Aerodynamic Centre and Pitching Moment

Correct answer: B (1 and 2 only). Statement 1 is the definition of aerodynamic center ($dC_M/d\alpha = 0$). Statement 2 is correct because positive camber generates an intrinsic nose-down couple at zero lift. Statement 3 is false: for subsonic flow, the aerodynamic centre is located at approximately the quarter-chord ($0.25c$), not the half-chord point.

Correct answer: C. Thin airfoil theory demonstrates that integrating the circulation across the chord yields a constant pitching moment about the quarter-chord point ($x_{\text{ac}} = c/4 = 0.25c$).

Correct answer: A. A symmetrical airfoil has identical upper and lower contours with zero camber. At zero lift ($\alpha = 0^\circ$), the pressure distribution is symmetrical, producing zero pitching couple ($C_{M0} = 0$). Because the aerodynamic centre has invariant moment, $C_{M,\text{ac}} = C_{M0} = 0$ for all linear angles of attack.

Correct answer: B. Positive camber creates a downward curvature of the mean camber line, producing accelerated flow and lower surface pressures over the forward upper contour and an aft downward force, resulting in a persistent nose-down (negative) pitching couple about the aerodynamic center ($C_{M,\text{ac}} < 0$).

Correct answer: C. Reflexing the trailing edge upward creates reverse local camber near the trailing edge, generating a downward aerodynamic force at the rear that produces an intrinsic nose-up pitching moment ($C_{M,\text{ac}} > 0$). This provides self-trimming longitudinal stability for flying wings without horizontal tails.

Correct answer: D. In supersonic linearized flow (Ackeret theory), the pressure distribution shifts from a forward-loaded parabolic distribution to an evenly distributed flat chordwise loading, moving the aerodynamic centre from the subsonic quarter-chord ($0.25c$) to the supersonic half-chord ($0.50c$).

Correct answer: A. As shock waves form and move aft across the upper wing surface in transonic flight, the aerodynamic centre moves rearward from $0.25c$ toward $0.50c$. Because the lift force now acts substantially further behind the CG, a severe uncommanded nose-down pitching moment is generated, known as Mach Tuck.

Correct answer: B. By definition, the Aerodynamic Centre is the point about which the moment coefficient does not vary with angle of attack or lift coefficient: $\frac{\partial C_{M,\text{ac}}}{\partial C_L} = 0$.

Correct answer: C. The Centre of Pressure (CP) is the point of application of the resultant aerodynamic force vector. Because the line of action passes directly through the CP, the moment arm is zero, meaning $M_{\text{cp}} = 0$ and $C_{M,\text{cp}} = 0$.

Correct answer: A. From $x_{\text{cp}} = 0.25c - c(C_{M,\text{ac}}/C_L)$, since $C_{M,\text{ac}}$ is negative, the second term is positive: $x_{\text{cp}} = 0.25c + c(|C_{M,\text{ac}}|/C_L)$. As $\alpha$ increases, $C_L$ increases, which decreases the second term, causing $x_{\text{cp}}$ to move forward towards $0.25c$.

Correct answer: D. Because $C_{M,\text{ac}} = 0$ for a symmetric airfoil, $x_{\text{cp}} = x_{\text{ac}} - c(0/C_L) = 0.25c$. The Centre of Pressure coincides exactly with the Aerodynamic Centre and does NOT move with angle of attack!

Correct answer: B. In the equation $x_{\text{cp}} = x_{\text{ac}} - c(C_{M,\text{ac}}/C_L)$, since $C_{M,\text{ac}} \ne 0$ for a cambered airfoil, dividing by $C_L \to 0$ causes $x_{\text{cp}} \to \infty$. This proves that a pure moment couple cannot be represented by a finite force at a finite point of application.

Correct answer: C. If a disturbance increases angle of attack ($\Delta \alpha > 0$), the CP moves forward. This forward movement increases the moment arm of the upward lift force ahead of the CG, amplifying the disturbance and pitching the aircraft further up. Hence, CP travel is inherently destabilizing.

Correct answer: A. $x_{\text{cp}} = x_{\text{ac}} - c(C_{M,\text{ac}}/C_L) = (0.25 \times 2.0) - 2.0 \times (-0.06 / 0.50) = 0.50 - 2.0 \times (-0.12) = 0.50 + 0.24 = 0.74\text{ m}$.

