Visual walkthrough — Area rule — Whitcomb's rule for transonic drag reduction

3.1.26 · D2 · Physics › Compressible Flow & Aerodynamics › Area rule — Whitcomb's rule for transonic drag reduction

Hum parent note, Slender-body theory, Shock waves and wave drag, aur Sears–Haack body ko finish line ke roop mein use karte hain.

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Step 0 — Speed ka measure jo humein chahiye: Mach number aur Mach angle

KYA HAI. Neeche ki har baat is par depend karti hai ki hum sound ki speed ke compare mein kitni tez ud rahe hain. Mach number define karo:

$$M_{\infty }=\frac{U_{\infty }}{a_{\infty }},$$

• $U_{\infty }$ ::: aircraft ki free-stream speed, yani hawa mein uski chaal.
• $a_{\infty }$ ::: uss undisturbed hawa mein sound ki speed jo door kahin hai.
• $M_{\infty }$ ::: unka ratio — hum kitne "sound-speeds" ki speed se ja rahe hain. $M_{\infty }=1$ bilkul sound ki speed hai; $M_{\infty }>1$ supersonic hai; $M_{\infty }\approx 1$ transonic hai (dekho Transonic flow).

KYUN. Jab $M_{\infty }>1$ hota hai, koi bhi disturbance aircraft se aage nahi nikal sakti, isliye har chote source ka influence ek peeche ki taraf swept cone — Mach cone — ke andar trapped ho jaata hai. Iska half-angle, Mach angle $\mu$, simple geometry se milta hai (dekho Mach angle and Mach cone):

$$\mu =\arcsin \left(\frac{1}{M_{\infty }}\right),\tan \mu =\frac{1}{\sqrt{M_{\infty }^2-1}}.$$

• $\mu$ ::: Mach cone ka tilt — $M_{\infty }=1$ se thoda upar steep (nearly $90^{\circ }$), aur $M_{\infty }$ badhne ke saath flat hota jaata hai.
• $\sqrt{M_{\infty }^2-1}$ ::: yeh crucial combination measure karta hai ki "hum kitne supersonic hain"; yeh neeche har jagah aata hai. Dhyan do ki yeh $M_{\infty }\to 1$ par zero ho jaata hai — ek warning flag jis par hum Step 6 mein wapas aate hain.

PICTURE. Ek source ke peeche Mach cone, half-angle $\mu$ ke saath, aur yeh kaise flat hota jaata hai jab $M_{\infty }$ badhta hai.

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Step 1 — "Cross-section" ka matlab asal mein kya hai

KYA HAI. Aircraft lo (ya koi bhi lamba patla body). Ise us arrow ke along point karo jise hum flight axis kehte hain. Ab us axis ke neeche glass ki ek flat sheet slide karo, jo hamesha uspe perpendicular rahe. Jahan glass body ko slice karta hai, wahan ek flat 2-D shape cut hoti hai — ek disc, oval, ya wing-plus-fuselage blob. Uss slice ka area ek number hota hai. Glass ki position ko $x$ kaho (nose se distance), aur slice area ko $A(x)$ kaho.

KYUN. Area rule mein jo bhi claim hai, woh sab is ek function $A(x)$ ke baare mein hai. Yeh kehne se pehle ki "drag is baat par depend karta hai ki $A$ kaise change hota hai," humein yeh crystal clear hona chahiye ki $A(x)$ sirf itna hai: ek distance $x$ chuno, slice karo, jo area cut hua woh measure karo.

PICTURE. Neeche red slice ek value of $x$ par hai; number $A(x)$ wahi red area hai. Slice ko slide karo aur tum poore body ko trace kar lete ho.

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Step 2 — Hawa sirf $A(x)$ "dekhti" hai (slender-body idea)

KYA HAI. Socho ki body thin hai — jo kitni bhi moti ho, usse kaafi zyada lambi. Jab hawa past flow karti hai, hawa ka har thin ring body ke around slide hone ke liye sirf utna hi baahir dhakela jaata hai jitna zaruri ho. Kitna baahir push? Exactly itna ki body ki badhti thickness ke liye jagah ban sake.

KYUN. Yahan woh key claim hai jise hum precise karenge: hawa ka ek ring yeh nahi jaanta ki jis width ke around use flow karna hai woh wing se aayi hai ya fuselage se. Woh sirf uss station par total area feel karta hai jise use dodge karna hai. Isliye jinke $A(x)$ same hai woh do aircraft hawa ko identically dhakelte hain — aur isliye same waves banate hain. Ek sentence mein Slender-body theory yahi hai.

