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

1. What is 3DOF, really?

WHAT we throw away: pitch/yaw/roll (3 rotational DOF). So "3DOF" = full 6DOF minus rotation.

HOW we set it up: Work in the velocity (wind) frame — resolve forces along the velocity vector (changes speed) and perpendicular to it (changes direction). This is the trick that makes the algebra clean.

2. Deriving the equations from scratch

We use a planar (2D vertical plane) flat-Earth model. Forces on the point mass:

• Thrust $T$ (along velocity, assuming zero angle of attack),
• Drag $D=\tfrac12\rho V^2 S C_D$ (opposes velocity),
• Weight $mg$ (straight down),
• (Lift $L$ perpendicular; often $0$ for a symmetric ballistic rocket).

Let $\gamma$ = flight-path angle = angle of velocity above horizontal.

Tangential (along velocity) — controls speed

Sum forces along $\vec V$. Thrust and drag act along the line; gravity's component along velocity is $-mg\sin\gamma$ (climbing loses speed):

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

Why this step? Project each force onto the unit velocity direction. Weight points down; the angle between "down" and "backward-along-velocity" gives the $\sin\gamma$.

Normal (perpendicular to velocity) — controls direction

The perpendicular acceleration is the centripetal term $mV\dot\gamma$. Only gravity's perpendicular component $-mg\cos\gamma$ (and lift $L$) act here:

$m V\dot\gamma = L - mg\cos\gamma$

For $L=0$: $\boxed{\,\dot\gamma = -\dfrac{g\cos\gamma}{V}\,}$

Why this step? Perpendicular to $\vec V$, "down" makes angle $\gamma$ with the normal, so its component is $g\cos\gamma$, and it curves the path downward (negative $\dot\gamma$).

Kinematics — controls position

Velocity components in the ground frame: $\boxed{\,\dot x = V\cos\gamma,\qquad \dot h = V\sin\gamma\,}$

These 4–5 coupled ODEs are integrated numerically for the trajectory.

3. Special cases (the 80/20 that gives most insight)

Vertical flight ($\gamma=90°$): $\dot V=\frac{T-D}{m}-g$, and $\dot\gamma=0$ — a vertical climb stays vertical. That's the ideal-rocket / Tsiolkovsky regime.

Gravity turn ($T$ along $V$, $L=0$): the only thing bending the path is gravity, via $\dot\gamma=-g\cos\gamma/V$. A tiny initial pitch-over lets gravity slowly rotate velocity toward horizontal — no wasted control effort. This is why real launches lean over just after liftoff.

Coast / ballistic ($T=0$): pure projectile-plus-drag; recovers parabola when $D=0$.

4. Worked examples

5. Common mistakes

6. Active recall

Connections

• Tsiolkovsky Rocket Equation — vertical/no-drag limit of $\dot V$
• Gravity Turn Trajectory — special case of the $\dot\gamma$ equation
• Drag and Atmospheric Models — supplies $D=\tfrac12\rho V^2 S C_D$
• 6DOF Rigid-Body Dynamics — adds rotational DOF removed here
• Flight-Path Angle and Velocity Frame — geometry behind $\gamma$
• Numerical Integration of Trajectories (RK4) — how these ODEs are solved

Concept Map

Hinglish (regional understanding)