EASA ATPL 081 Principles of Flight study materials
EASA ATPL 081 Principles of Flight study resources

EASA ATPL 081 Principles of Flight: Airflow and Aerofoils: Lift, Drag, Pressure Distribution, Boundary Layer, Separation and Stalls

EASA ATPL 081 Principles of Flight · Work through your exam, one topic at a time.

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Airflow and Aerofoils: Lift, Drag, Pressure Distribution, Boundary Layer, Separation and Stalls

This domain tests the section itself: how camber and angle of attack build the pressure distribution, how the boundary layer responds, and where lift ends. Items typically ask you to interpret CL/alpha curves and pressure distributions, and to break drag down into its components.

Lift and the lift curve

  • CL = L/(qS); lift is proportional to dynamic pressure, area and lift coefficient.
  • CL rises almost linearly with alpha at roughly 0.1 per degree for a two-dimensional section, until the critical alpha (typically 15–16°), where CLmax is reached and the section stalls.
  • Camber raises CL at zero alpha and makes the zero-lift angle negative; it does not remove the stall, it only moves it.

Pressure distribution

  • There is a suction peak near the leading edge on the upper surface, and the area between the upper and lower pressure distributions represents the lift.
  • The centre of pressure moves forward as alpha increases, approaching 25% chord, then jumps aft when the flow separates.
  • Pitching moment about the aerodynamic centre is roughly constant with alpha for a cambered section.

Boundary layer and separation

  • The layer is laminar near the leading edge and becomes turbulent after transition; a turbulent layer has more energy and resists the adverse pressure gradient longer.
  • Separation begins at the trailing edge and moves forward as alpha increases; full separation is the stall.
  • Profile drag = skin friction + form drag. Induced drag belongs to the finite wing, and wave drag appears at high Mach. Total drag is the sum.

Worked example

An aerofoil of area 16 m² flies at TAS 60 m/s at sea level with CL 0.40. q = 0.5 × 1.225 × 3600 = 2205 Pa, so L = 2205 × 16 × 0.40 = 14,112 N, about 14.1 kN, supporting roughly 1440 kg. If the aircraft actually weighs 1200 kg (11,772 N), the required CL is 11,772/(2205 × 16) = 0.33, reached at a smaller alpha. The same section, a different angle.

Traps

  • CL is not lift: at fixed CL, doubling speed quadruples lift.
  • The stall angle is fixed by geometry, not by weight or speed; weight changes only the speed at which that angle is reached.
  • A rough or iced leading edge lowers CLmax and reduces the stall angle, so Vs rises.
  • Induced drag does not exist on a two-dimensional section; quoting it in a section question loses the mark.
  • Reynolds number matters: section data taken at model scale does not transfer directly to full scale.