Worked examples — Area rule — Whitcomb's rule for transonic drag reduction

3.1.26 · D3 · Physics › Compressible Flow & Aerodynamics › Area rule — Whitcomb's rule for transonic drag reduction

Yeh page area rule ka stress-test hai. Parent note Area Rule ne aapko machinery di thi. Ab hum ise har kone mein push karte hain: smooth bodies, kinked bodies, degenerate (zero-length ya zero-area) bodies, perpendicular-cut $M\approx 1$ case aur tilted Mach-plane case, aur ek real aircraft. Agar koi scenario ho sakta hai, woh neeche worked out hai.

Shuru karne se pehle, ek reminder plain words mein, kyunki har example isi par tika hai:

Hume do aur quantities chahiye jo parent note ne use ki thi lekin is page ko plain words mein recall karni chahiye, kyunki Examples 4, 5 aur 7 inhi par depend karte hain:

Recall Parent se do quantities:

$q_{\infty }$ aur $V$ Dynamic pressure ::: $q_{\infty }=\frac{1}{2}{\rho }_{\infty }{U}_{\infty }^2$ — "push per unit area" jo moving air carry karti hai, $\text{Pa}={\text{N/m}}^2$ mein. Yahan ${\rho }_{\infty }$ free-stream air density hai aur $U_{\infty }$ flight speed. Body volume ::: $V={\int }_0^{\ell }A(x)dx$ — literally saari slice-areas ka sum length ke along, ${\text{m}}^3$ mein. Yeh woh hai jo body ko air mein "room banana" padta hai.

Neeche sab kuch inhi mein se ek hai, ya parent ka master integral:

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

Dynamic pressure ke saath likhne par, leading term $\frac{{\rho }_{\infty }{U}_{\infty }^2}{4\pi }=\frac{q_{\infty }}{2\pi }$ hota hai, kyunki $q_{\infty }=\frac{1}{2}{\rho }_{\infty }{U}_{\infty }^2$.

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The scenario matrix

"Har scenario" jo yeh topic aap par throw kar sakta hai, unhe is grid mein bhara hua socho. Baad ke har example ko us cell ke saath tag kiya gaya hai jise woh cover karta hai.

Cell	Kya cheez ise distinct banati hai	Covered by
C1 Smooth body, har jagah positive curvature	textbook Sears–Haack, $A^{\prime \prime }$ finite aur smooth	Ex 1
C2 $A(x)$ mein kink (slope jump)	$A^{\prime \prime }$ ek spike (impulse) ban jaata hai → drag blow up ho jaata hai	Ex 2
C3 Superposition: fuselage + wing bump	classic "Coke-bottle" fix, $A_{\text{tot}}=A_{\text{fus}}+A_{\text{wing}}$	Ex 3
C4 Degenerate: zero volume / zero length	limiting inputs — kya formula sane rehta hai?	Ex 4
C5 Scaling limit: $\ell \to$ large at fixed $V$	$D_w\propto V^2/{\ell }^4$ ka limiting behaviour	Ex 5
C6 Speed regime split: $M\approx 1$ vs $M>1$	perpendicular cut vs tilted Mach plane $\mu =\arcsin (1/M)$	Ex 6
C7 Real-world word problem	YF-102 → F-102A, percentage drag drop	Ex 7
C8 Exam twist: sign/units trap	kyun $D_w$ integral se positive aata hai minus ke bawajood	Ex 8

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Example 1 — C1: smooth Sears–Haack slice (positive-curvature baseline)

Forecast: Compute karne se pehle guess karo — kya maximum beech mein hai? Kya $A(0.25)$, $A_{\max }$ ke aadhe ke kareeb hai ya bahut kam?

Steps.

1. Maximum dhundho. Bracket $4\xi (1-\xi )$ $\xi$ mein ek downward parabola hai, jo peak karta hai jahan $\xi (1-\xi )$ sabse bada ho, yaani $\xi =\frac{1}{2}$. Wahan $4\cdot \frac{1}{2}\cdot \frac{1}{2}=1$, to $A=A_{\max }\cdot 1^{3/2}=1{\text{m}}^2$. Yeh step kyun? Area rule ek smooth, single-humped curve chahta hai; peak location batata hai ki sabse chodi fuselage kahan expect karo, aur Sears–Haack optimum ke liye yeh mid-body mein hona chahiye.

