Linearised supersonic flow and Ackeret theory
Ackeret linearised supersonic thin-airfoil theory: local Cp = 2θ/√(M² − 1), lift from angle only, wave drag from angle, camber and thickness, mid-chord aerodynamic centre.
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Why it matters
Ackeret's linearised theory gives lift, wave drag and moment of thin supersonic airfoils in closed form. It shows at a glance why supersonic wings are thin, why the aerodynamic centre jumps aft from quarter-chord to mid-chord when an aircraft goes supersonic, and why lift-to-drag ratio falls so sharply above Mach 1.
Key ideas
From the wave equation. For M∞ > 1 the linearised perturbation equation is λ²φxx − φyy = 0, λ = √(M∞² − 1). Its 2-D solution above the airfoil is φ = f(x − λy): disturbances are carried unchanged along left-running Mach lines and nothing travels upstream. Combining this with the tangency condition v' = V∞·dy/dx gives a purely local result: the pressure at a surface point depends only on the slope there.
Local pressure rule. Cp = 2θ/√(M∞² − 1), where θ is the local inclination of the surface to the free stream, positive when the surface faces into the flow (compression) and negative when it faces away (expansion). Compressions and expansions are treated alike — the main approximation compared with shock-expansion theory.
Decomposition. A thin airfoil at angle α can be split into a flat plate at α, a camber line and a symmetric thickness distribution. Because the theory is linear the effects add:
- Lift comes only from angle of attack:
cl = 4α/√(M∞² − 1). Camber and thickness add no lift (unlike subsonic flow, where camber gives lift). - Wave drag comes from all three:
cd = (4/√(M∞² − 1))·(α² + ⟨(dyc/dx)²⟩ + ⟨(dyt/dx)²⟩), where ⟨ ⟩ is the chordwise mean. Drag rises with the square of angle and thickness, so supersonic sections are thin (t/c ≈ 3–6%) and nearly uncambered. - Moment: for a flat plate the pressure difference is uniform along the chord, so the centre of pressure and the aerodynamic centre are at mid-chord:
cm,LE = −2α/√(M∞² − 1). In subsonic flow the aerodynamic centre is at quarter-chord, so going supersonic moves it aft and increases static stability (and trim drag).
Standard sections.
- Flat plate:
cd = 4α²/λ = cl·α. - Symmetric diamond (double wedge) of thickness ratio t/c: slopes ±t/c, so
cd = (4/λ)(α² + (t/c)²). - Symmetric biconvex (circular arc):
cd = (4/λ)(α² + (4/3)(t/c)²).
Lift-to-drag ratio. For a diamond, L/D = α/(α² + (t/c)²); it is maximum at α = t/c, where (L/D)max = 1/(2·t/c). Thin sections are essential — a 5% diamond can at best reach L/D = 10 before skin friction is added.
Validity. Thin sections, small angles, sharp leading edges, moderate supersonic Mach numbers: not close to M = 1 (where 1/λ → ∞) and not hypersonic (where M∞·θ is not small). Viscous drag must be added separately.
Formulas
λ = √(M∞² − 1)
Cp = 2θ/√(M∞² − 1) — θ in radians, + for compression.
cl = 4α/√(M∞² − 1)
cd,wave = (4/√(M∞² − 1))·[α² + ⟨(dyc/dx)²⟩ + ⟨(dyt/dx)²⟩]
cd = (4/√(M∞² − 1))·(α² + (t/c)²) — symmetric diamond; (4/3)(t/c)² replaces (t/c)² for a biconvex section.
cm,LE = −2α/√(M∞² − 1); x_cp = x_ac = c/2 (flat plate).
(L/D)max = 1/(2·t/c) at α = t/c (diamond, wave drag only).
Symbols: M∞ free-stream Mach number; θ local surface inclination (rad); α angle of attack (rad); yc, yt camber and thickness ordinates (m); x chordwise position (m); c chord (m); t/c thickness ratio; cl, cd, cm section coefficients (dimensionless).
Worked examples
Example 1 (standard). A flat plate is at α = 5° in a Mach 2 stream. Find cl, cd and the centre of pressure by Ackeret theory.
α = 5 × π/180 = 0.08727 rad;λ = √(4 − 1) = 1.732.cl = 4 × 0.08727/1.732 = 0.2015.cd = 4 × 0.08727²/1.732 = 0.01759.- The pressure difference is uniform, so the centre of pressure is at mid-chord.
Answer: cl ≈ 0.202, cd ≈ 0.0176, x_cp = 0.5c (shock-expansion theory gives 0.202 and 0.0177 here).
Example 2 (GATE level). A symmetric diamond airfoil with t/c = 0.06 flies at M∞ = 2.5 and α = 2°. Find cl, cd and L/D, and the angle and value of maximum L/D (wave drag only).
λ = √(6.25 − 1) = 2.2913;α = 0.03491 rad.cl = 4 × 0.03491/2.2913 = 0.0609.cd = (4/2.2913) × (0.03491² + 0.06²) = 1.7457 × (0.001219 + 0.0036) = 0.00841.L/D = 0.0609/0.00841 = 7.24.- Maximum L/D at
α = t/c = 0.06 rad = 3.44°, whereL/D = 1/(2 × 0.06) = 8.33.
Answer: cl ≈ 0.061, cd ≈ 0.0084, L/D ≈ 7.2; (L/D)max ≈ 8.3 at α ≈ 3.4°.
