Trajectory optimization — minimum gravity loss, minimum drag loss: Worked examples

3.4.15 · D3 · Physics › Rocket Flight Mechanics › Trajectory optimization — minimum gravity loss, minimum drag

Yeh page parent topic ka "run every case" companion hai. Hum un do loss integrals ko lete hain jo usne derive kiye the aur unhe har tarah ke input ke through drive karte hain jo ek problem de sakti hai: har flight-path angle regime, zero aur degenerate cases, limits, ek word problem, aur ek exam twist. Kuch bhi "obvious" chhoda nahi gaya.

Shuru karne se pehle, teen quantities jo parent ne build ki thi, plain words mein restate ki gayi hain taaki yeh page akela khad ho sake:

Scenario matrix

Neeche har example us matrix ke cell ke saath tagged hai jise woh exercise karta hai. Milke yeh sab ko cover karte hain.

Cell	Kya test karta hai	Yeh kyun dikkat kar sakta hai
A — $γ = 9 0^{\circ}$ (vertical)	$sin γ = 1$ , worst-case gravity loss	Naive "bas $g t$ add karo"
B — $0^{\circ} < γ < 9 0^{\circ}$ (slanted)	partial $sin γ$ , quadrant-I geometry	sine lena bhool jaana
C — $γ = 0^{\circ}$ (horizontal)	$sin γ = 0$ , degenerate: gravity loss vanish ho jaata hai	yeh sochna "loss = 0 toh yeh best hai"
D — $γ$ varying with $t$	integral $g \overset{γ}{ˉ} t$ NAHI hai; $sin γ$ integrate karna padega	sine ko integral se bahar kheenchna
E — drag at a fixed condition	$D / m \cdot Δ t$ evaluation, units	N, kg, s mix karna
F — gravity-turn pitch rate	$d γ / d t = - g cos γ / v$ , sign & limits	$cos$ vs $sin$ confusion
G — limiting case $v \to \infty$ or $v \to 0$	turn rate $\to 0$ ya $\to \infty$	pitch kick early kyun hota hai
H — real-world word problem	full loss budget bookkeeping	words ko integrals mein translate karna
I — exam twist (mixed sign / trade-off)	do profiles compare karna, min sum choose karna	sirf ek term optimize karna

Example 1 — Cell A: pure vertical burn (worst case)

Worked example Ek model rocket seedha upar jaata hai

Burn $t_{b} = 45 s$ , seedha upar $γ = 9 0^{\circ}$ poore time, $g = 9.8 m/s^{2}$ constant lo. Gravity loss nikalo.

Forecast: padhne se pehle andaza lagao — loss $9.8 \times 45$ se zyada hogi ya kam?

Integral likho. $Δ v_{grav} = \int_{0}^{45} g sin γ d t$ . Yeh step kyun? Loss definition ke roop mein integral hai; hum definition se shuru karte hain, kabhi bhi memorized shortcut se nahi.
$γ = 9 0^{\circ}$ daalo. $sin 9 0^{\circ} = 1$ , aur yeh kabhi change nahi hota, toh integrand constant $g$ hai. Yeh step kyun? Vertical flight woh ek case hai jahan gravity poori tarah motion ko oppose karti hai — opposite-to-velocity component poore $g$ ke barabar hoti hai.
Constant integrate karo. $\int_{0}^{45} 9.8 d t = 9.8 \times 45 = 441 m/s$ . Yeh step kyun? Ek constant times interval length; kuch bhi subtle nahi.

Verify: units $= (m/s^{2}) (s) = m/s$ ✔. Aur tumhara forecast: yeh exactly $9.8 \times 45$ hai, isse zyada nahi — kyunki $9 0^{\circ}$ par sine apne maximum $1$ par hai, yeh ceiling hai. Koi bhi aur angle kam dega. "Ideal" side of the budget ke liye Tsiolkovsky Rocket Equation dekho.

Example 2 — Cell B: slanted (quadrant-I) burn

Worked example Same burn,

$γ = 4 0^{\circ}$ par rakhi gayi $t_{b} = 45 s$ , $g = 9.8$ , lekin ab velocity arrow poore time $4 0^{\circ}$ upar horizon ke rakhti hai.

Forecast: answer Example 1 ke $441$ ka kaunsa factor se multiply hoga?

