Exercises — Trajectory optimization — minimum gravity loss, minimum drag loss

KYA: identify karo ki kaun sa term har extreme angle par maximise hota hai. Gravity loss integrand $g\sin \gamma$ hai. $\gamma =90^{\circ }$ par, $\sin 90^{\circ }=1$ — $\sin$ ki sabse badi value. Isliye gravity loss MAXIMUM hai. Seedha upar jaane par tum moti hawa se sabse jaldi bahar niklo aur speed low rakhte ho jab dense air mein ho, isliye drag loss (roughly) minimum hai. YEH AISA KYUN LAGTA HAI: figure s01 mein, jab path seedha upar point karta hai to down-arrow poori tarah path ke along pad jaata hai — poora weight motion se ladta hai.

KYA: har sentence ko uske integral ke physical thief tak trace karo. (a) drag loss $\Delta v_{\text{drag}}$. KYun: "air resistance" hi drag force $D=\frac{1}{2}\rho v^2{C}_{D}A$ hai; $m$ se divide karke time ke upar integrate karne par $\Delta v$ milta hai jo khayi gayi — exactly drag integral. (b) gravity loss $\Delta v_{\text{grav}}$. KYun: "gravity motion ke against kheenchti hai" along-path piece $g\sin \gamma$ hai (figure s01 mein pink arrow); uska time-integral gravity loss hai. (c) steering loss $\Delta v_{\text{steer}}$. KYun: thrust ko velocity se ek angle par aim karne par sirf $\cos (\text{angle})$ forward milta hai; waste fraction steering loss hai — dekho Steering losses and thrust vectoring.

KYA: ${\int }_0^{t_b}g\sin \gamma dt$ mein constant integrand ke saath plug karo. $\sin 90^{\circ }=1$, aur $g$ constant hai, isliye integral sirf $g\times t_b$ hai: $\Delta v_{\text{grav}}=9.7\times 45\times 1=436.5\text{m/s}$ KYun: ek constant value ko time ke upar integrate karna = value × time — height $9.7$ aur width $45$ ke rectangle ka area.

KYA: same formula, naya angle. $\Delta v_{\text{grav}}=9.7\times 45\times \sin 25^{\circ }=436.5\times 0.4226\approx 184.5\text{m/s}$ Factor $=\sin 25^{\circ }\approx 0.423$, matlab yeh vertical value ka lagbhag $42\%$ reh gaya. KYun: sirf $\sin \gamma$ badla. Shallower fly karne ka matlab hai gravity kam tumhari motion ke against jaati hai.

KYA: pehle drag force $D$ nikalo, phir deceleration $D/m$, phir time se multiply karo. $D=\frac{1}{2}\rho v^2{C}_{D}A=\frac{1}{2}(0.5)(400{)}^2(0.28)(1.2)=13440\text{N}$ $\Delta v_{\text{drag}}\approx \frac{D}{m}\Delta t=\frac{13440}{2.5\times 10^4}\times 15\approx 8.06\text{m/s}$ KYun: $D/m$ deceleration hai (Newton's 2nd law); $\Delta t$ ke upar roughly constant rakha jaaye to speed lost = deceleration × time. Dekho Aerodynamic Drag and Max-Q.

KYA: dono compute karo, phir interpret karo. $\Delta v_{\text{grav,A}}=9.7\times 50=485\text{m/s},\Delta v_{\text{grav,B}}=9.7\times 25=242.5\text{m/s}$ B half lose karta hai. KYun: gravity loss $\int g\sin \gamma dt$ gravity se lad kar bitaaye gaye time par depend karta hai, altitude gained par nahi. Jaldi jalo (zyada thrust-to-weight) aur lifting mein kam time spend karo ⇒ kam gravity loss. Yahi parent ki warning hai: gravity loss angle aur duration ke baare mein hai, height ke baare mein nahi.

KYA: hamare paas do competing terms ka sum hai — classic see-saw. Valley ka bottom dhundne ke liye hum poochhte hain "slope kahan zero hai?" Isliye hum differentiate karte hain: derivative $dL/d\gamma$ loss curve ka slope hai, aur minimum wahan baitha hai jahan slope negative se positive flip hota hai. $\frac{dL}{d\gamma }=K\cos \gamma +C\cdot 2\cos \gamma (-\sin \gamma )=K\cos \gamma -2C\sin \gamma \cos \gamma$ $\cos \gamma$ factor out karo ($\gamma <90^{\circ }$ ke liye nonzero): $\cos \gamma (K-2C\sin \gamma )=0\Rightarrow \sin \gamma =\frac{K}{2C}=\frac{400}{1800}=0.2222$ ${\gamma }_{\star }=\arcsin (0.2222)\approx 12.84^{\circ }$ KAISA DIKHTA HAI: figure s02 pink gravity term $K\sin \gamma$ ko angle ke saath badhte dikhata hai, blue drag proxy $C{\cos }^2\gamma$ ko angle ke saath girte dikhata hai, aur yellow total $L$. Stationary point ${\gamma }_{\star }$ dotted line se mark hai — lekin notice karo total curve wahan dip nahi karta; actually woh ek chhoti si hump hai. Hum isko endpoints check karke resolve karte hain.

