Visual walkthrough — Equations of motion — 3DOF point mass (trajectory analysis)

3.4.4 · D2 · Physics › Rocket Flight Mechanics › Equations of motion — 3DOF point mass (trajectory analysis)

Hum ek boxed result chase kar rahe hain, planar 3DOF set:

$\dot{V}=\frac{T-D}{m}-g\sin \gamma ,\dot{\gamma }=\frac{L-mg\cos \gamma }{mV},\dot{x}=V\cos \gamma ,\dot{h}=V\sin \gamma .$

Niche har step iska ek piece kamaata hai.

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Step 1 — Rocket ko ek arrow ki tarah draw karo

KYA. Rocket ki shape bhool jao. Use ek single dot se replace karo (centre of mass — woh ek point jo aise move karta hai jaise saari mass wahin baithi ho) aur ek arrow usse bahar nikaalta hua. Woh arrow hai velocity $\vec{V}$: uski length speed $V$ hai (dot kitne metres per second travel karta hai), aur uski direction woh hai jahan dot abhi ja raha hai.

KYU. Trajectory ke liye — range, altitude, burnout speed — hamen sirf dot kahan jaata hai se matlab hai, na ki body kaise ghoomti hai. Exactly yehi "3DOF" (3 Degrees Of Freedom: vertical plane mein position ke liye do, arrow ki speed–direction ke liye ek) ka matlab hai. Spin phek diya gaya hai.

PICTURE. Dot flat ground par baitha hai. Blue arrow $\vec{V}$ upar aur daayein point kar raha hai. Woh angle jo woh flat horizon se banata hai woh hai flight-path angle $\gamma$ (Greek letter "gamma"). $\gamma =0$ matlab level flying; $\gamma =90^{\circ }$ matlab seedha upar; aur — yeh abhi padh lo, baad mein use karenge — $\gamma <0$ matlab arrow horizon ke neeche point karta hai, yaani descending.

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Step 2 — Newton ka law ek vector statement hai

KYA. Newton ka second law kehta hai ki velocity ka arrow forces ki wajah se change hota hai jo dot ko push karte hain:

$m\frac{d\vec{V}}{dt}=\sum \vec{F}.$

• $m$ = rocket ka mass (kilograms) — ise accelerate karna kitna mushkil hai.
• $\frac{d\vec{V}}{dt}$ = "velocity arrow ki rate-of-change" — $\frac{d}{dt}$ ko padho "uske baad wali cheez, per second, kitni tezi se change ho rahi hai." Kyunki $\vec{V}$ ek arrow hai, uska change bhi ek arrow hai.
• $\sum \vec{F}$ = saare forces arrows ki tarah jod diye (thrust, drag, gravity — gravity matlab weight $mg$ jo upar define kiya).

KYU. Yeh ek line poori trajectory contain karti hai — lekin yeh stretching aur rotating ko ek arrow equation mein mila deti hai. Yeh awkward hai. Ek smart trick unhe suljhaati hai.

PICTURE. Teen force arrows usi dot se latke hain: thrust (yellow) forward, drag (pink) backward, weight (white) seedha neeche. Unka sum ek single net arrow hai — lekin ek unhelpful direction mein point karta hua.

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Step 3 — Magic split: arrow ke saath vs. arrow ke across

KYA. Horizontal/vertical ki jagah, hum do bilkul naye directions rakhte hain jo arrow ke saath travel karte hain:

• Tangential $\hat{t}$ — $\vec{V}$ ke saath point karta hai (jis taraf ja rahe hain).
• Normal $\hat{n}$ — $\vec{V}$ ke 90° across point karta hai. Hum iska orientation ek baar fix kar dete hain: $\hat{n}$ woh $\vec{V}$ hai jo 90° counter-clockwise rotate kiya gaya hai — toh jab arrow upar-aur-daayein point karta hai, $\hat{n}$ upar-aur-baayein point karta hai (path ki sky side ki taraf). Yeh choice matter karti hai: yeh across-axis par sab kuch ka sign decide karti hai.

Yeh pair velocity (wind) frame kehlata hai.

KYU. Yeh key fact hai, apni line deserve karta hai:

Arrow ke saath push sirf uski length (speed) change kar sakta hai. Arrow ke across push sirf use rotate kar sakta hai (heading change karta hai).

Toh har force ko $\hat{t}$ aur $\hat{n}$ par project karne se Newton ki ek vector equation do clean scalar equations mein split ho jaati hai — ek speed ke liye, ek turning ke liye. Yahi frame ke exist karne ki poori wajah hai. Humare chosen $\hat{n}$ ke saath, ek positive across-force arrow ko counter-clockwise rotate karta hai — yaani $\gamma$ badhata hai (use aur upar ki taraf ghuma deta hai).

