Area rule — Whitcomb's rule for transonic drag reduction: Exercises

3.1.26 · D4 · Physics › Compressible Flow & Aerodynamics › Area rule — Whitcomb's rule for transonic drag reduction

Climb karne se pehle, is chapter ke do main formulas ka ek reminder, simple words mein. Pehle, har symbol se milte hain:

Definition Is page par milne wale symbols

$x$ — flight axis ke saath distance (streamwise), nose se measure kiya gaya. Units: metres.
$ℓ$ — body ki total length, nose se tail tak. Toh $x$ , $0$ se $ℓ$ tak jaata hai.
$A (x)$ — station $x$ par cross-sectional area: yeh woh area hai jo tab milta hai jab aap body ko flow ke perpendicular plane se kaato. Units: $m^{2}$ .
$A^{'} (x) = \frac{d A}{d x}$ — area curve ki slope (downstream move karne par area kitni tezi se badhti hai). Units: $m^{2} / m = m$ .
$A^{''} (x) = \frac{d ^{2} A}{d x ^{2}}$ — area curve ki curvature (woh slope khud kitni sharply change hoti hai). Yahi drag ka driver hai. Units: $m^{2} / m^{2}$ (dimensionless).
$ρ_{\infty}$ — freestream air density: aircraft ke kaafi aage undisturbed air ka mass per unit volume. Subscript $\infty$ ka matlab hai "far upstream, undisturbed." Units: $kg/m^{3}$ .
$U_{\infty}$ — freestream velocity: us undisturbed air ki speed aircraft ke relative (yaani flight speed). Units: $m/s$ .
$V$ — body ka total volume. Units: $m^{3}$ .
$q_{\infty} = \frac{1}{2} ρ_{\infty} U_{\infty}^{2}$ — dynamic pressure, aanewali hawa ka "push per unit area." Units: $Pa$ .
$M$ — Mach number: flight speed ko local speed of sound se divide karo, $M = U_{\infty} / a$ (jahaan $a$ speed of sound hai). Yeh dimensionless hai. $M < 1$ = subsonic, $M \approx 1$ = transonic, $M > 1$ = supersonic. Dekho Transonic flow.
$μ$ — Mach angle: ek supersonic body ke peeche trail hone wale disturbances ke cone ka half-angle, $μ = arcsin (1/ M)$ se milta hai. Yeh ek angle hai (degrees ya radians mein). Dekho Mach angle and Mach cone.

Related ideas jo aap doosre tab mein open rakhna chahein: Sears–Haack body, Slender-body theory, Shock waves and wave drag, Mach angle and Mach cone, Drag breakdown: friction, induced, wave, Transonic flow, Prandtl–Glauert and compressibility corrections.

Level 1 — Recognition

Exercise 1.1 (L1)

Area rule ke according, transonic wave drag kis single physical quantity par depend karti hai, aur von Kármán–Sears integral ke andar woh single mathematical quantity kya hai jo drag ko "drive" karti hai — yeh ek-ek sentence mein batao.

Recall Solution

Physical quantity: total cross-sectional area ka longitudinal distribution, yaani $A (x)$ — yeh matter nahi karta ki woh area fuselage, wings, aur tail mein kaise split hai. Mathematical driver: second derivative $A^{''} (x)$ (area curve ki curvature). $A (x)$ mein abrupt changes / kinks $A^{''}$ ko spike karwa dete hain, jo double integral ko bada kar deta hai.

Exercise 1.2 (L1)

$M \approx 1$ par $A (x)$ measure karne ke liye cutting plane flow ke perpendicular hoti hai (yaad karo $M$ = flight speed ÷ speed of sound). $M > 1$ par iske jagah kya aata hai, aur woh kis angle par tilted hoti hai?

Recall Solution

$M > 1$ par aap body ko Mach planes se kaatte ho, jo Mach angle $μ$ par tilted hoti hain: $μ = arcsin (\frac{1}{M}),$ aur aap resulting area ko roll angle par average karte ho (supersonic / Jones area rule). Dekho Mach angle and Mach cone. Exactly $M = 1$ par, $arcsin (1) = 9 0^{\circ}$ , toh Mach plane flow ke perpendicular khadi ho jaati hai — dono rules wahan agree karte hain.

