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

We are chasing one boxed result, the planar 3DOF set:

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

Every step below earns one piece of this.

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Step 1 — Draw the rocket as one arrow

WHAT. Forget the shape of the rocket. Replace it with a single dot (the centre of mass — the one point that moves as if all the mass sat there) and one arrow sticking out of it. That arrow is the velocity $\vec{V}$: its length is the speed $V$ (how many metres per second the dot travels), and its direction is where the dot is heading right now.

WHY. For a trajectory — range, altitude, burnout speed — we only care about where the dot goes, not how the body spins. That is exactly what "3DOF" (3 Degrees Of Freedom: two for position in the vertical plane, one for the speed–direction of the arrow) means. Spin is thrown away.

PICTURE. The dot sits on flat ground. The blue arrow $\vec{V}$ points up and to the right. The angle it makes with the flat horizon is the flight-path angle $\gamma$ (Greek letter "gamma"). $\gamma =0$ means flying level; $\gamma =90^{\circ }$ means straight up; and — read this now, we'll use it later — $\gamma <0$ means the arrow points below the horizon, i.e. descending.

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Step 2 — Newton's law is a vector statement

WHAT. Newton's second law says the arrow of velocity changes because forces push the dot:

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

• $m$ = mass of the rocket (kilograms) — how hard it is to accelerate.
• $\frac{d\vec{V}}{dt}$ = the "rate-of-change of the velocity arrow" — read $\frac{d}{dt}$ as "how fast, per second, the thing after it is changing." Because $\vec{V}$ is an arrow, its change is also an arrow.
• $\sum \vec{F}$ = all the forces added together as arrows (thrust, drag, gravity — gravity being the weight $mg$ defined above).

WHY. This one line contains the whole trajectory — but it mixes stretching and rotating into one arrow equation. That is awkward. A smart trick untangles them.

PICTURE. Three force arrows hang off the same dot: thrust (yellow) forward, drag (pink) backward, weight (white) straight down. Their sum is a single net arrow — but pointing in an unhelpful direction.

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Step 3 — The magic split: along vs. across the arrow

WHAT. Instead of horizontal/vertical, we lay down two brand-new directions that travel with the arrow:

• Tangential $\hat{t}$ — points along $\vec{V}$ (the way we're going).
• Normal $\hat{n}$ — points 90° across $\vec{V}$. We fix its orientation once and for all: $\hat{n}$ is $\vec{V}$ rotated 90° counter-clockwise — so when the arrow points up-and-right, $\hat{n}$ points up-and-left (toward the sky side of the path). This choice matters: it decides the sign of everything on the across-axis.

This pair is called the velocity (wind) frame.

WHY. Here is the key fact, worth its own line:

A push along the arrow can only change its length (speed). A push across the arrow can only rotate it (change heading).

So projecting every force onto $\hat{t}$ and $\hat{n}$ splits Newton's one vector equation into two clean scalar equations — one for speed, one for turning. That is the whole reason this frame exists. With our chosen $\hat{n}$, a positive across-force rotates the arrow counter-clockwise — i.e. increases $\gamma$ (turns it more upward).

PICTURE. The blue velocity arrow, with $\hat{t}$ drawn on top of it and $\hat{n}$ perpendicular, pointing to the upper-left (counter-clockwise side). A little "speedometer" symbol sits on $\hat{t}$, a curved "steering" arrow on $\hat{n}$.

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Step 4 — Project the easy forces (thrust & drag)

WHAT. Two forces already lie on the path's line:

• Thrust $T$ (yellow) — the engine pushes along $\vec{V}$ (we assume the nose points into the wind, so no sideways thrust). It sits fully on $+\hat{t}$.
• Drag $D=\frac{1}{2}\rho V^{2}S{C}_D$ (pink) — air resistance always opposes motion, so it sits fully on $-\hat{t}$.

Here $\rho$ = air density, $S$ = reference area, $C_D$ = drag coefficient (see Drag and Atmospheric Models). For this derivation just treat $D$ as "a backward force whose size we know."

WHY. These two contribute nothing to the across-direction ($\hat{n}$), because they lie exactly on $\hat{t}$. So the only tricky force left is gravity — that's Step 5.

PICTURE. On the tangential axis: yellow $+T$ forward, pink $-D$ backward. On the normal axis: nothing from these two.

