Exercises — Thin airfoil theory — lift per unit span = πρV²(α + 2β - π c)

The theorem is $L^{\prime }=\rho V\Gamma .$

• $L^{\prime }$ — lift per unit span, units $\text{N/m}$.
• $\rho$ — density, ${\text{kg/m}}^3$.
• $V$ — speed, $\text{m/s}$.
• $\Gamma$ — circulation, ${\text{m}}^2/\text{s}$.

Check the units multiply correctly: $\frac{\text{kg}}{{\text{m}}^3}\cdot \frac{\text{m}}{\text{s}}\cdot \frac{{\text{m}}^2}{\text{s}}=\frac{\text{kg}\cdot \text{m}}{{\text{s}}^2}\cdot \frac{1}{\text{m}}=\frac{\text{N}}{\text{m}}.$ ✓ The lift is per metre of span, so the prime must never be dropped. See Kutta–Joukowski theorem.

Differentiate $c_{\ell }=2\pi \alpha$ with respect to $\alpha$: $\frac{d{c}_{\ell }}{d\alpha }=2\pi \approx 6.283\text{ per radian.}$ The "per radian" matters: $\alpha$ must be in radians because the derivation replaced $\sin \alpha$ by $\alpha$, an approximation only valid for small angles measured in radians.

Step 1 — convert the angle. $\alpha =4^{\circ }=4\times \frac{\pi }{180}=0.06981$ rad. Why: the formula assumes radians.

Step 2 — camber term is zero. $\beta =0$, so $\alpha +\frac{2\beta }{\pi }=0.06981$.

Step 3 — plug into the lift formula. $L^{\prime }=\pi \rho V^{2}c\left(\alpha +\frac{2\beta }{\pi }\right)=\pi (1.225)(60^2)(1.2)(0.06981).$ $L^{\prime }=\pi (1.225)(3600)(1.2)(0.06981)\approx 1161\text{N/m}.$ So each metre of span carries about $1161$ N.

Step 1 — camber contribution. $\frac{2\beta }{\pi }=\frac{2(0.04)}{\pi }=0.02546$.

Step 2 — total effective angle. $\alpha +\frac{2\beta }{\pi }=0.06981+0.02546=0.09527$.

Step 3 — lift. $L^{\prime }=\pi (1.225)(3600)(1.2)(0.09527)\approx 1585\text{N/m}.$

Step 4 — percentage increase. $\frac{1585-1161}{1161}\times 100\%\approx 36.5\%.$ Camber added $\sim 37\%$ lift at the same angle of attack — it shifts the whole curve up. See Lift coefficient and angle of attack.

(a) $c_{\ell }=2\pi (0.06981)+4(0.04)=0.4386+0.16=\mathrm{0.5986.}$

(b) Dynamic pressure part: $\frac{1}{2}\rho V^{2}c=0.5(1.225)(3600)(1.2)=2646$. $c_{\ell }=\frac{L^{\prime }}{2646}=\frac{1585}{2646}=\mathrm{0.599.}$ Both give $\approx 0.599$. ✓ They agree because $c_{\ell }=2\pi \left(\alpha +\frac{2\beta }{\pi }\right)=2\pi \alpha +4\beta$ is exactly the lift formula divided by the dynamic-pressure term.

Step 1 — set $L^{\prime }=0$. Lift vanishes when the bracket vanishes: $\alpha +\frac{2\beta }{\pi }=0\Rightarrow {\alpha }_{L=0}=-\frac{2\beta }{\pi }.$ Why this step: $\pi \rho V^{2}c>0$ always, so the only way to make $L^{\prime }=0$ is to kill the bracket.

Step 2 — plug in. ${\alpha }_{L=0}=-\frac{2(0.04)}{\pi }=-0.02546$ rad.

Step 3 — to degrees. $-0.02546\times \frac{180}{\pi }=-1.459^{\circ }.$

Physical meaning: a cambered wing already scoops air downward even when perfectly aligned with the flow. To produce zero net downward turning you must tilt the nose slightly down, hence a negative angle. Look at the shifted curve in the figure below — it crosses zero to the left of the origin.

Step 1 — recall which coefficients survive. From the parent derivation, $\Gamma =\pi cV\left(A_0+\frac{A_1}{2}\right).$ Only $A_0$ and $A_1$ enter, because integrating the sheet along the chord uses the orthogonality of the $\sin n\theta$ terms; every $A_n$ with $n\ge 2$ integrates to zero over $[0,\pi ]$. See Glauert's integral and Fourier coefficients.

Step 2 — plug in. $\Gamma =\pi (1.0)(40)\left(0.09+\frac{0.05}{2}\right)=\pi (40)(0.115)=14.45{\text{m}}^2/\text{s}.$

Step 3 — why $A_2$ is invisible to lift. $A_2$ (and higher) redistribute the vorticity along the chord — they change the moment (pitching) but their net contribution to total circulation, and hence total lift, cancels. So $A_2=0.03$ is simply irrelevant here.

Route A — Kutta–Joukowski. $L^{\prime }=\rho V\Gamma =(1.225)(40)(14.451)=708.1\text{N/m}.$

Route B — direct formula. $L^{\prime }=\pi \rho c{V}^2\left(A_0+\frac{A_1}{2}\right)=\pi (1.225)(1.0)(1600)(0.115)=708.1\text{N/m}.$ Both routes match, because Route B is Route A with $\Gamma =\pi cV(A_0+A_1/2)$ substituted.

Step 1 — change variable. With $x=\frac{c}{2}(1-\cos \theta )$, $1-\frac{2x}{c}=1-(1-\cos \theta )=\cos \theta ,\text{so}\frac{dz}{dx}=m\cos \theta .$ Why: the whole Glauert machinery lives in the $\theta$ variable, so we must rewrite the slope in $\theta$.

