Question bank — Reentry corridor — angle of attack constraints

${\alpha }_{\text{opt}}$ kahan se aata hai? (differentiation, step by step) Maximize karne wala ratio hai $\frac{L}{D}=\frac{C_{L\alpha }\alpha }{C_{D0}+k{C}_{L\alpha }^2{\alpha }^2}$. Differentiate kyun karte hain? Maximum ek flat spot hota hai: slope $d(L/D)/d\alpha$ wahan zero se guzarta hai. Numerator ko $u=C_{L\alpha }\alpha$ aur denominator ko $w=C_{D0}+k{C}_{L\alpha }^2{\alpha }^2$ mano. Quotient rule kehta hai $\frac{d}{d\alpha }\frac{u}{w}=\frac{u^{\prime }w-u{w}^{\prime }}{w^2}$, jahan $u^{\prime }=C_{L\alpha }$ aur $w^{\prime }=2k{C}_{L\alpha }^2\alpha$ hain. Toh derivative ka numerator hai $u^{\prime }w-u{w}^{\prime }=C_{L\alpha }(C_{D0}+k{C}_{L\alpha }^2{\alpha }^2)-\left(C_{L\alpha }\alpha \right)\left(2k{C}_{L\alpha }^2\alpha \right).$ Dono products expand karo: $=C_{L\alpha }{C}_{D0}+C_{L\alpha }k{C}_{L\alpha }^2{\alpha }^2-2k{C}_{L\alpha }^3{\alpha }^2=C_{L\alpha }{C}_{D0}+k{C}_{L\alpha }^3{\alpha }^2-2k{C}_{L\alpha }^3{\alpha }^2.$ Dono ${\alpha }^2$ terms combine ho jaate hain ($+1-2=-1$), bachta hai $C_{L\alpha }{C}_{D0}-k{C}_{L\alpha }^3{\alpha }^2$. Maximum ke liye yeh $=0$ chahiye; common $C_{L\alpha }$ divide kar do: $C_{D0}-k{C}_{L\alpha }^2{\alpha }^2=0\Rightarrow \boxed{C_{D0}=k{C}_{L\alpha }^2{\alpha }^2}.$ $\alpha$ ke liye solve karo (positive root lo, kyunki physical angle of attack positive hoti hai): $\boxed{{\alpha }_{\text{opt}}=\sqrt{\frac{C_{D0}}{k{C}_{L\alpha }^2}}=\frac{1}{C_{L\alpha }}\sqrt{\frac{C_{D0}}{k}}}$ Physically: peak wahan hai jahan fixed drag $C_{D0}$, lift-induced drag $k{C}_{L\alpha }^2{\alpha }^2$ ke barabar hoti hai — added lift aur added drag exactly trade off karte hain.

$\propto |\sin {\gamma }_E|$ aur heating $\propto \sqrt{\rho }{V}^3$ kyun hai Deceleration: drag deceleration $D/m=\frac{1}{2}\rho V^{2}A{C}_D/m$ hai (drag force ko mass $m$ se divide karne par acceleration milta hai). Jaise vehicle plunge karta hai, altitude rate $\dot{h}=V\sin \gamma$ se girti hai, isliye steepness $\sin {\gamma }_E$ control karti hai ki $\rho ={\rho }_0{e}^{-h/H}$ kitni tezi se per second thicken hoti hai. Ise exponential atmosphere ke through carry karne par (Allen–Eggers Ballistic Reentry result) single peak $\propto V_E^2|\sin {\gamma }_E|$ nikalta hai: $V_E^2$ dynamic pressure se, $|\sin {\gamma }_E|$ is baat se ki tum thicker hawa mein kitna sharply drive karte ho. Heating: nose par convective heating rate $\dot{q}\propto \sqrt{\rho }{V}^3$ scale hoti hai (dekho Aerodynamic Heating and Stanton Number) — $\sqrt{\rho }$ boundary-layer physics se aur $V^3$ isliye kyunki heat flux roughly (energy flux $\frac{1}{2}\rho V^2\cdot V$) Stanton number se modify hota hai; drag ke upar $V$ ki extra power heating ko zyada speed-sensitive limit banati hai.