Exercises — Airfoil aerodynamics — camber, chord, thickness

4-digit code ko aise padho: (camber%, kahan%, mota%) — digit 1, digit 2, digits 3–4.

• Digit 1 $=4$ → max camber $=4\%$ of chord $=0.04c$.
• Digit 2 $=4$ → LE se $40\%$ chord par hai $=0.40c$.
• Digits 3–4 $=15$ → max thickness $=15\%$ of chord $=0.15c$.

Symmetric ⟺ camber $=0$ ⟺ pehle do digits 00 hain. Sirf NACA 0018 00 se shuru hota hai, isliye woh symmetric hai (18% mota). Baaki dono ke pehle digits non-zero hain, toh unki mean camber line chord se door bow karti hai — woh cambered hain.

• Camber → (i) zero angle par lift (yeh flow ko neeche ki taraf pre-bend karta hai).
• Chord → (ii) reference length (saath hi $c_{\ell }$ mein denominator bhi hai).
• Thickness → (iii) structure + stall behaviour.

Har digit $c$ ka ek fraction hai, isliye bas multiply karo।

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

Radians kyun? Slope $2\pi$ "per radian" hai, isliye pehle angle convert karna zaroori hai। $\alpha =5^{\circ }\times \frac{\pi }{180}=0.08727\text{ rad}.$ $c_{\ell }=2\pi (0.08727-0)=\mathrm{0.5483.}$ Toh $c_{\ell }\approx 0.55$।

Dono angles radians mein convert karo: $\alpha =0,{\alpha }_{L0}=-2^{\circ }=-0.03491\text{ rad}.$ $c_{\ell }=2\pi (0-(-0.03491))=2\pi (0.03491)=\mathrm{0.2193.}$ Toh $c_{\ell }\approx 0.22>0$: chord level hote hue bhi positive lift — parent ke Section 3 mein describe kiya gaya camber ka fayda। Thin-Airfoil Theory se compare karo।

Arctan kyun? Slope $d{y}_c/dx$ "rise over run" hai — woh chhota right triangle jो camber line horizontal ke saath banati hai, uske dono legs ka ratio। $\tan \theta$ bilkul wahi ratio hai, isliye jo angle $0.10$ ke barabar tangent rakhe woh tilt hai: woh reverse question $\arctan$ answer karta hai।

$\theta =\arctan (0.10)=0.09967\text{ rad}=5.71^{\circ }.$ $\sin \theta =0.09950,\cos \theta =\mathrm{0.99504.}$ Approximation $\sin \theta$ ko $0$ se replace karti hai (error $\approx 0.0995$) aur $\cos \theta$ ko $1$ se (error $\approx 0.005$, yani $0.5\%$)। Toh $\cos \theta$ almost perfect hai, lekin $\sin \theta$ offset ek genuine $\sim 10\%$ sideways shift hai surface points ka — chhota, lekin zero nahi।

$\sin /\cos$ kyun? Flesh tilted skeleton ke perpendicular add hoti hai, unit normal $(-\sin \theta ,\cos \theta )$ ke along (figure dekho)। Us normal ka vertical component $\cos \theta$ hai, horizontal $\sin \theta$ hai।

$\sin \theta =0.09950$, $\cos \theta =0.99504$ ke saath: $y_U=0.03c+0.05c(0.99504)=0.07975c,$ $y_L=0.03c-0.05c(0.99504)=-0.01975c.$ Vertical gap $=y_U-y_L=2(0.05c)(0.99504)=0.09950c$। Naive chord-perpendicular value hoti $2y_t=0.10c$। Toh actual vertical gap $0.5\%$ chhota hai kyunki thickness tilted normal ke along laid out hai, seedha upar nahi। Yahi parent ke Section 2 ka "thin-airfoil coincidence" ($\theta \to 0$) hai।

