GEN 13A : Triangle of velocity, HDG, Track Drift

📄 Detailed Notes⚡ Cheat Sheet
 

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📄 Notes ⚡ Cheat Sheet
📄 Notes ⚡ Cheat Sheet

Q1. The relative bearing to a beacon is 090°R. Three minutes later, at a ground speed of 180 knots, it has changed to 135°R. What was the distance of the closest point of approach of the aircraft to the beacon?

9 NM –

  • Distance flown in 3 mins at 180 kts is 9 NM. The bearing change from 090R to 135R creates a 45-degree right triangle where the distance flown equals the distance at the beam (CPA).

    • Q2. Given the following: True track: 192° Magnetic variation: 7°E Drift angle: 5° left What is the magnetic heading required to maintain the given track?

    190° –

  • True Heading = Track (192) – Drift (Left is minus? No, Drift Left means Heading is Right of Track). TH = 192 + 5 = 197. Magnetic Heading = True (197) – Variation (7E) = 190.

    • Q3. An aircraft leaves at 0900UTC on a 250 NM journey with a planned ground speed of 115 knots. After 74 NM the aircraft is 1.5 minutes behind the planned schedule. What is the revised ETA at the destination?

    1115 –

  • Planned time for 74 NM is 38.6 min. Actual time is 40.1 min. Actual GS is 110.7 kts. Remaining 176 NM takes 95.4 min. Total time 2h 15m. ETA 1115.

    • Q4. You are flying at a True Mach No. of .82 in a SAT of -45°C. At 1000 hours you are 100 NM from the POL DME and your ETA at POL is 1012. ATC ask you to slow down to be at POL at 1016. What should your new TMN be if you reduce speed at 100 NM distance to go?

    M.61 –

  • Required time is 16 mins for 100 NM (GS 375). Old GS was 500 kts. Old TAS (M.82 @ -45C) was 483 kts. Wind is +17 kts. New TAS = 375 – 17 = 358 kts. M = 358/589 = 0.61.

    • Q5. Half way between two reporting points the navigation log gives the following information: TAS 360 kt, W/V 330°/80 kt, Compass heading 237°, Deviation on this heading -5°, Variation 19°W. What is the average ground speed for this leg?

    403 kt –

  • True Heading is 213°. Wind 330/80 gives a strong tailwind component. GS will exceed TAS significantly.

    • Q6. The distance between two waypoints is 200 NM. To calculate compass heading the pilot used 2°E magnetic variation instead of 2°W. Assuming that the forecast W/V applied, what will the off track distance be at the second waypoint?

    14 NM –

  • Total error is 4°. Using 1-in-60 rule: (4 * 200) / 60 = 13.3 NM.

    • Q7. Given: Runway direction 083°(M), Surface W/V 035/35 kt. Calculate the effective crosswind component.

    26 kt –

  • Angular difference is 48°. Crosswind = 35 * sin(48) ≈ 26 kts.

    • Q8. Given: True track 180° Drift 8°R Compass Heading 195° Deviation -2° Calculate the variation.

    21°W –

  • True Heading = 172°. Magnetic Heading = 193°. Variation = 172 – 193 = -21 (West).

    • Q9. During a low level flight 2 parallel roads are crossed at right angles by an aircraft. The time between these roads can be used to check the aircraft:

    ground speed –

  • Crossing fixed ground features at a known distance allows calculation of speed over the ground.

    • Q10. Given: TAS=375, Trk=335°(T), W/V=340°(T)/50. What is heading and Ground speed?

    336°(T) 326 –

  • Wind is 5° from the right. Headwind approx 50 kts. GS = 375 – 50 = 325. Drift is small to the left, so Heading > Track.

    • Q11. Heading is 156°(T), TAS is 320 knots, W/V is 130/45 and the Variation is 10°W. What is your magnetic track?

    170 –

  • Wind from left causes drift right. Track ≈ 160°T. Mag Track = 160 – (-10) = 170°M.

    • Q12. Given: True Track = 352, Variation = 11W, Deviation = -5, Drift = 8°R. What is Heading (C)?

    000°(C) –

  • TH = 344°. MH = 355°. Compass = 360° (000°).

    • Q13. Required course 045°(T), W/V = 190 /30, FL = 055 @ ISA, Variation = 15°W. CAS = 120 knots. What is mag heading and G/S?

    067°(M) 154 –

  • TAS ≈ 133 kts. Tailwind component adds to speed. Drift Correction is right. Mag Heading = TH + 15W.

    • Q14. Given: W/V 262/90 kt, Track 234° and TAS 305 kt, what are the Heading and Groundspeed?

    242° and 224 kt –

  • Wind from right requires heading right (242). Strong headwind reduces GS significantly (to 224).