Common mistakes
- Dropping the square root: the factor is
1/√(M∞² − 1), not1/(M∞² − 1). - Using degrees in
4α/λor4α²/λ. - Expecting camber to give lift in linearised supersonic flow — it only adds drag.
- Placing the aerodynamic centre at quarter-chord in supersonic flow; it is at mid-chord.
- Applying Ackeret near M = 1 or at hypersonic speeds.
- Forgetting the thickness term in wave drag for a real airfoil.
For GATE AE
Expect: cl, cd and cm of flat plates and diamond or biconvex sections; local Cp on a given panel; how cl and cd scale with α, t/c and M; the aerodynamic-centre shift from subsonic to supersonic; and (L/D)max for a diamond. Practise radian conversion and the decomposition into angle, camber and thickness contributions.
Quick check
- What is cl for a flat plate at α = 2° and M∞ = 3?
- Where is the aerodynamic centre of a thin airfoil in supersonic flow?
- Does camber change cl in Ackeret theory?
- How does wave drag of a diamond depend on t/c at α = 0?
- What is the local Cp on a surface inclined 0.05 rad into an M = 2 flow?
Answers: 1. 0.0494; 2. mid-chord; 3. no; 4. proportional to (t/c)²; 5. 0.0577.
Interview questions
All Compressible Aerodynamics interview questionsTry answering each one aloud before you open it.
1.What is linearised supersonic flow?Concept
Linearised supersonic flow refers to the simplification of the equations governing supersonic flow, assuming small perturbations in the flow field. This allows the nonlinear equations to be linearized, making them easier to solve. The linearization is valid when the flow disturbances are small compared to the free stream conditions.
2.Explain Ackeret theory in the context of supersonic flow.Concept
Ackeret theory is a linearized theory used to predict the lift and drag on thin airfoils in supersonic flow. It assumes small perturbations and uses potential flow theory to derive expressions for aerodynamic coefficients. The theory is particularly useful for calculating the pressure distribution over airfoils at supersonic speeds.
3.Why is linearised theory important in supersonic aerodynamics?Application
Linearised theory is important because it simplifies the complex nonlinear equations of supersonic flow, making them more tractable. This simplification allows engineers to predict aerodynamic forces and moments on bodies at supersonic speeds with reasonable accuracy, especially for small disturbances.
4.What are the limitations of Ackeret theory?Application
Ackeret theory is limited to thin airfoils and small angles of attack, where the flow disturbances are small. It does not account for shock waves or viscous effects, which can be significant in real-world supersonic flows. Therefore, its predictions may not be accurate for thicker airfoils or at higher angles of attack.
5.How does the Mach number affect the applicability of linearised supersonic flow theory?Application
The Mach number affects the applicability of linearised supersonic flow theory because the theory assumes small perturbations relative to the free stream. At higher Mach numbers, the flow disturbances can become more pronounced, potentially violating the assumptions of linearization. However, for moderate supersonic Mach numbers, the theory remains applicable.
6.What happens if the angle of attack is increased beyond the limits of linearised theory?Application
If the angle of attack is increased beyond the limits of linearised theory, the assumptions of small perturbations break down. This can lead to inaccurate predictions of lift and drag, as nonlinear effects, such as shock waves and flow separation, become significant. In such cases, more complex nonlinear theories or computational methods are needed.
7.Explain how Ackeret theory can be used to calculate the lift coefficient for a thin airfoil in supersonic flow.Application
Linearised supersonic theory gives the local pressure coefficient as Cp = 2θ/√(M∞² − 1), where θ is the local surface slope relative to the free stream. For a thin airfoil at angle α the lower surface has Cp ≈ +2α/√(M∞² − 1) and the upper surface −2α/√(M∞² − 1) (camber and thickness cancel between the surfaces), so integrating the difference over the chord gives cl = 4α/√(M∞² − 1). Lift is proportional to α and inversely proportional to √(M∞² − 1), not to M²; camber contributes no lift.
8.Calculate the lift coefficient for a thin airfoil at a Mach number of 2.0 and an angle of attack of 5 degrees using Ackeret theory.Numerical
Convert α to radians: 5° = 0.08727 rad. Ackeret theory gives cl = 4α/√(M∞² − 1) = 4 × 0.08727/√3 = 0.3491/1.732 ≈ 0.202. (Forgetting the square root gives 0.116, a common error.)
9.How does the Prandtl-Glauert factor relate to the factor that appears in linearised supersonic flow?Application
Both come from the coefficient (1 − M∞²) of the linearised potential equation. In subsonic flow it gives the Prandtl–Glauert factor 1/√(1 − M∞²), which scales incompressible results; in supersonic flow the equation becomes a wave equation and the corresponding factor is 1/√(M∞² − 1), as in Cp = 2θ/√(M∞² − 1) and cl = 4α/√(M∞² − 1). Supersonic results are not corrections of incompressible ones, and both factors blow up near M = 1 where linear theory fails.
10.Determine the wave-drag coefficient for a thin flat plate at a Mach number of 3.0 and an angle of attack of 2 degrees using Ackeret theory.Numerical
Convert α: 2° = 0.03491 rad. For a flat plate cd = 4α²/√(M∞² − 1) = 4 × 0.03491²/√8 = 0.004874/2.828 ≈ 0.00172. A real airfoil would add a thickness term and skin friction.
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