Exercise 1.3 (L1)

Ek-line reason ke saath true ya false batao: "Area ruling aircraft ki har tarah ki drag kam karti hai."

Recall Solution

False. Yeh sirf wave drag (shock waves se transonic/supersonic drag) kam karti hai. Skin-friction drag aur induced drag untouched rehte hain, aur wasp-waist low speed par thoda friction/structural penalty bhi add kar sakta hai. Dekho Drag breakdown: friction, induced, wave.

Level 2 — Application

Exercise 2.1 (L2)

Ek Sears–Haack body ka $A_{m a x} = 1.2 m^{2}$ aur length $ℓ = 10 m$ hai (yaad karo $ℓ$ nose-to-tail length hai). Iska area law hai: $A (x) = A_{m a x} [4 ξ (1 - ξ)]^{3/2}, ξ = x / ℓ .$ Mid-station $x = 5 m$ aur quarter-station $x = 2.5 m$ par area nikalo.

Recall Solution

$ξ = x / ℓ$ sirf body ke saath kitni door hai uska fraction hai (nose par 0, tail par 1). Mid-station $x = 5$ : $ξ = 0.5$ . Toh $4 ξ (1 - ξ) = 4 (0.5) (0.5) = 1$ , aur $1^{3/2} = 1$ , toh $A = 1.2 \times 1 = 1.20 m^{2} .$ Jaisa expected tha, area middle mein peak karti hai — wahan $A_{m a x}$ hota hai. Quarter-station $x = 2.5$ : $ξ = 0.25$ . Toh $4 (0.25) (0.75) = 0.75$ , aur $0.7 5^{3/2} = 0.6495$ , toh $A = 1.2 \times 0.6495 = 0.779 m^{2} .$

Exercise 2.2 (L2)

Ek fixed volume wala missile fixed speed aur altitude par fly kar raha hai. Ek engineer isse $ℓ_{1} = 3 m$ se $ℓ_{2} = 4.5 m$ tak stretch karta hai, same Sears–Haack shape aur same volume rakhte hue. Wave drag kis factor se change hogi?

Recall Solution

$V$ , $q_{\infty}$ fixed hone par $D_{w} \propto 1/ ℓ^{4}$ . Toh $\frac{D _{w, 2}}{D _{w, 1}} = (\frac{ℓ _{1}}{ℓ _{2}})^{4} = (\frac{3}{4.5})^{4} = (\frac{2}{3})^{4} = \frac{16}{81} \approx 0.198.$ Wave drag original ki lagbhag 20% tak girr jaati hai — sirf body ko 50% longer banane se factor-5 reduction. Isliye supersonic bodies lambi aur slender hoti hain.

Exercise 2.3 (L2)

$V = 3 m^{3}$ , $ℓ = 12 m$ , aur $q_{\infty} = 25, 000 Pa$ par fly kar rahe ek body ki Sears–Haack wave drag compute karo.

Recall Solution

$D_{w} = \frac{128 q _{\infty} V ^{2}}{π ℓ ^{4}} = \frac{128 \times 25000 \times 3 ^{2}}{π \times 1 2 ^{4}} .$ Numerator: $128 \times 25000 = 3.2 \times 1 0^{6}$ ; times $9 = 2.88 \times 1 0^{7}$ . Denominator: $π \times 20736 = 6.5145 \times 1 0^{4}$ . $D_{w} = \frac{2.88 \times 1 0 ^{7}}{6.5145 \times 1 0 ^{4}} \approx 442 N .$

Level 3 — Analysis

Exercise 3.1 (L3)

Do bodies ki same length aur same volume hai. Body A ki area curve mein ek kink hai (slope $A^{'} (x)$ ek station par jump karti hai). Body B ek smooth Sears–Haack body hai. Von Kármán–Sears integral use karke argue karo ki kis mein zyada wave drag hai aur kyun, aur kink par $A^{''}$ kya mathematical object ban jaata hai — yeh bhi batao.

