Advanced Aerospace interview questions

125 questions with answers (AER-301 to AER-425). Try each one aloud before reading the answer.

Incompressible Aerodynamics

AER-301 · An airliner weighing 600 kN with S = 120 m² cruises at 230 m/s where ρ = 0.38 kg/m³. What is its cruise C_L?

q = ½·0.38·230² ≈ 10,050 Pa. C_L = W/(q·S) = 600,000/(10,050 × 120) ≈ 0.50, a typical cruise value.

AER-302 · Find the induced drag coefficient for C_L = 0.5, AR = 9 and e = 0.85.

C_Di = C_L²/(π·e·AR) = 0.25/(π × 0.85 × 9) = 0.25/24.0 ≈ 0.0104, or about 104 drag counts.

AER-303 · What are the trade-offs of increasing wing aspect ratio?

Higher AR cuts induced drag and raises L/D, but increases root bending moment, wing weight, flexibility and flutter risk, reduces roll rate and internal fuel volume, and may exceed airport gate span limits (hence folding tips).

AER-304 · How do you choose wing taper ratio?

Taper moves spanwise loading toward elliptical and lowers root bending and weight. Too much taper raises tip loading and tip-stall risk, especially with sweep, so swept transport wings commonly use about 0.2–0.4 combined with washout.

AER-305 · Why is washout used, and what does it cost?

Washout (tip twisted nose-down) makes the root stall first, keeping aileron authority and giving benign stall. It moves loading inboard, so induced drag rises slightly off the design point.

AER-306 · Scenario: a light aircraft takes off with frost on the wing. What happens?

Surface roughness trips and thickens the boundary layer, cutting C_L,max and stall angle significantly and raising drag. The aircraft may stall below normal speed with stall warning not yet active. Wings must be clean before take-off.

AER-307 · Why is Reynolds number matching hard in wind tunnels, and how is it addressed?

Scale models at the same speed have far lower Re. Pressurised or cryogenic tunnels raise ρ/μ, and boundary-layer trips fix transition so scale effects can be corrected.

AER-308 · What is ground effect and how does it affect landing?

Within about one wingspan of the ground, the ground restricts downwash and the tip vortices, reducing induced drag and raising effective lift. Aircraft float in the flare if approach speed is too high.

AER-309 · How would you select a high-lift system for a new airliner?

Compare C_L,max gain against weight, complexity, noise and cost: slotted or Fowler flaps raise lift and area; slats add stall angle. Modern designs favour simpler single-slotted flaps with slats where field performance permits.

AER-310 · Find the sea-level stall speed for W = 50 kN, S = 20 m², C_L,max = 1.6.

V_s = √(2W/(ρ·S·C_L,max)) = √(100,000/(1.225 × 20 × 1.6)) = √2,551 ≈ 50.5 m/s (about 98 kt).

AER-311 · What are the challenges of natural laminar flow wings?

Laminar flow cuts skin friction, but it needs favourable pressure gradients, very smooth surfaces and low sweep to avoid crossflow instability. Insects, rain and steps trip it. Hybrid laminar flow control adds suction to overcome sweep limits.

Compressible & High-Speed Flow

AER-312 · Explain the transonic area rule.

Wave drag depends on how smoothly total cross-sectional area (fuselage plus wing, tail, nacelles) varies along the length. Fuselages are waisted where the wing joins ("coke-bottle") to keep the area distribution smooth and reduce transonic drag rise.

AER-313 · For a normal shock at M₁ = 2.0 (γ = 1.4), give the downstream values.

M₂ ≈ 0.577, p₂/p₁ = 4.5, T₂/T₁ ≈ 1.69, ρ₂/ρ₁ ≈ 2.67 and total pressure ratio p₀₂/p₀₁ ≈ 0.721, a 28% loss.

AER-314 · Mach 2 flow meets a 10° wedge. What are the shock angle and downstream Mach?

From the θ–β–M relation (weak solution), β ≈ 39.3° and M₂ ≈ 1.64, with pressure ratio about 1.7.

