Aspect ratio — effect on induced drag: Worked examples

3.1.23 · D3 · Physics › Compressible Flow & Aerodynamics › Aspect ratio — effect on induced drag

Yeh page ek drill hall hai. Parent note Aspect ratio — effect on induced drag ne woh ek formula build kiya jo is poore topic ko chalata hai:

$C_{D, i} = \frac{C _{L}^{2}}{π e A R}$

Yahan hum ise re-derive nahi karte — hum ise stress-test karte hain. Hum poochte hain: kya hota hai jab aapko har tarah ka input diya jaye? Bada span, zero span, zero lift, numbers ki ek table, ek real airplane, ek sneaky exam twist. End tak aapko koi aisa case nahi milna chahiye jo aapne pehle na dekha ho.

Kuch bhi shuru karne se pehle, yeh pakka karte hain ki har letter ek picture hai, na ki sirf ek squiggle.

Definition Paanch letters, har ek ek picture ki tarah

$C_{L}$ — lift coefficient. Socho yeh ek dial hai jo pilot ghoomata hai: nose upar karo, $C_{L}$ badhta hai. Iske koi units nahi hain. $C_{L} = 0$ matlab "wings zero lift produce kar rahe hain" (knife-edge ya exactly zero angle par gliding).
$A R$ — aspect ratio, $b^{2} / S$ , span squared divided by wing area. Picture karo "kitne chord-lengths wing ke along fit hote hain." Glider: lamba aur patla, $A R \sim 30$ . Fighter: chhota, $A R \sim 3$ .
$e$ — span efficiency, $0$ aur $1$ ke beech ki ek number. Yeh jawab deta hai "tumhara lift shape perfect elliptical ke kitna kareeb hai?" $e = 1$ perfect hai; ek plain rectangle $\approx 0.8$ hota hai.
$π$ — bas $3.14159 \dots$ , woh constant jo Lifting-line theory (Prandtl) mein elliptical integral se bahar aata hai.
$C_{D, i}$ — induced-drag coefficient, woh answer jo hum compute karte hain. Jitna bada = utna zyada drag jo purely ek finite wing par lift banana se paida hota hai.

Scenario matrix

Yeh har tarah ka case hai jo yeh topic aap par throw kar sakta hai. Neeche har worked example ko us cell ke saath tag kiya gaya hai jise woh fill karta hai.

Cell	Kya vary karta hai / degenerate input	Kaunsa sawaal answer karta hai	Example
A. Forward compute	ordinary $C_{L}, e, A R$	plug in karo, $C_{D, i}$ nikalo	Ex 1
B. $1/ A R$ scaling	sirf $A R$ badlo	span kaise payoff karta hai?	Ex 2
C. $C_{L}^{2}$ scaling	sirf $C_{L}$ badlo	slow flight kyun hurt karta hai?	Ex 3
D. $e$ effect	non-elliptical shape	kharab lift shape ki cost	Ex 4
E. Degenerate: $C_{L} = 0$	zero lift	kya koi induced drag hai?	Ex 5
F. Limit: $A R \to \infty$	infinite (2-D) wing	kya induced drag vanish ho jaata hai?	Ex 6
G. Backward solve	given $C_{D, i}$ , $A R$ nikalo	design/sizing problem	Ex 7
H. Real-world word problem	full drag polar, actual forces	Newtons mein fuel/thrust	Ex 8
I. Exam twist	ratio trap, dono inputs move karte hain	"cancel" mistakes se bacho	Ex 9

Hum saaton nau hit karenge.

Example 1 — Cell A: plain forward computation

Worked example Statement

Ek wing elliptical loading ( $e = 1$ ) ke saath $C_{L} = 0.8$ aur aspect ratio $A R = 6$ par fly karta hai. $C_{D, i}$ nikalo.

Forecast: Pehle order of magnitude guess karo. Lift coefficients around $1$ hote hain; $π \cdot 6 \approx 19$ . Toh $C_{D, i}$ roughly $0.64/19 \approx 0.03$ hona chahiye — kuch percent. Woh guess pakad ke rakho.

