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Top 25 ATPL Principles of Flight Questions (with Sourced Answers)

LAST UPDATED: 7 OCTOBER 2026

Published Oct 2026~14 min readEASA syllabusSubject 081

Principles of Flight (081) is one of the 13 EASA ATPL theory papers: 46 questions in 1 hour 30 minutes, passed at 75% like every other paper. It covers subsonic aerodynamics (lift, drag, the boundary layer, the stall and high-lift devices), stability and control, high-speed flight, operating limits and the flight envelope. Most questions test whether you understand why a figure changes, not the figure itself.

These 25 questions come from v1prep's ATPL Principles of Flight bank, with the four options, the correct answer marked, a short explanation and the page of the CAE Oxford 080 text it is taken from. Choose your answer before you read the one under it. For the other papers and how they are sat, see the EASA ATPL theory exams guide.

Sections: 5 questions each
  1. Lift & the aerofoil
  2. Drag & the boundary layer
  3. Stall & high-lift devices
  4. Stability & control
  5. High-speed flight & limitations

Lift & the aerofoil

Angle of attack, lift equation, ground effect
Question 1
The angle of attack (α, alpha, also called aerodynamic incidence) is the angle between:
  1. AThe chord line and the relative airflow ✓
  2. BThe wing root chord line and the longitudinal axis
  3. CThe mean camber line and the horizontal datum line
  4. DThe relative airflow and the aircraft's flight path direction
Answer: A. The angle of attack is the angle between the chord line and the relative airflow. It is not the angle of incidence, the fixed angle between the chord line and the longitudinal axis.Source: CAE Oxford 080, p. 53
Question 2
The basic lift equation is:
  1. AL = ½ ρ V · CL: no wing area term required in the formula
  2. BL = ρ V · CL · S: density × velocity × CL × wing area
  3. CL = ½ ρ V² · CL · S: the standard lift equation ✓
  4. DL = ρ V² · CD · S: using drag coefficient, not lift coefficient
Answer: C. L = ½ρV² × CL × S, where ρ is the air density, V the true airspeed, CL the lift coefficient (set by angle of attack and shape) and S the wing area. The ½ρV² term is the dynamic pressure.Source: CAE Oxford 080, p. 72
Question 3
The aerodynamic centre (AC) of an aerofoil at subsonic speeds (less than M 0.4) is located at approximately:
  1. AThe 25% chord point: regardless of camber, thickness, or angle of attack ✓
  2. BThe 50% chord point: the geometric midpoint of the chord line always
  3. CThe leading edge: coincident with the stagnation point at low AOA only
  4. DThe 75% chord point: near the trailing edge of the section in any case
Answer: A. Below about M 0.4 the aerodynamic centre is at the quarter-chord point of any aerofoil, whatever its camber, thickness or angle of attack. It is the point about which the pitching moment stays constant at normal angles of attack.Source: CAE Oxford 080, p. 60
Question 4
Wing tip vortices are caused by:
  1. AThe pressure difference between upper and lower surfaces, causing tip leakage ✓
  2. BIrregular flow disturbances caused by the engine nacelle vortex flow
  3. CVibrations of the aileron control surfaces during normal flight
  4. DHeat-driven turbulence in the upper-atmosphere boundary layer at altitude
Answer: A. The lower surface is at higher pressure than the upper. At the tips and trailing edge air leaks from below to above, which sets up a spanwise flow, towards the root on top and towards the tip underneath, and rolls up into the tip vortices.Source: CAE Oxford 080, p. 86
Question 5
Ground effect becomes significant when the wing is within approximately:
  1. AHalf the wingspan above the surface ✓
  2. BTwice the wingspan above the surface
  3. CFive times the wingspan above the surface
  4. DThe full wingspan above the surface: 100% wingspan height
Answer: A. Ground effect becomes significant within half a wingspan of the surface. It can be detected up to about one span, but only slightly: for a 40 m span, the reduction in induced drag is about 1.4% at 40 m, 23.5% at 10 m and 47.6% at 4 m.Source: CAE Oxford 080, p. 92