PICTURE. Neeche: same $x$ par do bahut alag bodies (ek fat tube, aur ek thin tube with wings) slice ki gayi hain. Agar unke red slice areas equal hain, toh hawa us station par same amount baahir push hoti hai.

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Step 3 — "Make room" ko ek source mein badalna ($f(x)$ define karna)

KYA HAI. "Body apne liye jagah banati hai" ko model karne ke liye, hum solid body ko tiny air-emitters ki ek line se replace karte hain jo axis ke neeche run karti hai. Har emitter, jise source kehte hain, hawa ko baahir blow karta hai. Strength $f(x)$ ka source axis ki per unit length, per second hawa ka ek volume baahir blow karta hai.

KYUN yeh tool. Hum ek weird solid shape ke around flow easily solve nahi kar sakte. Lekin ek point source se flow door — yeh ek simplest flows mein se ek hai (hawa baahri taraf spread hoti hai, symmetric). Axis ke along sources stack karke aur unki strengths choose karke, hum exactly woh outward-push reproduce kar sakte hain jo body demand karti hai — bina hard solid-boundary problem solve kiye. Yeh substitution Slender-body theory ki puri trick hai.

Source kitna strong hona chahiye? Length $dx$ ke ek slice par, body ka area $A$ se $A+dA$ ho jaata hai. Oncoming hawa speed $U_{\infty }$ (free-stream speed) par aati hai. Ek second mein, length $U_{\infty }$ ka hawa ka ek column pass hota hai; extra volume jitna use accommodate karne ke liye side mein dhakela jaana chahiye woh hai $U_{\infty }dA$. Per unit length yeh hai:

$$f(x)=U_{\infty }\frac{dA}{dx}.$$

Term by term:

• $f(x)$ ::: station $x$ par source strength — per second per unit axis-length emit ki gayi hawa ka volume.
• $U_{\infty }$ ::: oncoming hawa ki free-stream speed (subscript $\infty$ = "door, undisturbed").
• $\frac{dA}{dx}$ ::: distance ke saath area ka rate of change — "body yahan kitni tezi se moti (ya patli) ho rahi hai."

PICTURE. Jahan body moti hoti hai, $dA/dx>0$ → sources baahir blow karte hain (red, positive). Jahan slim hoti hai, $dA/dx<0$ → sources andar suck karte hain (negative source, "sink"). Jahan area constant hai, $f=0$: bilkul koi disturbance nahi.

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Step 4 — Waves momentum le jaati hain, aur lost momentum = drag

KYA HAI. Supersonic speed par har source hawa ko smoothly baahir nahi push karta — uska influence Step 0 ke Mach cones ke along trap ho jaata hai aur pressure waves / shocks ke roop mein radiate karta hai (dekho Shock waves and wave drag). Yeh waves body se peeche stream karti hain aur kabhi wapas nahi aatin. Woh momentum le jaati hain.

KYUN. Newton ke according drag woh rate hai jis par body hawa ko forward momentum kho deti hai. Agar waves peeche-ki-taraf momentum "debt" le jaati hain, toh body ko usse supply karna hota hai — aur wahi supply wave drag $D_w$ hai. Toh: zyada strong, zyada abrupt sources → zyada strong waves → zyada momentum bleed off → zyada drag.

PICTURE. Ek smoothly changing body (gentle sources) weak, spread-out waves shed karti hai. Ek body jisme achanak fattening hai (ek violent source spike) ek strong shock shed karta hai. Same hawa, bahut alag momentum loss.

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Step 5 — Abruptness curvature hai: $A^{\prime \prime }(x)$ ka entry

KYA HAI. "Source strength kitni abruptly change hoti hai" woh $f(x)$ ka rate of change hai. Kyunki $f=U_{\infty }dA/dx$, uska rate of change $U_{\infty }\frac{d}{dx}\left(\frac{dA}{dx}\right)$ hai. Beech wali cheez, "slope ka slope," second derivative hai, likha jaata hai:

$$A^{\prime \prime }(x)=\frac{d}{dx}\left(\frac{dA}{dx}\right)=\frac{d^{2}A}{d{x}^2}.$$

• $A^{\prime \prime }(x)$ ::: area curve ki curvature — slope $dA/dx$ khud kitni tezi se change ho raha hai.

