
Answer: 5400 NM
The longitudes are each others anti-meridian (both add up to 180°) so it will be a straight line over the nearer pole.
From position A to the pole (90°) = 90° - 59°34.1' = 30°25.9'
From the pole to position B = 90° - 30°25.9' = 59°34.1"
(30°25.9') + (59°34.1') = 90° latitude change
90° x 60 = 5400 NM
► An aircraft departing A (N40° 00 minutes E080° 00 minutes) 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?
Answer: N40° 00 minutes E064° 20 minutes
Departure = Ch long x Cos(lat)
Departure (distance) = 120kt (speed) x 6 (time) = 720 NM
Latitude = 40°
Ch Long = (Departure (720) / Cos(40) ) : 60
Ch Long = 939 : 60 = 15°40'
We started at 80°E and fly 270° (WEST direction) so 80°E - 15°40' = 65°20'E
► An aircraft flies the following rhumb line track and distances from position 04°00 minutes N 030° 00 minutes W; 600NM South, then 600NM East, then 600NM North, then 600NM West. The final position of the aircraft is:
Answer: 04° 00 minutes N 029° 58 minutes W
Start = 4N30W
We have a change in latitude towards the South of 600NM = (600NM : 60) 10°
Position 2 = 6S30W
We have a change in longitude towards the East of 600NM (= departure)
Ch Long = (600NM / Cos(6) ) : 60 = 10°3'
Position 3 = 6S19°57'W
We have a change in latitude towards the North of 600NM = (600NM : 60) 10°
Position 4 = 4N19°57'W
We have a change in longitude towards the West of 600NM (= departure)
Ch Long = (600NM / Cos(4) ) : 60 = 10°1'
Final position = 4°N 29°58'W
► A flight is to be made from A 49°S 180° E/W to B 58°S 180° E/W. The distance in kilometres from A to B is approximately:
Answer: 1000
Difference in latitude : 9°
9° x 60 = 540 NM
540NM x 1.852 = 1000 KM
► An aircraft at latitude 02° 20 minutes N tracks 180°(T) for 685 km. On completion of the flight the latitude will be:
Answer: 03° 50 minutes S
685km : 1.852 = 370NM
370NM : 60 = 6°10'
Start was 02°20N , we travel 6°10' South so new position will be 3°50'S
► Given: Position A N60 W020, position B N60 W021, position C N59 W020. What are, respectively the distances from A to B and from A to C?
Answer: 30 NM and 60 NM
1) From A to B
Departure = Ch long (1° x 60) x Cos(60)
Departure = 30 NM
2) From A to C
Ch in latitude = 1°
1° x 60 = 60 NM
► The sun moves from East to West at a speed of 15° longitude an hour. What ground speed will give you the opportunity to observe the sun due south at all times at 60°N?
Answer: 450kt
They want us to follow the sun from East to West at all times so Ch Long = 180°
Departure = Ch long (180 x 60) x Cos(60)
Departure = 5400 NM
15° in 1 hour means 180° in 12 hours
5400NM / 12H = 450kt
► An aircraft leaves 0°N/S 45°W and flies due South for 10 hours at a speed of 540kts. What is its position?
Answer: South Pole
10hours x 540kts = 5400NM (departure)
5400NM : 60 = 90° towards the South (end up at South Pole)
► An aircraft leaves point A (75°N 50°W) and flies due North. At the North Pole it flies due South along the meridian of 65°50minE until it reaches 75°N (point B). What is the total distance covered?
Answer: 1 800 NM
From A to North Pole = 90° - 75° = 15°
15° x 60 = 900 NM
From North Pole to B = 90° - 75° = 15°
15° x 60 = 900 NM
900 + 900 = 1800 NM
► Position A is (31°00minS 176°17minW)
Rhumb line track (T) from A to B is 270°
Initial great circle track (T) from A to B is 266.2°
The approximate position of B is:
Answer: 31°00minS 168°58minE
The difference between GCT (266.2) and RLT (270) = Convergence Angle (3.8°)
Convergence = 3.8° x 2 = 7.6
Convergence (7.6) = (Ch Long) x Sin(31°)
Change in Longitude = 14°45'
We started at A (176°17min WEST) and travel towards the WEST so position B will be (168°58'E)
176°17'W + 14°45' = 191°2'W (crossed the 180° Meridian)
192°2'W - 180° = 11° 2'
180° - 11.2' = 168° 58' E
Answer: 300 NM
Departure = Ch Long (10° x 60) x Cos(60°) = 300 NM
► The great circle distance between position A (59° 34,1 minutes N 008° 08.4 minutes E) and B (30° 25.9 minutes N 171° 51.6 minutes W) is:
Answer: 5400 NM
The longitudes are each others anti-meridian (both add up to 180°) so it will be a straight line over the nearer pole.