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Step 5 — Split gravity with a triangle (the heart of it)

WHAT. Weight $mg$ (white arrow) points straight down, not along the path. We must break it into an along-part and an across-part. Drop the weight arrow at the dot and build a right triangle whose long side is along $\hat{t}$ and short side along $\hat{n}$.

WHY a triangle, and why sine/cosine? A right triangle is the only honest way to ask "how much of this down-arrow points along the path?" Sine and cosine are exactly the two ratios that answer "what fraction of a length survives after tilting it by an angle." We use them because the question is a tilt question.

Look carefully at the geometry: the velocity is tilted up by $\gamma$ from horizontal, so "straight down" is tilted by $\gamma$ away from the backward-along-path direction. That single fact fixes both parts:

• Component along $\hat{t}$ (opposing the climb): $-mg\sin \gamma$.
• Component across $\hat{n}$: $-mg\cos \gamma$. (With our counter-clockwise $\hat{n}$ pointing up-left, "down" has a negative $\hat{n}$-component — hence the minus sign, which will curve the path earthward.)

PICTURE. The white weight arrow decomposed into a yellow along-piece ($mg\sin \gamma$, pointing backward down the path) and a blue across-piece ($mg\cos \gamma$, pointing toward the inside of the turn). The angle $\gamma$ is marked inside the triangle so you can read off which side gets sine.

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Step 6 — Assemble the tangential (speed) equation

WHAT. Add the three along-parts (thrust $+T$, drag $-D$, gravity $-mg\sin \gamma$) and set them equal to $m\dot{V}$:

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

Divide by $m$:

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

Term by term:

• $\dot{V}$ — how fast the speed grows each second.
• $\frac{T-D}{m}$ — net forward push per kilogram (engine minus air).
• $-g\sin \gamma$ — the slice of gravity that slows the climb; biggest when going straight up, zero when flying level.

WHY. This is the "throttle" equation: only along-forces appear, so it governs speed alone. Direction is nowhere in it. Clean, exactly as promised.

PICTURE. A number line for $\dot{V}$: yellow $+T/m$ pushing right, pink $-D/m$ and white $-g\sin \gamma$ pushing left; the leftover is $\dot{V}$.

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Step 7 — Assemble the normal (turning) equation

WHAT. Across the path, only gravity's $\hat{n}$-piece ($-mg\cos \gamma$) and any lift $L$ act. An across-force can't change the arrow's length; it can only rotate it, and the sideways (centripetal) acceleration needed to rotate an arrow of length $V$ at rate $\dot{\gamma }$ is $V\dot{\gamma }$. So the general turning equation, keeping lift, is:

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

What is $L$, and when do you keep it? Lift is an aerodynamic force perpendicular to the velocity (from wings, fins, or flying at an angle of attack). Along our counter-clockwise $\hat{n}$, positive $L$ points up-left and turns the arrow upward. Keep the $L$ term whenever the vehicle actually generates side-force (a winged launcher, a manoeuvring re-entry body). Set $L=0$ for a plain symmetric ballistic rocket flying nose-into-wind — then only gravity steers:

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

Term by term:

• $\dot{\gamma }$ — how fast the heading angle changes (turn rate); positive turns the arrow up, negative turns it down.
• $V\dot{\gamma }$ — sideways (centripetal) acceleration: a faster arrow needs more sideways force to turn the same amount, so $V$ multiplies $\dot{\gamma }$.
• $L$ — lift's turning share; $-g\cos \gamma$ — gravity's turning share, always pulling the arrow's tip down, hence its minus sign.

WHY. This is the "steering" equation: only across-forces appear, so it governs direction alone. With $L=0$ the minus sign says the path always curves earthward.

PICTURE. The velocity arrow at two instants a moment apart; gravity's blue across-piece nudges the tip downward, rotating $\gamma$ to a smaller value. The tiny angle swept is $\dot{\gamma }dt$.

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Step 8 — Turn the arrow into a position (kinematics)

WHAT. We know the arrow's length $V$ and tilt $\gamma$; where does the dot actually move? Split the velocity into its ground-shadow (downrange) and its rise (altitude) — another right triangle, this time of the velocity itself:

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

Term by term:

• $\dot{x}$ — downrange speed = shadow of the arrow on the flat ground = $V\cos \gamma$.
• $\dot{h}$ — climb speed = height of the arrow tip = $V\sin \gamma$. Note if $\gamma <0$ then $\sin \gamma <0$ and $\dot{h}<0$ — the vehicle loses altitude, exactly right for a descent.