Step 2 — compute $A_0$. $A_0=\alpha -\frac{1}{\pi }{\int }_0^{\pi }\frac{dz}{dx}d\theta =\alpha -\frac{1}{\pi }{\int }_0^{\pi }m\cos \theta d\theta .$ Since ${\int }_0^{\pi }\cos \theta d\theta =[\sin \theta {]}_0^{\pi }=0$, we get $A_0=\alpha .$

Step 3 — compute $A_1$. $A_1=\frac{2}{\pi }{\int }_0^{\pi }\frac{dz}{dx}\cos \theta d\theta =\frac{2m}{\pi }{\int }_0^{\pi }{\cos }^2\theta d\theta =\frac{2m}{\pi }\cdot \frac{\pi }{2}=m.$ (Used ${\int }_0^{\pi }{\cos }^2\theta d\theta =\pi /2$.)

Step 4 — assemble the lift bracket. $A_0+\frac{A_1}{2}=\alpha +\frac{m}{2}.$ Comparing with $\alpha +\frac{2\beta }{\pi }$ gives $\boxed{\frac{2\beta }{\pi }=\frac{m}{2}\Rightarrow \beta =\frac{\pi m}{4}.}$ So the abstract parameter $\beta$ is just $\frac{\pi }{4}$ times the maximum camber-line slope $m$.

Step 1 — angle to radians. $\alpha =3^{\circ }=0.05236$ rad.

Step 2 — lift bracket. $A_0+A_1/2=\alpha +m/2=0.05236+0.05=\mathrm{0.10236.}$

Step 3 — lift coefficient. $c_{\ell }=2\pi (A_0+A_1/2)=2\pi (0.10236)=\mathrm{0.6431.}$

Step 4 — lift per span. $L^{\prime }=\pi \rho V^{2}c(A_0+A_1/2)=\pi (1.225)(55^2)(1.4)(0.10236).$ $L^{\prime }=\pi (1.225)(3025)(1.4)(0.10236)\approx 1669\text{N/m}.$

Step 1 — total = per-span × span. $L=L^{\prime }\times b=1669\times 8=13352\text{N}\approx 13.4\text{kN}.$ Why: $L^{\prime }$ carries units N/m; multiplying by span (m) gives newtons. This is why the prime mattered all along.

Model. Thin airfoil theory says $c_{\ell }=2\pi \left(\alpha +\frac{2\beta }{\pi }\right)$, a straight line in $\alpha$. Two points fix the line.

Step 1 — convert angles. ${\alpha }_1=-2^{\circ }=-0.03491$ rad (where $c_{\ell }=0$); ${\alpha }_2=3^{\circ }=0.05236$ rad (where $c_{\ell }=0.55$).

Step 2 — slope from the two points. $\text{slope}=\frac{c_{\ell ,2}-c_{\ell ,1}}{{\alpha }_2-{\alpha }_1}=\frac{0.55-0}{0.05236-(-0.03491)}=\frac{0.55}{0.08727}=6.303\text{per rad}.$ Theory predicts $2\pi =6.283$. The measured $6.303$ is within $0.3\%$ — excellent agreement, confirming the slope really is $\approx 2\pi$ and camber did not change it.

Step 3 — camber from the zero-lift point. At $c_{\ell }=0$, $\alpha ={\alpha }_{L=0}=-\frac{2\beta }{\pi }$. $\frac{2\beta }{\pi }=-{\alpha }_{L=0}=-(-0.03491)=\mathrm{0.03491.}$ So the effective camber contributes $0.0349$ rad of "extra angle," equivalent to $\beta =\frac{\pi }{2}(0.03491)=0.05483$ rad.

Step 1 — incompressible value (from 5.1): $c_{\ell ,\text{incomp}}=0.55$.

Step 2 — the correction factor. $\sqrt{1-M_{\infty }^2}=\sqrt{1-0.36}=\sqrt{0.64}=\mathrm{0.8.}$ Why this tool: at higher speed air compresses, packing more pressure difference across the wing; Prandtl–Glauert scales the incompressible answer up by $1/\sqrt{1-M^2}$. It answers "how much does compressibility amplify my low-speed lift?"

Step 3 — corrected coefficient. $c_{\ell ,\text{comp}}=\frac{0.55}{0.8}=\mathrm{0.6875.}$ Compressibility boosted the lift coefficient by $25\%$. See Compressibility corrections (Prandtl–Glauert).

Step 1 — write the target. $c_{\ell }=2\pi \alpha +4\beta$ and (Problem 4.1) $4\beta =\pi m$, so equivalently $c_{\ell }=2\pi \alpha +\pi m$. Why: express everything in the design knob $m$.

Step 2 — angle to radians. $\alpha =2^{\circ }=0.03491$ rad.

Step 3 — solve for $m$. $0.80=2\pi (0.03491)+\pi m\Rightarrow 0.80=0.21932+\pi m.$ $\pi m=0.58068\Rightarrow m=\frac{0.58068}{\pi }=\mathrm{0.1848.}$ So the mean line needs a maximum slope of about $m\approx 0.185$. That is a fairly aggressive camber — sanity check: it puts the zero-lift angle at ${\alpha }_{L=0}=-m/2=-0.0924$ rad $=-5.3^{\circ }$, comfortably negative as expected for strong camber.

The lift bracket is ==$\alpha +\frac{2\beta }{\pi }$ and equals $\alpha +\frac{m}{2}$ for a parabolic camber line. Only $A_0$ and $A_1$ survive in the circulation integral. The lift-curve slope is $2\pi$ per radian and camber only shifts the curve, never the slope==. Prandtl–Glauert multiplies $c_{\ell }$ by ==$1/\sqrt{1-M^2}$==.