Drag & the boundary layer

Parasite, induced, VMD
Question 6
Parasite drag is further subdivided into:
  1. AWave drag, shock drag, and compressibility drag: applicable transonic only
  2. BInduced drag, lift-induced drag, and vortex drag: all the same component
  3. CSkin friction drag, form (pressure) drag, and interference drag ✓
  4. DLift drag, weight drag, and thrust drag: the four-force drags only
Answer: C. Skin friction drag, form (pressure) drag and interference drag. Skin friction and form drag together are called profile drag.Source: CAE Oxford 080, p. 111
Question 7
A laminar boundary layer compared to a turbulent boundary layer has:
  1. AHigher skin friction, more kinetic energy, and a greater tendency to remain attached
  2. BHigher mean speed, thicker depth, and resistant to adverse pressure gradient
  3. CIdentical drag and energy characteristics; the difference is purely visual
  4. DLower skin friction, lower kinetic energy, and a greater tendency to separate ✓
Answer: D. A laminar layer has lower skin friction but less kinetic energy, so it separates more readily. A turbulent layer has higher skin friction but more energy, and resists an adverse pressure gradient better.Source: CAE Oxford 080, p. 112
Question 8
Induced drag varies with airspeed approximately as:
  1. AV²: proportional to dynamic pressure, like parasite drag
  2. BConstant: independent of speed at all flight speeds and altitudes
  3. CV: linear with velocity in any flight regime
  4. D1/V²: decreases as the square of velocity ✓
Answer: D. Induced drag falls as the square of speed rises. At low speed a high CL is needed, the tip vortices are strong and induced drag is large; at high speed the reverse. Parasite drag rises with V², so total drag has a minimum, at VMD.Source: CAE Oxford 080, p. 117
Question 9
On the total drag vs airspeed curve (the 'drag curve'), VMD (minimum drag speed) coincides with:
  1. ABest L/D ratio, and induced drag = parasite drag ✓
  2. BMaximum L/D ratio at low altitude only: not at cruise altitude
  3. CThe stall speed of the aircraft: VMD is always at CLmax condition
  4. DVMO: the maximum operating speed at any flight altitude
Answer: A. At VMD induced drag equals parasite drag, total drag is at its minimum and the lift/drag ratio at its maximum. VMD is also the speed for the best glide range.Source: CAE Oxford 080, p. 124
Question 10
Below VMD ('back side of the drag curve'), the aircraft is:
  1. ASpeed stable: a gust slowing the aircraft causes drag to fall, restoring speed
  2. BSpeed unstable: slowing causes drag to rise, which slows aircraft further ✓
  3. CAlways in a stalled condition: VMD is always above the stall speed
  4. DSubject to control reversal: flight controls operate in reverse senses
Answer: B. Below VMD the drag curve has a negative gradient: a loss of speed makes induced drag rise steeply, which slows the aircraft further. Above VMD the gradient is positive and the aircraft is speed stable.Source: CAE Oxford 080, p. 128