WHY. These are what you integrate (add up over time) to draw the actual path on the board. Speed and heading are useless for plotting until projected onto ground and sky.

PICTURE. The velocity arrow as the hypotenuse; its horizontal shadow labelled $V\cos \gamma$, its vertical rise labelled $V\sin \gamma$, angle $\gamma$ marked at the base.

Recall Check the corner cases

Level flight $\gamma =0$ ::: $\dot{x}=V$, $\dot{h}=0$ — all downrange, no climbing. Straight up $\gamma =90^{\circ }$ ::: $\dot{x}=0$, $\dot{h}=V$ — all climb, no downrange; and $\dot{\gamma }=-g\cos 90^{\circ }/V=0$, so vertical stays vertical. Diving $\gamma =-30^{\circ }$ ::: $\dot{h}=V\sin (-30^{\circ })<0$ (losing height), and $-g\sin \gamma >0$ (gravity speeds it up). Coasting $T=0$, $D=0$ ::: $\dot{V}=-g\sin \gamma$, $\dot{\gamma }=-g\cos \gamma /V$ — recovers the plain projectile parabola.

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Step 9 — The mass shrinks (why acceleration explodes)

WHAT. The engine throws propellant out the back, so $m$ is not constant:

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

• $I_{sp}$ = specific impulse (how efficiently the engine uses fuel), measured in seconds.
• $g_0$ = the reference gravitational acceleration, the same $9.81{\text{m/s}}^2$ as the $g$ we defined at the top — but here it plays a purely bookkeeping role: it converts the engine's specific impulse (seconds) into an exhaust speed. It is written $g_0$ (with a subscript zero) by convention to flag "this is the frozen standard constant," never a local varying gravity. In our flat-Earth model $g$ and $g_0$ hold the identical number; they differ only in job, not value.
• The minus sign: mass only ever decreases while burning.

WHY. Look back at $\dot{V}=(T-D)/m-\dots$. As $m$ falls, the same thrust divides by a smaller number — so acceleration keeps rising, largest right at burnout. Forgetting this is the classic mistake. This links directly to the Tsiolkovsky Rocket Equation.

PICTURE. Mass bar draining left-to-right while the $\dot{V}$ arrow beside it grows taller — same thrust, lighter rocket, bigger kick.

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The one-picture summary

Everything above compressed into a single board: the velocity arrow with its $\hat{t}$/$\hat{n}$ frame (recall $\hat{n}$ is the counter-clockwise perpendicular), the three force arrows projected, and the four resulting equations flowing out — one for stretch ($\dot{V}$), one for rotate ($\dot{\gamma }$), two for move ($\dot{x},\dot{h}$).

Recall Feynman retelling — the whole walkthrough in plain words

Draw the rocket as a single dot with one arrow — that arrow is how fast and which way it's going. The tilt of the arrow above the horizon is $\gamma$: positive means climbing, negative means diving. Newton says forces bend and stretch that arrow, but if you measure forces along the arrow and across it (instead of up/down), the two effects come apart perfectly: along-forces only stretch it, across-forces only turn it. Thrust and drag already sit along the arrow. Gravity (a downward pull of size $mg$, where $g\approx 9.81$) points straight down, so we drop a right triangle and slice it: the sine-slice fights the climb (that's the speed rule, $\dot{V}=(T-D)/m-g\sin \gamma$ — and when diving it helps the speed), the cosine-slice curves the tip toward the ground (that's the turning rule, $\dot{\gamma }=-g\cos \gamma /V$, valid only while the rocket is actually moving, $V>0$; add a lift term $L$ if wings are involved). To find where the dot lands, we shadow the arrow onto the ground and the sky — $\dot{x}=V\cos \gamma$, $\dot{h}=V\sin \gamma$. Finally, the rocket keeps getting lighter as it burns fuel, so the same push accelerates it harder and harder. Four little rules, one moving arrow — the entire flight path.

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Connections

• Parent topic (Hinglish) — the note this deep-dive expands.
• Flight-Path Angle and Velocity Frame — the geometry of $\gamma$ and the $\hat{t},\hat{n}$ frame.
• Gravity Turn Trajectory — Step 7 as its own trajectory.
• Drag and Atmospheric Models — where $D$ comes from.
• Tsiolkovsky Rocket Equation — the vertical, drag-free limit and mass loss.
• 6DOF Rigid-Body Dynamics — restores the rotation we deleted in Step 1.
• Numerical Integration of Trajectories (RK4) — integrating these four ODEs.