Step 1 — Lift ko square karo. $C_{L}^{2} = 0. 8^{2} = 0.64$ Yeh step kyun? Formula ka numerator $C_{L}^{2}$ hai, $C_{L}$ nahi — lift "troublemaker" hai aur yeh squared enter karta hai. Pehle squaring karna arithmetic saaf rakhta hai.

Step 2 — Denominator banao. $π e A R = π \cdot 1 \cdot 6 = 18.85$ Yeh step kyun? $e = 1$ ke saath shape ideal hai, toh kuch extra denominator ko shrink nahi karta; yeh bas $π A R$ hai.

Step 3 — Divide karo. $C_{D, i} = \frac{0.64}{18.85} = 0.0340$ Yeh step kyun? Yeh definition hai, execute ki gayi. Numerator (harm) over denominator (span hamare liye kaam kar raha hai).

Verify: Hamare forecast " $\approx 0.03$ " se match karta hai. Units: saare inputs dimensionless hain, isliye $C_{D, i}$ dimensionless hai — coefficient ke liye correct. ✓

Example 2 — Cell B: span double karna ( $1/ A R$ payoff)

Worked example Statement

Ex 1 wali wing ( $C_{L} = 0.8$ , $e = 1$ ) lekin ab $A R = 12$ tak stretch ki gayi. $A R = 6$ ke answer se compare karo.

Forecast: Humne sirf $A R$ badla, aur woh akele bottom par baitha hai. Bottom double karne se answer half ho jaana chahiye. Guess $0.0170$ .

Step 1 — Numerator frozen rakho. $C_{L}^{2} = 0.64 (unchanged — humne C_{L} ko touch nahi kiya)$ Yeh step kyun? Jo variable change hua use isolate karna scaling visible banata hai. $C_{L}$ aur $e$ same hain, toh sirf denominator mein $A R$ move karta hai.

Step 2 — Denominator recompute karo. $π \cdot 1 \cdot 12 = 37.70$ Yeh step kyun? $A R$ ko $6$ se $12$ tak double karna $π A R$ ko exactly double karta hai.

Step 3 — Divide karo, phir ratio lo. $C_{D, i} = \frac{0.64}{37.70} = 0.0170, \frac{0.0170}{0.0340} = \frac{1}{2}$ Yeh step kyun? Ratio hi point hai: kyunki $C_{D, i} \propto 1/ A R$ , aspect ratio double = induced drag half. Figure mein blue aur yellow bars dekho — doosra exactly half-height hai.

Verify: $0.0170 \times 2 = 0.0340$ se $A R = 6$ wali value wapas milti hai. Consistent. ✓

Example 3 — Cell C: slow karna ( $C_{L}^{2}$ ka explosion)

Worked example Statement

Ek wing $A R = 10$ , $e = 1$ ke saath cruise mein $C_{L} = 0.4$ aur landing mein $C_{L} = 1.2$ par fly karti hai. $C_{D, i}$ compare karo.

Forecast: $C_{L}$ triple hota hai ( $0.4 \to 1.2$ ). Kyunki yeh squared enter karta hai, $3^{2} = 9$ times zyada drag guess karo.

Step 1 — Cruise case. $C_{D, i} = \frac{0. 4 ^{2}}{π \cdot 10} = \frac{0.16}{31.42} = 0.00509$ Yeh step kyun? Small-lift baseline establish karo taaki hum growth measure kar sakein.

Step 2 — Landing case. $C_{D, i} = \frac{1. 2 ^{2}}{π \cdot 10} = \frac{1.44}{31.42} = 0.0458$ Yeh step kyun? Same wing, same $π A R$ ; sirf numerator $0.16$ se $1.44$ tak swelled — ek factor of $9$ .

Step 3 — Ratio padho. $\frac{0.0458}{0.00509} = 9.0$ Yeh step kyun? Square law confirm karta hai. Figure mein parabola dikhata hai kyun: origin ke paas flat hai, lekin high $C_{L}$ par steeply shoot up karta hai.

Verify: $(1.2/0.4)^{2} = 3^{2} = 9$ , drag ratio se exactly match karta hai. Isliye slow, high-lift flight (takeoff/landing) induced drag se dominate hoti hai — Glide ratio & L/D max dekho. ✓

Example 4 — Cell D: non-elliptical shape ki price

Worked example Statement

Do wings, dono $A R = 8$ , dono $C_{L} = 0.7$ par. Wing X elliptical hai ( $e = 1$ ); Wing R ek plain rectangle hai ( $e = 0.8$ ). Rectangle kitne percent worse hai?