From position A to the pole (90°) = 90° - 59°34.1' = 30°25.9'
From the pole to position B = 90° - 30°25.9' = 59°34.1"
(30°25.9') + (59°34.1') = 90° latitude change
90° x 60 = 5400 NM
► An aircraft departing A (N40° 00 minutes E080° 00 minutes) 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?
Answer: N40° 00 minutes E064° 20 minutes
Departure = Ch long x Cos(lat)
Departure (distance) = 120kt (speed) x 6 (time) = 720 NM
Latitude = 40°
Ch Long = (Departure (720) / Cos(40) ) : 60
Ch Long = 939 : 60 = 15°40'
We started at 80°E and fly 270° (WEST direction) so 80°E - 15°40' = 65°20'E
► An aircraft flies the following rhumb line track and distances from position 04°00 minutes N 030° 00 minutes W; 600NM South, then 600NM East, then 600NM North, then 600NM West. The final position of the aircraft is:
Answer: 04° 00 minutes N 029° 58 minutes W
Start = 4N30W
We have a change in latitude towards the South of 600NM = (600NM : 60) 10°
Position 2 = 6S30W
We have a change in longitude towards the East of 600NM (= departure)
Ch Long = (600NM / Cos(6) ) : 60 = 10°3'
Position 3 = 6S19°57'W
We have a change in latitude towards the North of 600NM = (600NM : 60) 10°
Position 4 = 4N19°57'W
We have a change in longitude towards the West of 600NM (= departure)
Ch Long = (600NM / Cos(4) ) : 60 = 10°1'
Final position = 4°N 29°58'W
► A flight is to be made from A 49°S 180° E/W to B 58°S 180° E/W. The distance in kilometres from A to B is approximately:
Answer: 1000
Difference in latitude : 9°
9° x 60 = 540 NM
540NM x 1.852 = 1000 KM
► An aircraft at latitude 02° 20 minutes N tracks 180°(T) for 685 km. On completion of the flight the latitude will be:
Answer: 03° 50 minutes S
685km : 1.852 = 370NM
370NM : 60 = 6°10'
Start was 02°20N , we travel 6°10' South so new position will be 3°50'S
► Given: Position A N60 W020, position B N60 W021, position C N59 W020. What are, respectively the distances from A to B and from A to C?
Answer: 30 NM and 60 NM
1) From A to B
Departure = Ch long (1° x 60) x Cos(60)
Departure = 30 NM
2) From A to C
Ch in latitude = 1°
1° x 60 = 60 NM
► The sun moves from East to West at a speed of 15° longitude an hour. What ground speed will give you the opportunity to observe the sun due south at all times at 60°N?
Answer: 450kt
They want us to follow the sun from East to West at all times so Ch Long = 180°
Departure = Ch long (180 x 60) x Cos(60)
Departure = 5400 NM
15° in 1 hour means 180° in 12 hours
5400NM / 12H = 450kt
► An aircraft leaves 0°N/S 45°W and flies due South for 10 hours at a speed of 540kts. What is its position?
Answer: South Pole
10hours x 540kts = 5400NM (departure)
5400NM : 60 = 90° towards the South (end up at South Pole)
► An aircraft leaves point A (75°N 50°W) and flies due North. At the North Pole it flies due South along the meridian of 65°50minE until it reaches 75°N (point B). What is the total distance covered?
Answer: 1 800 NM
From A to North Pole = 90° - 75° = 15°
15° x 60 = 900 NM
From North Pole to B = 90° - 75° = 15°
15° x 60 = 900 NM
900 + 900 = 1800 NM
► Position A is (31°00minS 176°17minW)
Rhumb line track (T) from A to B is 270°
Initial great circle track (T) from A to B is 266.2°
The approximate position of B is:
Answer: 31°00minS 168°58minE
The difference between GCT (266.2) and RLT (270) = Convergence Angle (3.8°)
Convergence = 3.8° x 2 = 7.6
Convergence (7.6) = (Ch Long) x Sin(31°)
Change in Longitude = 14°45'
We started at A (176°17min WEST) and travel towards the WEST so position B will be (168°58'E)
176°17'W + 14°45' = 191°2'W (crossed the 180° Meridian)
192°2'W - 180° = 11° 2'
180° - 11.2' = 168° 58' E