KYUN yeh tool. Waves source strength mein changes se launch hoti hain, source strength se nahi. Ek body jo constant rate par moti hoti hai ($dA/dx$ constant, toh $A^{\prime \prime }=0$) koi nayi disturbance launch nahi karti — sab sources equal hain, downstream kuch change nahi hota. Sirf wahan jahan fattening-rate change hoti hai — jahan $A^{\prime \prime }\ne 0$ — fresh waves throw off hoti hain. Isliye second derivative, pehla nahi, drag driver hai.

PICTURE. Same body ke liye teen stacked curves: upar $A(x)$, beech mein uska slope $A^{\prime }(x)$, neeche uski curvature $A^{\prime \prime }(x)$ (red). Dhyan do ki $A(x)$ mein ek kink (ek wing bolt karne se) $A^{\prime }$ mein step aur $A^{\prime \prime }$ mein ek enormous spike ban jaata hai.

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Step 6 — Waves ko add karna: von Kármán–Sears integral

KYA HAI. Position $x_1$ par axis ka har tiny piece $\propto A^{\prime \prime }(x_1)$ strength ki wave launch karta hai. Woh wave apne Mach cone ke along peeche sweep karti hai aur $x_2$ par kisi doosre piece ki wave se interact karti hai (strength $\propto A^{\prime \prime }(x_2)$). Unka combined momentum loss dono strengths par aur do pieces kitne door hain $|x_1-x_2|$ par depend karta hai. Har pair of pieces ko sum karne par supersonic slender-body wave drag milta hai:

$$D_w=-\frac{{\rho }_{\infty }{U}_{\infty }^2}{4\pi }{\int }_0^{\ell }{\int }_0^{\ell }{A}^{\prime \prime }(x_1)A^{\prime \prime }(x_2)\ln \frac{|x_1-x_2|}{\ell }d{x}_{1}d{x}_2.$$

Symbol by symbol padhte hain:

• $D_w$ ::: wave drag jo hum compute kar rahe hain (ek force).
• ${\rho }_{\infty }$ ::: free-stream air density; $U_{\infty }$ ::: free-stream speed. Saath mein ${\rho }_{\infty }{U}_{\infty }^2$ dynamic pressure ka scale set karta hai — flow kitna "punchy" hai.
• $\frac{1}{4\pi }$ ::: flow mein spreading se aaya ek geometric constant (mat daro — yeh sirf bookkeeping hai).
• ${\int }_0^{\ell }{\int }_0^{\ell }\cdots d{x}_{1}d{x}_2$ ::: length ke along har pair of stations $(x_1,x_2)$ ka double sum — pehla integral $x_1$ choose karta hai, doosra $x_2$, aur hum sab combinations add karte hain.
• $A^{\prime \prime }(x_1)A^{\prime \prime }(x_2)$ ::: station $x_1$ ki wave ki strength times station $x_2$ ki wave ki strength — curvature ki ek self-correlation.
• $\ln \frac{|x_1-x_2|}{\ell }$ ::: separation divided by body length $\ell$ ka natural logarithm. $\ell$ se divide karne par argument ek pure number (dimensionless) ban jaata hai — tum "kitne metres" ka log nahi le sakte. Reference length ka choice $D_w$ ko sirf $(\int A^{\prime \prime }dx{)}^2=(A^{\prime }(\ell )-A^{\prime }(0){)}^2$ ke proportional ek term se shift karta hai, jo ek closed body (pointed nose aur tail) ke liye zero ho jaata hai, isliye answer well-defined hai.

KYUN exactly yeh form. Do sources linearized supersonic influence (ek Green's function) ke through interact karte hain. Mach cone geometry factor $\sqrt{M_{\infty }^2-1}$ ke through enter karta hai: cone ke along coordinates align karne ke baad yeh factor effective streamwise length mein absorb ho jaata hai, aur leading term ke liye yeh integral ki shape se drop out ho jaata hai — separation ka logarithm chhod ke. Isliye classic form $M_{\infty }$-free lagti hai jab geometry Mach-cone geometry hai.

PICTURE. Double integral ko square $[0,\ell ]\times [0,\ell ]$ par ek grid ke roop mein draw kiya gaya hai. Har cell ek pair $(x_1,x_2)$ hai; hum ise $A^{\prime \prime }(x_1)A^{\prime \prime }(x_2)$ se colour karte hain. Kink wali body bright hot cells (red) light up karti hai → bada total → bada drag. Smooth body sab jagah cool rehti hai.

