Exercises — Reentry corridor — angle of attack constraints

• $V$ ::: vehicle ki speed air ke through (m/s) — uske velocity vector ki length.
• $V_E$ ::: entry speed, yaani $V$ ki woh value jab vehicle pehli baar sensible atmosphere se milti hai ("interface" par). LEO return ke liye $\approx 7.8\text{ km/s}$; lunar return ke liye $\approx 11\text{ km/s}$.
• $m$ ::: vehicle ki mass (kg) — trajectory equations ke liye Newton's law $F=ma$ mein aata hai.
• $C_{L\alpha }$ ::: lift-curve slope ($/\text{rad}$) — angle of attack ke ek radian per kitne units lift coefficient $C_L$ milti hai, yaani $C_L\approx C_{L\alpha }\alpha$. Bada $C_{L\alpha }$ = shape zyada eagerly lift banata hai.
• $C_{D0}$ ::: zero-lift drag coefficient (dimensionless) — woh drag jo body ko $\alpha =0$ par bhi hoti hai, purely uski bluntness/shape se.
• $k$ ::: induced-drag factor (dimensionless) — $C_D\approx C_{D0}+k{C}_L^2$ mein penalty coefficient; yeh batata hai ki jab tum zyada lift maangto ho toh drag kitni tezi se badhti hai. (Sab Lift and Drag Coefficients se.)

Undershoot (woh steep boundary). Tezi se dive karne ka matlab hai ki vehicle dense low air mein pahunchti hai jab woh abhi bhi tezi se move kar rahi hoti hai, toh dynamic pressure $q=\frac{1}{2}\rho V^2$ aur drag dono saath spike karte hain. Undershoot limit boundary hai (g-load + heating). Overshoot skip-out boundary hai — woh shallow wali, jahan tum thin upper air se glance off karke dobara atmosphere chhodh dete ho.

• $\alpha$ lift $L$ aur drag $D$ dono ki size set karta hai ($C_L(\alpha )$ aur $C_D(\alpha )$ ke through).
• $\sigma$ (fixed-magnitude) lift ki direction set karta hai — kitna upar point karta hai, $L\cos \sigma$, versus sideways, $L\sin \sigma$.

$\alpha$ ko aero force ka throttle samjho aur $\sigma$ ko steering wheel jo use rotate karta hai.

Descent par $\gamma <0$: velocity vector local horizontal ke neeche point karta hai (horizon ke relative nose-down). $\gamma =0$ ka matlab hai tum exactly horizontally fly kar rahe ho — level flight, us instant mein na chadhna na girna. $\gamma >0$ ka matlab hoga path wapas upar curve ho raha hai (ek skip ki shuruat).

Parent ke boxed results use karo: ${\alpha }_{\text{opt}}=\frac{1}{C_{L\alpha }}\sqrt{\frac{C_{D0}}{k}}=\frac{1}{1.8}\sqrt{\frac{0.08}{0.4}}=\frac{1}{1.8}(0.4472)=0.2485\text{ rad}.$ Convert karo: $0.2485\times \frac{180}{\pi }\approx 14.2^{\circ }$. ${\left(\frac{L}{D}\right)}_{\max }=\frac{1}{2\sqrt{k{C}_{D0}}}=\frac{1}{2\sqrt{0.4\cdot 0.08}}=\frac{1}{2(0.1789)}=\mathrm{2.795.}$ KYUN peak exactly wahan baith hai (core idea). Neeche Figure s01 dekho. Yellow dotted line sirf numerator hai, $C_{L\alpha }\alpha$ — pure lift, seedhi line mein badhti. Blue curve puri $L/D$ hai. Chhote $\alpha$ par blue curve yellow line ke saath chalti hai: $\alpha$ add karna almost pure lift kharidta hai, kyunki drag denominator abhi constant $C_{D0}$ se dominate ho raha hai aur ${\alpha }^2$ term negligible hai. Jaise jaise $\alpha$ badhta hai, $k{C}_{L\alpha }^2{\alpha }^2$ drag term jaag uthta hai aur gains khaana shuru kar deta hai, toh blue curve peel off hoti hai aur bend ho jaati hai. Red dot exact balance point hai: ek aur unit $\alpha$ utna hi drag (denominator) add karta hai jitna lift (numerator), toh $L/D$ ka slope momentarily flat hai — woh flat top hi reason hai ki derivative ${\alpha }_{\text{opt}}$ par vanish hoti hai. Iske baye tum available lift waste kar rahe ho; iske daaye drag jeet jaata hai.

