Question bank — Finite wing theory — induced drag, Prandtl's lifting line

True — tips nahi hain toh trailing sheet nahi, downwash nahi, lift tilt nahi; yeh d'Alembert's paradox hai (ideal 2D flow mein saari drag gayab ho jaati hai).

False — induced drag woh kinetic energy hai jo trailing vortices mein ghoomti rehti hai, yeh ek pressure/lift effect hai jo zero viscosity par bhi zinda rehti hai; sirf friction drag ko viscosity chahiye.

False — yeh drag minimize karti hai ek given lift aur span ke liye downwash ko uniform banake ($\delta =0$); yeh lift maximize nahi karti, yeh wake energy cost minimize karti hai.

False — kyunki $C_{D,i}\propto C_L^{2}S/(\pi b^2)$ hai, fixed span par area (chord) badhane se induced drag badhti hai; sirf span help karta hai.

False — $\sin n\theta$ ki orthogonality sirf $A_1$ ko bachne deti hai: $C_L=\pi AR{A}_1$; higher harmonics drag add karte hain lekin lift nahi.

True — induced downwash geometric angle ka kuch hissa "kha" jaata hai, isliye aapko har unit $C_L$ ke liye zyada $\alpha$ chahiye: $a=a_0/(1+a_0/\pi AR)$.

False — $e=1/(1+\delta )$ jahan $\delta \ge 0$, isliye $e\le 1$, aur equality sirf elliptic (single-harmonic) case mein hoti hai.

True — $C_{D,i}\propto 1/AR$, isliye $AR\to 2AR$ karne se half induced drag milti hai; isliye gliders lambe aur patle hote hain (dekho Aspect ratio & wing design).

False — yeh isliye exist karte hain kyunki $\Gamma$ span ke along vary karta hai; Helmholtz ek filament ko fluid mein khatam karne se mana karta hai, isliye jahan bhi circulation change hota hai wahan vorticity $-d\Gamma /dy$ shed hoti hai.

Trailing filaments semi-infinite hote hain (wing se shuru hokar infinity tak jaate hain), isliye yeh $\Gamma /(4\pi r)$ hai — exactly woh factor jo downwash integral mein hai. $2\pi$ formula doubly-infinite line ke liye hai.

$A_2$ orthogonality ki wajah se lift mein zero contribute karta hai; yeh sirf $\delta$ badhata hai, induced drag add karta hai. Sirf $A_1$ hi $C_L$ set karta hai.

Induced drag inviscid hai, lift-coupled hai, aur wake ki swirl energy mein rehta hai — yeh surface skin-friction (profile) drag se bilkul alag hai.

Yeh subtract hota hai: ${\alpha }_{\text{eff}}=\alpha -{\alpha }_i$, kyunki downwash local flow ko neechay tilt karta hai, jisse section ko actually feel hone wala angle reduce ho jaata hai.

Yeh $C_L^2$ ke saath scale karta hai, isliye zero lift par ($C_L=0$) induced drag khatam ho jaata hai — yeh poora drag-due-to-lift hai; baat sirf itni hai ki ise viscosity nahi chahiye.

Har term $\sin n\theta$, $\theta =0,\pi$ (tips) par zero hoti hai, jo $\Gamma =0$ force karti hai wahan — yeh built in hai aur physically required hai, kyunki finite tip circulation ke liye infinite shed vortex ki zaroorat padegi.

Lift local relative wind ke perpendicular define hoti hai; downwash us wind ko ${\alpha }_i$ se neechay rotate karta hai, isliye abhi bhi perpendicular lift ${\alpha }_i$ se peeche rotate ho jaati hai, jisse uska rearward component ban jaata hai (Thin airfoil theory ke perpendicular-lift rule se).

Lift $\Gamma \cdot \sin \theta$ integrate karta hai, aur ${\int }_0^{\pi }\sin n\theta \sin \theta d\theta =0$ jab tak $n=1$ na ho; drag $\Gamma \cdot {\alpha }_i$ integrate karta hai jo Biot–Savart downwash ke through har mode se $n{A}_n^2$ laata hai.

Constant downwash ka matlab hai ki har strip ek jitna induced-angle "tax" bharta hai, jo ek fixed total lift ke liye Cauchy–Schwarz optimum hai; koi bhi spanwise variation ($\delta >0$) wasted wake energy hai.

Trailing vorticity ek lambe span par faili hoti hai, isliye downwash per unit lift kamzor hoti hai; $C_{D,i}\propto 1/AR=S/b^2$, aur yeh span $b$ (squared) hai jo dominate karta hai, area nahi.

$\delta ={\sum }_{n\ge 2}n(A_n/A_1{)}^2$ squares ka sum hai, isliye yeh $\ge 0$ hai; pure-$A_1$ elliptic shape se har departure sirf drag add karta hai, jo $e=1/(1+\delta)$ se capture hota hai.

Geometric angle ka kuch hissa induced downwash angle ${\alpha }_i$ ko cancel karne mein "kharach" ho jaata hai, isliye section ka effective angle reduce ho jaata hai; aapko geometrically ${\alpha }_i$ wapas add karna padta hai, isliye 3D slope $a<a_0$ hoti hai.

Ek constant $\Gamma$ jo tips par zero tak drop karni chahiye matlab hai $d\Gamma /dy\to -\infty$ wahan, jo infinitely strong tip vortex shed karti hai — physically impossible, isliye real loading zero tak taper karti hai (elliptic sabse smooth optimum hai).

$C_{D,i}=C_L^2/(\pi AR)\to 0$ aur $a=a_0/(1+a_0/\pi AR)\to a_0=2\pi$ — 2D drag-free, full-slope limit recover hoti hai, jo d'Alembert's paradox ke saath consistent hai.

Nahi — $A_1=0$ isliye $C_L=0$, aur $C_{D,i}=C_L^2/(\pi eAR)=0$; net lift nahi hai toh lift-carrying vortices shed karne ke liye koi spanwise loading nahi hai.

Nahi — $C_{D,i}=\pi AR\sum n{A}_n^2$, $A_n^2$ par depend karta hai, isliye yeh positive rehta hai; drag ko wake se thrust ke roop mein recover nahi kiya ja sakta, sirf $C_L$ ka sign badalta hai.

Lift zero hai (koi $A_1$ nahi), phir bhi $\delta ={\sum }_{n\ge 2}n{A}_n^2/A_1^2$ blow up hota hai / $C_{D,i}=\pi AR(2A_2^2)>0$ — pure induced drag bina lift ke, worst possible efficiency.

Identical — induced drag sirf span aur total lift par depend karta hai, chord par nahi: $C_{D,i}=L^2/(\frac{1}{2}\rho V^2\pi b^{2}q\dots )$ sirf $b$ aur lift par depend karta hai, isliye equal span ⇒ equal induced drag.