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Step 7 — Edge cases: constant, linear, aur kinked bodies

KYA HAI. Chaliye formula ko extreme cases par test karte hain taaki koi bhi scenario reader ko surprise na kare.

Case (a) — constant area (ek plain cylinder, $A(x)=\text{const}$). Tab $dA/dx=0$ aur $A^{\prime \prime }=0$ har jagah. Integrand hai $0\cdot 0$ → $D_w=0$. Ek parallel tube (is idealization mein) koi wave drag nahi banata: kuch bhi change nahi ho raha.

Case (b) — linearly growing area ($A(x)$ ek straight ramp). Tab $dA/dx=$ const, isliye $A^{\prime \prime }=0$ interior mein. Lekin do ends par slope achanak on aur off hota hai — kinks — $A^{\prime \prime }$ mein spikes produce karte hain. Drag poori tarah nose aur tail corners se aata hai. Lesson: corners round karo.

Case (c) — bolted-on wing (ek kink in $A$). Jaise Step 5 mein, kink → $A^{\prime \prime }$ mein ek delta-spike. Grid picture mein (Step 6) woh spike ek poori hot row aur column light up kar deta hai → large drag. Yahi villain hai jiske liye Coke-bottle fuselage exist karta hai.

KYUN teen dikhaye. Yeh poore landscape ko map karte hain: $A^{\prime \prime }=0$ free hai, corners thoda cost karte hain, kinks bahut cost karte hain. Design goal khud likh jaata hai: $A^{\prime \prime }$ spikes khatam karo; curvature ko dheere dheere spread karo.

PICTURE. Teen area curves side by side unke $A^{\prime \prime }$ ke saath neeche (red). Dekho $A^{\prime \prime }$ flat (free) se → do end-spikes (sasta) → ek tall interior spike (mehnga) tak jaata hai.

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Step 8 — Winner: smoothest possible curve = Sears–Haack

KYA HAI. Optimization question poocho: fixed length $\ell$ aur fixed volume $V={\int }_0^{\ell }A(x)dx$ (Step 1 se) wale sab area curves $A(x)$ mein se, kaun sa ek double integral ko smallest banata hai? Jawab (ek calculus-of-variations result) Sears–Haack body hai:

$$A(x)=A_{\max }[4\xi (1-\xi ){]}^{3/2},\xi =\frac{x}{\ell }\in [0,1].$$

• $\xi$ ::: fractional position, nose par $0$, tail par $1$.
• $A_{\max }$ ::: body par kahin bhi sabse bada cross-sectional area, $A_{\max }={\max }_{x}A(x)$. Yeh beech mein hota hai ($\xi =\frac{1}{2}$), jahan bracket $1$ ke barabar hota hai.
• $4\xi (1-\xi )$ ::: ek smooth hump, dono ends par zero, beech mein peak karta hai.
• power $3/2$ ::: nose/tail ko zero tak gently sharpen karta hai, taaki $A^{\prime \prime }$ kabhi spike na kare.

Fixed volume ke liye uska minimum wave drag hai:

$$D_w=\frac{128q_{\infty }{V}^2}{\pi {\ell }^4},q_{\infty }=\frac{1}{2}{\rho }_{\infty }{U}_{\infty }^2.$$

• $q_{\infty }$ ::: dynamic pressure, flow ka "push."
• $V={\int }_0^{\ell }A(x)dx$ ::: jo fixed total volume carry kiya ja raha hai.
• $V^2/{\ell }^4$ ::: scaling — zyada volume hurts (squared); zyada length enormously help karta hai (fourth power).

KYUN. Yeh concrete target hai jis par Whitcomb tujhe apna total $A(x)$ aim karne ko kehta hai. Fuselage wings ke paas precisely isliye pinch hota hai taaki fuselage+wing+tail ka sum ek lumpy ki jagah is gentle hump ko trace kare.

PICTURE. Sears–Haack area curve (red) ek lumpy "components-added" curve ke upar rakhi gayi hai. Same volume, same length — lekin smooth wali har jagah chhota $A^{\prime \prime }$ rakhti hai.