Recall Solution

Pehle, ek quick tool. Ek Dirac delta $δ (x - a)$ ek idealized "infinitely tall, infinitely thin spike" hai jo $x = a$ par baitha hai, itna narrow ki uska total area (integral) exactly $1$ hai. Yeh woh hota hai jo ek sudden step ke derivative jaisa dikhta hai: agar koi quantity instantly ek value se doosri par jump karti hai, toh uski rate of change everywhere zero hoti hai siwaay ek instant ke, jahaan woh ek spike hai — ek delta. (Yeh source distributions kahaan se aate hain, uske liye dekho Slender-body theory.)

Ab Body A ke derivatives step by step lete hain.

$A (x)$ mein ek kink hai ⇒ uski slope $A^{'} (x)$ us station par jump karti hai (ek value se doosri par step karti hai).
Ek step ka derivative ek delta hota hai: toh $A^{''} (x) = (jump size) \times δ (x - x_{k})$ kink station $x_{k}$ par. Curvature ek single delta spike hai.
Ise integral mein feed karo. Do deltas, ek $x_{1} = x_{k}$ par aur ek $x_{2} = x_{k}$ par, dono integrals ko single station $x_{k}$ par collapse kar dete hain, aur wahan kernel $ln ∣ x_{1} - x_{2} ∣$ ko $x_{1} = x_{2} = x_{k}$ par evaluate kiya jaata hai, yaani $ln ∣0∣ = ln 0 = - \infty$ .
Toh integral diverge karti hai: ek true kink infinite wave drag deta hai. Yeh honest, correct conclusion hai — $A (x)$ mein genuine slope-discontinuity physically realizable nahi hai precisely isliye kyunki woh infinite drag demand karega. Real "kinks" thodi rounded hoti hain, jo large-but-finite $A^{''}$ aur large-but-finite drag deti hain; corner jitna sharp hoga, drag utni hi infinity ke kareeb jaayegi.
Body B ka $A^{''}$ smooth aur bounded hai ⇒ chota, spread-out integrand ⇒ choti, finite drag. Sears–Haack exact minimizer hai. Conclusion: Body A (kink) mein enormously zyada wave drag hoti hai — idealized sharp-corner limit mein infinity tak diverge karti hai — same volume aur length hone ke bawajood. Size nahi balki $A (x)$ ki smoothness wave drag set karti hai.

"Off-diagonal" contributions ke baare mein ek note. Coincident point se door ( $x_{1} \neq = x_{2}$ ) kernel $ln ∣ x_{1} - x_{2} ∣$ perfectly finite hai, aur uske paas bhi log-singularity mild hai: yeh integrable hai. "Mild" kyun? Elementary integral yaad karo: $\int_{0}^{ε} ln x d x = [x ln x - x]_{0}^{ε} = ε ln ε - ε,$ jo ek finite number hai ( $x \to 0^{+}$ par $x ln x \to 0$ ). Toh $ln$ jab ek bounded, spread-out $A^{''}$ (Body B) ke against integrate hoti hai toh finite drag deti hai. Body A mein problem log se nahi balki saari curvature ko ek point (delta) mein squeeze karne se aati hai, jo kernel ko exactly uske singular point $ln 0$ par sample karne par majboor kar deti hai.

Figure dono area curves (left) aur Body A ki delta-spike $A^{''}$ aur Body B ki smooth, bounded $A^{''}$ ka contrast (right) dikhata hai.

Figure — Area rule — Whitcomb's rule for transonic drag reduction

Figure s01 — Left: equal volume aur length ke do bodies. Teal Sears–Haack curve smooth hai; orange curve mein $ξ = 0.4$ aur $ξ = 0.6$ par kinks (slope jumps) hain. Right: unki curvatures $A^{''}$ . Smooth body (teal) ka $A^{''}$ bounded aur gently-varying hai; kinks infinite delta spikes produce karte hain (orange arrows). Kyunki drag $A^{''}$ ke log-kernel ke against sampling se scale karti hai, spikes drag ko sharp-corner limit mein infinity tak drive karte hain.

Exercise 3.2 (L3)

Ek slender-body source strength $f (x) = U_{\infty} d A / d x$ hai (yaad karo $U_{\infty}$ flight speed hai). Exercise 2.1 ke Sears–Haack law ke liye, body ke saath source strength $f (x)$ zero kahaan hai, aur $A^{''} (x) = 0$ kahaan hai? Differentiation dikhao aur physically interpret karo.