2. $\xi =0.25$ evaluate karo. $4(0.25)(0.75)=0.75$, aur $0.75^{3/2}=0.6495$. To $A=0.6495{\text{m}}^2$. Yeh step kyun? Dikhata hai ki curve triangle nahi hai — ek quarter andar jaane par hi aap peak area ke $65\%$ par ho. Gentle, koi kink nahi.

3. Ends check karo. $\xi =0$ par: $4\cdot 0\cdot 1=0$, $0^{3/2}=0$ → $A=0$. Wahi $\xi =1$ par. Nose aur tail point tak taper ho jaate hain. Yeh step kyun? Yeh hamara pehla taste hai degenerate input ka — ends zero area ke legitimate cross-sections hain, aur formula unhe cleanly handle karta hai.

Verify: Units: $A_{\max }$ ${\text{m}}^2$ mein, bracket dimensionless hai, to $A$ ${\text{m}}^2$ mein hai. Peak value $1{\text{m}}^2$, $A_{\max }$ se exactly match karta hai ✓. Symmetry: $A(\xi )=A(1-\xi )$, to $A(0.75)=A(0.25)=0.6495{\text{m}}^2$ — ek fore-aft symmetric body, jaisa Sears–Haack hona chahiye ✓.

Neeche ka figure yeh area law plot karta hai. Blue curve dekho: yeh woh smooth, single hump hai jo area rule ko pasand hai — kahi koi corner nahi. Yellow dot step 1 ka mid-body peak mark karta hai; green dot step 2 ka $A(0.25)=0.6495{\text{m}}^2$ hai (already two-thirds upar, straight-line ramp nahi); do red dots step 3 ke degenerate zero-area nose aur tail hain. Left se right padho aur aap literally aircraft ke neeche ruler chalate ja rahe ho.

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Example 2 — C2: ek kink $A^{\prime \prime }$ ko spike mein badal deta hai

Forecast: Guess karo — kya $A^{\prime \prime }$ join par sirf "bada" hai, ya infinite hai?

Steps.

1. Pehla derivative $A^{\prime }$. $x<a$ ke liye, $A^{\prime }=k$ (constant slope). $x>a$ ke liye, $A^{\prime }=0$. To $A^{\prime }$, $x=a$ par $k$ se $0$ tak jump karta hai — ek step. Yeh step kyun? $A^{\prime }=dA/dx$ woh rate hai jis par slice grow karti hai. Cone constant rate se grow karta hai; cylinder bilkul nahi. Abrupt handover hi "kink" hai.

2. Doosra derivative $A^{\prime \prime }$. Ek step ka derivative ek Dirac impulse (delta spike $\delta$) hota hai. To $A^{\prime \prime }(x)=-k\delta (x-a)$: har jagah zero, siwan join par infinitely tall spike ke. Yeh step kyun? Parent ki chain kehti hai strong $A^{\prime \prime }$ ⇒ strong shock ⇒ drag. Impulse sabse violent possible $A^{\prime \prime }$ hai.

3. Drag integral mein daalo. $A^{\prime \prime }\propto \delta (x-a)$ ke saath, double integral $\ln |x_1-x_2|$ kernel se $x_1=x_2=a$ par takraata hai jahan $\ln 0=-\infty$. Self-correlation diverge ho jaata hai — linearized model literally unbounded drag predict karta hai slope discontinuity se. Yeh step kyun? Yeh woh mathematical scream hai jo kehta hai "apna area curve kabhi kink mat karo." Real air infinite drag produce nahi kar sakti, lekin woh ek strong shock produce karti hai, jo bilkul wahi hai jo hum avoid karna chahte hain.

