An aircraft performs a steady, coordinated level turn at a true airspeed of $V = 150\text{ m/s}$ with a bank angle $\phi = 60^\circ$. Taking $g = 9.81\text{ m/s}^2$, the turn radius $R$ (in meters) and load factor $n$ are closest to:

(A) $R = 1{,}324\text{ m}$ and $n = 2.0$

(B) $R = 2{,}293\text{ m}$ and $n = 1.15$

(C) $R = 1{,}324\text{ m}$ and $n = 1.41$

(D) $R = 2{,}648\text{ m}$ and $n = 2.0$

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles In a coordinated, level turn with bank angle $\phi$: 1. The lift vector balances both weight vertically and supplies the horizontal centripetal force: $$L \cos \phi = W \implies n = \frac{L}{W} = \frac{1}{\cos \phi}$$ 2. The horizontal centripetal balance is: $$L \sin \phi = \frac{W}{g} \frac{V^2}{R} \implies \tan \phi = \frac{V^2}{g R} \implies R = \frac{V^2}{g \tan \phi}$$ ### 2. Step-by-Step Calculation Given $\phi = 60^\circ$, $V = 150\text{ m/s}$, $g = 9.81\text{ m/s}^2$: 1. Load factor: $$n = \frac{1}{\cos(60^\circ)} = \frac{1}{0.5} = 2.0$$ 2. Turn radius $R$: $$\tan(60^\circ) = \sqrt{3} \approx 1.73205$$ $$R = \frac{(150)^2}{9.81 \times \sqrt{3}} = \frac{22{,}500}{9.81 \times 1.73205} = \frac{22{,}500}{16.9914} \approx 1{,}324.2\text{ m} \approx 1{,}324\text{ m}$$ ### 3. Distractor Forensics - **Option A:** Correct radius and load factor. - **Option B:** Uses $\sin(60^\circ)$ or incorrect trigonometric identity. - **Option C:** Inverted trigonometric functions. - **Option D:** Forgot factor of $\tan(60^\circ)$ in radius. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Always remember that in a level turn, load factor depends *only* on bank angle: $n = 1 / \cos \phi$. - **Standard Reference:** Anderson, J.D., *Introduction to Flight*, Chapter 6 (Turning Flight).

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