Stall & high-lift devices

Weight, bank, sweep, flaps, slats
Question 11
If the weight of an aircraft is reduced by 20% (e.g. fuel burn), the 1g stall speed (Vs1g) will:
  1. AIncrease by 20% in the same proportion as the weight change
  2. BRemain unchanged: Vs1g depends only on configuration, not weight
  3. CDecrease by 50%: direct proportionality at low altitude
  4. DDecrease by approximately 10%: Vs varies as √(weight) ✓
Answer: D. Stall speed varies with the square root of weight: √0.80 = 0.894, so the 1g stall speed falls by about 10%. The stalling angle of attack does not change.Source: CAE Oxford 080, p. 169
Question 12
If an aircraft's 1g stall speed is 150 kt CAS, the stall speed in a level 60° bank turn is approximately:
  1. A150 kt: bank angle does not affect stall speed
  2. BAbout 178 kt: calculated as Vs × √(1/cos 45°) = 150 × 1.19
  3. CAbout 212 kt: calculated as Vs × √(1/cos 60°) = 150 × 1.41 ✓
  4. DAbout 130 kt: bank angles below 90° reduce stall speed
Answer: C. Stall speed in a level turn = Vs × √(1/cos φ). At 60° of bank, √(1/0.5) = 1.414, so 150 × 1.414 ≈ 212 kt CAS. At 45° it would be about 178 kt.Source: CAE Oxford 080, p. 172
Question 13
When a swept-back wing begins to stall, the natural tendency is for the aircraft to:
  1. APitch nose-down, just like a rectangular wing aircraft
  2. BYaw violently at constant pitch and roll attitude in any condition
  3. CRoll inverted at constant pitch attitude in any flight regime
  4. DPitch up: tip stalls first, moving the CP forward ✓
Answer: D. The tips stall first, so lift is lost outboard and the centre of pressure moves forward towards the root, giving a nose-up moment. With lift now concentrated inboard, more downwash reaches the tailplane and adds to it. It can lead to a deep stall.Source: CAE Oxford 080, p. 160
Question 14
Lowering trailing-edge flaps in flight will affect the stall speed and stall AOA as follows:
  1. AStall speed decreases, stall AOA reduced ✓
  2. BStall speed increases, stall AOA increases: flap deployment penalises both
  3. CStall speed unchanged, stall AOA unchanged: flap is purely a drag device
  4. DStall speed decreases, stall AOA increases: flap raises critical AOA
Answer: A. Flaps raise CLmax, so the stall speed falls. The stalling angle of attack, measured against the basic section's chord line, is reduced because the flap increases the camber. Leading-edge slats act differently: they delay separation and raise the stalling angle.Source: CAE Oxford 080, p. 173
Question 15
When leading-edge slats are deployed, the stalling angle of attack:
  1. ADecreases: slats reduce critical AOA
  2. BDrops to zero: slats prevent stall at any AOA
  3. CRemains unchanged: slats only affect drag, not stall AOA
  4. DIncreases significantly ✓
Answer: D. Slats re-energise the boundary layer and delay separation to a higher angle of attack: typically about 25°, against about 16° for the basic section.Source: CAE Oxford 080, p. 218

Stability & control

Static, dynamic, lateral, adverse yaw
Question 16
An aircraft displaced from a trimmed condition that initially tends to return to equilibrium has:
  1. APositive static stability ✓
  2. BNeutral static stability: neither returning nor diverging from disturbance
  3. CNegative static stability: the most desirable for fighter aircraft only
  4. DDynamic instability: independent of static behaviour at any altitude
Answer: A. Positive static stability is an initial tendency to return to equilibrium after a disturbance. Most transport aircraft are designed to be positively stable in pitch and yaw and close to neutral in roll.Source: CAE Oxford 080, p. 241
Question 17
The aft CG limit on an aircraft is set to ensure:
  1. AMaximum controllability for aerobatic manoeuvres at any altitude
  2. BEqual weight distribution between forward and aft cargo holds
  3. CA minimum degree of static longitudinal stability ✓
  4. DMinimum trim drag at cruise speed and altitude
Answer: C. The aft limit keeps a minimum of static longitudinal stability: the CG always stays some distance forward of the neutral point. The forward limit is set by controllability.Source: CAE Oxford 080, p. 251
Question 18
An undamped oscillation (constant amplitude with time) demonstrates:
  1. ANegative static, positive dynamic stability: convergent oscillation at any altitude
  2. BNo stability information: undamped behaviour is independent of stability on any type
  3. CNeutral static stability and divergent dynamic stability
  4. DPositive static stability and neutral dynamic stability ✓
Answer: D. The initial tendency to return shows positive static stability, but with no damping the amplitude never decreases: neutral dynamic stability. Positive dynamic stability needs damping.Source: CAE Oxford 080, p. 284
Question 19
Dutch roll is a coupled lateral-directional oscillation that occurs when:
  1. ADirectional stability is much larger than dihedral effect
  2. BDihedral effect is large compared to directional stability ✓
  3. CBoth stabilities are zero at any altitude or condition
  4. DAircraft has any swept wing at any altitude
Answer: B. Dutch roll appears when dihedral effect is large compared with directional stability: the aircraft rolls strongly away from the sideslip but yaws back slowly, giving a coupled rolling and yawing oscillation. It is worse on swept wings at low speed and is damped by the yaw damper.Source: CAE Oxford 080, p. 309
Question 20
Adverse aileron yaw arises because:
  1. AThe up-going aileron causes increased induced drag on its wing
  2. BThe fin produces a destabilising side-force
  3. CRudder cables are not connected to ailerons
  4. DThe up-going wing has more induced drag, yawing opposite to roll ✓
Answer: D. The down-going aileron raises lift, and with it induced drag, on the up-going wing, while the other wing has less of both. The drag difference yaws the aircraft away from the direction of roll.Source: CAE Oxford 080, p. 342