Correct answer: B. From the leading edge ($x_{\text{ref}} = 0$): $x_{\text{cp}}/c = 0 - (C_{M,\text{LE}}/C_L) = -(-0.15 / 0.60) = +0.25$. Thus the Centre of Pressure is located at $25\%$ of chord ($0.25c$).

Correct answer: D (0.0065). Formula: $C_{M,\text{cg}} = C_{M,\text{ac}} + C_L((x_{\text{cg}} - x_{\text{ac}})/c)$. Given $C_{M,\text{ac}} = -0.016$, $C_L = 0.45$, and $(x_{\text{cg}} - x_{\text{ac}})/c = +0.05$. Therefore: $C_{M,\text{cg}} = -0.016 + (0.45 \times 0.05) = -0.016 + 0.0225 = +0.0065$.

Correct answer: B (0.50). Setting $C_{M,\text{cg}} = 0$: $0 = C_{M,\text{ac}} + C_L((x_{\text{cg}} - x_{\text{ac}})/c) \implies 0 = -0.05 + C_L(0.35 - 0.25) \implies 0.10 C_L = 0.05 \implies C_L = 0.05 / 0.10 = 0.50$.

Correct answer: C. When $x_{\text{cg}} = x_{\text{ac}}$, the moment arm $(x_{\text{cg}} - x_{\text{ac}}) = 0$. The lift force produces zero moment about the CG, meaning $C_{M,\text{cg}} = C_{M,\text{ac}} = \text{constant}$ regardless of $C_L$ or $\alpha$.

Correct answer: A. Differentiating with respect to $C_L$: $\frac{dC_{M,\text{cg}}}{dC_L} = \frac{x_{\text{cg}} - x_{\text{ac}}}{c}$. When $x_{\text{cg}} < x_{\text{ac}}$, this slope is negative ($\frac{dC_{M,\text{cg}}}{dC_L} < 0$). An increase in lift produces a stabilizing nose-down restoring moment.

Correct answer: B. For stability, CG must be ahead of AC ($x_{\text{cg}} < x_{\text{ac}}$), which means $\frac{x_{\text{cg}} - x_{\text{ac}}}{c} < 0$. To trim at $C_{M,\text{cg}} = 0$: $0 = C_{M,\text{ac}} + C_L \frac{x_{\text{cg}} - x_{\text{ac}}}{c} \implies C_L \left|\frac{x_{\text{cg}} - x_{\text{ac}}}{c}\right| = C_{M,\text{ac}}$. For positive $C_L$, $C_{M,\text{ac}}$ MUST be positive! This is why tailless aircraft use reflexed airfoils or trailing-edge elevons deflected upward.

Correct answer: C. $C_{M,\text{cg}} = C_{M,\text{ac}} + C_L((x_{\text{cg}} - x_{\text{ac}})/c) = -0.04 + 0.80 \times (+0.08) = -0.04 + 0.064 = +0.024$.

Correct answer: D. $C_{M,0.20c} = C_{M,\text{ac}} + C_L((0.20c - 0.25c)/c) \implies -0.02 = C_{M,\text{ac}} + 0.40(-0.05) \implies -0.02 = C_{M,\text{ac}} - 0.02 \dots$ Wait: $-0.02 = C_{M,\text{ac}} - 0.02 \implies C_{M,\text{ac}} = 0$. Let's re-verify: $C_{M,\text{ref}} = C_{M,\text{ac}} + C_L((x_{\text{ref}} - x_{\text{ac}})/c) \implies C_{M,\text{ac}} = C_{M,\text{ref}} - C_L((x_{\text{ref}} - x_{\text{ac}})/c) = -0.02 - 0.40(0.20 - 0.25) = -0.02 - (-0.02) = 0$. For a cambered airfoil with $C_{M,\text{ac}} = -0.04$: $C_{M,0.20c} = -0.04 + 0.40(-0.05) = -0.06$. Here if $C_{M,0.20c} = -0.06$, $C_{M,\text{ac}} = -0.04$.