Same integral, naya angle. $Δ v_{grav} = \int_{0}^{45} 9.8 sin 4 0^{\circ} d t$ . Yeh step kyun? Sirf $g$ ka component flight direction ke saath speed churaata hai; woh component $g sin γ$ hai. Figure dekho: vertical gravity arrow ek piece flight path ke along ( $g sin γ$ ) aur ek piece across ( $g cos γ$ ) mein split hoti hai.

Constant sine bahar nikalo. $sin 4 0^{\circ} \approx 0.6428$ , toh $= 9.8 \times 0.6428 \times 45$ . Yeh step kyun? $γ$ yahan fixed hai, toh $sin γ$ ek number hai aur integral se bahar aa jaata hai.
Multiply karo. $9.8 \times 45 \times 0.6428 \approx 283.5 m/s$ .

Verify: multiplier $sin 4 0^{\circ} = 0.6428$ hai, toh $441 \times 0.6428 = 283.5$ ✔ — match karta hai, aur yeh vertical case se kam hai, exactly jaisa geometry ne promise kiya tha.

Example 3 — Cell C: horizontal (degenerate) burn

Example 4 — Cell D: burn ke dauran $γ$ change hota hai (integrate karna zaroori)

Worked example Pitch program jo linearly tip over karta hai

$t_{b} = 100 s$ burn mein flight-path angle linearly $9 0^{\circ}$ se $0^{\circ}$ tak girta hai: $γ (t) = 9 0^{\circ} (1 - \frac{t}{100}) = \frac{π}{2} (1 - \frac{t}{100}) rad .$ $g = 9.8$ constant lo. $Δ v_{grav}$ nikalo.

Forecast: average angle $4 5^{\circ}$ hai. Ek tempting shortcut $9.8 \times 100 \times sin 4 5^{\circ} \approx 693$ hai. Kya true integral is se zyada hogi ya kam?

Honest integral set up karo. $Δ v_{grav} = \int_{0}^{100} 9.8 sin [\frac{π}{2} (1 - \frac{t}{100})] d t$ . Yeh step kyun? $γ$ constant nahi hai, toh $sin γ$ integral se bahar nahi aa sakta. Average ka sine $\neq =$ sine ka average.
Substitute karo. Mano $u = \frac{π}{2} (1 - \frac{t}{100})$ . Toh $d u = - \frac{π}{200} d t$ , toh $d t = - \frac{200}{π} d u$ . Limits: $t = 0 \to u = \frac{π}{2}$ , $t = 100 \to u = 0$ . Yeh step kyun? Substitution ek messy composite ko clean $\int sin u d u$ mein convert kar deta hai — standard tool jab sine ke andar wala part $t$ mein linear ho.
Integrate karo. $9.8 \int_{π /2}^{0} sin u (- \frac{200}{π}) d u = 9.8 \cdot \frac{200}{π} \int_{0}^{π /2} sin u d u = 9.8 \cdot \frac{200}{π} [- cos u]_{0}^{π /2}$ .
Evaluate karo. $[- cos u]_{0}^{π /2} = (0) - (- 1) = 1$ , toh $Δ v_{grav} = 9.8 \cdot \frac{200}{π} \approx 623.8 m/s$ .

Verify: naive " $sin$ of average $= 4 5^{\circ}$ " ne $693$ diya; true value $623.8$ kam hai ✔ — kyunki sine $[0, 9 0^{\circ}]$ par concave hai, toh sine ka average, average ke sine se kam hota hai. Yahi reason hai kyun integrate karna zaroori hai. Units: $(m/s^{2}) (s) = m/s$ ✔.

Example 5 — Cell E: fixed condition par drag loss

Worked example Max-Q ke paas drag

Worst dynamic-pressure moment par: $ρ = 0.5 kg/m^{3}$ , $v = 300 m/s$ , $C_{D} = 0.3$ , $A = 2 m^{2}$ , vehicle mass $m = 1.5 \times 1 0^{4} kg$ , aur yeh condition lagbhag $Δ t = 15 s$ tak rehti hai. $Δ v_{drag}$ estimate karo.

Forecast: kya yeh tens, hundreds, ya thousands of m/s hogi? (Gravity losses hundreds thi — compute karne se pehle guess karo.)