PICTURE. Blue velocity arrow, jiske upar $\hat{t}$ drawn hai aur $\hat{n}$ perpendicular, upper-left ki taraf point karta hua (counter-clockwise side). Ek chhota "speedometer" symbol $\hat{t}$ par baitha hai, ek curved "steering" arrow $\hat{n}$ par.

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Step 4 — Easy forces project karo (thrust & drag)

KYA. Do forces pehle se path ki line par hain:

• Thrust $T$ (yellow) — engine $\vec{V}$ ke saath push karta hai (hum assume karte hain ki naak wind mein point kar rahi hai, toh koi sideways thrust nahi). Yeh poori tarah $+\hat{t}$ par baitha hai.
• Drag $D=\frac{1}{2}\rho V^{2}S{C}_D$ (pink) — air resistance hamesha motion oppose karti hai, toh yeh poori tarah $-\hat{t}$ par baitha hai.

Yahan $\rho$ = air density, $S$ = reference area, $C_D$ = drag coefficient (dekho Drag and Atmospheric Models). Is derivation ke liye bas $D$ ko "ek backward force jiski size hume pata hai" maano.

KYU. Yeh across-direction ($\hat{n}$) mein kuch contribute nahi karte, kyunki woh exactly $\hat{t}$ par hain. Toh abhi sirf ek tricky force gravity bachi hai — woh Step 5 hai.

PICTURE. Tangential axis par: yellow $+T$ forward, pink $-D$ backward. Normal axis par: in donon se kuch nahi.

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Step 5 — Ek triangle se gravity split karo (iska heart)

KYA. Weight $mg$ (white arrow) seedha neeche point karta hai, path ke saath nahi. Hume ise along-part aur across-part mein todna hoga. Dot par weight arrow dalo aur ek right triangle banao jiska lamba side $\hat{t}$ ke saath ho aur chhota side $\hat{n}$ ke saath.

Triangle kyun, aur sine/cosine kyun? Ek right triangle hi ek honest tarika hai yeh poochne ka ki "is down-arrow ka kitna hissa path ke saath point karta hai?" Sine aur cosine exactly woh do ratios hain jo jawab dete hain "ek angle se tilt karne ke baad length ka kaunsa fraction bachta hai." Hum inhe isliye use karte hain kyunki sawaal hi ek tilt sawaal hai.

Geometry dhyan se dekho: velocity $\gamma$ se horizontal se upar tilted hai, toh "seedha neeche" backward-along-path direction se $\gamma$ door tilted hai. Woh ek fact donon parts fix kar deta hai:

• Component along $\hat{t}$ (climb oppose karta hua): $-mg\sin \gamma$.
• Component across $\hat{n}$: $-mg\cos \gamma$. (Humare counter-clockwise $\hat{n}$ ke saath jo upper-left point karta hai, "down" ka ek negative $\hat{n}$-component hai — isliye minus sign, jo path ko earthward curve karega.)

PICTURE. White weight arrow ek yellow along-piece ($mg\sin \gamma$, path ke neeche backward point karta hua) aur ek blue across-piece ($mg\cos \gamma$, turn ke inside ki taraf point karta hua) mein decompose hua. Angle $\gamma$ triangle ke andar marked hai taaki aap padh sako kaunsa side sine leta hai.

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Step 6 — Tangential (speed) equation assemble karo

KYA. Teen along-parts jodo (thrust $+T$, drag $-D$, gravity $-mg\sin \gamma$) aur unhe $m\dot{V}$ ke barabar rakho:

$m\dot{V}=T-D-mg\sin \gamma .$

$m$ se divide karo:

$\boxed{\dot{V}=\frac{T-D}{m}-g\sin \gamma }$

Term by term:

• $\dot{V}$ — speed har second kitni tezi se badhti hai.
• $\frac{T-D}{m}$ — net forward push per kilogram (engine minus air).
• $-g\sin \gamma$ — gravity ka woh slice jo climb slow karta hai; seedha upar jaate waqt sabse bada, level flying mein zero.

KYU. Yeh "throttle" equation hai: sirf along-forces appear karte hain, toh yeh speed akela govern karta hai. Direction isme kahin nahi. Clean, exactly jaisa promise kiya tha.