AER-315 · How do you design an inlet for Mach 2.2 flight?

A simple pitot inlet's single normal shock loses too much total pressure above about Mach 1.6. Use external or mixed compression with ramps or a cone to create several oblique shocks before a weak terminal normal shock, with variable geometry and boundary-layer bleed to stabilise the shock system.

AER-316 · A nozzle has A_e/A_t = 4. What is the supersonic exit Mach number?

Solving A/A* = (1/M)[(2/(γ+1))(1 + (γ−1)/2·M²)]^((γ+1)/(2(γ−1))) with γ = 1.4 gives M_e ≈ 2.94.

AER-317 · What makes hypersonic flow different from supersonic flow?

Thin shock layers close to the body, entropy layers, strong viscous interaction, real-gas effects (vibrational excitation, dissociation, ionisation), extreme heating, and Mach number independence of some coefficients.

AER-318 · Scenario: a transonic aircraft shows buffet near M 0.84 at high C_L. What is happening and what are the options?

Shock oscillation and shock-induced separation on the upper surface cause buffet. Options: reduce Mach or C_L (lower altitude), adjust airfoil (weaker shock), add vortex generators, or adjust twist. Certification requires a buffet margin, typically 1.3 g.

AER-319 · What is a waverider?

A hypersonic shape designed so its attached shock sits on the leading edges, trapping high-pressure flow under the body for a high L/D at a single design Mach number.

AER-320 · How can sonic boom be reduced?

Long, slender shaping and careful lift and volume distribution that keep shocks from coalescing into a strong N-wave, giving a weaker "thump". Lower weight and higher altitude also help.

AER-321 · Mach 2 flow turns away by 10° around a corner. What is the new Mach number?

ν(2.0) = 26.38°. New ν = 36.38°, which corresponds to M ≈ 2.38 from Prandtl–Meyer tables. Pressure falls isentropically.

AER-322 · Compute choked mass flow for A* = 0.01 m², p₀ = 500 kPa, T₀ = 300 K.

For air, ṁ = 0.0404·A*·p₀/√T₀ = 0.0404 × 0.01 × 500,000/17.32 ≈ 11.7 kg/s.

Gas Turbine Propulsion

AER-323 · What is ideal Brayton thermal efficiency for a pressure ratio of 30?

η = 1 − 30^(−0.286) = 1 − 1/2.64 ≈ 0.62, or 62%. Real engines are lower because of component losses.

AER-324 · Find propulsive efficiency for V₀ = 250 m/s and V_j = 350 m/s.

η_p = 2/(1 + V_j/V₀) = 2/(1 + 1.4) ≈ 0.83.

AER-325 · An engine swallows 100 kg/s at 250 m/s and exhausts at 600 m/s. Estimate net thrust.

Ignoring fuel flow and pressure thrust, F = ṁ(V_j − V₀) = 100 × 350 = 35 kN.

AER-326 · What are the trade-offs of very high bypass ratio?

Better propulsive efficiency and lower jet noise, but larger fan diameter means more nacelle weight and drag, ground-clearance and wing-integration problems, and lower specific thrust. Geared fans and short nacelles help.

AER-327 · How would you choose an engine for a supersonic cruise aircraft?

Use low bypass ratio or a turbojet for high specific thrust and small frontal area, with variable inlet and convergent-divergent nozzle. Supercruise designs avoid afterburning at cruise; variable-cycle engines adapt bypass to flight phase.

AER-328 · What does a compressor map show?

Pressure ratio versus corrected mass flow with corrected speed lines, efficiency contours, the surge line and the engine operating line. Surge margin is the distance between operating and surge lines.

AER-329 · Scenario: an engine's EGT margin shrinks over several thousand hours. What are the likely causes and actions?

Blade erosion, tip clearance growth, seal wear and fouling reduce component efficiencies, so the engine must run hotter for the same thrust. Trend monitoring, borescope inspection, on-wing washing and shop-visit restoration are used.

AER-330 · What does the Euler turbomachine equation state?