Verify: Sanity via smoothing. Corner ko small length $\epsilon$ par round karo: to $A^{\prime }$, $\epsilon$ par ramp down karta hai jump ke bajaye, to $A^{\prime \prime }\sim k/\epsilon$ (bada lekin finite). Jaise $\epsilon \to 0$, $A^{\prime \prime }\to \infty$ — delta limit confirm karta hai aur confirm karta hai ki drag unbounded grow karti hai jaise corner sharpen hota hai ✓. Units check: $[A^{\prime }]={\text{m}}^2/\text{m}=\text{m}$, $[A^{\prime \prime }]={\text{m}}^2/{\text{m}}^2=$ dimensionless density, consistent hai $k$ ke saath jo $\text{m}$ units rakhta hai ✓.

Neeche ke teen stacked panels yeh story top-to-bottom batate hain. Top (blue): $A(x)$ ramp up hota hai phir flatten ho jaata hai — yellow arrow corner ko point karta hai. Middle (green): iska slope $A^{\prime }$ abruptly $k$ se $0$ tak jump karta hai (step 1). Bottom (red): second derivative $A^{\prime \prime }$ har jagah zero hai siwan tall red arrow ke — step 2 ka delta spike. Teeno panels mein $x=a$ par ek vertical line trace karo aur dekho kaise $A$ mein barely-visible corner $A^{\prime \prime }$ mein ek infinite spike ban jaata hai.

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Example 3 — C3: superposition, Coke-bottle cancellation

Forecast: Guess karo — kya fuselage ko exactly wing bump ke barabar, zyada, ya kam dip karna chahiye?

Steps.

1. Total likho. $A_{\text{tot}}(x)=A_{\text{fus}}(x)+A_{\text{wing}}(x)$. Area rule sirf is sum ki care karta hai, split ki nahi. Yeh step kyun? Yeh ek line mein poora theorem hai — identical $A_{\text{tot}}$ wale do shapes ka identical slender-body wave drag hota hai.

2. Smoothness demand karo. Wing station par koi bump na ho, iske liye require karo ki $A_{\text{tot}}$ wahan surrounding tube value $1.00{\text{m}}^2$ ke barabar ho: $A_{\text{fus}}^{\text{pinched}}+0.30=1.00\Rightarrow A_{\text{fus}}^{\text{pinched}}=0.70{\text{m}}^2.$ Yeh step kyun? Fuselage ko exactly wing ka added area shed karna hai — $0.30{\text{m}}^2$ ka dip. Hump cancel karo, uski curvature $A^{\prime \prime }$ cancel karo, us drag ko cancel karo jo usne cause ki.

3. Pinch interpret karo. Fuselage cross-section $1.00$ se $0.70{\text{m}}^2$ tak girta hai, yaani us station par $30\%$ area reduction — visible "wasp waist." Yeh step kyun? Abstract cancellation ko physical Coke-bottle shape se jodata hai jo parent note describe karta hai.

Verify: $A_{\text{tot}}=0.70+0.30=1.00{\text{m}}^2$ — wing region mein flat, to $A_{\text{tot}}^{\prime }=0$ aur $A_{\text{tot}}^{\prime \prime }=0$ wahan. Koi curvature spike nahi, koi wing-induced shock nahi ✓. Fractional pinch $0.30/1.00=30\%$, era ke "~25–30% drag reduction" numbers se order of magnitude mein match karta hai ✓.

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Example 4 — C4: degenerate inputs (zero volume, zero length)

Forecast: Guess karo — kaun sa degenerate case zero drag deta hai aur kaun sa blow up karta hai?

Steps.

0. $128/\pi$ kahan se aata hai? Yeh magic nahi hai — yeh woh hai jo aapko parent ne flag kiya hua calculus-of-variations minimization karne ke baad milta hai. Sears–Haack area law $A(x)=A_{\max }[4\xi (1-\xi ){]}^{3/2}$ ko von Kármán–Sears double integral mein feed karo aur evaluate karo. Do pieces combine hote hain: (i) us specific shape ka volume $V=\frac{3\pi }{16}{A}_{\max }\ell$ nikalta hai, to $A_{\max }=\frac{16V}{3\pi \ell }$; (ii) is smooth curve ke liye $\iint A^{\prime \prime }{A}^{\prime \prime }\ln |\cdot |$ integral ek pure number times $A_{\max }^2/{\ell }^2$ evaluate hota hai. $A_{\max }$ ko $V$ ke terms mein substitute karo aur leading $\frac{\rho U^2}{4\pi }=\frac{q_{\infty }}{2\pi }$ collect karo, exactly $\frac{128q_{\infty }{V}^2}{\pi {\ell }^4}$ milta hai. To $128/\pi$ is optimal shape ka fingerprint hai, kuch aur nahi. Yeh step kyun? Learners ko dekhna chahiye ki constant minimization se earned hota hai, upar se nahi diya jaata; har doosra body $q_{\infty }{V}^2/{\ell }^4$ ke aage bada number deta hai.

