Visual walkthrough — Pitch program — open-loop pitch-over

3.4.14 · D2 · Physics › Rocket Flight Mechanics › Pitch program — open-loop pitch-over

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Step 1 — "Flight-path angle" kya hota hai? Velocity arrow draw karo

KYA HAI. Ek rocket hawaon mein kisi jagah hai, kisi direction mein ja raha hai. Hum uski velocity ko ek arrow ki tarah draw karte hain: length = kitni tez (isko $v$, yaani speed kehte hain), direction = abhi motion kahan ja rahi hai isi waqt.

KYUN. "Turning" ki baat karne se pehle humein ek number chahiye jo bataye ki motion kitni tilted hai. Woh number hai flight-path angle, likha jaata hai $\gamma$ (Greek letter "gamma"). Yeh angle hai velocity arrow aur flat horizon ke beech ka.

PICTURE. Figure s01 dekho. Horizontal dashed line zameen/horizon hai. Blue arrow velocity $v$ hai. Unke beech ka pale-yellow wedge $\gamma$ hai.

• $\gamma =90^{\circ }$ → arrow seedha upar point karta hai (pad se abhi abhi uthaa).
• $\gamma =0^{\circ }$ → arrow sideways point karta hai (orbital, yahi goal hai).
• $\gamma$ time ke saath chhota hona = "rocket pitch over kar raha hai."

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Step 2 — $\dot{\gamma }$ kya hai? Dot matlab "rate of change"

KYA HAI. Kisi letter ke upar dot lagate hain matlab "woh letter har second mein kitni tez change ho raha hai." Toh $\dot{\gamma }$ (bolo "gamma-dot") = flight-path angle har second mein kitne degrees (ya radians) change hota hai.

KYUN. Poori ascent ek kahani hai $\gamma$ ki, jo $90^{\circ }$ se $0^{\circ }$ tak slide karti hai. Us slide ki speed — $\dot{\gamma }$ — exactly wahi hai jo ek pitch program control karta hai. Agar $\dot{\gamma }$ bahut bada ho toh hum turn over karke dive karte hain; bahut chhota ho toh hum kabhi horizontal nahi jaate.

PICTURE. Figure s02 mein same blue arrow ek second ke antar par do instants mein dikhaaya gaya hai. Yellow wedge chhota ho gaya: woh shrinkage, per second, hi $\dot{\gamma }$ hai. Kyunki $\gamma$ decrease ho raha hai, $\dot{\gamma }$ ek negative number hai — yeh sign final formula mein wapas aayega, toh abhi yaad rakh lo.

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Step 3 — Rocket par forces, aur kaun sa bend kar sakta hai path ko

KYA HAI. Teen forces hain hamare point-mass rocket par: thrust $T$ (engine ka push, body ke saath), drag $D$ (air resistance, backward), aur weight $mg$ (gravity, seedha neeche). Yahan $m$ = mass, $g$ = gravity ka pull per kilogram ($\approx 9.8{\text{m/s}}^2$).

KYUN. Ek ideal gravity turn mein body exactly velocity ke saath point karta hai — angle of attack $\alpha \approx 0$. Matlab thrust aur drag dono arrow ke saath lie karte hain. Arrow ke saath wali force sirf use speed up ya slow down kar sakti hai — woh use bend nahi kar sakti. Sirf woh force jo arrow ke sideways ho, use turn kar sakti hai. Gravity seedha neeche point karti hai, toh uska ek hissa tilted arrow ke sideways hota hai. Gravity akela turner hai.

PICTURE. Figure s03 mein velocity arrow $\gamma$ angle par tilted hai, $T$ aur $D$ uske saath draw hain (chalk-blue aur pink), aur weight $mg$ seedha neeche point kar raha hai (yellow). Note karo ki weight hi akela ek aisa arrow hai jo motion ke saath aligned nahi hai.

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Step 4 — Gravity ko velocity ke "saath" aur "across" mein split karo

KYA HAI. Hum downward weight $mg$ ko do pieces mein todte hain, tilted velocity arrow ke relative measure karke:

• woh hissa jo arrow ke saath hai (climbing ko oppose karta hai): $mg\sin \gamma$,
• woh hissa jo arrow ke across hai (use bend karta hai): $mg\cos \gamma$.

