Q1. 1> An aircraft takes off from A (20N 30E) and follows a great circle track to B (30 N 60E). The great circle track at A is:
13 degree less than B. – Movement is Eastbound in Northern Hemisphere. Rule: NH Eastbound -> Track Increases. Therefore, Track A must be less than Track B. Difference = Convergency = $dLong times sin(text{Mean Lat}) = 30 times sin(25) approx 12.7^circ$ (round to 13°).
Q2. 2> An aircraft takes off from A (60N 175E) and follows a great circle track to B (40 N 173 W). The great circle track at A is
9 degree less than B. – Movement is Eastbound (crossing 180°). In NH, Eastbound tracks Increase. Convergency = $12 times sin(50) approx 9.2^circ$.
Q3. 3> An aircraft takes off from A (60N 30E) and follows a great circle track to B (40 N 60E). The great circle track at A as compared to the rhumb line track
11.5 degree less. – The difference between GC and RL tracks is the Conversion Angle (CA). CA = $0.5 times dLong times sin(text{Mean Lat}) = 0.5 times 30 times sin(50) approx 11.5^circ$. In NH Eastbound, Initial GC < RL.
Q4. 4> The autopilot is coupled to the Inertial Navigation System steering from A (20S 30E) to B (20S 20W). The great circle track at A is … than the rhumb line track
8.5 degree less. – Movement is Westbound (30E to 20W) in Southern Hemisphere. CA = $0.5 times 50 times sin(20) approx 8.55^circ$. In SH Westbound, Initial GC < RL.
Q5. 5> An aircraft takes off from A (42N 167W) and follows a great circle track to B (38 N 175 W). The great circle at B as compared to A……………
decreases by 5 degree – Movement is Westbound in NH. Rule: NH Westbound -> Track Decreases. Change = Convergency = $8 times sin(40) approx 5.1^circ$.
Q6. 6> An aircraft takes off from A (68 S 010E) and follows a great circle track to B (62 S 017E). On track A to B great circle track:
decreases by 6 degree – Movement is Eastbound in SH. Rule: SH Eastbound -> Track Decreases. Change = Convergency = $7 times sin(65) approx 6.3^circ$.
Q7. 7> An aircraft autopilot is engaged to route A (40 N 173 W) to B (40 N 157W) to C (40 N 140 W). As the aircraft crosses B the great circle track:
increases by 10 degree – Movement is Eastbound in NH (173W -> 157W). Track Increases. Convergency (A to B) = $16 times sin(40) approx 10.3^circ$.
Q8. 8> An aircraft autopilot is engaged to route A (40 S 163 E) to B (40 S 158 W) to C (40 S 140 W). As the aircraft crosses B the great circle track:
decreases by 25 degree – Movement is Eastbound in SH. Track Decreases. Total dLong (A to C) = 17+22=39. Conv = $39 times sin(40) approx 25^circ$. (Note: Question likely asks for total change or change at a specific point relative to start, context implies full route or significant leg).
Q9. 9> An aircraft autopilot is engaged to route A (20N 167 E) to B (20 S 157 W). On track A to B the great circle track:
first increases then decreases. – The route crosses the Equator (NH to SH). In NH Eastbound, track increases. In SH Eastbound, track decreases.
Q10. 10> A great circle track joins position A (59 S 178 W) and B (61 S 168 E). What is the difference between the great circle track at A and B?
increases by 12 degree – Movement is Westbound (178W to 168E is West across 180). In SH Westbound, track Increases. Conv = $14 times sin(60) approx 12.1^circ$.
Q11. 11> The difference between the initial and final great circle track on route A (30N 20 E) to B is 30 degrees. Given: Rhumb line track A to B is 090. find the location of B.
30 N 80 E – Convergency = 30°. Lat = 30°. $dLong = 30 / sin(30) = 60^circ$. A is 20E + 60 = 80E.
Q12. 12> The difference between the initial and final great circle track on route A (30 S 173 E) to B is 10 degrees. Given: Rhumb line track A to B is 270. find the location of B.
30 S 153 E – Convergency = 10°. Lat = 30°. $dLong = 10 / sin(30) = 20^circ$. A is 173E. RL is 270 (West). 173E – 20 = 153E.
Q13. 13> On route A to B great circle track increases by 18 degrees. given Rhumb line track A to B is 090. mean latitude is 30 degrees and the longitude of A is 20 E. find B.
30 N 56 E – Track Increases + Eastbound (090) = Northern Hemisphere. Conv = 18. $dLong = 18 / sin(30) = 36^circ$. A=20E + 36 = 56E.
Q14. 14> On route A to B great circle track decreases by 26 degrees. given Rhumb line track A to B is 270. mean latitude is 60 degrees and the longitude of A is 160 E. find B.
60N 130 E – Track Decreases + Westbound (270) = Northern Hemisphere. Conv = 26. $dLong = 26 / sin(60) approx 30^circ$. A=160E – 30 = 130E.
Q15. 15> On route A to B great circle track decreases by 19 degrees. Given: mean latitude is 60 N and longitude of A is 40 E. find the longitude of B?
18 E – Track Decreases + Northern Hemisphere (60N) = Westbound. Conv = 19. $dLong = 19 / sin(60) approx 22^circ$. A=40E – 22 = 18E.
Q16. 16> Given: The autopilot is coupled to the Inertial Navigation System steering from Waypoint 1 at 60 N 020W, to Waypoint 2 at 000W, to Waypoint 3 at 60 N 020 E. What is the approximate latitude at waypoint 2 is
61.5 N – Waypoint 2 is the Vertex (midpoint longitude). $tan(text{LatV}) = tan(60) / cos(20) approx 1.84$. $arctan(1.84) approx 61.5^circ$.
Q17. 17> Given: The autopilot is coupled to the Inertial Navigation System steering from Waypoint 1 at 30 N 020W, to Waypoint 2 at 040W, to Waypoint 3 at 30 N 060 W. What is the approximate latitude at waypoint 2 is:
31 31 00 N – Vertex Lat. dLong to vertex = 20°. $tan(text{LatV}) = tan(30) / cos(20) approx 0.614$. $arctan(0.614) approx 31.55^circ$ (31°33’N). Closest is 31 31 N.
Q18. 18> Given: The autopilot is coupled to the Inertial Navigation System steering from Waypoint 1 at 20 S 120W, to Waypoint 2 at 130W, to Waypoint 3 at 20 S 140 W. What is the approximate latitude at waypoint 2 is:
20 17 00 S – Vertex Lat. dLong = 10°. $tan(text{LatV}) = tan(20) / cos(10) approx 0.369$. $arctan(0.369) approx 20.29^circ$ (20°17’S).
Q19. 19> Given: The autopilot is coupled to the Inertial Navigation System steering from Waypoint 1 at 45 S 160W, to Waypoint 2 at 45S 170 E. What is the approximate latitude at when passing 175 W.
45 59 00 S – 175W is the midpoint (Vertex) between 160W and 170E. dLong = 15°. $tan(text{LatV}) = tan(45) / cos(15) approx 1.035$. $arctan(1.035) approx 45.99^circ$ (45°59’S).
Q20. 20> Given: The autopilot is coupled to the Inertial Navigation System steering from Waypoint 1 at 15 N 020W, to Waypoint 2 at 15S 020 E. What is the approximate latitude at when passing 000 E/W?
00 00 00 N – The Great Circle crosses the Equator at the midpoint of longitude difference (000 E/W) since start/end latitudes are symmetric (15N/15S).