1. Zero-volume body ($V=0$). Tab $D_w=\frac{128q_{\infty }\cdot 0}{\pi {\ell }^4}=0$. Yeh step kyun? Zero volume wala body ek needle/streamline hai jiska koi cross-section nahi jo air ko dhakele. Koi displaced volume nahi ⇒ koi source strength nahi $f=U_{\infty }{A}^{\prime }=0$ ⇒ koi waves nahi ⇒ koi wave drag nahi. Formula agree karta hai.

2. Vanishing length ($\ell \to 0$, $V$ fixed). $D_w\propto \frac{1}{{\ell }^4}\to \infty$. Yeh step kyun? Ek fixed volume ko zero length mein cramming karna area ko ek tiny distance par spike hone par force karta hai — enormous $A^{\prime \prime }$. Formula correctly predict karta hai unbounded wave drag: aap ek fat body ko infinitely short nahi bana sakte aur slender nahi reh sakte.

3. Validity ki boundary. Case (b) exactly wahan hai jahan slender-body theory khatam hoti hai: isne assume kiya tha thickness $\ll$ length. $1/{\ell }^4$ growth woh theory hai jo politely apni apni edge par blow up karti hai. Yeh step kyun? Har model ka ek domain hota hai; use name karna "har scenario cover karne" ka hissa hai.

Verify: $D_w=128q_{\infty }{V}^2/(\pi {\ell }^4)$ ka dimensional check: $[q_{\infty }]=\text{Pa}={\text{N/m}}^2$, $[V^2]={\text{m}}^6$, $[{\ell }^4]={\text{m}}^4$, to $[D_w]=\frac{\text{N}}{{\text{m}}^2}\cdot \frac{{\text{m}}^6}{{\text{m}}^4}=\text{N}$ — ek force ✓. Monotonic: $D_w$ badhta hai jaise $V$ badhta hai ya $\ell$ ghatta hai, dono physically correct hain ✓.

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Example 5 — C5: length-scaling limit

Forecast: $2\times$ length case ke liye drag ratio guess karo, aur yeh ki $1\%$ target ke liye roughly $2\times$, $3\times$, ya $4\times$ lamba body chahiye.

Steps.

1. Double-length drag ratio. $\frac{D_{w,2}}{D_{w,1}}=\frac{{\ell }_1^4}{{\ell }_2^4}={\left(\frac{1}{2}\right)}^4=\frac{1}{16}=0.0625$. Yeh step kyun? $D_w\propto V^2/{\ell }^4$ mein $V,q_{\infty }$ fixed hain to sirf ${\ell }^{-4}$ factor bachta hai. Length double ⇒ $2^4=16$-fold drop.

2. $1\%$ target ke liye solve karo. Chahiye $\frac{{\ell }_1^4}{{\ell }_3^4}=0.01$, to ${\left(\frac{{\ell }_3}{{\ell }_1}\right)}^4=100$, milta hai $\frac{{\ell }_3}{{\ell }_1}=100^{1/4}=\sqrt{10}\approx 3.162$. Yeh step kyun? Steep payoff aur uski limit dikhata hai: slenderness huge drag cuts deta hai, lekin fourth-power ka matlab hai sirf ~$3.16\times$ length chahiye $100\times$ drag cut ke liye — geometry length ko dramatically reward karta hai.