Specific work Δh₀ = U·ΔC_θ, blade speed times the change in tangential (whirl) velocity. Velocity triangles relate this to blade angles.

AER-331 · Why do axial compressor stages often use about 50% reaction?

Equal static pressure rise in rotor and stator gives symmetric velocity triangles, balancing diffusion loading and limiting separation in both rows.

AER-332 · What are the pros and cons of open-rotor engines?

Very high effective bypass ratio gives notable fuel savings over turbofans, but noise, blade-off containment, certification, installation and cruise Mach limits are significant challenges.

AER-333 · How is NOx controlled in modern combustors?

NOx rises strongly with flame temperature and residence time. Rich-quench-lean and lean-premixed staged combustors limit peak flame temperature, trading against stability and higher pressure ratios.

Rocket Propulsion

AER-334 · State the Tsiolkovsky rocket equation and its implication.

Δv = Isp·g₀·ln(m₀/m_f). Δv grows only logarithmically with mass ratio, so reaching orbit needs huge propellant fractions. This is why rockets use staging.

AER-335 · Find Δv for Isp = 300 s and a mass ratio of 5.

Δv = 300 × 9.81 × ln 5 = 2,943 × 1.609 ≈ 4.74 km/s.

AER-336 · A 1,000 kg (dry) spacecraft with Isp = 320 s needs Δv = 1.5 km/s. How much propellant?

m₀/m_f = e^(1500/(320×9.81)) = e^0.478 ≈ 1.613, so propellant ≈ 613 kg.

AER-337 · Why is single-stage-to-orbit so difficult?

Orbit needs roughly 9–10 km/s including losses. With Isp 350 s and dry mass 10% of lift-off mass, ideal Δv = 350 × 9.81 × ln 10 ≈ 7.9 km/s, before any payload. Staging discards dead mass and fixes this.

AER-338 · How can nozzle losses across altitude be reduced?

Bell nozzles are optimum at only one pressure. Options: lower-expansion first-stage nozzles, extendable nozzle skirts for upper stages, dual-bell nozzles, or aerospikes, which adapt to ambient pressure.

AER-339 · Engine: ṁ = 250 kg/s, v_e = 3,000 m/s, p_e = 60 kPa, A_e = 1 m². Find sea-level and vacuum thrust.

Sea level: 750 kN + (60 − 101.3) kPa × 1 m² ≈ 709 kN. Vacuum: 750 + 60 = 810 kN.

AER-340 · An ion thruster has P = 5 kW, efficiency 0.6 and Isp = 2,000 s. Estimate thrust.

F = 2ηP/v_e = 2 × 0.6 × 5,000/19,620 ≈ 0.31 N. Thrust is power-limited, so high Isp means low thrust.

AER-341 · Why is there an optimum Isp for electric propulsion missions?

Raising Isp cuts propellant mass but, at fixed thrust, needs more power and a heavier power system. Total mass is minimised at an Isp that depends on mission Δv, burn time and specific power.

AER-342 · How do you choose propellants for an upper stage?

LOX/LH₂ gives the highest chemical Isp but low density, large tanks and boil-off. Kerosene or methane give denser tanks. Storable hypergolics suit long missions and multiple restarts despite lower Isp.

AER-343 · What are gravity and drag losses in launch?

Gravity loss is ∫g·sin γ dt from thrusting against gravity; drag loss is ∫(D/m) dt. Together with steering losses they typically add roughly 1.5–2 km/s to orbital velocity for a LEO launch.

AER-344 · Scenario: X-ray inspection finds a crack in a solid motor grain. Why is this serious?

The crack adds burning surface, raising chamber pressure (via r = a·pⁿ) beyond case design limits, which can cause burst. The motor must be rejected or repaired and reinspected.

Flight Mechanics & Stability

AER-345 · Find the neutral point and static margin: h_ac,wb = 0.25, V_H = 0.6, η_t = 0.9, a_t/a = 0.7, dε/dα = 0.4, CG at 0.30 MAC.

h_n = 0.25 + 0.6 × 0.9 × 0.7 × (1 − 0.4) = 0.25 + 0.227 = 0.477. Static margin = 0.477 − 0.30 ≈ 0.18, or 18% MAC.