PICTURE. $\dot{V}$ ke liye ek number line: yellow $+T/m$ daayein dhakelta hai, pink $-D/m$ aur white $-g\sin \gamma$ baayein; jo bachta hai woh $\dot{V}$ hai.

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Step 7 — Normal (turning) equation assemble karo

KYA. Path ke across, sirf gravity ka $\hat{n}$-piece ($-mg\cos \gamma$) aur koi bhi lift $L$ act karta hai. Ek across-force arrow ki length change nahi kar sakta; woh sirf use rotate kar sakta hai, aur sideways (centripetal) acceleration jo $V$ length ke arrow ko $\dot{\gamma }$ rate se rotate karne ke liye chahiye woh $V\dot{\gamma }$ hai. Toh lift ke saath general turning equation hai:

$mV\dot{\gamma }=L-mg\cos \gamma .$

$L$ kya hai, aur kab rakhte hain? Lift ek aerodynamic force hai velocity ke perpendicular (wings, fins, ya angle of attack par fly karne se). Humare counter-clockwise $\hat{n}$ ke saath, positive $L$ up-left point karta hai aur arrow ko upward turn karta hai. $L$ term tabhi rakho jab vehicle actually side-force generate karta ho (ek winged launcher, ek manoeuvring re-entry body). $L=0$ set karo ek plain symmetric ballistic rocket ke liye jo nose-into-wind fly kar raha ho — phir sirf gravity steer karti hai:

$\boxed{\dot{\gamma }=\frac{L-mg\cos \gamma }{mV}\stackrel{L=0}{\to }\dot{\gamma }=-\frac{g\cos \gamma }{V}}$

Term by term:

• $\dot{\gamma }$ — heading angle kitni tezi se change hota hai (turn rate); positive arrow ko upar ghuma deta hai, negative neeche.
• $V\dot{\gamma }$ — sideways (centripetal) acceleration: ek tezi se jaata arrow utni hi sideways force se utna hi ghoomne ke liye, toh $V$, $\dot{\gamma }$ ko multiply karta hai.
• $L$ — lift ka turning share; $-g\cos \gamma$ — gravity ka turning share, hamesha arrow ki tip ko neeche kheenchta hai, isliye minus sign.

KYU. Yeh "steering" equation hai: sirf across-forces appear karte hain, toh yeh direction akela govern karta hai. $L=0$ ke saath minus sign kehta hai path hamesha earthward curve karti hai.

PICTURE. Velocity arrow do instants par ek moment apart; gravity ka blue across-piece tip ko neeche nudge karta hai, $\gamma$ ko ek chhoti si value par rotate karta hua. Swept hua tiny angle hai $\dot{\gamma }dt$.

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Step 8 — Arrow ko position mein badlo (kinematics)

KYA. Hume arrow ki length $V$ aur tilt $\gamma$ pata hai; dot actually kahan move karta hai? Velocity ko uski ground-shadow (downrange) aur uski rise (altitude) mein split karo — is baar velocity ki khud ek aur right triangle:

$\boxed{\dot{x}=V\cos \gamma ,\dot{h}=V\sin \gamma }$

Term by term:

• $\dot{x}$ — downrange speed = arrow ki flat ground par shadow = $V\cos \gamma$.
• $\dot{h}$ — climb speed = arrow tip ki height = $V\sin \gamma$. Note karo agar $\gamma <0$ toh $\sin \gamma <0$ aur $\dot{h}<0$ — vehicle altitude lose karta hai, descent ke liye exactly sahi.

KYU. Inhe hi aap integrate karte ho (time ke saath jod dete ho) board par actual path draw karne ke liye. Speed aur heading plot karne ke liye useless hain jab tak ground aur sky par project na ho.

PICTURE. Velocity arrow hypotenuse ki tarah; uski horizontal shadow labelled $V\cos \gamma$, uski vertical rise labelled $V\sin \gamma$, angle $\gamma$ base par marked.

Recall Corner cases check karo

Level flight $\gamma =0$ ::: $\dot{x}=V$, $\dot{h}=0$ — sab downrange, koi climbing nahi. Straight up $\gamma =90^{\circ }$ ::: $\dot{x}=0$, $\dot{h}=V$ — sab climb, koi downrange nahi; aur $\dot{\gamma }=-g\cos 90^{\circ }/V=0$, toh vertical vertical rehta hai. Diving $\gamma =-30^{\circ }$ ::: $\dot{h}=V\sin (-30^{\circ })<0$ (height khona), aur $-g\sin \gamma >0$ (gravity ise speed up karti hai). Coasting $T=0$, $D=0$ ::: $\dot{V}=-g\sin \gamma$, $\dot{\gamma }=-g\cos \gamma /V$ — plain projectile parabola recover ho jaata hai.