Recall Solution

Shape function $g (ξ) = [4 ξ (1 - ξ)]^{3/2} = [4 ξ - 4 ξ^{2}]^{3/2}$ likho, toh $A = A_{m a x} g (ξ)$ . Jahaan $f = 0$ : $f = U_{\infty} A^{'} (x)$ , aur $A^{'} = 0$ jahaan $g^{'} (ξ) = 0$ . $g^{'} (ξ) = \frac{3}{2} [4 ξ - 4 ξ^{2}]^{1/2} (4 - 8 ξ) .$ Bracket $(4 - 8 ξ) = 0$ se $ξ = 0.5$ milta hai; square-root factor endpoints $ξ = 0, 1$ deta hai. Toh maximum mid-body par $ξ = 0.5$ par baithta hai, jahaan $A^{'} = 0 \Rightarrow f = 0$ . Physical meaning: sabse chaudi station par body na grow ho rahi hai na shrink — toh woh streamtube volume inject ya remove nahi karti — koi source, koi sink nahi. Jahaan $A^{''} = 0$ (inflection points): $g^{'}$ ko phir differentiate karo. $u = 4 ξ - 4 ξ^{2}$ likhne par $u^{'} = 4 - 8 ξ$ , $g^{''} (ξ) = \frac{3}{2} [\frac{1}{2} u^{- 1/2} (u^{'})^{2} + u^{1/2} (- 8)] = \frac{3}{4} u^{- 1/2} [(4 - 8 ξ)^{2} - 16 u] .$ Bracket ko zero karo ( $u^{- 1/2}$ factor sirf pointed ends par blow up karta hai): $(4 - 8 ξ)^{2} - 16 (4 ξ - 4 ξ^{2}) = 0.$ Expand karo: $(4 - 8 ξ)^{2} = 16 - 64 ξ + 64 ξ^{2}$ aur $16 (4 ξ - 4 ξ^{2}) = 64 ξ - 64 ξ^{2}$ , toh $16 - 64 ξ + 64 ξ^{2} - 64 ξ + 64 ξ^{2} = 128 ξ^{2} - 128 ξ + 16 = 0 \Rightarrow 8 ξ^{2} - 8 ξ + 1 = 0.$ Quadratic formula: $ξ = \frac{8 \pm 64 - 32}{16} = \frac{8 \pm 32}{16} = \frac{1}{2} (1 \pm \frac{1}{2})$ , yaani $ξ \approx 0.1464 aur ξ \approx 0.8536.$ Yeh inflection points hain, jahaan area curve concave-up (ends ke paas) se concave-down (middle ke paas) switch karti hai. Yahan curve "seedhi" hoti hai, $D_{w}$ mein kam local curvature contribute karti hai.

Level 4 — Synthesis

Exercise 4.1 (L4)

Ek straight fuselage ka area $A_{fus} (x) = 0.8 m^{2}$ hai (ek section ke saath constant). Mid-body par ek wing add ki jaati hai, jo ek bump contribute karti hai: $A_{wing} (x) = 0.5 exp [- (\frac{x - x _{0}}{w})^{2}] m^{2} .$ Yahan $x_{0}$ woh streamwise station hai jahaan wing area peak karti hai (uska centre, metres mein) aur $w$ woh width parameter hai jo batata hai bump kitna spread-out hai (chota $w$ = narrow, sharp hump; bada $w$ = gentle, wide hump — units metres). $x_{0} = 5 m$ , $w = 1.2 m$ lo. (a) Sketch karo ki total area $A_{tot} = A_{fus} + A_{wing}$ ka kya hoga. (b) Ek fuselage dip $Δ A_{fus} (x)$ design karo jo ek smooth total restore kare. (c) Area-rule language mein explain karo ki yeh "Coke-bottling" kyun kaam karti hai.