Woh tool jo hum use karte hain (aur yeh kahan se aata hai). Peak-deceleration formula ${\left|\frac{dV}{dt}\right|}_{\max }=\frac{V_E^2|\sin {\gamma }_E|}{2eH}$ simply diya nahi gaya — yeh neeche problem L4.1 mein scratch se derive kiya gaya hai (Allen–Eggers; dekho Allen–Eggers Ballistic Reentry bhi). Abhi ise ready tool ki tarah use karo: yeh kehta hai ki ek purely ballistic (no-lift) capsule jo worst deceleration feel karta hai woh sirf entry speed squared, kitna steeply dive karta hai ($\sin {\gamma }_E$), atmosphere ki scale height $H$, aur constant $e$ par depend karta hai. Numerator: $(7800{)}^2\sin 5^{\circ }=6.084\times 10^7\times 0.08716=5.303\times 10^6$. Denominator: $2(2.71828)(7000)=3.806\times 10^4$. ${\left|\frac{dV}{dt}\right|}_{\max }=\frac{5.303\times 10^6}{3.806\times 10^4}=139.3{\text{ m/s}}^2=\frac{139.3}{9.81}\approx 14.2g.$ Kyun $e$ aata hai: density $e^{h/H}$ ki tarah badhti hai jabki speed ghatti hai; unka product $\rho V^2$ exactly ek baar peak karta hai, aur peak value usi crossover se $1/e$ ka factor carry karti hai (L4.1 mein prove kiya hai).

$\frac{\Delta {\gamma }_B}{\Delta {\gamma }_A}=\frac{(L/D{)}_B}{(L/D{)}_A}=\frac{1.5}{0.3}=5.$ B ka corridor 5× wider hai. Zyada lift tumhe ek steeper dive se bhi pull out karne deti hai (undershoot dodge karo) aur ek shallow mein bhi dig in karne deti hai (overshoot dodge karo) — yeh dono boundaries ek saath widen karti hai.

WHAT karte hain: quotient differentiate karo aur zero set karo. $f(\alpha )=\frac{N}{D}$ lo jahan $N=C_{L\alpha }\alpha$, $D=C_{D0}+k{C}_{L\alpha }^2{\alpha }^2$. Quotient rule: $f^{\prime }=\frac{N^{\prime }D-N{D}^{\prime }}{D^2}$. $f^{\prime }$ ka sign numerator ka sign hai $N^{\prime }D-N{D}^{\prime }=C_{L\alpha }(C_{D0}+k{C}_{L\alpha }^2{\alpha }^2)-C_{L\alpha }\alpha (2k{C}_{L\alpha }^2\alpha )=C_{L\alpha }(C_{D0}-k{C}_{L\alpha }^2{\alpha }^2).$ Zero set karo: $C_{D0}=k{C}_{L\alpha }^2{\alpha }^2\Rightarrow {\alpha }_{\text{opt}}=\frac{1}{C_{L\alpha }}\sqrt{C_{D0}/k}$. ✓ Kyun yeh maximum hai: bracket $C_{D0}-k{C}_{L\alpha }^2{\alpha }^2$ chhote $\alpha$ ke liye positive hai (function badhta hai) aur bade $\alpha$ ke liye negative hai (function girta hai). Badhna-phir-girna ⇒ ek peak. Wapas plug karo: optimum par $D=C_{D0}+C_{D0}=2C_{D0}$, aur $N=C_{L\alpha }\cdot \frac{1}{C_{L\alpha }}\sqrt{C_{D0}/k}=\sqrt{C_{D0}/k}$, toh ${\left(\frac{L}{D}\right)}_{\max }=\frac{\sqrt{C_{D0}/k}}{2C_{D0}}=\frac{1}{2\sqrt{k{C}_{D0}}}.✓$