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Step 9 — Mass shrink karta hai (isliye acceleration explode karta hai)

KYA. Engine propellant peechhe phenkta hai, toh $m$ constant nahi hai:

$\dot{m}=-\frac{T}{I_{sp}{g}_0}.$

• $I_{sp}$ = specific impulse (engine fuel kitni efficiently use karta hai), seconds mein measured.
• $g_0$ = reference gravitational acceleration, wahi same $9.81{\text{m/s}}^2$ jo humne upar define kiya — lekin yahan yeh purely bookkeeping role play karta hai: engine ki specific impulse (seconds) ko exhaust speed mein convert karta hai. Yeh convention se $g_0$ (subscript zero ke saath) likha jaata hai taaki flag ho "yeh frozen standard constant hai," kabhi koi local varying gravity nahi. Humare flat-Earth model mein $g$ aur $g_0$ identical number hold karte hain; woh sirf job mein differ karte hain, value mein nahi.
• Minus sign: mass sirf tab decrease hoti hai jab burn ho raha ho.

KYU. Wapas dekho $\dot{V}=(T-D)/m-\dots$ par. Jaise $m$ girti hai, wahi thrust ek chhote number se divide hota hai — toh acceleration badhta rehta hai, burnout ke waqt sabse zyada. Yeh bhool jaana classic mistake hai. Yeh directly Tsiolkovsky Rocket Equation se link karta hai.

PICTURE. Mass bar left-to-right drain hota hua jabki $\dot{V}$ arrow paas mein taller hota jaata hai — wahi thrust, halka rocket, bada kick.

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Ek-picture summary

Upar sab kuch ek single board mein compress kiya: velocity arrow apne $\hat{t}$/$\hat{n}$ frame ke saath (yaad raho $\hat{n}$ counter-clockwise perpendicular hai), teen force arrows projected, aur char resulting equations baahir flow karte hue — ek stretch ($\dot{V}$) ke liye, ek rotate ($\dot{\gamma }$) ke liye, do move ($\dot{x},\dot{h}$) ke liye.

Recall Feynman retelling — poora walkthrough plain words mein

Rocket ko ek single dot ki tarah draw karo ek arrow ke saath — woh arrow hai kitni tezi se aur kis taraf ja raha hai. Arrow ka horizon ke upar tilt hai $\gamma$: positive matlab climbing, negative matlab diving. Newton kehta hai forces us arrow ko bend aur stretch karte hain, lekin agar aap forces arrow ke saath aur across measure karo (upar/neeche ki jagah), toh donon effects perfectly alag ho jaate hain: along-forces sirf stretch karti hain, across-forces sirf turn karti hain. Thrust aur drag pehle se arrow ke saath baithe hain. Gravity (ek downward pull $mg$ size ka, jahan $g\approx 9.81$) seedha neeche point karti hai, toh hum ek right triangle drop karte hain aur ise slice karte hain: sine-slice climb se ladti hai (yeh speed rule hai, $\dot{V}=(T-D)/m-g\sin \gamma$ — aur diving mein yeh speed help karti hai), cosine-slice tip ko ground ki taraf curve karti hai (yeh turning rule hai, $\dot{\gamma }=-g\cos \gamma /V$, sirf tabhi valid jab rocket actually move kar raha ho, $V>0$; ek lift term $L$ add karo agar wings involved hain). Yeh jaanne ke liye ki dot kahan land karti hai, arrow ko ground aur sky par shadow karo — $\dot{x}=V\cos \gamma$, $\dot{h}=V\sin \gamma$. Finally, rocket fuel burn karte hue halka hota rehta hai, toh wahi push ise zyada se zyada accelerate karta hai. Char chhote rules, ek moving arrow — poora flight path.

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Connections

• Parent topic (Hinglish) — woh note jise yeh deep-dive expand karta hai.
• Flight-Path Angle and Velocity Frame — $\gamma$ aur $\hat{t},\hat{n}$ frame ki geometry.
• Gravity Turn Trajectory — Step 7 apni trajectory ki tarah.
• Drag and Atmospheric Models — $D$ kahan se aata hai.
• Tsiolkovsky Rocket Equation — vertical, drag-free limit aur mass loss.
• 6DOF Rigid-Body Dynamics — woh rotation restore karta hai jo hum Step 1 mein delete kar diya tha.
• Numerical Integration of Trajectories (RK4) — in char ODEs ko integrate karna.