High-speed flight & limitations

MCRIT, sweep, Mach tuck, VA
Question 21
The critical Mach number (MCRIT) of an aircraft is defined as:
  1. AThe free-stream Mach number at which local airflow first reaches M 1.0 ✓
  2. BThe Mach number at which shockwaves first appear on the leading edge in any condition
  3. CThe maximum operating Mach number set by the manufacturer on every aircraft and altitude
  4. DThe Mach number at which the aircraft itself first exceeds the speed of sound globally
Answer: A. MCRIT is the free-stream Mach number at which the local flow first reaches M 1.0 somewhere on the aircraft, typically near the point of maximum thickness on the wing upper surface. Above it, supersonic regions and shock waves appear.Source: CAE Oxford 080, p. 414
Question 22
The principal benefit of wing sweepback for high-speed flight is to:
  1. AReduce structural weight by 30%: a swept spar carries less bending load than a straight wing
  2. BImprove low-speed handling: sweep raises CLmax and delays the stall to a higher angle of attack
  3. CEliminate stall behaviour entirely: a swept wing is unable to reach its stalling angle of attack
  4. DIncrease MCRIT: the velocity component perpendicular to the LE is less than free-stream (V × cos Λ) ✓
Answer: D. Only the velocity component perpendicular to the leading edge (V × cos Λ) sets the pressure distribution, so a swept wing sees a lower effective speed than the free stream and MCRIT rises. In effect the section behaves as if it were thinner.Source: CAE Oxford 080, p. 434
Question 23
Mach tuck (high-speed tuck under) is the:
  1. ANose-down moment from rearward CP shift and reduced tail downwash ✓
  2. BTendency of the aircraft to pitch up rapidly above MCRIT at any weight
  3. CSudden increase in pitch authority above MCRIT
  4. DStable cruise condition above MCRIT
Answer: A. Above MCRIT the centre of pressure moves aft and the downwash at the tail decreases, so the tailplane carries less download: both give a nose-down moment, and a further Mach increase makes it worse. A Mach trim system counters it.Source: CAE Oxford 080, p. 426
Question 24
The 'coffin corner' on a buffet onset chart represents the altitude where:
  1. ALow-speed (stall) buffet boundary intersects high-speed (Mach) buffet boundary: only one speed can be flown ✓
  2. BThe maximum certified ceiling for the type is reached, set by pressurisation limits rather than by the buffet boundaries
  3. CThe maximum operating speed VMO becomes equal to MMO, leaving a single airspeed limit for the rest of the climb
  4. DStall speed VS rises to equal the long-range cruise speed, so the aircraft can no longer be climbed
Answer: A. At the coffin corner the low-speed (stall) buffet boundary meets the high-speed (Mach) buffet boundary, so only one speed can be flown without buffet.Source: CAE Oxford 080, p. 431
Question 25
VA (design manoeuvring speed) is best defined as:
  1. AThe minimum speed at which the aircraft can be flown straight-and-level at any aircraft mass
  2. BThe maximum operating speed for the aircraft
  3. CThe highest speed for sudden full nose-up elevator without exceeding the limit load factor ✓
  4. DThe speed at which buffet first occurs
Answer: C. VA is the highest speed at which sudden full nose-up elevator can be applied without exceeding the design limit load factor; below it, the wing stalls first. VA = Vs1g × √n, where the CLmax line meets the limit load line on the V-n diagram.Source: CAE Oxford 080, p. 462

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