Exercises — Airfoil aerodynamics — camber, chord, thickness

The 4-digit code reads (camber%, where%, fat%) — digit 1, digit 2, digits 3–4.

• Digit 1 $=4$ → max camber $=4\%$ of chord $=0.04c$.
• Digit 2 $=4$ → located at $40\%$ chord $=0.40c$ from the LE.
• Digits 3–4 $=15$ → max thickness $=15\%$ of chord $=0.15c$.

Symmetric ⟺ camber $=0$ ⟺ the first two digits are 00. Only NACA 0018 starts with 00, so it is the symmetric one (18% thick). The other two have non-zero first digits, so their mean camber line bows away from the chord — they are cambered.

• Camber → (i) lift at zero angle (it pre-bends the flow downward).
• Chord → (ii) the reference length (also the denominator in $c_{\ell }$).
• Thickness → (iii) structure + stall behaviour.

Each digit is a fraction of $c$, so just multiply.

• (a) camber $=0.02\times 1.5=0.03\text{ m}=30\text{ mm}$.
• (b) position $=0.40\times 1.5=0.60\text{ m}$ from the LE.
• (c) thickness $=0.12\times 1.5=0.18\text{ m}=180\text{ mm}$.

Why radians? The slope $2\pi$ is "per radian", so we must convert the angle first. $\alpha =5^{\circ }\times \frac{\pi }{180}=0.08727\text{ rad}.$ $c_{\ell }=2\pi (0.08727-0)=\mathrm{0.5483.}$ So $c_{\ell }\approx 0.55$.

Convert both angles to radians: $\alpha =0,{\alpha }_{L0}=-2^{\circ }=-0.03491\text{ rad}.$ $c_{\ell }=2\pi (0-(-0.03491))=2\pi (0.03491)=\mathrm{0.2193.}$ So $c_{\ell }\approx 0.22>0$: positive lift with the chord level — the camber payoff described in the parent's Section 3. Compare with Thin-Airfoil Theory.

Why arctan? The slope $d{y}_c/dx$ is "rise over run" — the ratio of the two legs of the little right triangle the camber line makes with the horizontal (see figure). $\tan \theta$ is exactly that ratio, so the angle whose tangent equals $0.10$ is the tilt: that inverse question is what $\arctan$ answers.

$\theta =\arctan (0.10)=0.09967\text{ rad}=5.71^{\circ }.$ $\sin \theta =0.09950,\cos \theta =\mathrm{0.99504.}$ The approximation replaces $\sin \theta$ by $0$ (error $\approx 0.0995$) and $\cos \theta$ by $1$ (error $\approx 0.005$, i.e. $0.5\%$). So $\cos \theta$ is almost perfect, but the $\sin \theta$ offset is a genuine $\sim 10\%$ sideways shift of the surface points — small, but not zero.

Why $\sin /\cos$? The flesh is added perpendicular to the tilted skeleton, along the unit normal $(-\sin \theta ,\cos \theta )$ (see figure). The vertical component of that normal is $\cos \theta$, the horizontal is $\sin \theta$.

With $\sin \theta =0.09950$, $\cos \theta =0.99504$: $y_U=0.03c+0.05c(0.99504)=0.07975c,$ $y_L=0.03c-0.05c(0.99504)=-0.01975c.$ Vertical gap $=y_U-y_L=2(0.05c)(0.99504)=0.09950c$. The naive chord-perpendicular value would be $2y_t=0.10c$. So the true vertical gap is $0.5\%$ smaller because the thickness is laid out along the tilted normal, not straight up. This is exactly the "thin-airfoil coincidence" ($\theta \to 0$) from the parent's Section 2.

Convert: $4^{\circ }=0.06981$ rad, $8^{\circ }=0.13963$ rad, ${\alpha }_{L0}=-3^{\circ }=-0.05236$ rad.

At $\alpha =4^{\circ }$: $c_{\ell ,A}=2\pi (0.06981)=0.4386,c_{\ell ,B}=2\pi (0.06981+0.05236)=\mathrm{0.7676.}$ Difference $=2\pi (0.05236)=0.3290$.

At $\alpha =8^{\circ }$: $c_{\ell ,A}=2\pi (0.13963)=0.8773,c_{\ell ,B}=2\pi (0.13963+0.05236)=\mathrm{1.2063.}$ Difference $=2\pi (0.05236)=0.3290$ again.