Recall Solution

(a) Ek flat line mein Gaussian hump add karne se $A_{tot}$ mein ek bump aata hai: yeh $0.8 m^{2}$ par flat start karta hai, $x_{0} = 5$ m par $0.8 + 0.5 = 1.3 m^{2}$ ki peak tak jaata hai, phir $0.8 m^{2}$ par wapis aata hai. Yeh single rise-and-fall large positive-then-negative curvature $A^{''}$ ka ek region create karta hai — exactly wahi jo von Kármán–Sears integral punish karta hai. Aapke sketch mein middle mein ek smooth bump wali horizontal line honi chahiye (neeche figure ka teal-plus-plum → orange "hump" panel). (b) Fuselage ko same hump subtract karke pinch karo: $Δ A_{fus} (x) = - 0.5 exp [- (\frac{x - x _{0}}{w})^{2}] .$ Tab $A_{tot} = A_{fus} + Δ A_{fus} + A_{wing} = 0.8 m^{2},$ phir se flat — $A^{''} = 0$ har jagah (figure, right panel). (Practice mein aap iska saara hissa subtract nahi kar sakte — fuselage payload ke liye kaafi bada rehna chahiye — lekin aap jitna structurally possible ho utni curvature cancel karte ho.) (c) Kyunki $D_{w}$ sirf $A_{tot} (x)$ par depend karta hai, wing ke hump ko fuselage dip se cancel karna $A_{tot}^{''} \to 0$ kar deta hai aur wave drag ko smooth-body minimum ki taraf drive karta hai. Yahi Whitcomb ki insight hai aur "wasp-waist" ki origin hai (dekho Sears–Haack body).

Neeche teen panels left→middle→right chalte hain: alag parts, bura naive total, aur achha area-ruled total.

Figure s02 — Teen panels mein Coke-bottling. Left ("Parts"): constant fuselage area (teal, $0.8 m^{2}$ ) aur wing ka Gaussian bump (plum) $x_{0} = 5$ m par centred, width $w = 1.2$ m ke saath. Middle ("Naive total, bad"): simply unhe add karne se $1.3 m^{2}$ par peaking hump milta hai — badi curvature $A^{''}$ , high wave drag (orange). Right ("Area-ruled total, good"): fuselage ko ek equal, opposite dip (dashed teal) se pinch karna hump cancel kar deta hai, $0.8 m^{2}$ par flat total deta hai (orange) jisme $A^{''} = 0$ hai — minimal wave drag.

Exercise 4.2 (L4)

Ek aircraft $M = 1.4$ par supersonic cruise ke liye area-rule ho raha hai. (a) Mach angle $μ$ nikalo. (b) $μ$ ke terms mein explain karo ki aapko jo area smooth karni hai woh $M \approx 1$ case se kaise alag hai, aur "averaging over roll" ka kya matlab hai.

Recall Solution

(a) Mach angle ek supersonic body ke disturbances ke cone ka half-angle hai: $μ = arcsin (\frac{1}{M}) = arcsin (\frac{1}{1.4}) = arcsin (0.7143) = 45.5 8^{\circ} .$ (b) Flow ke perpendicular planes se body kaatne ki jagah, aap Mach planes se kaatte ho jo axis se $μ = 45. 6^{\circ}$ par tilted hoti hain. Har aisi plane Mach cone ki tangent hoti hai, toh woh capture karti hai ki $M > 1$ par disturbances actually kaise propagate hote hain (dekho Mach angle and Mach cone). Kyunki ek Mach plane ko flight axis ke around kisi bhi roll direction mein tilt kiya ja sakta hai, aap kai roll angles ke liye "cut area" compute karte ho aur unhe average karte ho — flow ki wave drag us roll-averaged area distribution par respond karti hai, jise aap phir pehle ki tarah smooth karte ho.

Level 5 — Mastery

Exercise 5.1 (L5)

Von Kármán–Sears integral se dimensional/scaling analysis karke length-scaling law $D_{w} \propto V^{2} / ℓ^{4}$ prove karo, bina double integral evaluate kiye. Assume karo ki shape fixed hai (sirf $ℓ$ aur overall area scale change hote hain).

