Figure 30: Wind component chart
FAA-CT-8080-5H · Flight & Ground Instructor supplement · asked on the Flight Instructor test
2 FAA questions that refer to this figure
Question 1 · Flight Instructor
(Refer to FAA-CT-8080-5H, Figure 30.) Using a maximum demonstrated crosswind component equal to 0.2 VSO, what is a pilot able to determine? VSO: 60 kts Landing Rwy: 12 Wind: 150° at 20 kts
- A.Headwind component exceeds recommended limits.
- B.Crosswind component is within safe limits.
- C.Maximum demonstrated crosswind component is exceeded.
Show answer and explanation
Correct: B. Crosswind component is within safe limits.
Runway 12 is 120 degrees and the wind is from 150, a 30-degree angle, so the crosswind component is 20 x sin 30 = 10 knots (the chart gives the same). The maximum demonstrated crosswind is 0.2 x 60 = 12 knots, so the crosswind is within limits; the headwind component (about 17 knots) is not a limit.
Open this question’s pageQuestion 2 · Flight Instructor
(Refer to FAA-CT-8080-5H, Figure 30.) What is the crosswind component for a landing on Runway 18 if the tower reports the wind as 220° at 25 knots?
- A.16 knots.
- B.19 knots.
- C.22 knots.
Show answer and explanation
Correct: A. 16 knots.
The wind from 220 degrees is 40 degrees off runway 18 (180 degrees). The crosswind component is 25 x sin 40 = 25 x 0.64 = about 16 knots, which the crosswind component chart also gives for a 40-degree angle and 25 knots. 19 and 22 knots would be the headwind component (25 x cos 40 = 19) and a larger angle.
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