Question

An aircraft is flying in the north direction at a velocity of 60.5m/s under cross wind from the east to west of 5m/s. If the value of Cnβ=0.02/deg, where. Find sideslip angle β.

(a) -4.72°

(b) -5°

(c) 4.7°

(d) 3.18°

I had been asked this question by my school teacher while I was bunking the class.

My enquiry is from Lateral-Directional Static Stability and Control in portion Stability, Control, and Handling Qualities of Aircraft Design

Answer 1

The correct choice is (a) -4.72°

Easy explanation: Given, North velocity n = 60.5m/s, East to west velocity e = 5m/s.

For given cross wind condition sideslip angle β is given by,

β = -arctan (e/n) = -arctan (5/60.5) = -4.72°.