Centre of pressure and aerodynamic centre

Centre of pressure versus aerodynamic centre: definitions, how the CP moves with lift, locating the AC from moment data, and links to stability.

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Why it matters

To place the wing on the fuselage, size the tail and locate the centre of gravity, you need to know where the aerodynamic force acts and how the pitching moment changes with angle of attack. The centre of pressure answers the first question but wanders with α; the aerodynamic centre gives a fixed reference point that makes stability analysis simple.

Key ideas

Force plus moment, or force alone. The distributed pressure and shear on an airfoil can be replaced by the resultant force acting at any chosen point together with the moment about that point. Choosing a different point changes the moment, not the force.

Centre of pressure (CP). The point on the chord about which the resultant aerodynamic moment is zero, so the whole load can be represented by the force alone. For small α (lift ≈ normal force) its distance from the leading edge is x_cp = −M′_LE/N′ ≈ −M′_LE/L′.

  • For a cambered airfoil with positive camber, c_m,ac is negative (nose-down), and x_cp = x_ac − c_m,ac·c/c_l. As c_l falls towards zero, x_cp runs far aft (to infinity at zero lift); as c_l increases, the CP moves forward towards the aerodynamic centre. This makes the CP an awkward reference for analysis.
  • For a symmetric airfoil, c_m,ac = 0 and the CP coincides with the aerodynamic centre at the quarter chord for all α in the linear range.

Aerodynamic centre (AC). The point about which the pitching-moment coefficient is independent of angle of attack (in the linear, unstalled range). Thin airfoil theory places it at the quarter chord in incompressible flow; real airfoils have x_ac within a few percent of 0.25c. In supersonic flow it moves to about mid-chord.

Why the AC exists. Write the moment about a point x as c_m,x = c_m,ac + c_l·(x − x_ac)/c. Since c_l varies linearly with α and c_m,ac is constant, c_m,x varies linearly with α at every point except x_ac. Experimentally you locate the AC by measuring c_m about any convenient point (often c/4) at two or more α and finding where the slope vanishes.

Link to stability. For a complete aircraft the analogous point is the neutral point. If the centre of gravity is ahead of it, a nose-up disturbance increases lift aft of the CG and produces a restoring nose-down moment — static longitudinal stability. Because c_m,ac of a cambered wing is negative, a tail (or reflexed camber on a flying wing) is needed to trim.

Assumptions. Small α (L ≈ N), attached flow, and moments taken about points on the chord line. Near stall the AC concept breaks down because c_l is no longer linear in α.

Formulas

  • x_cp = −M′_LE / N′ — distance of CP from the LE (m); M′_LE nose-up positive (N·m/m), N′ normal force per span (N/m).
  • x_cp/c = x_ac/c − c_m,ac/c_l — CP from AC data; with x_ac = c/4 this is x_cp/c = 1/4 − c_m,c/4/c_l.
  • c_m,x = c_m,ac + c_l·(x − x_ac)/c — moment coefficient about any chord point x; nose-up positive.
  • x_ac/c = 1/4 − dc_m,c/4/dc_l — AC from measured quarter-chord moments; equivalently x_ac/c = 1/4 − m₀/a₀ with m₀ = dc_m,c/4/dα and a₀ = dc_l/dα (per same angle unit).
  • M′_ac = c_m,ac·q∞·c² (airfoil, N·m/m) and M_ac = C_M,ac·q∞·S·c̄ (wing, N·m).

Worked examples

Example 1 (standard). A NACA 2412 section of chord 1.2 m has c_m,ac = −0.053 (data-book value) with the AC at the quarter chord. Find the CP location at c_l = 0.4 and at c_l = 1.0.

  1. x_cp/c = 0.25 − c_m,ac/c_l.
  2. c_l = 0.4: x_cp/c = 0.25 + 0.053/0.4 = 0.25 + 0.1325 = 0.3825, so x_cp = 0.3825 × 1.2 = 0.459 m.
  3. c_l = 1.0: x_cp/c = 0.25 + 0.053 = 0.303, so x_cp = 0.364 m.

Answer: x_cp = 0.459 m (0.383c) at c_l = 0.4 and 0.364 m (0.303c) at c_l = 1.0 — the CP moves forward as lift increases.

Example 2 (GATE level). Wind-tunnel tests on an airfoil give c_l = 0.2, c_m,c/4 = −0.05 at α = −2°, and c_l = 1.0, c_m,c/4 = −0.03 at α = 6°. Find x_ac, c_m,ac and x_cp at α = 6°.

  1. Slope: dc_m,c/4/dc_l = (−0.03 − (−0.05))/(1.0 − 0.2) = 0.02/0.8 = 0.025.
  2. x_ac/c = 0.25 − 0.025 = 0.225.
  3. c_m,ac = c_m,c/4 + c_l(x_ac − 0.25c)/c = −0.05 + 0.2 × (−0.025) = −0.055. Check with the second point: −0.03 + 1.0 × (−0.025) = −0.055 ✓.
  4. At α = 6°: x_cp/c = x_ac/c − c_m,ac/c_l = 0.225 + 0.055/1.0 = 0.280. Check from the quarter chord: 0.25 − (−0.03)/1.0 = 0.280 ✓.

