Q1. An aircraft passes position A (60°00’N 120°00’W) on route to position B (60°00’N140°30’W).What is the great circle track on departure from A?
279° – NH Westbound along 60N. RL=270. CA = 0.5 * 20.5 * sin(60) = 8.9°. Initial GC = 270 + 8.9 = 278.9°.
Q2. An aeroplane flies from A (59°S 142°W) to B (61°S 148°W) with a TAS of 480 kt.The autopilot is engaged and coupled with an Inertial Navigation System in which AB track is active.On route AB, the true track:
increases by 5° – SH Westbound -> Track Increases. Conv = dLong * sin(MeanLat) = 6 * sin(60) ≈ 5.2°.
Q3. The angle between the true great-circle track and the true rhumb-line track joining the following points: A (60°S 165°W) B (60°S 177°E) at the place of departure A, is:
7.8° – Conversion Angle = 0.5 * dLong * sin(Lat). dLong = 18°. CA = 0.5 * 18 * 0.866 = 7.8°.
Q4. 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 – SH Westbound across 180. Track Increases. Conv = 14 * sin(60) ≈ 12°.
Q5. In the Northern Hemisphere the rhumb line track from position A to B is 230°, the convergency is 6° and the difference in longitude is 10°.What is the initial rhumb line track from B to A?
050° – Rhumb lines have constant bearing (on sphere/mercator). Track B to A is reciprocal of A to B. 230 – 180 = 50.
Q6. Rhumb line track B (35N 150 W) to A (45N 160E) is 285 degrees.Find the Great-circle track at A on route A-B?
89 – RL A->B = 285-180=105. CA = 0.5 * 50 * sin(40) = 16. NH Eastbound Initial GC < RL. 105 – 16 = 89.
Q7. The Great Circle bearing of ‘B’ (70°S 060°E), from ‘A’ (70°S 030°W), is approximately:
135°(T) – RL=090. CA=42.3. SH Eastbound GC > RL. 132.3 is closest to 135.
Q8. 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. – NH Eastbound -> Track Increases. A < B. Difference = Conv = 13°.
Q9. Position A is (31°00’S, 176°17’W) 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:
(31°00’S, 168°58’E) – CA=3.8 -> dLong=14.75°. A is 3.7° from 180. Remaining 11.05° past 180E -> 168.97°E.
Q10. Given: Waypoint 1. 60°S 030°W Waypoint 2. 60°S 020°WWhat will be the approximate latitude shown on the display unit of an inertial navigation system at longitude 025°W?
060°06’S – Vertex Latitude calculation. Midpoint of Great Circle between two points at 60S. tan(Lv) = tan(60)/cos(5).
Q11. “A” latitude is 00°N/S “B” is located at 33°N 101°E. True track (great circle) from “A” to “B”, at “B”, is 090°.The initial true track of the great circle at “A” is:
057° – Using Napier’s/Vertex rules. sin(TrackA)*cos(LatA) = sin(TrackB)*cos(LatB). sin(TA) = cos(33) = 0.838. TA=57°.
Q12. An aircraft is in the position (86°N, 020°E). When following a rhumb line track of 085°(T) it will:
fly via a spiral to the North Pole – Constant track < 90 deg spirals into the pole.
Q13. A great circle track joins position A (59°S 141°W) and B (61°S 148°W).What is the difference between the great circle track at A and B?
It increases by 6° – SH Westbound -> Track Increases. Conv = 7 * sin(60) ≈ 6°.
Q14. Given : A is N55° 000° B is N54° E010° The average true course of the great circle is 100°.The true course of the rhumb line at point A is:
100° – The Rhumb Line track is approximately the average of the initial and final Great Circle tracks. Thus RL = 100°.
Q15. 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 – SH Eastbound -> Track Decreases. Conv = 7 * sin(65) ≈ 6.3°.