KYUN. Humne $\sin$ aur $\cos$ use kiya — kyun yahi? Kyunki $\sin$ aur $\cos$ exactly woh tools hain jo jawaab dete hain "ek seedha-neeche arrow is doosri direction mein kitna point karta hai?" Woh weight ki do axes par shadow-lengths hain jo humein chahiye. Hum inhe choose karte hain, tan ya kuch aur nahi, kyunki humein projections chahiye, aur sin/cos hi ek right triangle ke projection ratios hain.

PICTURE. Figure s04 mein woh right triangle draw hai jo weight $mg$ (down-arrow) aur velocity ke saath rotate hue do axes se banta hai. Along-piece arrow ke saath chipka hai; across-piece usse perpendicular bahar nikal raha hai. Triangle ke andar ka angle $\gamma$ hai (yeh copy over hota hai kyunki horizon aur vertical dono ek jaisa $\gamma$ se rotate hote hain).

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Step 5 — Newton across the path: turning = centripetal-jaisi bending

KYA HAI. $v$ length wala velocity arrow jo $\dot{\gamma }$ rate se swing karta hai, uski tip sideways $v\dot{\gamma }$ speed se move karti hai. Newton kehta hai: (mass)×(woh sideways acceleration) = (across force). Akela across force $mg\cos \gamma$ hai, neeche point karta hua, jo $\gamma$ ko reduce karta hai. Toh: $mv\dot{\gamma }=-mg\cos \gamma$

Minus kyun? Gravity ka across-piece neeche point karta hai aur neeche matlab $\gamma$ shrink hona (Step 2 ne bataya ki shrinking = negative $\dot{\gamma }$). Toh right side negative honi chahiye — hum minus explicitly likhte hain.

Term by term:

• $m$ — rocket ki mass (redirect karna kitna mushkil hai).
• $v\dot{\gamma }$ — velocity arrow ki tip sideways kitni tez swing karti hai.
• $mg\cos \gamma$ — Step 4 se gravity ka across-component.
• $-$ — kyunki woh gravity component $\gamma$ ko neeche ki taraf bend karta hai.

PICTURE. Figure s05 mein velocity arrow ek second mein ek chhota sa angle sweep karta dikhaaya gaya hai; tip jo chhota arc trace karti hai uski length $v\dot{\gamma }$ hai, aur yellow across-force $mg\cos \gamma$ us arc ke saath point karta hai, tip ko neeche kheench raha hai.

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Step 6 — Mass cancel karo: gravity-turn law appear hoti hai

KYA HAI. Mass $m$ is equation ke dono sides par hai: $mv\dot{\gamma }=-mg\cos \gamma .$ Dono sides ko $m$ se divide karo (allowed hai — $m$ kabhi zero nahi hota), phir $v$ se divide karo: $\boxed{\dot{\gamma }=-\frac{g\cos \gamma }{v}}$

Yeh kyun beautiful hai. Mass gaayab ho gaya. Ek bhaari rocket aur ek halka rocket same speed aur angle par same rate se turn karte hain. Turn sirf teen cheezein set karti hain: gravity $g$, current angle $\gamma$, aur current speed $v$.

Term by term:

• $g$ — zyada gravity → zyada tez turn.
• $\cos \gamma$ — vertical ke paas ($\gamma \to 90^{\circ }$) yeh ~0 hai, toh turn barely-there hai; horizontal ke paas yeh ~1 hai, turning sabse strong hai.
• $v$ denominator mein — tez rockets slowly turn karte hain; slow, low rockets quickly turn karte hain. (Yahi hai why speed matters flip: zyada inertia, kam bendable.)
• $-$ — arrow hamesha neeche tip kar raha hota hai.

PICTURE. Figure s06 mein $\dot{\gamma }$ vs $\gamma$ plot hai: ek curve jo $90^{\circ }$ par flat (near zero) hai aur $0^{\circ }$ ke paas steepest hai, teen sample rockets alag $v$ par jo faster $v$ ke liye slower turns dikhate hain.