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

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

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

Gap har $\alpha$ par constant ($0.329$) hai। Geometrically: dono lift lines ka slope $2\pi$ same hai; camber sirf B ki line ko left shift karta hai $|{\alpha }_{L0}|$ se। Fixed slope ki ek straight line ko horizontally shift karo toh fixed vertical gap milta hai — yahi camber ka poora role hai।

Slope kyun check karein? Ek smooth curve ka maximum wahan hota hai jahan uski tangent flat hoti hai — slope $=0$। Derivative exactly yahi sawaal answer karti hai: "yahan $y_c$ kitni tezi se badal rahi hai?" Peak par jawab zero hota hai।

Front branch use karo ($0\le x\le p$), $m=0.02$, $p=0.40$, toh $m/p^2=0.02/0.16=0.125$।

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

(b) Slope: $\frac{d{y}_c}{dx}=0.125(2p-2x)=0.125(0.80-2x)$। $x=0.40$ par: $\frac{d{y}_c}{dx}=0.125(0.80-0.80)=0.✓$ Dono branches $x=p$ par agree karte hain aur $x=0.40c$ par height $0.02c$ ka flat peak dete hain — "2" aur "4" digits, geometry se recover kiye। NACA Airfoil Series dekho।

Front branch slope: $d{y}_c/dx=0.125(0.80-2x)$। $x=0$ par: $\frac{d{y}_c}{dx}=0.125(0.80)=\mathrm{0.10.}$ $\theta =\arctan (0.10)=5.71^{\circ }.$ Yeh mean line ka sabse steep point hai (4-digit airfoil ke liye LE par slope sabse badi hoti hai)। Yahan bhi $\theta \approx 5.7^{\circ }$ se $\cos \theta =0.995$ milta hai, toh thin-airfoil approximation thickness direction mein sirf $0.5\%$ off hai — safe hai। Downstream, slope sirf decrease hoti hai ($x=p$ par $0$ tak), isliye approximation wahan aur bhi better hai।

Change of variable $\hat{x}=\frac{1}{2}(1-\cos \theta )$ kyun? Yeh chord $\hat{x}\in [0,1]$ ko $\theta \in [0,\pi ]$ par map karta hai aur LE aur TE ke paas points cluster karta hai jahan physics sharp hai; yeh vortex-sheet integral ko Fourier-cosine form mein bhi badal deta hai jo cleanly integrate ho jaata hai। Yeh standard Thin-Airfoil Theory substitution hai।

Step 1 — camber line ka slope। $z=4m\hat{x}(1-\hat{x})=4m(\hat{x}-{\hat{x}}^2)$ ke saath: $\frac{dz}{d\hat{x}}=4m(1-2\hat{x}).$

Step 2 — $\theta$ mein rewrite karo। Kyunki $\hat{x}=\frac{1}{2}(1-\cos \theta )$, humein milta hai $1-2\hat{x}=1-(1-\cos \theta )=\cos \theta$। Toh: $\frac{dz}{d\hat{x}}=4m\cos \theta .$

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

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

Hmm — carefully recompute karo: $-\frac{4m}{\pi }\cdot \frac{\pi }{2}=-2m$। Toh ${\alpha }_{L0}=-2m$ radians। (Prompt mein "$-m\pi$" ek trap hai — mistake box dekho।)

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

Toh yeh strongly-cambered plate lift karta rehta hai jab tak use almost $4.6^{\circ }$ nose-down nahi tilt kiya jaata — bada camber, bada shift, exactly parent ke Section 3 ka trend।

$\alpha =0$ par: $c_{\ell }=2\pi (0-{\alpha }_{L0})=2\pi (2m)=4\pi m$। $0.5$ ke barabar set karo: $4\pi m=0.5\Rightarrow m=\frac{0.5}{4\pi }=0.03979\approx \mathrm{0.0398.}$ Toh $m\approx 3.98\%\approx 4\%$ camber — remarkably close to a real NACA 4-something airfoil, yahi wajah hai ki moderately-cambered wings zero tilt ke paas aaram se cruise karti hain। Lift and Drag Coefficients ke saath lift value cross-check karo।