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An airfoil (British spelling: aerofoil) is the cross-sectional profile of a lifting surface, such as a wing or blade. Its chord is the straight segment from leading edge to trailing edge, with length $c$; its mean camber line runs midway between its upper and lower surfaces, and its thickness measures their separation. Camber is the departure of that mean line from the chord. The angle of attack describes the profile's orientation relative to the incoming flow. These independent choices let a designer alter the distribution of pressure, lift and drag, rather than simply enlarging the surface. ^[ref]

The distinction between an airfoil and a finite wing matters. A two-dimensional section model assumes no variation along span and predicts lift per unit span $L^{\prime }$. A wing of span $b$ has a spanwise distribution $L^{\prime }(z)$ and total lift $L={\int }_{-b/2}^{b/2}{L}^{\prime }(z)dz$. Its tip flow and trailing vortices generate a downward induced velocity, called downwash, which changes the local flow seen by the sections. Thus one cannot generally multiply the isolated two-dimensional $L^{\prime }$ by $b$. The wing aspect ratio $b^2/S$ measures its span relative to its reference area and helps characterize these three-dimensional effects. ^[ref]

For a thin section with small camber slopes and small angle of attack, thin airfoil theory predictswhere both angles are in radians and ${\alpha }_{L=0}$ is the zero-lift angle of attack. The model assumes two-dimensional, steady flow, incompressible flow, inviscid flow, an attached outer flow, and smooth departure at a sharp trailing edge (the Kutta condition). This last condition selects the circulation used by the Kutta–Joukowski theorem; with counterclockwise circulation $\Gamma$ positive and ${\mathbf{U}}_{\infty }=U_{\infty }{\mathbf{e}}_x$, that theorem gives $L^{\prime }=-{\rho }_{\infty }{U}_{\infty }\Gamma$ for positive lift towards $+{\mathbf{e}}_y$. The beauty is that a detailed velocity field collapses to a simple force law, while the geometry enters through ${\alpha }_{L=0}$. ^{[ref] [ref]}

A symmetric airfoil has ${\alpha }_{L=0}=0$ in this model: at $\alpha =4{}^{\circ }=\pi /45$ the predicted lift coefficient is $c_l\simeq 0.439$, and reversing $\alpha$ reverses lift. A stipulated cambered section with ${\alpha }_{L=0}=-2{}^{\circ }$ instead predicts $c_l\simeq 0.219$ at zero angle of attack. These are ideal calculations, not measurements or guarantees for a particular profile. Air parcels on the upper and lower surfaces have no obligation to arrive together at the trailing edge; the pressure and velocity field must satisfy the flow problem. A symmetric lifting section already makes an unequal-path-length explanation inadequate. ^[ref]

Real airfoils have a viscous boundary layer, even when inviscid flow approximates the outer motion. At sufficiently demanding incidence, flow separation can become extensive and the lift coefficient departs from its approximately linear law, reaches a maximum and may fall: this is aerodynamic stall. Lift does not generally vanish at aerodynamic stall. The onset depends on profile, Reynolds number, surface condition and motion history; $2\pi (\alpha -{\alpha }_{L=0})$ predicts neither a universal stall angle nor post-stall behavior. The pressure/balance experiment in lift makes this limit observable by measuring changes in the surface pressure distribution alongside the total force. ^[ref]

The Magnus effect is a transverse force on a spinning body in relative motion through a fluid. Rotation changes the surrounding circulation, pressure distribution and wake. In a two-dimensional cylinder model, the Kutta–Joukowski theorem gives force per unit span ${\mathbf{F}}^{\prime }=\rho \mathbf{U}\times (\Gamma \hat{\mathbf{z}})$, with relative fluid velocity $\mathbf{U}$ and circulation $\Gamma$ positive anticlockwise viewed from $+z$. Real viscosity and boundary-layer behavior help determine the circulation; impermeability alone does not determine it from spin rate.

With polar unit vectors ${\mathbf{e}}_r=(\cos \theta ,\sin \theta ,0)$ and ${\mathbf{e}}_{\theta }=(-\sin \theta ,\cos \theta ,0)$, the second term is $\Gamma {\mathbf{e}}_{\theta }/(2\pi r)$. It is tangent to circles, independent of $z$, and has zero divergence and zero vorticity away from the axis. Its singular axis carries a line vortex; adding the constant velocity field gives a uniform flow plus that line vortex.

The counterclockwise circulation around a positively oriented loop is $\oint \mathbf{u}\cdot d\mathbf{r}$. For a radius-$R$ circle, the uniform flow contributes zero and the line vortex contributesAny loop winding once around the axis has the same circulation, by Stokes theorem in the nonsingular intervening region.

On $C_R$, $\mathbf{n}={\mathbf{e}}_r$, $\mathbf{u}\cdot \mathbf{n}=U\cos \theta$, and $dl=Rd\theta$. ThusIn fact this equality holds for every $R>0$ for the stated field, since the remaining vector integral is $(0,\pi ,0)$.

