Discussion
For an aeroplane with \( \omega_{n} = 4 \ \mathrm{rad/s} \) and a normal load factor sensitivity of \( n/\alpha = 20 \) per radian, the control anticipation parameter \( \mathrm{CAP} = \omega_{n}^{2}/(n/\alpha) \) is
Answer: C
Substituting, \( \mathrm{CAP} = 16/20 = 0.8 \), which sits inside the Level 1 band of roughly 0.28 to 3.6 for a manoeuvring aeroplane. The parameter exists because pitch frequency alone does not tell a pilot how the aeroplane will feel: what matters is how much initial pitch acceleration arrives for the load factor eventually delivered. 1.25 inverts the ratio. 3.2 divides the frequency rather than its square by \( n/\alpha \) and then scales wrongly. 0.2 divides by \( \omega_n^2 \) twice. A large CAP feels abrupt, a small one sluggish.
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