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Step 7 — Edge cases: vertical start, horizontal end, zero speed

KYA HAI. Humein har corner check karna hai taaki reader ko kabhi surprise na mile.

1. Seedha upar, $\gamma =90^{\circ }$: $\cos 90^{\circ }=0\Rightarrow \dot{\gamma }=0$. Rocket apne aap kabhi turn nahi karega. Isliye deliberately vertical se ek kick ki zaroorat hai — tumhe manually ek chhota $\cos \gamma \ne 0$ banana padta hai process shuru karne ke liye.
2. Kick ke thodi der baad, $\gamma =89^{\circ }$: $\cos 89^{\circ }\approx 0.017$, bahut chhota par non-zero → ek slow turn shuru hoti hai, jo $\cos \gamma$ badhata hai, jo turn badhata hai: self-amplifying.
3. Horizontal, $\gamma =0^{\circ }$: $\cos 0^{\circ }=1$ → maximum turn rate. Agar engine yahan bhi jal raha hai, gravity nose ko horizon ke neeche kheenchti rehti hai — aapko cut off karna hoga warna dive ho jaate ho.
4. Zero speed, $v\to 0$: formula blow up kar jaata hai ($÷0$). Physically yeh liftoff ka instant hai jab real speed exist nahi karti — gravity-turn model tab tak apply nahi hota jab tak $v$ safely positive na ho. Isliye kick tower clear karne ke baad hoti hai, pad par nahi.

PICTURE. Figure s07 ek four-panel strip hai: frozen $90^{\circ }$ rocket, tiny-kick $89^{\circ }$ rocket, max-turn $0^{\circ }$ rocket, aur $v\to 0$ "undefined" warning.

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Step 8 — Picture par numbers (worked checks)

Recall "1/v" claim khud check karo

$v=60$ par turn rate divided by $v=120$ par turn rate ::: $2$ ke barabar hota hai — $v$ half karne se $\dot{\gamma }$ double ho jaata hai.

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Ek-picture summary

Figure s08 poori kahani stack karta hai: tilted velocity arrow angle $\gamma$ ke saath, weight along ($mg\sin \gamma$) aur across ($mg\cos \gamma$) mein split, mass cancel hota hua, aur boxed result $\dot{\gamma }=-g\cos \gamma /v$ — neeche teen special angles ke saath mark kiya gaya ($90^{\circ }$ stuck, $89^{\circ }$ nudge, $0^{\circ }$ max).

Recall Feynman retelling — ise ek kahani ki tarah bolo

Ek rocket velocity arrow ke saath fly karta hai jo zameen se $\gamma$ angle par tilted hai. Us arrow ko turn karne ke liye tumhe ek force chahiye jo usse sideways point kare. Engine aur air-drag dono arrow ke saath point karte hain, toh woh sirf ise lamba ya chhota kar sakte hain — bend nahi kar sakte. Gravity, lekin, hamesha seedha neeche point karti hai, aur jab arrow tilted hota hai, gravity ka ek hissa arrow ke across jhuk jaata hai. Woh across-part, $mg\cos \gamma$, akela path ko bend karta hai. Newton kehta hai mass times arrow ka sideways swing ($mv\dot{\gamma }$) us across-force ke barabar hai, aur kyunki yeh arrow ko neeche bend karta hai, hum minus sign lagate hain. Mass dono sides se cancel ho jaata hai — light aur heavy rockets same turn karte hain — aur milta hai $\dot{\gamma }=-g\cos \gamma /v$. Jab rocket bilkul vertical hota hai, $\cos 90^{\circ }=0$, toh kuch turn nahi hota: isliye hum ise thoda vertical se kick karte hain process jagaane ke liye. Ek baar tilt hua, turning tilt badhata hai, jo turning badhata hai — ek gentle avalanche jo rocket ko seedha-upar se sideways le jaata hai, sab gravity ne kiya, koi steering nahi chahiye.

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Related: Gravity turn trajectory · Gravity loss and steering loss · Thrust-to-weight ratio · Attitude control and thrust vectoring (gimbal) · Closed-loop ascent guidance (PEG / IGM)