Correct answer: A. When angle of attack changes by $\Delta \alpha$, additional lift $\Delta L$ is generated across the chord. The centroid of this incremental lift distribution acts exactly at the aerodynamic centre ($0.25c$). Because $\Delta L$ acts with zero moment arm about $0.25c$, it produces zero additional pitching moment about the AC ($\Delta M_{\text{ac}} = 0$).

Correct answer: B. The horizontal tail volume ratio is $V_H = \frac{S_t l_t}{S \bar{c}}$, where $S_t$ is tail area, $l_t$ is tail moment arm, $S$ is wing area, and $\bar{c}$ is mean aerodynamic chord.

Correct answer: C (0.50). $V_H = \frac{S_t l_t}{S \bar{c}} = \frac{4.0 \times 5.0}{20.0 \times 2.0} = \frac{20}{40} = 0.50$.

Correct answer: D. An aircraft must satisfy two criteria: (1) $C_{M0} > 0$ (positive zero-lift pitching moment to enable trim at positive lift $C_L > 0$), and (2) $\frac{dC_M}{d\alpha} < 0$ (negative slope, so that an increase in angle of attack generates a restoring nose-down moment).

Correct answer: A. The neutral point is the aerodynamic centre of the complete aircraft (wing + body + tail). When the CG is placed at the neutral point, the restoring moment derivative is zero ($\frac{dC_M}{d\alpha} = 0$), giving neutral static stability.

Correct answer: B. Static margin is the nondimensional distance between the neutral point and the CG: $K_n = \frac{x_{\text{np}} - x_{\text{cg}}}{\bar{c}} = -\frac{dC_M}{dC_L}$. Positive static margin means the CG is ahead of the neutral point.

Correct answer: C (+10%). $K_n = \frac{x_{\text{np}} - x_{\text{cg}}}{c} = \frac{0.42c - 0.32c}{c} = +0.10 = +10\%$. Because $K_n > 0$, the aircraft is statically stable.

Correct answer: A. Because the wing possesses a nose-down pitching couple ($C_{M,\text{ac}} < 0$) and the CG is typically located ahead of the wing CP/neutral point, a downward load on the tail is required to produce a balancing nose-up moment to achieve longitudinal trim ($C_{M,\text{cg}} = 0$).

Correct answer: D. Because the tail pushes downward, the main wing must generate lift equal to the weight PLUS the tail downforce ($L_{\text{wing}} = W + |L_t|$). This additional required lift increases induced drag, known as trim drag.

Correct answer: B. Loading the CG excessively forward creates a huge nose-down moment arm. At landing, dynamic pressure is low and flap deployment creates strong nose-down pitching. If the CG is beyond the forward limit, the elevator lacks sufficient aerodynamic power to hold the nose up for flare touchdown.

Correct answer: A. When CG moves aft near or beyond the neutral point, static margin approaches zero or becomes negative. Stick force per g drops dramatically, the aircraft becomes hyper-sensitive in pitch, and upon stalling, it diverges violently nose-up into an unrecoverable deep stall or spin.

Correct answer: C. Swept wings have outboard tanks positioned substantially aft of inboard/center tanks. Burning wing fuel causes significant chordwise CG movement, which modern fuel management systems automate (sometimes transferring fuel to a tail trim tank to reduce trim drag in cruise).

Correct answer: D. In a T-tail aircraft at high angles of attack, the turbulent separated wake of the stalled main wing flows directly over the high-mounted horizontal tail. The elevator loses dynamic pressure and pitching control, locking the aircraft into a continuous nose-up deep stall.

Correct answer: A. To balance a nose-down wing pitching moment, a canard mounted ahead of the CG produces an upward lift force ($L_c > 0$), so $L_{\text{total}} = L_{\text{wing}} + L_{\text{canard}}$. This avoids the downward trim load penalty of conventional aft tails.

Correct answer: B. Static stability requires $\frac{dC_M}{d\alpha} < 0$. A positive slope ($\frac{dC_M}{d\alpha} > 0$) means any upward gust triggers a further pitch-up moment, leading to rapid diverge in pitch. The CG is located aft of the neutral point.

Correct answer: C. The Mach Trim System is an automated flight control subsystem that senses increasing Mach number and progressively repositions the variable-incidence horizontal stabilizer nose-up to counteract Mach tuck and maintain positive control column force gradients.