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Ek-picture summary

Is page par jo kuch bhi hai woh compressed hai: ek speed choose karo (Mach $M_{\infty }$) → body ko slice karo → $A(x)$ milta hai (uska volume $V=\int Adx$) → uska slope $A^{\prime }$ → uski curvature $A^{\prime \prime }$ → $A^{\prime \prime }$ ke pairs ko double integral mein daalo → $D_w$ nikalta hai. Smooth $A(x)$ ⇒ tiny $A^{\prime \prime }$ ⇒ tiny drag; kink ⇒ spike ⇒ big drag.

Recall Feynman retelling — seedhi bhaasha mein wapas bolo

Pehle fix karo tum kitni tez ud rahe ho: Mach number $M_{\infty }$, tumhari speed sound ki speed par. $M_{\infty }=1$ se upar jo bhi disturb karo woh tumse aage nahi nikal sakta, isliye har little push ek backward cone — Mach cone — ke andar trap ho jaata hai, Mach angle $\mu$ par tila hua. Ab plane ko ek arrow ke along point karo. Glass ki ek sheet arrow ke neeche slide karo; jahan yeh plane ko cut kare, area measure karo. Isse ek curve milti hai: area versus distance, $A(x)$; sab slices add karo aur volume milta hai $V=\int Adx$. Hawa jo rush karti hai use baahri taraf dhakela jaana chahiye exactly utna ki body ke mote hone ke liye jagah bane — isliye main body ko chhote blowers ki ek row se replace karta hoon jinka strength yeh hai ki area kitni tezi se badh raha hai. Sab ek jitna blow karne wale blowers koi waves nahi banate; tabhi waves throw off hoti hain jab blowing rate change hoti hai. "Blowing rate kitni tezi se change hoti hai" meri area curve ki curvature hai, $A^{\prime \prime }$. Woh waves apne Mach cones ke along peeche race karti hain aur kabhi nahi lauttein, momentum le jaati hain — aur lost momentum exactly drag hai. Total karne ke liye main har point ki wave ko har doosre point ki wave se interact karaata hoon: yahi double integral hai, unke scaled separation ke log se weighted (length $\ell$ se scale kiya taaki log ke andar plain number ho). Yeh $A^{\prime \prime }$ ke alawa kisi par depend nahi karta, jo $A(x)$ ke alawa kisi se nahi aata — isliye do bilkul alag dikhne wale aircraft jinka area curve same ho unka wave drag same hoga. Yeh clean formula supersonic slender-body limit hai; bilkul $M_{\infty }\approx 1$ par linear theory strain karti hai, lekin design lesson survive karta hai: kam drag chahiye, smooth area curve chahiye, aur sabse smooth wali Sears–Haack shape hai. Isliye designers fuselage ko wahan pinch karte hain jahan wings area add karti hain. Aur kyunki drag $1/{\ell }^4$ scale karta hai, poori cheez ko lambi banana sabse bada win hai.

Recall Quick self-test

Ek constant-area cylinder ka yahaan zero wave drag kyun hai? ::: Kyunki $A^{\prime \prime }=0$ har jagah hai, isliye integral mein har term zero hai — flow mein kuch bhi change nahi ho raha. $A^{\prime \prime }$ kyun, $A$ ya $A^{\prime }$ kyun nahi? ::: Waves source strength mein changes se launch hoti hain; source strength $\propto A^{\prime }$ hai, isliye uska change $\propto A^{\prime \prime }$ hai — curvature. Double integral physically kya represent karta hai? ::: Har station ki wave har doosre station ki wave se interact karti hai, unke (length-scaled) separation ke $\ln$ se weighted — curvature ki ek self-correlation. Sears–Haack drag formula mein $V$ kya hai? ::: Body ka total volume, $V={\int }_0^{\ell }A(x)dx$. $\ln$ ke andar $\ell$ se divide kyun karte hain? ::: Logarithm ko dimensionless argument chahiye; $|x_1-x_2|/\ell$ ek pure number hai, aur reference length ka choice closed body ke liye cancel ho jaata hai. Kya boxed integral $M_{\infty }=1$ par exact hai? ::: Nahi — yeh linearized supersonic ($M_{\infty }>1$) slender-body result hai; transonic $M_{\infty }\approx 1$ par linear theory break down karti hai, haalaanki smooth-$A(x)$ design rule phir bhi hold karti hai.

Related: Transonic flow, Prandtl–Glauert and compressibility corrections, Mach angle and Mach cone.