The gap is constant ($0.329$) at every $\alpha$. Geometrically: both lift lines have the same slope $2\pi$; camber only shifts B's line left by $|{\alpha }_{L0}|$. A horizontal shift of a straight line of fixed slope produces a fixed vertical gap — that is the whole role of camber.

Why check the slope? A maximum of a smooth curve is where its tangent is flat — slope $=0$. That is precisely the question the derivative answers: "how fast is $y_c$ changing right here?" At the peak the answer is zero.

Use the front branch ($0\le x\le p$), $m=0.02$, $p=0.40$, so $m/p^2=0.02/0.16=0.125$.

(a) Value at $x=0.40$: $y_c=0.125(2(0.40)(0.40)-0.40^2)=0.125(0.32-0.16)=0.125(0.16)=\mathrm{0.02.}✓$

(b) Slope: $\frac{d{y}_c}{dx}=0.125(2p-2x)=0.125(0.80-2x)$. At $x=0.40$: $\frac{d{y}_c}{dx}=0.125(0.80-0.80)=0.✓$ Both branches agree at $x=p$ and give a flat peak of height $0.02c$ at $x=0.40c$ — the "2" and "4" digits, recovered from the geometry. See NACA Airfoil Series.

Front branch slope: $d{y}_c/dx=0.125(0.80-2x)$. At $x=0$: $\frac{d{y}_c}{dx}=0.125(0.80)=\mathrm{0.10.}$ $\theta =\arctan (0.10)=5.71^{\circ }.$ This is the steepest point of the mean line (slope is largest at the LE for a 4-digit airfoil). Even here $\theta \approx 5.7^{\circ }$ gives $\cos \theta =0.995$, so the thin-airfoil approximation is off by only $0.5\%$ in the thickness direction — safe. Downstream, the slope only decreases (reaching $0$ at $x=p$), so the approximation is even better there.

Why the change of variable $\hat{x}=\frac{1}{2}(1-\cos \theta )$? It maps the chord $\hat{x}\in [0,1]$ onto $\theta \in [0,\pi ]$ and clusters points near the LE and TE where the physics is sharp; it also turns the vortex-sheet integral into a Fourier-cosine form that integrates cleanly. This is the standard Thin-Airfoil Theory substitution.

Step 1 — slope of the camber line. With $z=4m\hat{x}(1-\hat{x})=4m(\hat{x}-{\hat{x}}^2)$, $\frac{dz}{d\hat{x}}=4m(1-2\hat{x}).$

Step 2 — rewrite in $\theta$. Since $\hat{x}=\frac{1}{2}(1-\cos \theta )$, we get $1-2\hat{x}=1-(1-\cos \theta )=\cos \theta$. So $\frac{dz}{d\hat{x}}=4m\cos \theta .$

Step 3 — plug into the integral. ${\alpha }_{L0}=-\frac{1}{\pi }{\int }_0^{\pi }4m\cos \theta (\cos \theta -1)d\theta =-\frac{4m}{\pi }{\int }_0^{\pi }({\cos }^2\theta -\cos \theta )d\theta .$

Step 4 — do the two standard integrals. ${\int }_0^{\pi }{\cos }^2\theta d\theta =\frac{\pi }{2},{\int }_0^{\pi }\cos \theta d\theta =0.$ Therefore ${\alpha }_{L0}=-\frac{4m}{\pi }\left(\frac{\pi }{2}-0\right)=-2m.$

Hmm — recompute carefully: $-\frac{4m}{\pi }\cdot \frac{\pi }{2}=-2m$. So ${\alpha }_{L0}=-2m$ radians. (The "$-m\pi$" in the prompt is the trap — see the mistake box.)

Numeric. For $m=0.04$: ${\alpha }_{L0}=-2(0.04)=-0.08\text{ rad}=-0.08\times \frac{180}{\pi }=-4.58^{\circ }.$

So this strongly-cambered plate lifts until tilted nearly $4.6^{\circ }$ nose-down — big camber, big shift, exactly the trend of the parent's Section 3.

At $\alpha =0$: $c_{\ell }=2\pi (0-{\alpha }_{L0})=2\pi (2m)=4\pi m$. Set equal to $0.5$: $4\pi m=0.5\Rightarrow m=\frac{0.5}{4\pi }=0.03979\approx \mathrm{0.0398.}$ So $m\approx 3.98\%\approx 4\%$ camber — remarkably close to a real NACA 4-something airfoil, which is exactly why moderately-cambered wings cruise happily near zero tilt. Cross-check the lift value with Lift and Drag Coefficients.