Candidates par $L$ evaluate karo:

• $\gamma =0^{\circ }$: $L=0+900(1)=900$
• ${\gamma }_{\star }=12.84^{\circ }$: $L=400(0.2222)+900(0.9506)=88.9+855.6=944.4$
• $\gamma =90^{\circ }$: $L=400(1)+0=400$ Isliye stationary point is particular toy model ka ek maximum hai, aur interval par true minimum endpoint $\gamma =90^{\circ }$ par hai. Lesson: hamesha endpoints check karo — zero-derivative point ek hilltop ho sakta hai, valley nahi. (Real ascent ${\cos }^2$ proxy sirf low-altitude segment ke liye use karta hai; poori flight mein physical minimum genuinely interior hota hai. Yahan hum tumhari calculus rigour test karte hain.)

KYA: opening formula box se turn-rate law $\frac{d\gamma }{dt}=-\frac{g\cos \gamma }{v}$ apply karo. $\frac{d\gamma }{dt}=-\frac{9.8\cos 75^{\circ }}{250}=-\frac{9.8(0.2588)}{250}=-0.01014\text{rad/s}$ Convert karo: $-0.01014\times \frac{180}{\pi }\approx -0.581^{\circ }/\text{s}$. KYun: sirf cross-path gravity $g\cos \gamma$ trajectory curve karta hai, free mein. Turn slow hai kyunki $v$ already moderate hai — isliye initial pitch kick wahan plant karna zaroori hai jab $v$ chhota ho.

KYA: bookkeeping identity rearrange karo. $\Delta v_{\text{ideal}}=\Delta v_{\text{orbit}}+\Delta v_{\text{grav}}+\Delta v_{\text{drag}}+\Delta v_{\text{steer}}$ $\Delta v_{\text{grav}}=9100-7800-90-40=1170\text{m/s}$ KYun: Tsiolkovsky ka ideal $\Delta v$ useful orbital speed aur teen choron mein split hota hai. Ek unknown ke liye solve karo. Note karo gravity loss ($1170$) drag ($90$) se bahut zyada hai — topic ka central fact.

KYA & KYun: turn-rate equation $\gamma$ aur $t$ ko saath bandhta hai. Angle change se time paane ke liye, hum do variables ko opposite sides par separate karte hain — yahi "separation of variables" ka poora point hai: saare $\gamma$ left par collect karo, saare $t$ right par, phir dono ko independently integrate karo. $\frac{d\gamma }{-\cos \gamma }=\frac{g}{v}dt\Rightarrow -{\int }_{{\gamma }_0}^{{\gamma }_1}\sec \gamma d\gamma =\frac{g}{v}{\int }_0^{t}dt$ $-[\ln |\sec \gamma +\tan \gamma |{]}_{89^{\circ }}^{45^{\circ }}=\frac{g}{v}t$ Har end compute karo (trig ke liye angles radians mein, values unitless hain):

• $45^{\circ }$ par: $\sec 45^{\circ }+\tan 45^{\circ }=1.4142+1=2.4142$, $\ln =0.8814$
• $89^{\circ }$ par: $\sec 89^{\circ }+\tan 89^{\circ }=57.30+57.29=114.59$, $\ln =4.7413$ $-(0.8814-4.7413)=3.8599=\frac{9.7}{300}t$ $t=3.8599\times \frac{300}{9.7}\approx 119.4\text{s}$ KAISA DIKHTA HAI (figure s03): yellow $\gamma (t)$ curve $89^{\circ }$ ke paas nearly flat hai (turn rate $\propto \cos \gamma$ vertical ke kareeb tiny hai) phir steep hota jaata hai jaise $\gamma$ blue $45^{\circ }$ target line ki taraf girta hai. Woh flat start isliye hai kyun kick early shuru karni padti hai — vertical ke kareeb rehte hue vehicle mudta hi nahi hai.

KYA: constants ke saath steering-loss integral mein plug karo. $\Delta v_{\text{steer}}=\frac{T}{m}(1-\cos \alpha )\Delta t=20(1-\cos 6^{\circ })40$ $\cos 6^{\circ }=0.99452$, isliye $1-\cos 6^{\circ }=0.005478$: $\Delta v_{\text{steer}}=20\times 0.005478\times 40\approx 4.38\text{m/s}$ KYun: thrust ko off-velocity $\alpha$ par aim karne par sirf $\cos \alpha$ forward milta hai; lost fraction $(1-\cos \alpha )$ pure waste hai. Gravity turn $\alpha =0$ rakhta hai aur kuch nahi pay karta — yahi uski elegance hai (dekho Steering losses and thrust vectoring). Chhota $\alpha$ yahan sasta lagta hai, lekin ek lambi forced turn mein yeh accumulate karta hai, aur bada $\alpha$ loss ko ${\alpha }^2$ ki tarah badhata hai (kyunki $1-\cos \alpha \approx {\alpha }^2/2$).

Ek rocket ka ek fuel budget hota hai ($\Delta v_{\text{ideal}}$). Teen chor ise nibble karte hain: gravity (sabse bada, ~1 km/s, jitna zyada vertical aur jitna lambe time burn karo utna worse), drag (chhota, moti hawa mein fast hone par worst), aur steering (jab engine sideways point karo to worst). Clever trick gravity turn hai: shuru mein thoda tip over karo, phir gravity khud tumhari velocity horizontal swing kar deti hai free mein — zero steering pay karo, sirf itna time vertical spend karo ki drag wall se bacho.