Pitch program — open-loop pitch-over: Worked examples

3.4.14 · D3 · Physics › Rocket Flight Mechanics › Pitch program — open-loop pitch-over

Yeh page parent note ke gravity-turn pitch program ka case-by-case drill hai. Hum ek hi master law lete hain aur use har corner mein push karte hain: har angle regime, do degenerate endpoints ( $γ = 9 0^{\circ}$ seedha upar, $γ = 0^{\circ}$ horizontal), turn rate ki sign, ek real launch word-problem, aur ek exam-style twist.

Ek hi tool jo hum poore waqt use karenge:

Scenario matrix

Is topic mein jo bhi situation aa sakti hai woh inn cells mein se kisi ek mein fit hoti hai. Neeche ke worked examples mein har ek ke saath uski cell ka tag lagaya gaya hai.

Cell	Kya special hai	Example
A. Near-vertical ( $γ \to 9 0^{\circ}$ )	$cos γ \to 0$ , toh $\overset{γ}{˙} \to 0$ — muskil se murta hai	Ex 1
B. Mid-turn ( $γ \approx 4 5^{\circ}$ )	$cos γ$ bada — sabse tez turning	Ex 2
C. Near-horizontal ( $γ \to 0^{\circ}$ )	$cos γ \to 1$ , lekin $v$ bahut bada — turn phir slow ho jaata hai	Ex 3
D. Speed dependence ( $γ$ fix, $v$ vary)	$\overset{γ}{˙} \propto 1/ v$ — inverse law	Ex 4
E. Degenerate: exactly vertical ( $γ = 9 0^{\circ}$ )	$\overset{γ}{˙} = 0$ — "stuck" case, kick kyun zaroori hai	Ex 5
F. Degenerate: exactly horizontal ( $γ = 0^{\circ}$ )	maximum $cos γ = 1$ ; burnout instant check	Ex 6
G. Real-world word problem	units convert karo, ek time window mein turn dhundho	Ex 7
H. Exam twist: sign / integration	time eliminate karo, $d v / d γ$ , over-vs-under-kick	Ex 8

Figure — Pitch program — open-loop pitch-over

Upar ka figure poori turn mein $cos γ$ plot karta hai. Notice karo ki dono ends (vertical aur horizontal) woh jagah hain jahan raw turning force sabse chhoti aur sabse badi hoti hai — lekin actual turn rate $v$ se bhi divide hoti hai, isliye real turning pehle bunched hoti hai (cell A→B) jab $v$ abhi chhota hota hai.

Worked examples

Ex 1 — Cell A: near-vertical, high speed

Worked example Kick ke baad turn rate

$v = 120 m/s$ , $γ = 8 9^{\circ}$ , $g = 9.8 m/s^{2}$ . $\overset{γ}{˙}$ dhundho rad/s aur degrees per second dono mein.

Forecast: pehle guess karo — kya turn ek degree per second ka bada fraction hoga, ya bahut chhota?

$cos 8 9^{\circ}$ compute karo. $cos 8 9^{\circ} = 0.017452$ . Yeh step kyun? Formula ko gravity ka sideways fraction chahiye; $8 9^{\circ}$ par almost saari gravity seedha path ke along wapas kheenchti hai, sirf ek chhoti si sliver sideways kheenchti hai.
Plug in karo. $\overset{γ}{˙} = - \frac{9.8 \times 0.017452}{120} = - 0.0014253 rad/s$ . Yeh step kyun? Seedha ek law ka use; minus "hamesha neeche jhukta hai" sign rakhta hai.
Degrees mein convert karo. $180/ π = 57.2958$ se multiply karo: $- 0.0014253 \times 57.2958 = - 0.0816 6^{\circ} / s$ . Yeh step kyun? Insaan degrees padhte hain; rad/s formula ki natural unit hai.

Verify karo: units — $[m/s^{2}] / [m/s] = 1/ s$ , jo rad/s hai. ✓ Aur answer bahut chhota hai (0.08°/s), jo cell A ke promise se match karta hai ki near-vertical rocket muskil se jhukta hai — yeh prove karta hai ki tumhe deliberately kick dena zaroori hai.

Ex 2 — Cell B: mid-turn, sabse tez bending

Ex 3 — Cell C: near-horizontal, bahut tez

Ex 4 — Cell D: inverse-speed law

Ex 5 — Cell E: degenerate vertical case

Worked example Bilkul seedha upar — kya yeh kabhi jhukta hai?

Ek rocket perfectly vertical uthta hai: $γ = 9 0^{\circ}$ exactly, koi bhi speed $v$ . $\overset{γ}{˙}$ dhundho.

Forecast: bina hawa, bina kick ke, kya yeh kabhi apne aap tip over karega?

$cos 9 0^{\circ} = 0$ . Kyun? Exactly vertical par, gravity seedha flight path ke neeche point karti hai — yeh rocket ko slow karti hai lekin use turn karne ke liye zero sideways component hai.
Plug in karo. $\overset{γ}{˙} = - \frac{g \times 0}{v} = 0$ kisi bhi $v$ ke liye. Kyun? Zero numerator poori cheez ko zero kar deta hai.