AER-346 · What is relaxed static stability and why use it?

Placing the CG near or behind the neutral point cuts trim drag and tail size and improves agility. The bare airframe may be unstable, so a fly-by-wire system must provide artificial stability.

AER-347 · How is the horizontal tail sized initially?

Using the tail volume coefficient V_H = S_t·l_t/(S·c̄) from similar aircraft, then checked against forward CG rotation and flare authority, aft CG stability and trim limits.

AER-348 · What determines rudder size on a twin-engine aircraft?

Minimum control speed with the critical engine inoperative (V_MC): the rudder must balance the asymmetric thrust yawing moment at low speed, plus crosswind landing and spin recovery requirements.

AER-349 · Scenario: cargo is loaded so CG is behind the aft limit. What can happen?

Reduced or negative static margin, light pitch forces and possible pitch-up and loss of control at rotation or in climb. Tail tipping on the ground is also possible.

AER-350 · What is deep stall?

On T-tail aircraft at very high angle of attack, the wing wake blankets the horizontal tail, removing nose-down authority and possibly locking the aircraft in a stable stalled state. Stick pushers prevent entry.

AER-351 · Why do swept wings tend to pitch up at the stall?

Spanwise flow thickens the outboard boundary layer, so tips stall first. Since swept tips are aft of the CG, loss of tip lift moves the centre of pressure forward, giving a nose-up moment. Fences, notches and washout reduce it.

AER-352 · Estimate phugoid period at 200 m/s.

Lanchester's approximation T ≈ π·√2·V/g = 4.443 × 200/9.81 ≈ 91 s. Its damping ratio is roughly 1/(√2·L/D), so it is very lightly damped.

AER-353 · Stability augmentation vs control augmentation?

A stability augmentation system (SAS) feeds back rates to add damping, like a yaw damper. A control augmentation system (CAS) also shapes the response to pilot commands, such as pitch-rate or g command.

AER-354 · What is inertia coupling?

In rapid rolls by slender, fuselage-heavy aircraft, gyroscopic and inertial terms couple roll into pitch and yaw, potentially causing divergence. It is managed with roll-rate limits and larger tails.

AER-355 · How is a linear aircraft model used for stability analysis?

The equations of motion are linearised about trim into state-space form ẋ = Ax + Bu. Eigenvalues of A give frequency and damping of each mode, which are compared with handling-quality criteria.

Aircraft Performance

AER-356 · Breguet: V = 240 m/s, L/D = 18, TSFC = 0.6 /h, W₀/W₁ = 1.3. Find range.

c = 0.6/3,600 = 1.667×10⁻⁴ s⁻¹. R = (V/c)(L/D)·ln(1.3) = 1.44×10⁶ × 18 × 0.262 ≈ 6,800 km.

AER-357 · How does a 10% weight increase affect take-off distance?

Ground roll scales roughly with W²/(ρ·S·C_L,max·T), so distance rises by about 1.1² ≈ 21%, more if thrust is reduced at altitude or temperature.

AER-358 · T = 120 kN, D = 60 kN, W = 600 kN, V = 150 m/s. Find rate of climb and gradient.

RoC = (T − D)·V/W = 60,000 × 150/600,000 = 15 m/s (about 2,950 ft/min). Climb gradient = (T − D)/W = 10%.

AER-359 · Why do airliners step climb?

Optimum altitude rises as weight falls. Since ATC assigns discrete levels, aircraft climb in steps to stay near the optimum altitude and best specific range.

AER-360 · What is the cost index?

The ratio of time-related costs to fuel cost, entered in the FMS. A low index gives slow, fuel-saving speeds; a high index gives faster speeds.

AER-361 · How far can an aircraft with L/D = 17 glide from 10 km altitude in still air?

Distance = height × L/D = 10 × 17 = 170 km, flown at best-glide speed.

AER-362 · How are landing distance requirements set for jet transports?

