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

1. What is the physical setup?

HOW the geometry talks: the trailing sheet of strength $-\,d\Gamma/dy$ behaves like a row of semi-infinite vortex lines. Each contributes a downwash by Biot–Savart, and we sum (integrate) them.

2. Deriving the downwash (from Biot–Savart)

Formula Downwash integral

A semi-infinite vortex filament of strength $d\Gamma = \left(\dfrac{d\Gamma}{d\eta}\right)d\eta$ starting at spanwise station $\eta$ induces at point $y$ a downwash: $dw = \frac{(d\Gamma/d\eta)\,d\eta}{4\pi (y-\eta)}$ Why the $4\pi(y-\eta)$ ? Biot–Savart for a semi-infinite straight filament at perpendicular distance $r$ gives $\frac{\Gamma}{4\pi r}$ (half of the $\frac{\Gamma}{2\pi r}$ for an infinite line). Here $r=|y-\eta|$ . Integrating over the whole span: $\boxed{\,w(y)=\frac{1}{4\pi}\int_{-b/2}^{+b/2}\frac{(d\Gamma/d\eta)}{y-\eta}\,d\eta\,}$

3. The fundamental equation of lifting-line theory

Formula Monoplane / lifting-line equation

The effective angle of attack at each station is the geometric one minus the induced one: $\alpha_{\text{eff}}(y)=\alpha(y)-\alpha_i(y).$ Thin-airfoil theory says local lift coefficient $c_\ell = 2\pi(\alpha_{\text{eff}}-\alpha_{L=0})$ , and Kutta–Joukowski says $c_\ell=\dfrac{2\Gamma}{V_\infty c(y)}$ . Equating and substituting $w$ : $\boxed{\;\alpha(y)=\frac{\Gamma(y)}{\pi V_\infty c(y)}+\alpha_{L=0}(y)+\frac{1}{4\pi V_\infty}\int_{-b/2}^{b/2}\frac{d\Gamma/d\eta}{y-\eta}\,d\eta\;}$ This integro-differential equation determines $\Gamma(y)$ for any planform.

HOW we solve it: substitute $y=-\tfrac{b}{2}\cos\theta$ and a Fourier sine series $\Gamma(\theta)=2bV_\infty\sum_{n=1}^{\infty}A_n\sin(n\theta).$ This automatically satisfies $\Gamma=0$ at the tips ( $\theta=0,\pi$ ). The $A_n$ are found by collocation.

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

4. Lift and induced drag in terms of $A_n$

5. Worked examples

6. Common mistakes (Steel-manned)

7. Active recall

Recall Quick self-test (hide answers)

What kind of drag does a 2D infinite wing have in inviscid flow? → None (d'Alembert).
What physically creates induced drag? → Trailing tip vortices / downwash tilting lift back.
Which $\Gamma$ distribution minimizes induced drag? → Elliptic, $C_{D,i}=C_L^2/(\pi AR)$ .
Why only $A_1$ in lift? → Orthogonality of $\sin n\theta$ over $[0,\pi]$ .

Recall Feynman: explain to a 12-year-old

Imagine pushing a wide tray flat through bathwater. Near the edges, water sneaks around from the high-pressure bottom to the low-pressure top, making little spinning tornadoes that trail behind. Making those tornadoes costs energy, and that energy "drag" is felt as the wing being slightly pulled back. A longer, thinner tray makes the edge-tornadoes weaker compared to all the good lifting in the middle — so long skinny wings (like a glider's) waste less energy. The perfect sharing of lift, smallest near the tips and biggest in the middle in an oval shape, makes the smallest tornadoes possible.

8. Connections

Kutta–Joukowski theorem — gives $L'=\rho V_\infty\Gamma$ used per strip.
Thin airfoil theory — supplies local lift slope $2\pi$ and $\alpha_{L=0}$ .
Biot–Savart law — basis of the downwash integral.
Helmholtz vortex theorems — why trailing vorticity must be shed.
d'Alembert's paradox — why a 2D wing has zero drag.
Aspect ratio & wing design — practical consequence for gliders/airliners.
Oswald efficiency factor — drag polar $C_D=C_{D,0}+C_L^2/(\pi e\,AR)$ .

