Trajectory optimization — minimum gravity loss, minimum drag loss: Visual walkthrough

3.4.15 · D2 · Physics › Rocket Flight Mechanics › Trajectory optimization — minimum gravity loss, minimum drag

Step 0 — Woh ek arrow jis par sab kuch tika hai

KYA HAI. Ek rocket kisi ek instant mein move kar raha hai. Uski motion ek arrow hai: uski ek length hai (kitni tez — speed $v$ , metres per second mein naapi jaati hai) aur ek direction hai (woh kis taraf point karta hai).

KYUN. Jo bhi loss hum compute karenge woh sab is baare mein hai ki yeh arrow ground ke relative kis direction mein point karta hai. Isliye hume pehle us direction ka naam rakhna hoga, kisi bhi formula se pehle.

PICTURE. Figure dekho. Magenta arrow velocity hai. Dashed orange line local horizon hai — rocket ke bilkul neeche flat ground. Arrow aur horizon ke beech ka angle poore show ka star hai.

Do words jinpar hum depend karenge, usi picture par define kiye gaye hain:

Step 1 — Gravity ko "slow karta hai" aur "bend karta hai" mein split karo

KYA HAI. Gravity hamesha seedha neeche point karti hai, strength $g$ ke saath (ground ke paas lagbhag $9.8$ metres-per-second-per-second). Hum us neeche-arrow ko do pieces mein split karte hain: ek velocity ke saath, ek uske aarpaar.

KYUN. Tumhari motion ke saath ek force tumhari speed badlata hai. Tumhari motion ke aarpaar ek force tumhari direction badalta hai, speed nahi. Gravity dono karti hai, ek proportion mein jo bilkul $γ$ se set hoti hai. Hume unhe alag karna hoga kyunki woh do alag effects cause karte hain (ek speed loss, aur ek free turn).

PICTURE. Figure mein navy arrow gravity hai jo neeche point kar raha hai. Violet arrow uska piece hai jo path ke upar backwards point karta hai — woh wala part hai jo slow karta hai. Orange arrow uska piece hai jo path ke aarpaar point karta hai — woh part bend karta hai.

Yahan woh geometry ka piece hai jis par poora page tika hai. Down-arrow ko dono directions par drop karo:

Step 2 — Path ke saath Newton ka law likho

KYA HAI. Velocity arrow ke saath kaam karne wale har force ko add karo aur total ko mass times speed kitni tezi se badlti hai ke barabar set karo.

KYUN. Newton ka second law kehta hai (ek direction mein net force) = (mass) × (us direction mein acceleration). Path ke saath, "acceleration" bas $\frac{d v}{d t}$ hai — woh rate jis par speed number $v$ badhta hai. Yahi woh quantity hai jis ki hume ultimately parwah hai: yeh orbital speed ban jaati hai.

PICTURE. Velocity line par teen forces rehte hain: thrust $T$ aage push karta hai (magenta), drag $D$ peeche push karta hai (violet), aur gravity-along piece $m g sin γ$ peeche push karta hai (navy).

Step 3 — Mass se divide karo taaki har term ek acceleration ho

KYA HAI. Har term ko mass $m$ se divide karo.

KYUN. Hum $Δ v$ (ek speed budget) par end karna chahte hain, aur speed change acceleration × time se aata hai. Force ko mass se divide karna use acceleration mein badal deta hai (Newton phir se: $a = F / m$ ). Ab right side par har term "yeh effect har second kitne m/s deta ya leta hai" hai.

Step 4 — Poore burn par add karo (integrate karo)

KYA HAI. Ignition ( $t = 0$ ) se burnout ( $t = t_{b}$ ) tak har instant par har term ko sum karo. Woh summing-over-time operation integral $\int_{0}^{t_{b}} (\dots) d t$ hai.

KYUN. Ek instant par acceleration ek tiny speed change deta hai; total speed change woh saare tiny bits hain jo burn par add hote hain. Integral bas "ek continuously changing quantity add karo" hai. Har term integrate hokar ek total- $Δ v$ contribution mein badal jaata hai.

PICTURE. Figure mein har acceleration curve ko uske neeche shaded area ke roop mein dikhaya gaya hai — area hi integral hai, woh $Δ v$ jo us term ne contribute kiya.

Rearrange karo, yeh woh bookkeeping equation hai jo parent note ne promise ki thi: $Δ v_{achieved} = Δ v_{ideal} - Δ v_{drag} - Δ v_{grav}$

Step 5 — Do integrands padho: $γ$ dono kyun nahi jeet sakta

KYA HAI. Dono loss integrands ko ghoor kar dekho aur pucho: jab mein pitch $γ$ badlaata hoon, har ek kis taraf jaata hai?

