Discussion
An aeroplane has \( L_p = -2.5 \ \mathrm{s^{-1}} \) and an aileron input giving \( L_{\delta a}\delta_a = 0.60 \ \mathrm{s^{-2}} \). Its steady roll rate is
Answer: D
In the steady roll the aileron moment balances the damping moment:
\[ \begin{aligned} p_{ss} &= \frac{L_{\delta a}\delta_a}{-L_p} \\ &= \frac{0.60}{2.5} = 0.24 \ \mathrm{rad/s} \end{aligned} \]
which is 0.24 rad/s, about 14 degrees per second. 1.50 rad/s multiplies where it should divide. 0.42 rad/s inverts the ratio of the two numbers. 0.06 rad/s divides by ten as well. The relation shows why roll performance and roll damping pull against each other: the damping that makes the response crisp also limits the rate it settles at.
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