    • Q15. Given: Magnetic heading 311° Drift is 10° left Relative bearing of NDB 270. What is the magnetic bearing of the NDB measured from the aircraft?

    221° –

  • QDM = MH + RB. 311 + 270 = 581. 581 – 360 = 221°.

    • Q16. Given: For take-off an aircraft requires a headwind component of at least 10 kt and has a cross-wind limit of 35 kt. The angle between the wind direction and the runway is 45°. Calculate the maximum and minimum allowable wind speeds.

    15 kt and 50 kt –

  • Min Wind = 10 / cos(45) = 15 kt. Max Wind = 35 / sin(45) = 50 kt.

    • Q17. Given the following: Magnetic heading: 060° Magnetic variation: 8°W Drift angle: 4° right What is the true track?

    056° –

  • TH = 052°. Track = TH + Drift (4R) = 056°.

    • Q18. Given Magnetic Heading: 015M, Variation: -5, Deviation +4. True Heading is:

    10 –

  • True = Magnetic + Variation (-5). 15 – 5 = 10.

    • Q19. X Y Z 30 NM 20 NM. ATA X is 1420. ETA Y is 1447. ATA Y is 1450. What is new ETA Z?

    1510 –

  • Speed is 1 NM/min (30 NM in 30 mins). Next 20 NM takes 20 mins. 1450 + 20 = 1510.

    • Q20. An aircraft is at FL140 with an IAS of 210 and a true OAT of -5°C. The wind component is -35 knots. When the aircraft is at 150 NM from a reporting point, ATC request the crew to lose 5 minutes by the time they get to the beacon. How much do they need to reduce IAS?

    20 knots –

  • The calculation requires reducing ground speed to extend flight time, which corresponds to an IAS reduction of approximately 20 knots.

    • Q21. On a particular take-off, you can accept up to 10 knots tailwind. The runway QDM is 047, the variation is 17°E and the ATIS gives the wind direction as 210. What is the maximum wind strength you can accept?

    11 knots –

  • The tailwind component is W * cos(angle). Resolving for W with a 10 kt limit gives 11 kts.

    • Q22. Pressure Altitude is 28 000 feet, OAT=-45°C Mach No=0.46, W/V=270/85, Track=200°T. What is the drift and groundspeed?

    17L/228 knots –

  • Wind from the right (270 vs 200) causes Left drift. High crosswind creates significant drift (17°). Headwind reduces TAS to GS (228).

    • Q23. Track = 090°(T), TAS=460 knots, W/V = 360°(T) / 100, Variation = 12°E, Deviation = -2. What is compass heading and ground speed?

    067° 450 knots –

  • Wind from Left causes Right Drift (Heading Left). TH ≈ 077°. Compass = TH – Var – Dev ≈ 067°.

    • Q24. Given, Compass Heading: 25, Deviation: +8, Variation: -9. True Heading is…….

    24 –

  • MH = 33. TH = 33 – 9 = 24.

    • Q25. Given: Heading 165(M), Variation 25W, Drift 10°R G/S 360 knots. At ‘A’ your relative bearing to an NDB is 325R. Five minutes later, at ‘B’, the relative bearing is 280(R). What is the true bearing and distance from ‘B’ to the NDB?

    060°(T) 30 NM –

  • Double angle (325R to 280R is 45° change). Distance to station equals distance flown (5 mins @ 360kts = 30 NM).

    • Q26. [A]——-30 NM——-[B]——–20 NM——–[C] ATA A is 1010. ETA B is 1030. ETA C is 1043. ATA B is 1027. What is revised ETA C?

    1038 –

  • The aircraft is gaining time (17 mins actual vs 20 planned). Applying this ratio to the next leg yields ETA 1038.

    • Q27. The wind velocity is 359/25. An aircraft is heading 180 at a TAS of 198 knots. (All directions are True). What is its track and ground speed?

    180, 223 –

  • Direct tailwind maintains track 180 and increases speed by wind speed (198 + 25 = 223).

    • Q28. An island is observed to be 15°(T) to the left. The aircraft heading is 120° (M), variation 17°(W). The bearing (°T) from the aircraft to the island is:

    088 –

  • TH = 103°. Bearing is 15° Left = 088°.

    • Q29. G/S = 240 knots, Distance go = 530 NM. What is time to go?

    2 h 12 m –

  • Time = 530 / 240 = 2.2 hours (2h 12m).

    • Q30. Given: Track 198°, Heading 184°, TAS 427 kt and GS 453kt, what are the W/V and Drift Angle?

    087°/109kt and 14°S –

  • Heading 184 to Track 198 is Drift Right (Starboard). GS > TAS implies tailwind. Wind from 087 fits these vectors.
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