Recall Solution

Area law ko nondimensional form mein likho. $ξ = x / ℓ \in [0, 1]$ lo aur $A (x) = S g (ξ)$ , jahaan $g$ fixed shape function hai (dimensionless, family ke saare members ke liye same) aur $S$ peak-area scale hai (units $m^{2}$ ). Step 1 — $A^{''}$ rewrite karo. Kyunki $x = ℓ ξ$ , har $d / d x = \frac{1}{ℓ} d / d ξ$ . Isliye $A^{''} (x) = \frac{S}{ℓ ^{2}} g^{''} (ξ) .$ Kyun: $x$ mein do derivatives do $1/ ℓ$ factors pull out karti hain. Step 2 — integral mein substitute karo. Variables change karo $x_{i} = ℓ ξ_{i}$ , toh $d x_{1} d x_{2} = ℓ^{2} d ξ_{1} d ξ_{2}$ , aur $ln ∣ x_{1} - x_{2} ∣ = ln ℓ + ln ∣ ξ_{1} - ξ_{2} ∣$ . $ln ℓ$ term $\iint g^{''} g^{''} d ξ_{1} d ξ_{2}$ ke against integrate hoti hai; lekin $\int_{0}^{1} g^{''} (ξ) d ξ = g^{'} (1) - g^{'} (0) = 0$ ek closed body ke liye (pointed ends, zero slope), toh $ln ℓ$ piece vanish ho jaata hai. Jo bachta hai: $D_{w} = - \frac{ρ _{\infty} U _{\infty}^{2}}{4 π} \cdot (A^{''})^{2} \frac{S ^{2}}{ℓ ^{4}} \cdot d x_{1} d x_{2} ℓ^{2} \iint g^{''} (ξ_{1}) g^{''} (ξ_{2}) ln ∣ ξ_{1} - ξ_{2} ∣ d ξ_{1} d ξ_{2} .$ Last integral ek pure number $I$ hai jo shape se fixed hai. Toh $D_{w} = - \frac{ρ _{\infty} U _{\infty}^{2}}{4 π} I \frac{S ^{2}}{ℓ ^{2}} = (const) q_{\infty} \frac{S ^{2}}{ℓ ^{2}} .$ Step 3 — volume laao. Volume $V = \int_{0}^{ℓ} A d x = S ℓ \int_{0}^{1} g d ξ = c S ℓ$ jahaan $c$ ek shape constant hai. Isliye $S = V / (c ℓ)$ aur $S^{2} = V^{2} / (c^{2} ℓ^{2})$ . Substitute karo: $D_{w} \propto q_{\infty} \frac{S ^{2}}{ℓ ^{2}} = q_{\infty} \frac{V ^{2}}{c ^{2} ℓ ^{2} \cdot ℓ ^{2}} \propto \frac{q _{\infty} V ^{2}}{ℓ ^{4}} . ■$ Precise Sears–Haack constant $128/ π$ hai, jo $D_{w} = 128 q_{\infty} V^{2} / (π ℓ^{4})$ deta hai, parent note se consistent.

Exercise 5.2 (L5)

Historical/engineering synthesis: YF-102 $M = 1$ exceed nahi kar sakta tha; F-102A mein area-rule karne ke baad woh kar sakta tha. Causal chain ko paanch links mein explain karo, har ek "X ⇒ Y" form mein. Phir estimate karo: agar area-ruling ne peak $A^{''}$ ko factor of 2 se cut kiya (aur integral roughly $(A_{peak}^{''})^{2}$ ki tarah scale karta hai ek fixed region par), toh wave drag kis factor se giregi?

Recall Solution

Causal chain:

Wing mid-body par add hui ⇒ total $A (x)$ mein ek hump.
Hump ⇒ us station par bada $A^{''}$ (sharp curvature).
Bada $A^{''}$ ⇒ strong local flow acceleration/deceleration ⇒ strong shock waves (dekho Shock waves and wave drag).
Strong shocks ⇒ $M = 1$ ke paas badi wave drag ("transonic drag rise," dekho Transonic flow).
Woh drag wall ⇒ available thrust YF-102 ko $M = 1$ ke through push nahi kar sakta tha. Area-ruling $A (x)$ ko flatten karta hai ⇒ chota $A^{''}$ ⇒ weak shocks ⇒ drag wall neeche ⇒ F-102A through punch karta hai.

Estimate: agar $D_{w} \sim (A_{peak}^{''})^{2}$ aur $A_{peak}^{''}$ half ho jaaye, toh $\frac{D _{w, new}}{D _{w, old}} = (\frac{1}{2})^{2} = \frac{1}{4} .$ Toh wave drag apni purani value ki lagbhag 25% tak girr jaati hai — Whitcomb ne wind tunnel mein jo ~25–30% transonic drag reductions measure kiye the unse consistent.