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DGCA Air Navigation · 286 questions

Wind Triangle & Drift: DGCA Air Navigation Questions and Answers

286 practice questions on Wind Triangle & Drift with answers and explanations, plus the key concepts and formulas, from the TrueHeading question bank.

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About this topic

Use this page to revise Wind Triangle & Drift for the DGCA Air Navigation paper. Below are the key ideas first, then 12 practice questions with answers and explanations, picked from the 286 questions in the TrueHeading bank for this topic.

In short: Heading and TAS, track and ground speed, wind direction and speed.

Open the full set in the student zone to answer every question, track your accuracy and take timed mock tests.

Key concepts: Wind Triangle & Drift

8 ideas to know cold.

What are the six quantities in the wind triangle?

Heading and TAS, track and ground speed, wind direction and speed.

  • The vectors close: heading + wind = track
  • Drift is the angle between heading and track
  • WCA (wind correction angle) is the heading change needed to hold track: WCA = −drift
  • Wind direction is always where it blows from
WCAHeading + TASWindTrack + Ground speedWind blows along the orange vectorPoint the nose upwind by the WCA to hold track

How do you calculate the wind correction angle?

Use the sine rule on the wind triangle.

sin⁡(WCA)=WTAS sin⁡(angle between track and wind)\sin(\text{WCA})=\frac{W}{TAS}\,\sin(\text{angle between track and wind})
  • Small angles: WCA ≈ 60 × crosswind ÷ TAS (degrees)
  • Wind from the right means crab right, so heading is to the right of track

How do you find headwind and crosswind components?

Resolve the wind against the runway or track.

Crosswind=Vsin⁡θ\text{Crosswind}=V\sin\theta
Headwind=Vcos⁡θ\text{Headwind}=V\cos\theta
  • θ is the angle between wind direction and runway heading (or track)
  • Example: 30 kt at 40° off: crosswind 19 kt, headwind 23 kt
θWindCrosswind = V sin θHeadwind= V cos θRWYheading

The clock method for crosswind in your head?

Use fractions of the wind speed by angle off the runway:

  • 15° → ¼ of the wind speed
  • 30° → ½
  • 45° → ¾
  • 60° or more → all of it

Tip: For headwind use the opposite fractions: 60°→ ½, 30°→ ⅞ roughly.

GS from TAS and wind: a quick estimate?

Ground speed ≈ TAS − headwind component (or + tailwind).

  • Direct headwind of 30 kt at TAS 150 kt gives GS 120 kt
  • A pure crosswind changes GS very little, mainly the drift

How do you find drift and groundspeed from a wind triangle?

Draw heading/TAS from the start, add the wind vector, and the closing line is the track/groundspeed.

sin⁡(WCA)=WVsin⁡(wind angle)\sin(\text{WCA})=\dfrac{W}{V}\sin(\text{wind angle})
  • Drift = angle between heading and track
  • Drift right: track is to the right of heading
  • Correct for wind: heading = track − drift (drift right → steer left of track)

Tip: Wind from the right blows you to the left of heading (drift left); the correction is to steer into the wind.

What is the maximum drift angle for a given wind and TAS?

Maximum drift occurs when the wind is perpendicular to the track (a pure crosswind).

Driftmax=sin⁡−1(WTAS)\text{Drift}_{max}=\sin^{-1}\left(\dfrac{W}{TAS}\right)
  • Maximum drift ≈ arcsin(W / TAS)
  • Example: W = 30 kt, TAS = 120 kt → about 14.5°

How do you use the 1-in-60 rule for a quick wind correction?

Drift (°) ≈ crosswind component × 60 / TAS.

Drift≈60×xwindTAS\text{Drift}\approx\dfrac{60\times\text{xwind}}{TAS}
  • Crosswind = wind speed × sin(wind angle)
  • Example: 40 kt wind at 30° to track, TAS 120 kt: xw = 20; drift = 20 × 60/120 = 10°

Practice questions: Wind Triangle & Drift

Tap “Show answer” after you have tried each one.

Q1Runway 16 (heading 160°). The wind is 200°/40 kt. The crosswind component is approximately:

  1. 26 kt
  2. 22 kt
  3. 36 kt
  4. 28 kt
Show answer

Answer: A. 26 kt

Crosswind = 40 × sin(40°) = 26 kt.