This is the advective part of the momentum flux. To obtain the force, include pressure. For steady irrotational flow, the Bernoulli equation givesThe constant terms integrate to zero against $\mathbf{n}$; the remaining pressure contribution is $\oint p\mathbf{n}dl=\rho \mathbf{\Gamma }\times \mathbf{U}/2$. The total outward momentum flux plus pressure term is therefore $\rho \mathbf{\Gamma }\times \mathbf{U}$, the force of the obstacle on the fluid. The opposite force, of fluid on the obstacle per unit span, isThis is the Kutta–Joukowski theorem, with positive $\Gamma$ defined counterclockwise: positive $U$ and $\Gamma$ give force in the negative $y$ direction. For an actual two-dimensional aerofoil, faster-decaying far-field terms do not change this limit. The displayed advective integral supplies half the lift; pressure supplies the other half.

Differentiate the velocity potential in polar coordinates:At $r=a$ the normal velocity is zero. At large radius these components tend to $U\cos \theta ,-U\sin \theta$, the polar components of uniform flow in the positive $x$ direction. The circulation around any concentric circle is ${\int }_0^{2\pi }{u}_{\theta }rd\theta =\kappa$. Although the potential's circulation term is multivalued, its velocity is single-valued and irrotational away from the excluded axis.

For steady inviscid constant-density flow, Bernoulli equation gives $p+\rho |u{|}^2/2=p_{\infty }+\rho U^2/2$. On the cylinder, $u_{\theta }=-2U\sin \theta +\kappa /(2\pi a)$, soIn the displayed force integral, every constant or ${\sin }^2\theta$ term has zero cosine and sine first moments. The cross term in pressure is $\rho U\kappa \sin \theta /(\pi a)$. Therefore, using exactly the positive-sign convention printed,per unit axial length, since ${\int }_0^{2\pi }{\sin }^2\theta d\theta =\pi$.

With the usual normal directed outward from the cylinder into the fluid, pressure exerts traction $-pn$, so the physical force of the fluid on the cylinder is instead $\boxed{(0,-\rho U\kappa )}$. The positive integral printed in the question has the opposite sign. This makes the convention explicit and is consistent with the Kutta–Joukowski theorem for positive counterclockwise circulation.

For an incompressible flow, $\nabla \cdot \mathbf{u}=0$. If the flow is also irrotational, then locally $\mathbf{u}=\nabla \varphi$ for a velocity potential, and therefore

The boundary conditions areThe fluid region is not simply connected, so a potential may change by the constant $\kappa$ after one circuit while its gradient remains single-valued. Superposing uniform flow, the cylinder doublet, and the circulation gives the potential flow around a circular cylinder with circulationIts velocity components are

On $r=a$, $u_r=0$ and, with $\lambda =\kappa /(4\pi Ua)$,The Bernoulli equation gives $p=p_{\infty }+\rho (U^2-u_{\theta }^2)/2$. The pressure force per unit length on the cylinder isThe $x$ component vanishes by symmetry, while ${\int }_0^{2\pi }{\sin }^2\theta d\theta =\pi$ givesin agreement with the Kutta–Joukowski theorem.

A stagnation point satisfies $u_r=u_{\theta }=0$. On the cylinder this means $\sin \theta =\lambda$. Thus $|\lambda |<1$ gives two surface stagnation points, $|\lambda |=1$ gives one coincident surface point, and $|\lambda |>1$ gives none on the surface. Away from the cylinder, $u_r=0$ requires $\cos \theta =0$. Solving $u_{\theta }=0$ then gives one physical exterior root when $|\lambda |>1$:At $|\lambda |=1$ this root lies on $r=a$ and agrees with the single surface point.

The Magnus effect is the transverse force produced when a spinning body moves relative to a fluid. Rotation changes the flow and circulation around the body; the resulting asymmetric pressure and wake deflect the fluid, with the opposite transverse force on the body. Examples include the curved flight of spinning balls and the lift of rotating-cylinder sails.

A useful quantitative model is steady two-dimensional incompressible flow past a cylinder of radius $a$, with fluid velocity $\mathbf{U}=U\hat{\mathbf{x}}$ far away. Let circulation $\Gamma$ be positive anticlockwise as viewed from $+z$. The potential-flow velocity at the cylinder surface isThe Bernoulli equation outside the boundary layer givesIntegrating the pressure force around the cylinder, per unit span, yieldsThus the Kutta–Joukowski theorem givesThe sign here refers to the fluid velocity relative to the cylinder. For a body moving through otherwise stationary fluid, that relative fluid velocity points opposite to the body's translational velocity.

Real viscosity and the no-slip boundary condition transmit the surface rotation to the fluid and help set the circulation and separation pattern. Pure inviscid flow with only an impermeability condition does not determine $\Gamma$ from the cylinder's angular speed. Therefore the circulation must not be assigned a universal spin-rate formula without additional assumptions. The cylinder result illustrates the lift mechanism; quantitative forces on a finite spinning ball also depend on its geometry, Reynolds number, rotation rate, surface roughness and wake.