Under typical rules, the dry demonstrated landing distance must fit within 60% of the landing distance available (a factor of 1.67), with a further 15% added for wet runways.

AER-363 · Scenario: departing a hot-and-high airport. What limits performance and what can be done?

Low density cuts thrust and lift, raising true airspeed, take-off distance and reducing climb gradient. Reduce take-off weight, pick optimal flap, use full thrust, avoid the heat of the day, or use a longer runway.

AER-364 · Speed for maximum endurance in a propeller aircraft?

Minimum power occurs at maximum C_L^1.5/C_D, where induced drag is three times parasite drag, at about 0.76 times the minimum-drag speed.

AER-365 · Power required: W = 10 kN, L/D = 12, V = 60 m/s.

D = W/(L/D) = 833 N, so P = D·V = 833 × 60 = 50 kW. Shaft power must be greater after propeller efficiency.

Aircraft Structures

AER-366 · Find minimum fuselage skin thickness for Δp = 60 kPa, r = 2 m and allowable hoop stress 100 MPa.

t = Δp·r/σ = 60,000 × 2/100×10⁶ = 1.2 mm. Fatigue, damage tolerance and fittings usually raise the real thickness.

AER-367 · A semi-span of 15 m carries 300 kN of uniformly distributed lift. What is the root bending moment?

The resultant acts at mid semi-span, so M = 300 kN × 7.5 m = 2.25 MN·m. Elliptic loading would move the centroid inboard to about 0.42 of semi-span and lower it.

AER-368 · A pinned strut of length 1 m has E = 70 GPa and I = 2×10⁻⁷ m⁴. Find its Euler buckling load.

P_cr = π²EI/L² = 9.87 × 70×10⁹ × 2×10⁻⁷/1 ≈ 138 kN.

AER-369 · Scenario: an inspection finds cracks at several fasteners in a fuselage lap joint. What do you do?

Treat it as possible multi-site damage, which can link up and lower residual strength sharply. Assess residual strength, repair (e.g., doubler or skin replacement), check other aircraft of similar age and adjust the inspection programme.

AER-370 · What are the trade-offs of a composite versus aluminium wing?

Composites save weight, resist fatigue and corrosion, and allow aeroelastic tailoring. They cost more, need lightning protection, have impact damage and hot-wet knockdowns, need different repair and inspection methods, and are hard to change after design.

AER-371 · How are cutouts such as doors and windows designed?

With rounded corners, edge frames, doublers and surrounding structure to redistribute load and lower stress concentration. The de Havilland Comet accidents showed the cost of fatigue cracking at cutout corners.

AER-372 · Why are engine pylons designed with fuse pins?

So that under extreme overload, such as a seizure or ground impact, the engine separates cleanly rather than tearing the wing box or fuel tank.

AER-373 · How is structural strength demonstrated for certification?

Analysis supported by test: coupon, element and component testing (the building-block approach), a full-scale static test to ultimate load, and full-scale fatigue tests to support damage tolerance and inspection intervals.

AER-374 · K_IC = 30 MPa·√m, σ = 150 MPa, Y = 1. Find the critical crack length.

a_c = (1/π)(K_IC/(Y·σ))² = (1/π)(0.2)² ≈ 0.0127 m, or about 12.7 mm.

AER-375 · How do you minimise weight in a stiffened panel design?

Size skins and stringers to balance failure modes (buckling, crippling, material strength), respect minimum gauge and damage tolerance, use optimisation such as sizing or topology methods, and verify with FEA and test.

Aerospace Materials

AER-376 · Why use composites in aircraft, and what are the downsides?

They offer high strength-to-weight and stiffness-to-weight, fatigue and corrosion resistance and tailorable properties. They are harder to inspect and repair, sensitive to impact and hot-wet conditions, need lightning protection and cost more to certify.

AER-377 · How do you select a material for a high-pressure turbine blade?

Required: creep and rupture strength at high temperature, oxidation and hot-corrosion resistance, thermal fatigue resistance and castability. The usual answer is a single-crystal nickel superalloy with internal cooling, a bond coat and a ceramic TBC.

