FAA Private Pilot Helicopter (PRH)Navigation and AirspaceHard
A helicopter pilot plans a flight on a true course of 180° with a true airspeed of 100 knots. The wind is reported from 270° at 20 knots, a direct crosswind. Using the wind triangle, approximately what wind correction angle and resulting groundspeed should the pilot expect?
- A12° correction to the left; groundspeed 120 knots
- B12° correction to the right; groundspeed 98 knots
- C20° correction to the right; groundspeed 80 knots
- D6° correction to the left; groundspeed 100 knots
Show answer & explanationAnswer & explanation
Correct answer: B. 12° correction to the right; groundspeed 98 knots
With a wind directly from the west (270°) blowing across a southbound course (180°), the wind pushes the aircraft east. The wind correction angle equals arcsin(20/100) ≈ 11.5°, rounded to about 12° to the right (west) to compensate. Groundspeed equals sqrt(TAS² − crosswind²) = sqrt(100² − 20²) = sqrt(9600) ≈ 98 knots.
Why the other options are wrong
- A. Groundspeed cannot exceed TAS with a pure crosswind, and the correction direction is also wrong.
- C. 20° correction overstates the required angle for these wind and airspeed values.
- D. 6° is too small a correction for a 20-knot direct crosswind against 100 knots TAS.
Wind Triangle - Direct Crosswind
When wind blows perpendicular to the intended course, it requires a wind correction angle calculated using the sine relationship between wind speed and true airspeed.
- WCA (degrees) ≈ arcsin(crosswind component / TAS)
- Groundspeed = sqrt(TAS² − crosswind component²) for a pure crosswind
- Correct into the wind (turn toward the direction the wind is coming from)
Memory trick: Turn your nose toward the wind's source, like leaning into a push.