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

3.4.4 · Physics › Rocket Flight Mechanics

1. 3DOF actually hai kya?

KYA CHHODH DETE HAIN: pitch/yaw/roll (3 rotational DOF). Toh "3DOF" = full 6DOF minus rotation.

KAISE SET UP KARTE HAIN: Velocity (wind) frame mein kaam karo — forces ko resolve karo velocity vector ke saath (speed change karta hai) aur uske perpendicular (direction change karta hai). Yahi trick hai jo algebra ko saaf rakhti hai.

2. Equations ko scratch se derive karna

Hum ek planar (2D vertical plane) flat-Earth model use karte hain. Point mass par forces:

• Thrust $T$ (velocity ke saath, assuming zero angle of attack),
• Drag $D=\frac{1}{2}\rho V^{2}S{C}_D$ (velocity ko oppose karta hai),
• Weight $mg$ (seedha neeche),
• (Lift $L$ perpendicular; aksar $0$ hota hai ek symmetric ballistic rocket ke liye).

$\gamma$ = flight-path angle = horizontal se upar velocity ka angle.

Tangential (velocity ke saath) — speed control karta hai

$\vec{V}$ ke saath forces ka sum. Thrust aur drag line ke saath act karte hain; velocity ke saath gravity ka component $-mg\sin \gamma$ hai (climb karne se speed kam hoti hai):

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

Yeh step kyun? Har force ko unit velocity direction par project karo. Weight neeche point karta hai; "neeche" aur "velocity-ke-saath-peeche" ke beech ka angle $\sin \gamma$ deta hai.

Normal (velocity ke perpendicular) — direction control karta hai

Perpendicular acceleration centripetal term $mV\dot{\gamma }$ hai. Sirf gravity ka perpendicular component $-mg\cos \gamma$ (aur lift $L$) yahan act karta hai:

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

$L=0$ ke liye: $\boxed{\dot{\gamma }=-\frac{g\cos \gamma }{V}}$

Yeh step kyun? $\vec{V}$ ke perpendicular, "neeche" normal ke saath angle $\gamma$ banata hai, toh uska component $g\cos \gamma$ hai, aur yeh path ko neecha karta hai (negative $\dot{\gamma }$).

Kinematics — position control karta hai

Ground frame mein velocity components: $\boxed{\dot{x}=V\cos \gamma ,\dot{h}=V\sin \gamma }$

Yeh 4–5 coupled ODEs trajectory ke liye numerically integrate ki jaati hain.

3. Special cases (80/20 jo sabse zyada insight deta hai)

Vertical flight ($\gamma =90°$): $\dot{V}=\frac{T-D}{m}-g$, aur $\dot{\gamma }=0$ — ek vertical climb vertical rehta hai. Yahi ideal-rocket / Tsiolkovsky regime hai.

Gravity turn ($T$, $V$ ke saath, $L=0$): path ko sirf gravity bend karta hai, $\dot{\gamma }=-g\cos \gamma /V$ ke zariye. Ek choti si initial pitch-over gravity ko velocity ko horizontal ki taraf slowly rotate karne deti hai — koi control effort waste nahi. Isliye real launches liftoff ke turant baad lean over karti hain.

Coast / ballistic ($T=0$): pure projectile-plus-drag; $D=0$ hone par parabola recover ho jaata hai.

4. Worked examples

5. Common mistakes

6. Active recall

Connections

• Tsiolkovsky Rocket Equation — $\dot{V}$ ka vertical/no-drag limit
• Gravity Turn Trajectory — $\dot{\gamma }$ equation ka special case
• Drag and Atmospheric Models — $D=\frac{1}{2}\rho V^{2}S{C}_D$ supply karta hai
• 6DOF Rigid-Body Dynamics — yahan hataaye gaye rotational DOF add karta hai
• Flight-Path Angle and Velocity Frame — $\gamma$ ke peeche ki geometry
• Numerical Integration of Trajectories (RK4) — in ODEs ko kaise solve kiya jaata hai

Concept Map