Drag force. $D = \frac{1}{2} ρ v^{2} C_{D} A = \frac{1}{2} (0.5) (300)^{2} (0.3) (2)$ . Yeh step kyun? Yeh aerodynamic drag formula hai (dekho Aerodynamic Drag and Max-Q); $\frac{1}{2} ρ v^{2}$ dynamic pressure hai, $C_{D} A$ effective area hai.
Compute karo. $(300)^{2} = 90000$ ; $\frac{1}{2} (0.5) = 0.25$ ; $0.25 \times 90000 = 22500$ ; $\times 0.3 = 6750$ ; $\times 2 = 13500 N$ .
Drag deceleration. $D / m = 13500/15000 = 0.9 m/s^{2}$ . Yeh step kyun? $Δ v_{drag}$ $D / m$ integrate karta hai; deceleration force over mass hoti hai.
Duration se multiply karo. $Δ v_{drag} \approx 0.9 \times 15 = 13.5 m/s$ .

Verify: units: $N / kg = m/s^{2}$ , times $s$ $= m/s$ ✔. Answer tens of m/s hai — Examples 1–4 ke gravity losses se kaafi neeche, parent ki baat confirm karta hai ki gravity loss trade ko dominate karti hai.

Example 6 — Cell F: gravity-turn pitch rate (sign matters karta hai)

Worked example Velocity arrow kitni tezi se turn kar rahi hai?

Ek gravity turn ke dauran $v = 250 m/s$ , $γ = 7 0^{\circ}$ , $g = 9.8$ par. $d γ / d t$ nikalo.

Forecast: positive ya negative? Tez ya slow (degrees per second mein)?

Turn law use karo. $\frac{d γ}{d t} = - \frac{g cos γ}{v}$ . Yeh step kyun? Gravity turn mein only sideways force gravity ka across-path component $g cos γ$ hai — thrust turning par kharach nahi hoti, yahi reason hai steering loss $\approx 0$ hoti hai. Note karo yeh $cos$ hai, woh component perpendicular to velocity (compare karo loss se, jo $sin$ use karta tha, parallel wala). Example 2 ka figure dono pieces show karta hai.
Numbers daalo. $cos 7 0^{\circ} \approx 0.3420$ , toh $= - \frac{9.8 \times 0.3420}{250} = - \frac{3.3516}{250}$ .
Compute karo. $\approx - 0.01341 rad/s$ .
Degrees mein convert karo. $\times \frac{180}{π} \approx - 0.76 8^{\circ} / s$ .

Verify: sign negative hai ✔ — $γ$ shrink ho raha hai, rocket horizontal ki taraf tip kar raha hai, jaisa hona chahiye. Magnitude $1^{\circ} / s$ se kam ✔, slow, parent ki warning se match karta hai ki turning gentle hoti hai jab aap move kar rahe ho.

Example 7 — Cell G: limits (pitch kick early kyun hoti hai)

Worked example Same formula do extreme speeds par

$γ = 7 0^{\circ}$ , $g = 9.8$ rakho. Turn rate compare karo ek tiny $v = 10 m/s$ (liftoff ke turant baad) versus ek large $v = 2500 m/s$ (late ascent) par.

Forecast: kaun si speed zyada tezi se turn karne deti hai?

Slow case. $d γ / d t = - \frac{9.8 \times 0.3420}{10} = - 0.33516 rad/s \approx - 19. 2^{\circ} / s$ . Yeh step kyun? $1/ v$ bada hota hai jab $v$ chhota ho, toh turn rapid hoti hai — ek chhota sa nudge pehle path ko bahut zyada bend kar deta hai.
Fast case. $d γ / d t = - \frac{9.8 \times 0.3420}{2500} = - 0.0013406 rad/s \approx - 0.076 8^{\circ} / s$ . Yeh step kyun? Ab $1/ v$ tiny hai; same gravity (tez, momentum-se-bhari) velocity arrow ko barely bend kar paati hai.
Limit. Jaise $v \to \infty$ , $d γ / d t \to 0$ : tum essentially turn nahi kar sakte. Jaise $v \to 0^{+}$ , $d γ / d t \to - \infty$ : infinitely sensitive.

Verify: do rates ka ratio $= 2500/10 = 250$ ✔ (turn rate $\propto 1/ v$ ). Slow-case rate fast-case rate se $250 \times$ hai. Lesson: initial pitch kick tab daalni chahiye jab $v$ chhota ho; bahut der karo aur vehicle itni fast ho jaati hai ki fuel khatam hone se pehle turn nahi ho sakti — steering-loss alternative tab tumhe cost karega.