KYUN. Yeh ek hi nazar mein poora optimization problem hai. Pitch $γ$ ek single knob hai jo dono integrals ke andar baitha hai — unhe opposite directions mein push karta hai.

PICTURE. $γ$ ke against do curves: gravity loss flat jaane par girati hai; drag loss flat jaane par badhti hai (flat matlab neeche ghane air mein tez). Unka sum beech mein ek minimum wala bowl hai.

Knob position	$sin γ$	Gravity loss	Drag loss
Vertical, $γ \to 9 0^{\circ}$	$\to 1$	bada	chhota (ghane air se jaldi nikal jaata hai)
Flat, $γ \to 0^{\circ}$	$\to 0$	chhota	bada (neeche $ρ v^{2}$ bahut zyada)

Step 6 — Edge cases: formula ko uski extremes par check karo

KYA HAI. $γ$ ko uske dono limits aur ek degenerate burn par push karo, aur confirm karo ki formula kuch sane karta hai.

KYUN. Jis derivation par tum trust karte ho use apne corners survive karne chahiye. Agar koi bhi extreme nonsense deta, toh humara koi sign ya sine galat hota.

PICTURE. Teen mini-rockets: seedha upar, flat, aur ek zero-length burn.

Seedha upar, $γ = 9 0^{\circ}$ : $sin 9 0^{\circ} = 1$ , isliye $Δ v_{grav} = \int g d t$ — poori gravity poore burn mein tumse ladhti hai. Worst-case gravity loss. ✔ intuition se match karta hai.
Flat, $γ = 0^{\circ}$ : $sin 0^{\circ} = 0$ , isliye $Δ v_{grav} = 0$ . Gravity koi speed nahi leta — lekin Step 1 ka across piece $g cos 0^{\circ} = g$ ab full strength par hai, tumhe ground mein curve kar raha hai. Formula sahi kehta hai ki gravity slow karna band ho gayi aur sirf bend karna shuru ho gayi. ✔
Zero burn time, $t_{b} = 0$ : har integral ek empty interval par hai, isliye saare terms $0$ hain. No burn, no gain, no loss. ✔ Bookkeeping degenerate limit par bhi self-consistent hai.

Ek-picture summary

Ek velocity arrow, ek neeche-pointing gravity, angle $γ$ se ek slow-you piece ( $sin$ ) aur ek bend-you piece ( $cos$ ) mein split; Newton path ke saath, mass se divide kiya, burn par sum kiya — loss bookkeeping nikal aata hai, aur single knob $γ$ dono losses ko ek see-saw par trade karta hai jiska bottom gravity turn hai.

Recall Feynman: poora walkthrough plain words mein

Woh arrow banao jis par rocket move karta hai. Gravity seedha neeche point karti hai; us down-arrow ko us part mein slice karo jo arrow ke against line up hai aur us part mein jo uske sideways hai. Against-part $g sin γ$ hai aur speed churata hai; sideways-part $g cos γ$ hai aur bas tumhara path free mein bend karta hai. Arrow ke saath Newton ka law likho: thrust aage, drag aur gravity-along peeche. Weight se divide karo taaki sab kuch "speed gained per second" ho, phir burn ke har second ko add karo. Thrust total fuel ka ideal $Δ v$ hai; drag total aur gravity total woh do losses hain jo tumne pay kiye lekin rakhi nahi. Angle $γ$ dono losses ke andar chhupta hai aur unhe opposite ways mein dhakelta hai — steep gravity par maarta hai, flat drag par maarta hai — isliye sabse sasti climb sirf itna tip over karti hai ki gravity khud turning kar sake.

Recall Quick self-test

Gravity loss sine kyun hai cosine kyun nahi? ::: Kyunki gravity ka along-path (speed-robbing) piece $g sin γ$ hai — yeh max hai jab vertical ho ( $sin 9 0^{\circ} = 1$ ) aur zero jab flat ho ( $sin 0^{\circ} = 0$ ), exactly match karta hai ki gravity motion ko kaise oppose karti hai. Newton ke law ko $m$ se divide karne se kya hota hai? ::: Yeh har force ko ek acceleration (m/s per s) mein badal deta hai, isliye time par integrate karne se directly ek $Δ v$ budget milta hai. $γ = 0^{\circ}$ par gravity loss zero hai — kya gravity kuch nahi kar rahi? ::: Nahi: uska across-path piece $g cos γ$ full strength par hai, trajectory bend kar raha hai (tumhe ground ki taraf curve kar raha hai). Woh bas tumhe slow nahi kar raha.