Question

For a conventional fixed-wing aircraft in a 360\(^\circ\) inverted vertical loop maneuver, what is the load factor ($n$) at the topmost point of the loop? Assume the flight to be steady at the topmost point.

• A. $n = 1$
• B. $n < 1$
• C. $n = -1$
• D. $n > -1$

Hint:

At the top of an inverted loop, lift acts downward and must help provide centripetal force; hence $n$ is negative but greater than $-1$.

Answer 1

The Correct Option is D

At the top of an inverted loop, the aircraft is upside down, and the pilot experiences a downward (negative) load factor. The forces acting are:
- Lift ($L$) acting downward (towards the center of the loop) because the aircraft is inverted.
- Weight ($W$) acting downward.
For steady flight at the top of the loop, the centripetal force requirement is: \[ L + W = \frac{m V^2}{R} \] The load factor is: \[ n = \frac{L}{W} \] Since $L$ is downward, $n$ is negative. However, the aircraft still requires centripetal force to stay in the loop. Therefore: \[ L = \frac{m V^2}{R} - W \] Thus: \[ n = \frac{L}{W} = \frac{mV^2/R}{W} - 1 = \frac{V^2}{gR} - 1 \] Because the aircraft must still generate sufficient downward force to stay in the loop: \[ \frac{V^2}{gR} > 0 \] So: \[ n > -1 \] Why not the other options?
- (A) $n=1$: Wrong, because at the top the lift is downward and cannot equal positive weight.
- (B) $n<1$: Too vague; needs sign.
- (C) $n=-1$: Only true if $V^2/R = 0$ (impossible).
Thus, the only correct option is (D).