3. Limit padhna. Jaise $\ell \to \infty$ fixed $V$ par, $D_w\to 0$: ideal slender needle. Real limits (weight, structure, friction) rokate hain, lekin wave drag akela khatam ho jaata hai. Yeh step kyun? Limiting behaviour cell explicitly name karta hai.

Verify: $2^4=16$ ⇒ ratio $1/16=0.0625$ ✓. $(\sqrt{10}{)}^4=10^2=100$ ⇒ drag ratio $1/100=0.01$ ✓. Dono $D_w\propto {\ell }^{-4}$ ke consistent hain.

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Example 6 — C6: perpendicular cut vs tilted Mach-plane cut

Forecast: $M=2$ par $\mu$ guess karo — $45^{\circ }$ se bada ya chhota? Aur kya tilt equivalent body ko $x$ mein lamba ya chhota dikhata hai?

Steps.

1. Angle kyun hota hai? Transonic flow mein $M\approx 1$ par, disturbances almost straight sideways spread hote hain, to wave jo "footprint" feel karta hai woh ordinary perpendicular cross-section hai. $M=1$ ke upar, har disturbance ek Mach cone mein sweep back ho jaata hai; jo plane wave actually "dekhta" hai woh tilted hai. Yeh step kyun? Tool switch justify karta hai — hum physics nahi badal rahe, hum cutting plane badal rahe hain taaki match kare ki signals kaise propagate karte hain.

2. $M=2$ par $\mu$ compute karo. $\mu =\arcsin (1/M)=\arcsin (0.5)=30^{\circ }$. Yeh step kyun? $\arcsin$ jawab deta hai "kaun sa angle is sine ka hai?" Yahan $\sin \mu =1/M=1/2$, aur woh angle jiska sine $\frac{1}{2}$ hai $30^{\circ }$ hai. $45^{\circ }$ se chhota, to high $M$ par cone narrower hai aur cut flow direction ki taraf zyada tilted hai.

3. Area law par effect. Cut ko tilt karna aur roll angle par average karna area contributions ko $x$ ke along stretch aur smear karta hai (ek station ka area aage/peeche project hota hai). Equivalent body lamba aur uska area curve smoother hota hai — yahi reason hai ki supersonic designs ko choose kiye hue Mach number ke liye area-rule kiya jaata hai, sirf $M=1$ ke liye nahi. Yeh step kyun? Geometric tilt ko ek alag $A(x)$ se connect karta hai, hence alag $D_w$ — ek alag supersonic rule ka poora reason.

Verify: $M\to 1^{+}$ deta hai $\mu =\arcsin (1)=90^{\circ }$ — Mach plane perpendicular ho jaata hai, smoothly case (a) recover karta hai ✓. $M\to \infty$ deta hai $\mu =\arcsin (0)=0^{\circ }$ — cone axis par flatten ho jaata hai ✓. $M=2$ par, $\sin (30^{\circ })=0.5=1/M$ ✓.

Neeche ka curve Mach angle $\mu$ ko Mach number $M$ ke against plot karta hai. Blue curve ko left se follow karo: $M=1$ ke kareeb yeh $90^{\circ }$ se start karta hai (green arrow) — yeh case (a) ka perpendicular cut hai, to transonic rule sirf supersonic rule ka endpoint hai. Yellow dot hamare answer $\mu =30^{\circ }$ at $M=2$ ko mark karta hai, red dashed $45^{\circ }$ line ke neeche baitha, confirm karta hai ki cone narrow hota hai jaise tez fly karo. Aage right padte jaao, $\mu$ $0^{\circ }$ ki taraf girta rehta hai, hypersonic limit jahan disturbances almost axis ke along hote hain.

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Example 7 — C7: real-world word problem (YF-102 → F-102A)

Forecast: Guess karo — kya total drag drop $28\%$ se bada hai ya chhota, aur kyun?

Steps.

1. Naya wave drag. $28\%$ hatao: $C_{D,w}^{\text{new}}=0.0200\times (1-0.28)=0.0200\times 0.72=0.0144$. Yeh step kyun? Area ruling sirf wave-drag piece par attack karta hai (parent mistake #3), to hum sirf us term ko scale karte hain aur baaki ko chhodte hain.