Verify karo: $\overset{γ}{˙} = 0$ ka matlab angle kabhi nahi badlega — ek perfectly vertical rocket hamesha vertical rahega aur kabhi orbit nahi pahunchega. Yeh mathematical proof hai ki pitch kick mandatory hai: tumhe manually $γ = 9 0^{\circ}$ symmetry todna hoga. Yeh equation ka ek sach mein degenerate fixed point hai. ✓

Ex 6 — Cell F: degenerate horizontal case

Ex 7 — Cell G: real-world word problem

Worked example Ek time window mein angle change

Kick ke baad ek rocket roughly $v \approx 150 m/s$ aur $γ \approx 8 5^{\circ}$ (lagbhag constant) 10-second window ke liye rakhta hai. Approximately kitne degrees flight-path angle us window mein girta hai? (Chhote window par $\overset{γ}{˙}$ ko constant maano.)

Forecast: kuch tenths of a degree, kuch degrees, ya tens of degrees?

Instantaneous rate. $cos 8 5^{\circ} = 0.087156$ , toh $\overset{γ}{˙} = - \frac{9.8 \times 0.087156}{150} = - 0.0056942 rad/s$ . Kyun? Master law se per-second turn rate nikalo.
°/s mein. $- 0.0056942 \times 57.2958 = - 0.3262 3^{\circ} / s$ . Kyun? Convert karo taaki answer degrees mein padha ja sake.
Window se multiply karo. $Δ γ \approx \overset{γ}{˙} \times Δ t = - 0.32623 \times 10 = - 3.262 3^{\circ}$ . Yeh step kyun? Chhote window par jahan $\overset{γ}{˙}$ roughly constant hai, angle change ≈ rate × time (curve ka straight-line approximation).

Verify karo: rocket 10 s mein ~85° se ~81.7° tak jhukta hai — ek gentle few-degree bend, bilkul "ek baar kick, phir coast" waali picture. Agar humne galti se $v = 15$ m/s use kiya hota toh hum wild 32° turn predict karte, toh units aur magnitude sanity-check result ko verify karte hain. ✓

Ex 8 — Cell H: exam twist, time eliminate karo aur kick judge karo

Worked example Degree turn per speed gain (drag-free)

Parent note ke time-eliminated form ke saath constant thrust-to-weight $n = a / g = 2$ , drag neglected: $\frac{d v}{d γ} = \frac{v ( n - sin γ )}{- cos γ}$ . Is slope ko $γ = 6 0^{\circ}$ , $v = 300 m/s$ par evaluate karo, aur batao uski sign ka matlab kya hai.

Forecast: jab rocket turn karta hai (γ ghatta hai), kya yeh yahan speed up ho raha hai ya slow down?

Values. $sin 6 0^{\circ} = 0.866025$ , $cos 6 0^{\circ} = 0.5$ . Kyun? Slope formula ko current angle par dono chahiye.
Numerator. $v (n - sin γ) = 300 (2 - 0.866025) = 300 \times 1.133975 = 340.1924$ . Kyun? $n - sin γ$ thrust-along-path minus gravity-along-path hai; yahan thrust jeet raha hai.
$- cos γ$ se divide karo. $\frac{340.1924}{- 0.5} = - 680.3849$ (m/s per rad of $γ$ ). Kyun? $d v / d γ$ complete karta hai.
Sign interpret karo. $γ$ turn ke dauran ghatta hai ( $d γ < 0$ ), aur $d v / d γ < 0$ , toh $d v = (d v / d γ) d γ > 0$ : rocket turn karte waqt speed up hota hai. Kyun? Do negatives multiply hokar positive $Δ v$ dete hain.

Verify karo: $- 680.3849$ per rad = $- 680.3849/57.2958 = - 11.875$ m/s per degree of turn. Jab thrust gravity-along-path term se zyada hai ( $n > sin γ$ ), vehicle jhukते waqt accelerate karta hai — physically correct ek healthy ascent ke liye. Agar $n < sin γ$ hota toh sign flip hota aur rocket speed khota (ek under-powered, gravity-loss-heavy climb). ✓

Recall Kaun sa cell "stuck" wala hai, aur kyun?

Cell E, $γ = 9 0^{\circ}$ exactly ::: $cos 9 0^{\circ} = 0$ se $\overset{γ}{˙} = 0$ hota hai, toh ek perfectly vertical rocket kabhi nahi jhukta — pitch kick ko yeh symmetry todna hi padta hai.

Recall Sabse tez

turn rate horizontal par kyun NAHI hoti (jahan $cos γ$ sabse bada hai)? Kyunki $\overset{γ}{˙} \propto 1/ v$ ::: jab tak $γ$ chhota hota hai, $v$ bahut bada ho jaata hai, aur denominator mein badi speed maximum $cos γ$ ko overwhelm kar deti hai.

Dekho bhi: Gravity turn trajectory · Angle of attack and dynamic pressure (max-Q) · Thrust-to-weight ratio · Gravity loss and steering loss · Tsiolkovsky rocket equation · Closed-loop ascent guidance (PEG / IGM) · Attitude control and thrust vectoring (gimbal).