Forecast: $e$ bottom par baitha hai. Ise $1.0$ se $0.8$ tak drop karna ek chhote number se divide karta hai, toh drag badhti hai. $1/0.8 = 1.25$ , toh guess karo +25%.

Step 1 — Elliptical wing. $C_{D, i} = \frac{0.49}{π \cdot 1 \cdot 8} = \frac{0.49}{25.13} = 0.0195$ Yeh step kyun? $e = 1$ benchmark hai — is $C_{L}$ aur $A R$ ke liye physically possible least drag (Elliptical lift distribution minimum).

Step 2 — Rectangular wing. $C_{D, i} = \frac{0.49}{π \cdot 0.8 \cdot 8} = \frac{0.49}{20.11} = 0.0244$ Yeh step kyun? Rectangle ka lift tips ki taraf zyada pile up karta hai, jisse stronger tip vortices bante hain; $e = 0.8$ us penalty ko encode karta hai.

Step 3 — Percent increase. $\frac{0.0244 - 0.0195}{0.0195} = 0.25 = 25%$ Yeh step kyun? Do numbers ko design-relevant statement mein badalta hai: ek kharab lift shape ek quarter zyada induced drag karti hai. Isliye designers ellipse fake karne ke liye taper aur washout use karte hain.

Verify: $0.0195/0.8 = 0.0244$ exactly — $e = 0.8$ se divide karna hi $1.25$ se multiply karna hai. ✓

Example 5 — Cell E: degenerate case $C_{L} = 0$

Worked example Statement

Ek wing ( $A R = 7$ , $e = 0.85$ ) exactly zero lift par fly karti hai (zero-lift angle par gliding, ya knife-edge). Uska induced drag kya hai?

Forecast: Koi lift nahi matlab kuch bhi backward tilt nahi hota. Guess karo exactly zero.

Step 1 — Formula mein $C_{L} = 0$ plug karo. $C_{D, i} = \frac{0 ^{2}}{π \cdot 0.85 \cdot 7} = \frac{0}{18.69} = 0$ Yeh step kyun? Kyunki numerator $C_{L}^{2}$ hai, wahan zero poori cheez ko kill kar deta hai chahe $A R$ ya $e$ kitna bhi bada ho.

Step 2 — Physics ka sanity-check karo. Yeh step kyun? Zero lift matlab tips ke paas upar aur neeche equal pressure ⇒ kuch bhi leak nahi hota ⇒ koi trailing vortices nahi ⇒ koi downwash nahi ⇒ lift (joaise bhi zero hai) kisi cheez se back tilt nahi hoti. Parent note ki chain simply collapse ho jaati hai.

Verify: $C_{D, i} = 0$ . Note karo wing mein abhi bhi Parasite drag hai — skin friction aur form drag ko lift ki zaroorat nahi. Induced drag ek akaila term hai jo zero lift par vanish hota hai. Yeh directly is galati ko refute karta hai ki "induced drag sirf shape par depend karta hai." ✓

Example 6 — Cell F: limit $A R \to \infty$ (infinite 2-D wing)

Worked example Statement

Ek fixed lift lo ( $C_{L} = 1.0$ , $e = 1$ ) aur span ko bina bound ke badhne do: $A R = 10, 100, 1000, \to \infty$ . $C_{D, i}$ kya approach karta hai?

Forecast: $A R$ bottom par hai; ise infinity tak push karna fraction ko zero tak drive karna chahiye.

Step 1 — Trend tabulate karo. $A R = 10 : \frac{1}{π \cdot 10} = 0.0318, A R = 100 : 0.00318, A R = 1000 : 0.000318$ Yeh step kyun? $A R$ mein har 10× increase drag ko 10 se divide karta hai — ek clean $1/ A R$ decay jo tum zero ki taraf march karte dekh sakte ho.