AER-378 · How do you select a thermal protection system for re-entry?

Match heat flux and total heat load. Reusable silica tiles and blankets suit moderate heating. Reinforced carbon-carbon suits hot leading edges. Ablators such as PICA suit very high flux from lunar or planetary return.

AER-379 · What are hot-wet knockdowns for composites?

Moisture absorption plus high temperature lower the matrix glass transition temperature and matrix-dominated properties, such as compression and shear. Design allowables must use the worst environmental condition, with Tg margin above the maximum operating temperature.

AER-380 · How are CFRP structures protected from lightning?

Conductive surface layers such as expanded copper or bronze mesh, bonded and sealed fasteners, bonding straps and fuel-tank spark prevention measures.

AER-381 · Which NDT method would you use for composite delaminations?

Ultrasonic inspection (pulse-echo or C-scan) is the primary method. Thermography is fast for large areas and tap testing gives a quick field check. Eddy current suits surface cracks in metals rather than composite delamination.

AER-382 · Estimate longitudinal modulus for V_f = 0.6, E_f = 230 GPa, E_m = 3.5 GPa.

Rule of mixtures: E₁ = V_f·E_f + V_m·E_m = 0.6 × 230 + 0.4 × 3.5 ≈ 139 GPa.

AER-383 · What material index is used for a light, stiff panel in bending?

For a plate of fixed stiffness with free thickness, minimise mass by maximising E^(1/3)/ρ. This favours low-density materials and explains the value of sandwich and composite panels.

AER-384 · Why is 7075-T6 replaced by T73 or T76 tempers in some parts?

T6 is susceptible to stress corrosion cracking and exfoliation. Overaged T73/T76 tempers greatly improve SCC resistance at some loss of strength.

AER-385 · What are the challenges of qualifying additively manufactured parts?

Porosity, anisotropy, residual stress, surface roughness and process variability. Qualification requires process control, powder management, HIP and machining, NDT such as CT, and statistically based allowables.

Orbital Mechanics

AER-386 · What is a Hohmann transfer? Compute Δv from a 300 km LEO to GEO (coplanar).

The most fuel-efficient two-impulse transfer between coplanar circular orbits. r₁ = 6,678 km, r₂ = 42,164 km: Δv₁ ≈ 10.15 − 7.73 = 2.43 km/s and Δv₂ ≈ 3.07 − 1.61 = 1.47 km/s, total ≈ 3.9 km/s, taking about 5.3 hours.

AER-387 · Why combine plane change with the GTO apogee burn from 28.5° inclination?

Velocity is lowest at apogee (≈1.61 km/s). Combining the 28.5° plane change with circularisation gives Δv = √(1.61² + 3.07² − 2·1.61·3.07·cos 28.5°) ≈ 1.83 km/s, far less than separate burns (≈3.0 km/s).

AER-388 · When is a bi-elliptic transfer better than a Hohmann transfer?

When the ratio of final to initial orbit radius exceeds about 11.94 (and always above about 15.58). It saves Δv but takes much longer.

AER-389 · How is a sun-synchronous orbit designed?

Choose altitude and inclination so J2 nodal regression equals +0.9856°/day, the Sun's apparent motion. For about 800 km this requires an inclination near 98.6°.

AER-390 · How does a chaser spacecraft catch up with a target ahead in the same orbit?

It burns retrograde to drop into a lower, faster phasing orbit, then burns prograde to rejoin. Relative motion is modelled with the Clohessy–Wiltshire (Hill) equations.

AER-391 · What are the trade-offs in LEO constellation design?

Lower altitude gives lower latency and launch cost but more satellites for coverage, more drag and shorter life. Walker patterns set planes, phasing and inclination. Debris and collision avoidance grow with number.

AER-392 · Describe a Hohmann transfer to Mars.

Earth departure into a heliocentric ellipse touching both orbits, taking about 259 days. Launch windows repeat every synodic period of about 26 months.

AER-393 · How do you design end-of-life disposal for a LEO satellite?