2. Naya total. Friction+induced unchanged hai $0.0300$ par. To $C_D^{\text{new}}=0.0300+0.0144=0.0444$. Yeh step kyun? Drag breakdown additive hai; aapne sirf woh component edit kiya jo area ruling touch karta hai, phir woh parts re-add karo jo yeh touch nahi kar sakti.

3. Percentage total drop. $\frac{0.0500-0.0444}{0.0500}=\frac{0.0056}{0.0500}=0.112=11.2\%$. Yeh step kyun? Real-world lesson dikhata hai: $28\%$ wave-drag cut ek chhota total cut ban jaata hai kyunki friction aur induced drag unchanged ride karte hain — phir bhi $11.2\%$ F-102A ko Mach 1 ke paar push karne ke liye kaafi tha jab un-ruled YF-102 nahi kar sakta tha.

Verify: $0.02\times 0.72=0.0144$ ✓; $0.03+0.0144=0.0444$ ✓; $(0.05-0.0444)/0.05=0.112$ ✓. Total drop ($11.2\%$) kam hai wave-drag drop ($28\%$) se exactly isliye kyunki wave drag total ka sirf $0.02/0.05=40\%$ hai, aur $0.28\times 0.40=0.112$ ✓.

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Example 8 — C8: exam twist — minus sign positive drag kyun deta hai

Forecast: Guess karo — kareeb stations ya dur wale stations $\ln$ kernel ke through zyada drag contribute karte hain?

Steps.

1. $\ln$ ka sign handle karo. $|x_1-x_2|<1$ ke liye (body ki apni length units mein separation measure karte hue), $\ln |x_1-x_2|<0$. Leading minus $-\frac{{\rho }_{\infty }{U}_{\infty }^2}{4\pi }$ se multiply karo aur un close-station pairs se positive contribution milta hai. Yeh step kyun? Scary minus sign bookkeeping kar raha hai, thrust produce nahi kar raha; negative log ke saath combined yeh $D_w>0$ deta hai jaisa physics demand karta hai.

2. Do curvature impulses se probe karo. $A^{\prime \prime }$ ko do spikes model karo; kernel value $\propto -\ln (\text{separation})$ hai. Separation $s$ par: weight $-\ln s$. $2s$ par: weight $-\ln (2s)=-\ln s-\ln 2$. Kyunki $-\ln 2\approx -0.6931$, $2s$ case ka chhota $-\ln$ weight hai. Yeh step kyun? Yeh dikhata hai ki curvature ko spread out karna (badi separation) drag lower karta hai — mathematical reason ki smooth, gently varying $A(x)$ wins, aur abrupt back-to-back area changes lose karte hain.

3. Design se jodo. Closely bunched area changes (kink, ya wing area jahan fuselage already fat hai) log kernel ke through strongly correlate karte hain ⇒ high drag; ek body jo apne area changes puri length par spread karta hai self-correlation minimize karta hai. Yeh step kyun? Area-rule design principle ko integral ke sign structure se directly recover karta hai.

Verify: $-\ln 2=-0.6931$; to separation $s$ ke relative, $2s$ tak doubling (positive-drag) weight se $0.6931$ subtract karta hai — wider spacing par kam drag ✓. Sign check: ${\rho }_{\infty },U_{\infty }^2>0$ ke saath aur interacting region par typical negative $\ln$, $-(\text{neg})=\text{positive}$, to $D_w>0$ ✓.

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Recall Kya har cell cover hui?

C1 Ex1 ::: smooth Sears–Haack baseline C2 Ex2 ::: kink → delta-spike $A^{\prime \prime }$ C3 Ex3 ::: fuselage+wing superposition (Coke-bottle) C4 Ex4 ::: zero-volume aur zero-length degenerates, plus $128/\pi$ prefactor ka origin C5 Ex5 ::: ${\ell }^{-4}$ scaling limit C6 Ex6 ::: perpendicular vs Mach-plane cut, $\mu =\arcsin (1/M)$ C7 Ex7 ::: real F-102A drag-count word problem C8 Ex8 ::: sign/units exam twist