Step 2 — Formal limit lo. $lim_{A R \to \infty} \frac{C _{L}^{2}}{π e A R} = 0$ Yeh step kyun? Ek infinite wing ke koi tips nahi hote. Koi tips nahi ⇒ koi leakage nahi ⇒ koi vortices nahi ⇒ flow truly 2-D hai. Induced drag purely ek 3-D, tip-born effect hai, toh 2-D mein yeh exactly zero hona chahiye — aur algebra agree karta hai.

Verify: Curve $1/ (π A R)$ axis ko hug karta hai lekin kabhi cross nahi karta — induced drag zero approach karta hai, "2-D airfoil ka koi induced drag nahi" se match karta hai. ✓

Example 7 — Cell G: aapko jo wing chahiye uske liye backward solve karna

Worked example Statement

Ek designer chahta hai ki induced drag $C_{D, i} = 0.012$ se worse na ho jabki $C_{L} = 0.9$ par fly karo expected $e = 0.9$ ke saath. Minimum aspect ratio kya hai?

Forecast: Hum formula invert kar rahe hain. Kyunki drag $A R$ badhne par girta hai, ek chhota $A R$ hai jo just target meet karta hai. Mid-teens number expect karo.

Step 1 — Formula ko $A R$ ke liye rearrange karo. $C_{D, i} = \frac{C _{L}^{2}}{π e A R} ⟹ A R = \frac{C _{L}^{2}}{π e C _{D, i}}$ Yeh step kyun? $A R$ denominator mein daba hua tha; dono sides ko $A R$ se multiply karo aur $C_{D, i}$ se divide karo use bahar nikalne ke liye. Ab yeh ek explicit design equation hai.

Step 2 — Plug in karo. $A R = \frac{0. 9 ^{2}}{π \cdot 0.9 \cdot 0.012} = \frac{0.81}{0.03393} = 23.87$ Yeh step kyun? Teen jaane-maane numbers ka direct substitution.

Step 3 — "Minimum" interpret karo. Yeh step kyun? $A R = 23.9$ target exactly hit karta hai. Koi bhi bada $A R$ kam drag deta hai (better), isliye minimum acceptable aspect ratio $\approx 24$ hai — glider territory.

Verify: $A R = 23.87$ ko forward formula mein feed karo: $0.81/ (π \cdot 0.9 \cdot 23.87) = 0.012$ . Round-trip close ho jaata hai. ✓

Example 8 — Cell H: ek real-world word problem (Newtons, sirf coefficients nahi)

Worked example Statement

Ek drone $V = 30 m/s$ par cruise karta hai, air density $ρ = 1.2 kg/m^{3}$ , wing area $S = 2.0 m^{2}$ , span $b = 6.0 m$ , $e = 0.9$ . Uska weight $W = 250 N$ hai aur steady level flight mein lift weight ke barabar hoti hai. Induced-drag force $D_{i}$ Newtons mein nikalo.

Forecast: Cruise mein induced drag usually weight ka kuch percent hota hai. Kuch Newtons guess karo.

Step 1 — Dynamic pressure $q$ . $q = \frac{1}{2} ρ V^{2} = \frac{1}{2} (1.2) (3 0^{2}) = 540 Pa$ Yeh step kyun? Coefficients $q S$ ke through forces mein badal jaate hain; hume pehle $q$ chahiye. Units: $kg\cdotm^{- 3} \cdot m^{2} s^{- 2} = Pa$ . ✓

Step 2 — Level flight se lift coefficient. $L = W = 250 N \Rightarrow C_{L} = \frac{L}{q S} = \frac{250}{540 \cdot 2.0} = 0.2315$ Yeh step kyun? Hume $C_{L}$ given nahi hai; steady flight ( $L = W$ ) hume deta hai. Yeh forces aur coefficients ke beech classic word-problem link hai.

Step 3 — Aspect ratio. $A R = \frac{b ^{2}}{S} = \frac{6. 0 ^{2}}{2.0} = 18$ Yeh step kyun? Formula ko $A R$ chahiye, aur hume raw geometry ( $b$ aur $S$ ) di gayi hai — toh ise build karo.

Step 4 — Induced-drag coefficient. $C_{D, i} = \frac{C _{L}^{2}}{π e A R} = \frac{0.231 5 ^{2}}{π \cdot 0.9 \cdot 18} = \frac{0.05360}{50.89} = 0.001053$ Yeh step kyun? Ab har ingredient haath mein hai; yeh forward compute (Cell A) real numbers par apply hai.

