a small blimp has an air speed of 24km/h on a heading of 42°. The wind's speed is 9 km/h and its direction is 312°. Find the ground speed and drift angle of the blimp.

Answer 1

If R is the resultant vector, then
R = (24cos42,24sin42) + (9cos312,9sin312)
= (17.8355, 16.059) + (6.02218, -6.6883)
= (23.85766, 9.3707)

magnitude = √(23.85766^2 + (9.3707)^2) = appr 25.63 km/h
direction: Ø = tan^-1(9.3707/23.85766) = appr 21.44°

or

make a sketch and see that the two vectors are at right angles
to each other, so
R^2 = 24^2 + 9^2
R = appr 25.63, same as before

larger angle of right-angled triangle:
tan A = 24/9
A = 69.44°
direction angle = 69.44 + 312 = 381.44° which is coterminal with 21.44°
the same as before.

So if the heading was 42° and the actual direction is 21.44, what is the drift ?

Answer 2

All angles are measured CW from +Y-axis.

Vb + 9[312o] = 24km[42o].
Vb + 9*sin312+i9*Cos312 = 24*sin42 + i24*Cos42,
Vb - 6.69 + 6.02i = 16.1 + 17.8i,
Vb = 22.79 + 11.78i = 25.7km[62.7o]. = Velocity and direction of the blimp(gnd. speed).

Tan A = X/Y = 22.79/11.78.

Drift = 62.7 - 42. = 20.7o.