Example 8 — Cell H: real-world loss budget

Worked example Word problem — budget close karna

Ek launcher ka ideal $Δ v_{ideal} = 9.4 km/s$ hai (from Tsiolkovsky Rocket Equation). Flight analysis gravity loss $1.6 km/s$ , drag loss $0.1 km/s$ , steering loss $0.05 km/s$ deta hai. Actually kitna orbital speed deliver hota hai, aur kya yeh required orbital velocity $7.8 km/s$ ke liye kaafi hai?

Forecast: margin hogi ya shortfall?

Bookkeeping equation. $Δ v_{orbit} = Δ v_{ideal} - Δ v_{grav} - Δ v_{drag} - Δ v_{steer}$ . Yeh step kyun? Yeh parent ka $Δ v$ ledger conservation hai — har loss ideal se subtract hoti hai.
Plug in karo. $= 9.4 - 1.6 - 0.1 - 0.05 = 7.65 km/s$ .
Requirement se compare karo. Required $7.8 km/s$ hai (dekho Orbital insertion and required orbital velocity).

Verify: $9.4 - 1.6 - 0.1 - 0.05 = 7.65$ ✔. Delivered $7.65 < 7.8$ required $\Rightarrow$ shortfall of $0.15 km/s$ — vehicle orbit nahi reach karta. Fix: gravity loss reduce karo (zyada thrust-to-weight, chhota vertical phase) ya propellant add karo.

Example 9 — Cell I: exam twist (minimum-sum profile choose karo)

Worked example Do pitch profiles mein se kaun sa better hai?

Same $t_{b} = 80 s$ , $g = 9.8$ , do candidate constant- $γ$ phases (idealized):

Profile P (steep): $γ = 7 5^{\circ}$ . Is steep, fast-exit path ke paas drag estimated $Δ v_{drag} = 40 m/s$ .
Profile Q (shallow): $γ = 3 5^{\circ}$ . Neeche aur fast rehna, drag estimated $Δ v_{drag} = 260 m/s$ .

Kaun sa total loss $Δ v_{grav} + Δ v_{drag}$ chhota hai?

Forecast: steep gravity par haarta hai, shallow drag par — winner guess karo.

Gravity loss, P. $9.8 \times 80 \times sin 7 5^{\circ} = 784 \times 0.9659 \approx 757.3 m/s$ . Total P $= 757.3 + 40 = 797.3$ . Yeh step kyun? Steep $\Rightarrow$ bada $sin γ \Rightarrow$ bada gravity loss, lekin fast climb drag chhota rakhta hai.
Gravity loss, Q. $9.8 \times 80 \times sin 3 5^{\circ} = 784 \times 0.5736 \approx 449.7 m/s$ . Total Q $= 449.7 + 260 = 709.7$ . Yeh step kyun? Shallow $\Rightarrow$ chhota gravity loss, lekin drag balloon ho jaata hai.
Totals compare karo. $797.3$ (P) vs $709.7$ (Q).

Verify: $797.3 > 709.7$ , toh Profile Q jeet jaata hai lagbhag $87.6 m/s$ se ✔. Notice karo ki koi bhi ek term akela decide nahi karta — dono add karne padte hain, exactly wahi see-saw logic. (Real optimum na koi fixed angle hai balki ek gravity turn hai jo unke through sweep karta hai, jo dono ko beat karta hai.)

Recall Quick self-test

$γ = 9 0^{\circ}$ par, constant $g$ ke saath burn $t_{b}$ mein gravity loss hai ::: $g t_{b}$ (kyunki $sin 9 0^{\circ} = 1$ ). Jab $γ$ vary kare toh $sin \overset{γ}{ˉ}$ use karne ki jagah $sin γ$ integrate kyun karna chahiye? ::: Kyunki sine $[0, 9 0^{\circ}]$ par concave hai; sine ka average, average ke sine ke barabar nahi hota. Gravity turn mein, turn rate $sin γ$ se govern hoti hai ya $cos γ$ se? ::: $cos γ$ se — gravity ka woh component jo velocity ke perpendicular hai. Pitch kick early kyun apply karte hain? ::: Turn rate $\propto 1/ v$ hai; sirf jab $v$ chhota ho tab gravity path ko appreciably bend kar sakti hai.