Step 5 — Force mein wapas convert karo. $D_{i} = C_{D, i} q S = 0.001053 \cdot 540 \cdot 2.0 = 1.14 N$ Yeh step kyun? Sawaal force in Newtons maangta tha, isliye hum coefficient ko usi $q S$ se undo karte hain. Units: dimensionless $\times$ Pa $\times m^{2} = N$ . ✓

Verify: $D_{i} = 1.14 N$ weight ka $1.14/250 = 0.46%$ hai — chhota, exactly jaisa high- $A R$ ( $= 18$ ) cruise drone ke liye expect kiya tha. Yeh tiny induced drag precisely wahi reason hai kyun long-span drones itne efficiently loiter karte hain. ✓

Example 9 — Cell I: exam twist (terms ko "cancel" mat hone do)

Worked example Statement

Trap question: "Ek wing ka aspect ratio double kiya jaata hai aur uska lift coefficient double kiya jaata hai. Kyunki ek top par hai aur ek bottom par, induced drag unchanged rehta hai, sahi?"

Forecast: Suspicious — $C_{L}$ squared enter karta hai lekin $A R$ first power mein enter karta hai. Unhe cancel nahi karna chahiye. Guess karo: drag actually double hoti hai.

Step 1 — Before aur after symbolically likhо. $C_{D, i}^{old} = \frac{C _{L}^{2}}{π e A R}, C_{D, i}^{new} = \frac{( 2 C _{L} ) ^{2}}{π e ( 2 A R )}$ Yeh step kyun? $2 C_{L}$ aur $2 A R$ substitute karna algebra ko honest rakhta hai instead of "top vs bottom" eyeballing karne ke.

Step 2 — Ratio simplify karo. $\frac{C _{D, i}^{new}}{C _{D, i}^{old}} = \frac{4 C _{L}^{2}}{2 A R} \cdot \frac{A R}{C _{L}^{2}} = \frac{4}{2} = 2$ Yeh step kyun? $4$ doubled $C_{L}$ ko square karne se aata hai; $2$ doubled $A R$ se. Ye dete hain $4/2 = 2$ , na ki $1$ . Square law jeetta hai.

Step 3 — Corrected answer state karo. Yeh step kyun? Induced drag double hoti hai, same nahi rehti. Trap yeh bhoolne par rely karta hai ki $C_{L}$ squared hai.

Verify (concrete numbers): Start karo $C_{L} = 0.5, A R = 8, e = 1$ : $C_{D, i} = 0.25/ (8 π) = 0.009947$ . Ab $C_{L} = 1.0, A R = 16$ : $C_{D, i} = 1.0/ (16 π) = 0.019894$ . Ratio $= 2.000$ . ✓

Recall Self-test: inme se har ek kaunsi cell hai?

"Zero lift par wing" ::: Cell E (degenerate $C_{L} = 0$ → $C_{D, i} = 0$ ) "Ek target drag given hai, span nikalo" ::: Cell G (backward solve) " $C_{L}$ triple, drag ×9" ::: Cell C ( $C_{L}^{2}$ scaling) "Infinite/2-D wing" ::: Cell F ( $A R \to \infty$ , drag → 0) "Rectangle vs ellipse" ::: Cell D (span-efficiency $e$ effect)

Connections

Aspect ratio — effect on induced drag — parent formula jo har example use karta hai
Lifting-line theory (Prandtl) — jahan $C_{D, i} = C_{L}^{2} / (π A R)$ paida hota hai
Trailing vortices & downwash — Cell E aur F ke peeche ki physics
Elliptical lift distribution — Cell D mein $e = 1$ benchmark
Drag polar — Example 8 ki forces $C_{D} = C_{D, 0} + C_{L}^{2} / (π e A R)$ ke andar rehti hain
Parasite drag — woh term jo zero lift par vanish nahi hota (Cell E)
Wingtip devices (winglets) — Cell G mein effective $A R$ / $e$ raise karne ka ek tarika
Glide ratio & L/D max — kyun Cell C ka $C_{L}^{2}$ growth low speed par matter karta hai