$\dot{\gamma }$ tab sabse zyada negative hoga jab $L\cos \sigma$ sabse zyada negative ho, yaani $\cos \sigma =-1$, toh $\boxed{\sigma =180^{\circ }}.$ $\sigma =180^{\circ }$ par lift seedha neeche point karti hai, gravity mein add hokar path ko atmosphere mein wapas bend karti hai. Kyun: bank lift vector ko rotate karta hai bina uski length change kiye; $\cos \sigma$ $+1$ (lift fully up) se $0$ (lift fully sideways) se $-1$ (lift fully down) tak jaata hai. "Entry ke liye commit karo" = $180^{\circ }$ tak roll karo. $\frac{m{V}^2}{R+h}\cos \gamma$ term (centrifugal, curved Earth ride karne se) aur gravity yahan fixed hain — sirf lift term hamare control mein hai.

Heating mein $V^3$ hai; deceleration mein $V^2$ hai. Toh fractional speed increase $\delta$ heating ko $(1+\delta {)}^3$ se multiply karta hai lekin deceleration ko sirf $(1+\delta {)}^2$ se. Heating zyada speed-sensitive hai. Consequence: lunar/interplanetary return ke liye ($V_E\approx 11\text{ km/s}$) corridor mainly heat shield se squeeze hota hai, crew g-tolerance se nahi; low-Earth-orbit return ke liye ($V_E\approx 7.8\text{ km/s}$) g-load zyada even competitor hai. Faster returns "heating-limited" hote hain. (Dekho Aerodynamic Heating and Stanton Number.)

WHAT: hum woh altitude chahte hain jahan $\rho V^2$ sabse bada ho. $u=e^{-h/H}$ likho toh $\rho ={\rho }_{0}u$ aur $B=\frac{{\rho }_{0}H}{2\beta |\sin {\gamma }_E|}$ lo, jisse $V=V_E{e}^{-Bu}$ milta hai, isliye $\rho V^2={\rho }_{0}u{V}_E^2{e}^{-2Bu}.$ WHY: deceleration magnitude $\propto \rho V^2$, toh iska peak hi peak deceleration hai. $u$ ke respect mein differentiate karo ($h$ ka monotone function): $\frac{d}{du}\left(u{e}^{-2Bu}\right)=e^{-2Bu}(1-2Bu)=0\Rightarrow u_{\text{pk}}=\frac{1}{2B}$. Phir ${\rho }_{\text{pk}}={\rho }_0{u}_{\text{pk}}=\frac{{\rho }_0}{2B}=\frac{{\rho }_0}{2}\cdot \frac{2\beta |\sin {\gamma }_E|}{{\rho }_{0}H}=\frac{\beta |\sin {\gamma }_E|}{H}.$ ✓ $u_{\text{pk}}$ par: exponent $2B{u}_{\text{pk}}=1$, toh $V_{\text{pk}}=V_E{e}^{-1/2}$, aur ${\left|\dot{V}\right|}_{\max }=\frac{{\rho }_{\text{pk}}{V}_{\text{pk}}^2}{2\beta }=\frac{1}{2\beta }\cdot \frac{\beta |\sin {\gamma }_E|}{H}\cdot V_E^2{e}^{-1}=\frac{V_E^2|\sin {\gamma }_E|}{2eH}.✓$ WHAT yeh dikhta hai (neeche Figure s02 padho). Blue curve density $\rho$ hai — jaise tum girte ho yeh steadily badhti hai (right-se-left, kyunki altitude ghatti hai). Yellow curve $V^2$ hai — upar yeh abhi bhi full entry speed ke paas hai, lekin neeche yeh almost kuch nahi reh jaata. Deceleration $\propto \rho V^2$ (red curve) ek badhti cheez aur ek ghatti cheez ka product hai, toh iska ek single hump hona zaroori hai: upar chhota hai (koi air nahi) aur neeche chhota hai (koi speed nahi), beech mein peak karta hai. Green dashed line woh peak mark karta hai; algebra ise exactly wahan pin karta hai jahan speed $V_E/\sqrt{e}$ tak drop ho gayi hai (yaani $2Bu=1$) — precisely yahi reason hai ki final formula mein $1/e$ factor aata hai.