Answer: x_ac = 0.225c, c_m,ac = −0.055, x_cp = 0.280c at α = 6°.

Common mistakes

  • Treating the CP as a fixed point. It moves with c_l unless the section is symmetric.
  • Saying "the AC is the point where the moment is zero". That is the CP; at the AC the moment is constant, generally not zero.
  • Getting the sign of the transfer term wrong. Write c_m,x = c_m,ac + c_l(x − x_ac)/c and check a simple case: a point behind the AC sees a nose-up contribution from lift.
  • Using x_cp/c = 0.25 − c_m,ac/c_l when the AC is not at the quarter chord.
  • Applying the AC idea beyond stall where c_l is non-linear.

For GATE AE

Expect: locating the CP from c_m,ac and c_l; finding x_ac and c_m,ac from two measured (c_l, c_m) pairs about the quarter chord or the leading edge; transferring moments between points; and conceptual MCQs on CP movement with α and on symmetric versus cambered sections. Always note the sign convention used in the question before substituting.

Quick check

  1. Where is the CP of a symmetric airfoil at small α?
  2. For c_m,ac = −0.04 and c_l = 0.5 with the AC at c/4, where is the CP?
  3. Is the moment about the AC zero?
  4. What happens to the CP of a positively cambered airfoil as c_l → 0?

Answers: 1. At the quarter chord (it coincides with the AC). 2. x_cp = 0.25c + 0.08c = 0.33c. 3. No — it is constant with α, and negative for positive camber. 4. It moves aft without limit.

Interview questions

All Incompressible Aerodynamics interview questions

Try answering each one aloud before you open it.

  1. 1.What is the centre of pressure in aerodynamics?Concept

    The centre of pressure is the point on the chord about which the resultant aerodynamic moment is zero, so the distributed pressure and shear loading can be replaced by the resultant force acting there alone. It is found from x_cp = −M′_LE/N′. For a cambered airfoil it moves with angle of attack, which is why the aerodynamic centre is usually preferred as a reference point.

  2. 2.Define the aerodynamic centre of an airfoil.Concept

    The aerodynamic centre is a point on the chord of an airfoil where the pitching moment coefficient is constant with changes in the angle of attack. For subsonic flow, it is typically located at the quarter-chord point, or 25% of the chord length from the leading edge.

  3. 3.Explain the difference between the centre of pressure and the aerodynamic centre.Concept

    The centre of pressure is the point where the total aerodynamic force acts, and it can shift with changes in the angle of attack. The aerodynamic centre, however, is a fixed point on the airfoil where the pitching moment remains constant regardless of the angle of attack. This makes the aerodynamic centre more useful for stability analysis.

  4. 4.Why is the aerodynamic centre important in aircraft design?Application

    The aerodynamic centre is crucial in aircraft design because it simplifies the analysis of aerodynamic forces and moments. Since the pitching moment about the aerodynamic centre is constant, it allows engineers to predict the stability and control characteristics of the aircraft more easily.

  5. 5.What happens to the centre of pressure as the angle of attack increases?Application

    For a positively cambered airfoil c_m,ac is negative and x_cp/c = x_ac/c − c_m,ac/c_l, so as α and c_l increase the centre of pressure moves forward towards the aerodynamic centre; as c_l tends to zero it moves far aft. For a symmetric airfoil c_m,ac = 0 and the CP stays at the quarter chord throughout the linear range. This travel is why stability analysis uses the fixed aerodynamic centre instead.

  6. 6.How does the location of the aerodynamic centre affect aircraft stability?Application

    For the whole aircraft the equivalent point is the neutral point. If the centre of gravity lies ahead of it, an increase in angle of attack adds lift behind the CG and produces a nose-down restoring moment, so the aircraft is statically stable in pitch; if the CG is behind it the disturbance grows. The distance between CG and neutral point, as a fraction of the mean aerodynamic chord, is the static margin.

  7. 7.Why is the quarter-chord point often used as the aerodynamic centre in subsonic flow?Application

    The quarter-chord point is often used as the aerodynamic centre in subsonic flow because, for many airfoils, the pitching moment about this point remains nearly constant with changes in the angle of attack. This simplifies the analysis and design of the airfoil.

  8. 8.Calculate the location of the aerodynamic centre for a symmetric airfoil with a chord length of 2 meters.Numerical

    For a symmetric airfoil in subsonic flow, the aerodynamic centre is typically located at the quarter-chord point. Therefore, the location of the aerodynamic centre is 0.25 × 2 m = 0.5 meters from the leading edge.

  9. 9.If the centre of pressure is located at 0.3c and the aerodynamic centre is at 0.25c, what is the implication for the pitching moment?Application

    Since x_cp/c = x_ac/c − c_m,ac/c_l, a CP 0.05c behind the AC means c_m,ac = −0.05·c_l at that condition, i.e. a nose-down moment about the AC, which tells you the airfoil has positive camber. The lift acting 0.05c aft of the AC produces exactly this nose-down moment. A tail or reflex camber is needed to trim it.

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