Flashcards

Define induced drag.

Drag-due-to-lift from trailing vortices: downwash tilts the local lift vector rearward; exists even in inviscid flow.

State Prandtl's downwash integral.

w(y)=\frac{1}{4\pi}\int_{-b/2}^{b/2}\frac{d\Gamma/d\eta}{y-\eta}\,d\eta

Why factor

4\pi

not

2\pi

in trailing filament downwash?

Trailing filaments are semi-infinite, giving

\Gamma/(4\pi r)

, half the infinite-line value.

C_L

in terms of Fourier coefficients?

C_L=\pi\,AR\,A_1

(only the first coefficient).

C_{D,i}

in terms of Fourier coefficients?

C_{D,i}=\pi\,AR\sum_{n=1}^\infty nA_n^2=\frac{C_L^2}{\pi AR}(1+\delta)

Which lift distribution minimizes induced drag and why?

Elliptic (

\Gamma=\Gamma_0\sin\theta

); only

A_1\neq0

\delta=0

, constant downwash.

Minimum induced drag formula?

C_{D,i}=C_L^2/(\pi AR)

for an elliptic wing.

Define span efficiency

e

e=1/(1+\delta)\le1

;

C_{D,i}=C_L^2/(\pi e AR)

;

e=1

for elliptic.

Finite-wing lift slope (elliptic)?

a=a_0/(1+a_0/(\pi AR))

with

a_0=2\pi

Why does

C_{D,i}

depend on span not just area?

C_{D,i}=C_L^2 S/(\pi b^2)

; spreading vorticity over larger span weakens downwash.

Why only

A_1

contributes to lift?

Orthogonality:

\int_0^\pi\sin n\theta\sin\theta\,d\theta=0

for

n\neq1

Aspect ratio definition?

AR=b^2/S

(span squared over planform area).

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, ek infinite (2D) wing ka inviscid flow mein koi drag hota hi nahi — ye d'Alembert ka paradox hai. Lekin real wing finite hoti hai, uske tips par neeche ka high pressure air upar low pressure side ki taraf ghoom jaata hai, jisse peeche trailing vortices (chakkar khaate tornado) ban jaate hain. Ye vortices wing ke peeche air ko neeche dhakelte hain — isko downwash kehte hain.

Ab downwash ka kamaal: local flow ab sirf $V_\infty$ nahi, balki $V_\infty$ + thoda neeche ki velocity hai. Isse effective angle thoda kam ho jaata hai (induced angle $\alpha_i$ ), aur jo Lift banta hai wo local flow ke perpendicular hota hai, yaani thoda peeche ki taraf tilt ho jaata hai. Is tilt ka jo peeche wala component hai, wahi induced drag hai. Yaad rakho: ye friction wala drag NAHI hai — ye lift ke kaaran paida hota hai, even inviscid flow mein.

Prandtl ne ise solve karne ke liye lifting line banayi: span ke saath circulation $\Gamma(y)$ ko ek Fourier sine series $\sum A_n\sin n\theta$ se likha. Result bahut elegant hai: $C_L=\pi\,AR\,A_1$ — sirf $A_1$ lift deta hai. Aur $C_{D,i}=\pi AR\sum n A_n^2=\frac{C_L^2}{\pi AR}(1+\delta)$ — saare harmonics drag dete hain. Isliye sabse kam induced drag tab milta hai jab sirf $A_1$ ho, yaani elliptic lift distribution, jismein $C_{D,i}=C_L^2/(\pi AR)$ .

Practical baat: $C_{D,i}\propto 1/AR$ , isliye lamba patla wing (glider jaisa) kam induced drag deta hai. Yaad rakho span important hai, area nahi — kyunki $C_{D,i}=C_L^2 S/(\pi b^2)$ . Exam mein bas yeh teen cheezein pakad lo: downwash kyun banta hai, elliptic kyun best hai, aur $AR$ kyun matter karta hai.