Question

A pilot wishes to fly on a course of 290° with an air speed of 300 knots when the wind blows from 224° at 18 knots. Find the drift angle to the nearest hundredth of a degree.

Answer 1

Explanation:

The resulting course vector (magnitude 300, angle = 160° with the positive x axis (= 290° bearing)) with components (300cos160,300sin160) is the sum of the vector of the actual flight vector and the wind vector.

The wind vector has magnitude 18 and direction 46° with the positive x-axis ( it blows from - 134° with the positive x axis (= 224° bearing) into direction 46° with the positive x-axis). So its components are (18cos46,18sin46).

The course that the plane has to steer is :

(300cos160,300sin160)-(18cos46,18sin46)

=(300cos160 - 18cos46, 300sin160 - 18sin46)

=(-294.41, 89.6579)

The magnitude of this steering vector is = 307.749 knots and its direction will be:

`tan^(-1)=(89.6579)/(294.41)=16.94^{{\circ}}` with the negative x-axis (286.94° bearing).

Thus, the drift angle will be=290° - 286.94° = 3.06°.