1.

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»

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°.



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