Plan a deorbit burn or drag device to meet the deorbit time rule (e.g., FCC's 5-year rule for US-licensed LEO satellites), passivate batteries and tanks, and assess casualty risk for uncontrolled re-entry.

AER-394 · What Δv is needed to escape from a 300 km circular orbit?

v_esc = √2 × 7.73 ≈ 10.93 km/s, so a single prograde burn needs about 3.2 km/s.

AER-395 · What is the station-keeping budget in GEO?

North–south control against luni-solar inclination drift dominates, at roughly 45–50 m/s per year. East–west control against Earth's triaxiality needs only a few m/s per year.

Spacecraft Systems

AER-396 · Size a LEO solar array: 1 kW load, 60 min sunlight, 35 min eclipse, path efficiencies 0.85 (daylight) and 0.65 (eclipse).

P_sa = (P·T_e/X_e + P·T_d/X_d)/T_d = (1,000×35/0.65 + 1,000×60/0.85)/60 ≈ 2.07 kW at end of life.

AER-397 · Size the battery for the same 1 kW, 35-minute eclipse at 30% DoD and 90% efficiency.

E = P·t_e/(DoD·η) = 1,000 × 0.583 h/(0.3 × 0.9) ≈ 2.16 kWh.

AER-398 · Size a radiator to reject 500 W at 290 K with ε = 0.85 (ignore environmental heating).

q = ε·σ·T⁴ = 0.85 × 5.67×10⁻⁸ × 290⁴ ≈ 341 W/m², so A ≈ 1.47 m². Real designs add margin for solar and Earth infrared input.

AER-399 · How do you size a reaction wheel's momentum capacity?

Integrate disturbance torques (gravity gradient, solar pressure, drag, magnetic) over the unloading interval. A secular 10⁻⁵ N·m over one day accumulates 0.86 N·m·s, plus slew and cyclic needs and margin.

AER-400 · Find the minimum time for a 90° slew with I = 500 kg·m² and maximum torque 0.2 N·m (bang-bang).

t = 2√(θ·I/τ) = 2√(1.571 × 500/0.2) ≈ 125 s.

AER-401 · Compute free-space path loss at 2 GHz over 1,000 km.

FSPL(dB) = 20log₁₀(d_km) + 20log₁₀(f_MHz) + 32.44 = 60 + 66.0 + 32.44 ≈ 158.5 dB.

AER-402 · How do you design spacecraft safe mode?

On a detected fault, shed non-essential loads, point arrays to the Sun with a simple, robust attitude law and independent sensors, keep thermal survival and a low-rate command link, and wait for ground action.

AER-403 · Hot vs cold redundancy in spacecraft?

Hot spares run continuously for instant switchover but use power and age. Cold spares stay off until needed, saving power and life but needing switch time. Cross-strapping lets units pair with either side.

AER-404 · Scenario: a payload runs hotter every year. Why, and what can be done?

UV, contamination and atomic oxygen raise radiator coating solar absorptivity, increasing absorbed heat. Size radiators for end-of-life properties, use stable coatings such as OSRs, and add heater or louver margin.

AER-405 · Why carry mass and power margins through spacecraft design?

Estimates grow as designs mature. Margins are set by maturity, larger early and reduced as hardware is measured, to avoid exceeding launch capability or power late in the programme.

Avionics, Navigation & Control

AER-406 · How is a fly-by-wire system made fault tolerant?

Triplex or quadruplex computers and sensors with voting and monitoring, dissimilar hardware and software in command and monitor lanes, separate power and hydraulic channels, and degraded control laws.

AER-407 · Loosely vs tightly coupled INS/GNSS integration?

Loosely coupled fuses GNSS position and velocity solutions with INS. Tightly coupled fuses raw pseudoranges, so it works with fewer than four satellites and detects faults better, at more complexity.

AER-408 · Scenario: GNSS jamming is reported along a route. What mitigations exist?

Revert to INS, DME/DME or VOR navigation; use multi-constellation, multi-frequency receivers, controlled-reception-pattern antennas and spoofing detection by cross-checking with inertial and air data.

