Wind tunnel
Change the shape and angle of an airfoil, then watch the flow respond. Explore how pressure and lift change together.
Set the flow
Raise or lower the leading edge.
78 knots in this example.
Maximum mean-line height, as a fraction of chord.
- Section lift coefficient
- Cₗ
- Lift per meter of span
- N/m
- Air density
- 1.225kg/m³
Two-dimensional ideal flow · 1 m chord · incompressible air
Surface pressure
Upper surface Lower surfaceNegative Cₚ means pressure below the freestream value.
What to notice
Start with the symmetric airfoil at zero angle. Its upper and lower pressure distributions match, so the net lift is zero.
Add camber or increase the angle of attack. The pressure distribution changes, and the force no longer cancels. Doubling the speed multiplies the predicted lift by four.
The flow is computed with a Joukowski transformation and the Kutta condition. The lift readout and pressure plot come from the same solution.
The model
This experiment maps flow around a circle into flow around an airfoil using the Joukowski transformation. The Kutta condition sets the circulation so the flow leaves the sharp trailing edge smoothly.
Surface pressure follows from the local velocity through Bernoulli’s equation. Lift per unit span follows from the Kutta–Joukowsky theorem. The airfoil’s chord is fixed at one meter, and the air density is fixed at 1.225 kg/m³.
What it leaves out
This is a steady, two-dimensional, inviscid, incompressible solution. It has no boundary layer, flow separation, stall, compressibility, or finite-wing effects. It is useful for exploring ideal flow, rather than predicting the performance of a particular aircraft.
For the underlying ideas, see NASA Glenn’s conformal-mapping explanation and MIT’s Kutta–Joukowsky lecture.