Peak deceleration $=8g=78.48{\text{ m/s}}^2$ set karo aur $\sin {\gamma }_E$ solve karo: $|\sin {\gamma }_E|=\frac{2eH\cdot 8g}{V_E^2}=\frac{2(2.71828)(7000)(78.48)}{(7600{)}^2}=\frac{2.987\times 10^6}{5.776\times 10^7}=\mathrm{0.05171.}$ $|{\gamma }_E|=\arcsin (0.05171)=2.964^{\circ }\approx 2.96^{\circ }.$ Toh $\approx -3.0^{\circ }$ se steeper kuch bhi $8g$ crew limit exceed karta hai — yeh is vehicle ke liye undershoot edge hai.

Corridor width: $\frac{\Delta {\gamma }_B}{\Delta {\gamma }_A}=\frac{1.4}{0.35}=4.0$ — Option B ka corridor 4× wider hai, toh guidance errors aur atmospheric uncertainty kahin zyada forgiving hain. Deceleration: zyada lift B ko ek lofted, gentle trajectory fly karne deta hai, energy loss ko zyada time mein spread karta hai ⇒ lower peak $g$ aur lower peak heating rate (halanki possibly bada total heat load kyunki yeh zyada der tak garam rehta hai — Aerodynamic Heating and Stanton Number). Ki price: ek lifting body ko zyada complex, heavy, asymmetric heat shield aur active guidance chahiye (Bank Angle Modulation and Guidance); blunt capsule lighter, simpler, self-stabilizing, aur cheaper hai. Verdict: ek razor-thin lunar-return corridor ke liye, B ka 4× margin worth hai mass ke liye — lekin capsules (Apollo, Orion) historically isliye jeete kyunki active bank modulation ne ek simple, robust shape se enough corridor diya.

(a) ${\left|\dot{V}\right|}_{\max }=\frac{V_E^2|\sin {\gamma }_E|}{2eH}=\frac{(11000{)}^2\sin 6.5^{\circ }}{2(2.71828)(7200)}=\frac{1.21\times 10^8\times 0.11320}{3.914\times 10^4}=\frac{1.370\times 10^7}{3.914\times 10^4}=349.9{\text{ m/s}}^2.$ In $g$: $349.9/9.81=35.7g$. (b) $35.7g>12g$ — haan, yeh exceed karta hai $12g$ limit ko. (Historically Apollo practice mein ~7g ke paas peak karta tha kyunki vehicle lifts karta hai aur lofted profile fly karta hai — yeh ballistic estimate worst-case, purely-ballistic upper bound hai, aur exactly isliye unhone use kiya $L/D\approx 0.3$ lift ise shave down karne ke liye.) (c) $12g$ par steepest angle: formula ko $12g=117.72{\text{ m/s}}^2$ ke equal set karo aur $\sin {\gamma }_E$ solve karo: $|\sin {\gamma }_E|=\frac{2eH(12g)}{V_E^2}=\frac{2(2.71828)(7200)(117.72)}{(11000{)}^2}=\frac{4.609\times 10^6}{1.21\times 10^8}=0.03809,$ $|{\gamma }_E|=\arcsin (0.03809)=2.183^{\circ }\approx 2.18^{\circ }.$ Isko samjho: ek purely ballistic path par even $6.5^{\circ }$ bahut zyada steep hai — lift + bank modulation ka pura point yahi hai ki ek survivable corridor exist ho sake jahan ballistics akele kehta hai "nahi." Ek crewed vehicle simply is steep se bina active lift ke enter nahi kar sakta trajectory loft karne aur deceleration spread karne ke liye.