Q2Runway 07 (heading 070°). The wind is 350°/20 kt. The crosswind component is approximately:

  1. 12 kt
  2. 28 kt
  3. 20 kt
  4. 25 kt
Show answer

Answer: C. 20 kt

Crosswind = 20 × sin(80°) = 20 kt.

Q3Runway 33 (heading 330°). The wind is 010°/25 kt. The headwind component is approximately:

  1. 19 kt
  2. 31 kt
  3. 24 kt
  4. 14 kt
Show answer

Answer: A. 19 kt

Headwind = 25 × cos 40° = 19 kt.

Q4TAS 240 kt, required true track 320°, wind 200°(T)/50 kt. The true heading to steer is approximately:

  1. 304°
  2. 300°
  3. 310°
  4. 316°
Show answer

Answer: C. 310°

WCA = asin(50/240 × sin(200−320)) = -10.4°, so heading ≈ 310°.

Q5TAS 160 kt, required true track 130°, wind 330°(T)/40 kt. The groundspeed is approximately:

  1. 197 kt
  2. 217 kt
  3. 227 kt
  4. 118 kt
Show answer

Answer: A. 197 kt

GS = TAS × cos WCA − W × cos(wind angle) ≈ 197 kt.

Q6True heading 310°, drift 10° right. The true track made good is:

  1. 317°
  2. 330°
  3. 320°
  4. 324°
Show answer

Answer: C. 320°

Drift right: track = heading + 10° = 320°.

Q7An aircraft is to make good a true track of 060°. TAS is 320 kt and the wind is 160°/55 kt. What are the true heading and ground speed?

  1. 050° / 325 kt
  2. 050° / 315 kt
  3. 070° / 325 kt
  4. 070° / 310 kt
Show answer

Answer: C. 070° / 325 kt

WCA = asin(55/320 × sin(100°)) = +9.7° (wind from the right, so turn right into it). Heading = 060° + 9.7° ≈ 070°. GS = TAS × cos WCA − wind component along track ≈ 325 kt.

Q8An aircraft is to make good a true track of 315°. TAS is 250 kt and the wind is 240°/15 kt. What are the true heading and ground speed?

  1. 318° / 254 kt
  2. 312° / 246 kt
  3. 312° / 254 kt
  4. 318° / 246 kt
Show answer

Answer: B. 312° / 246 kt

WCA = asin(15/250 × sin(75°)) = −3.3° (wind from the left, so turn left into it). Heading = 315° − 3.3° ≈ 312°. GS = TAS × cos WCA − wind component along track ≈ 246 kt.

Q9TAS 430 kt, true track 270°, wind 160°/70 kt. The ground speed is closest to:

  1. 461 kt
  2. 360 kt
  3. 449 kt
  4. 500 kt
Show answer

Answer: C. 449 kt

Wind angle to track = 110°. WCA ≈ −8.8°. GS = 430·cos(8.8°) − 70·cos(110°) ≈ 449 kt.

Q10You wish to fly a true track of 275° at TAS 280 kt. The wind is 000°/25 kt. The true heading to steer is:

  1. 275°
  2. 280°
  3. 270°
  4. 290°
Show answer

Answer: B. 280°

The wind comes from the right of the track, so the heading is to the right of the track by the WCA of 5.1°. Heading ≈ 280°.

Q11TAS 360 kt, track 255°, wind 210°/55 kt. To maintain the track the aircraft heading must be about:

  1. 6° to the right of track
  2. 12° to the left of track
  3. 2° to the right of track
  4. 6° to the left of track
Show answer

Answer: D. 6° to the left of track

sin WCA = (wind speed / TAS) × sin(wind angle) → WCA ≈ 6°. Steer into the wind, i.e. to the left of the track.

Q12Runway 28 (magnetic 280°). Wind 320°/25 kt. The component along the runway on take-off is:

  1. 16 kt headwind
  2. 16 kt tailwind
  3. 19 kt tailwind
  4. 19 kt headwind
Show answer

Answer: D. 19 kt headwind

Angle between runway heading and wind = 40°. Along-runway component = 25 × cos 40° ≈ 19 kt, a headwind.

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