AER-409 · How is a pitch-attitude autopilot structured?

An inner pitch-rate damping loop, a pitch-attitude loop around it, and outer altitude or flight-path loops, each designed with bandwidth well separated from the next.

AER-410 · Why is gain scheduling used in flight control?

Aircraft dynamics change strongly with dynamic pressure, Mach and configuration. Gains are designed at many points and interpolated using scheduled variables, with stability checked between points.

AER-411 · What stability margins are typically required in flight control design?

A common guideline is at least 6 dB gain margin and 45° phase margin per loop, checked with time delays, actuator dynamics and model uncertainty.

AER-412 · Scenario: pitot probes ice at cruise and airspeed becomes unreliable. What should the systems and crew do?

The autopilot may disconnect and protections may degrade. Crews fly a memorised pitch and thrust setting and cross-check GNSS ground speed and angle of attack. Designs add probe heating, dissimilar sources and synthetic airspeed.

AER-413 · Why is time and space partitioning needed in IMA?

Several applications of different DALs share a processor. Partitioning (e.g., ARINC 653) ensures a fault or overrun in one cannot corrupt memory or steal time from another.

AER-414 · An unaided accelerometer has 1 mg bias. Estimate position error after 10 minutes.

x = ½·b·t² = 0.5 × 0.0098 × 600² ≈ 1.8 km, ignoring Schuler oscillation, which bounds horizontal errors over its 84.4-minute period.

AER-415 · What does DAL A software require beyond lower levels?

The most objectives, the most performed with independence, and structural coverage down to modified condition/decision coverage (MC/DC).

Aeroelasticity, Certification & Safety

AER-416 · Explain classical bending–torsion flutter.

As speed rises, aerodynamic coupling draws the bending and torsion frequencies together. Their phase lets the airflow do net positive work on the structure each cycle, so damping becomes negative and amplitude grows until failure. It can be catastrophic.

AER-417 · How is a flutter flight test conducted?

Expand speed in small steps, excite the structure with control pulses or exciters, measure damping and frequency, and extrapolate damping trends before each step. Stop if damping falls toward zero.

AER-418 · Scenario: after repainting, an aileron flutters in flight. Why?

Added paint mass behind the hinge upset the mass balance, shifting the surface CG aft and lowering flutter speed. Control surfaces must be re-balanced after painting or repair.

AER-419 · What sets divergence speed for a typical section?

q_D = K_θ/(e·S·a), where K_θ is torsional stiffness, e is the distance the aerodynamic centre lies ahead of the elastic axis, S area and a lift slope. Moving the elastic axis forward raises q_D.

AER-420 · What is active flutter suppression and load alleviation?

Control surfaces driven by sensors add aerodynamic damping or redistribute lift to reduce gust and manoeuvre loads, allowing lighter or more flexible wings.

AER-421 · What is whirl flutter?

An instability of propeller or rotor and nacelle where gyroscopic and aerodynamic forces drive a whirling motion of the engine mount. It is critical in turboprops and tiltrotors.

AER-422 · How is compliance with airworthiness requirements shown?

A certification basis and compliance plan list each requirement with its means of compliance: analysis, ground or flight test, inspection, similarity or simulation, approved by the authority.

AER-423 · How would you show a catastrophic failure condition is extremely improbable?

No single failure may cause it, regardless of probability. Quantify with fault tree analysis to below 1×10⁻⁹ per flight hour and use common cause analysis (zonal, particular risks, common mode) to show independence.

AER-424 · What is limit of validity (LOV)?

The flight-cycle or hour limit, supported by tests and analysis, up to which widespread fatigue damage will not occur in the airframe. Aircraft cannot operate beyond it without further evidence.

AER-425 · What lesson did the 737 MAX MCAS accidents teach about safety assessment?

A function relying on a single AoA sensor could repeatedly command nose-down trim. The safety assessment's assumptions about failure effects and pilot response were inadequate. Redundant sensor inputs, limited authority and validated human-factors assumptions are essential.

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