6DOF equations — translational (Newton), rotational (Euler's equations): Worked examples

Before anything: the symbols we reuse, in plain words.

Definition The cast of symbols (all defined once, here)

$m$ — the rocket's mass (how much stuff), in kilograms.
$v = (u, v, w)$ — velocity of the center of mass (the balance point) written in body axes: $u$ = forward speed, $v$ = sideways speed, $w$ = up/down speed. Units m/s.
$ω = (p, q, r)$ — how fast the rocket twists: $p$ = roll (barrel-roll like a log), $q$ = pitch (nose up/down), $r$ = yaw (nose left/right). Units rad/s.
$F = (F_{x}, F_{y}, F_{z})$ — total external force (push), in newtons.
$M = (M_{x}, M_{y}, M_{z})$ — total external moment/torque (twisting effort), in N·m.
$I_{x}, I_{y}, I_{z}$ — the moments of inertia about the three body axes: "how hard is it to get this axis spinning". Units kg·m². Built in Inertia tensor and principal axes.
$I_{t}$ — the transverse moment of inertia of an axisymmetric body: when the two "side" axes are identical we write $I_{y} = I_{z} = I_{t}$ (one shared number instead of two). Units kg·m².
$ℓ$ — a lever arm: the perpendicular distance from the center of mass to the line of a force. Multiplying a force by its lever arm gives a moment, $M = F ℓ$ . Units metres.
The symbol $\times$ is the cross product — it takes two arrows and returns a third arrow perpendicular to both, whose length is "how much they fail to be parallel".

The two master equations we will apply again and again (from the parent):

The scenario matrix

Each row is one class of situation. The last column names the worked example that hits it.

#	Case class	What makes it special	Covered by
A	Zero rotation ( $ω = 0$ )	all cross terms vanish → plain 1D Newton	Ex 1
B	Pure turn, constant body speed	$\dot{v} = 0$ but real acceleration exists	Ex 2
C	Sign of a coupling term ( $q r > 0$ vs $< 0$ )	which way the gyroscopic torque points	Ex 3
D	Symmetric spinner ( $I_{y} = I_{z}$ )	steady coning, oscillation frequency	Ex 4
E	Fully asymmetric, torque-free ( $I_{x} \neq = I_{y} \neq = I_{z}$ )	intermediate-axis instability	Ex 5
F	Degenerate inertia (sphere $I_{x} = I_{y} = I_{z}$ )	all coupling dies, $\dot{ω}$ from torque only	Ex 6
G	Decoupled single-axis maneuver	one moment, clean angular accel	Ex 7
H	Word problem (real thruster + gravity)	build $F, M$ from a story	Ex 8
I	Limiting / exam twist (mass → variable)	why naive $d (m v) / d t$ is wrong	Ex 9

Ex 1 — Case A: zero rotation (sanity floor)

Worked example Vertical climb, no spin

A rocket rises straight up. Body- $x$ points up. Thrust $T = 30, 000$ N up, weight $m g$ with $m = 2000$ kg, $g = 9.8$ m/s². No rotation: $p = q = r = 0$ , and $v = w = 0$ . Find the forward acceleration $\overset{u}{˙}$ .

Forecast: guess before reading — should the answer just be $(T - m g) / m$ ? Why or why not?

Write Newton- $x$ : $F_{x} = m (\overset{u}{˙} + q w - r v)$ . Why this step? It is the only equation containing $\overset{u}{˙}$ , our unknown.
Set $q = r = 0$ : the terms $q w$ and $r v$ die, leaving $F_{x} = m \overset{u}{˙}$ . Why this step? With no rotation there is no transport correction — the body frame and inertial frame agree on acceleration.
The net force is $F_{x} = T - m g = 30000 - 2000 (9.8) = 10400$ N. Why this step? Both forces act along body- $x$ ; up is positive, weight is negative, so we just add the two vertical forces with their signs.
Solve: $\overset{u}{˙} = F_{x} / m = 10400/2000 = 5.2$ m/s². Why this step? Divide the net force by mass — the final rearrangement of $F_{x} = m \overset{u}{˙}$ .

Verify: units $N / kg = m/s^{2}$ ✓. If $T = m g = 19600$ N then $\overset{u}{˙} = 0$ (hovering) — matches intuition. This is the "floor" every richer case must reduce to.

Ex 2 — Case B: pure turn, constant body speed

Worked example Coordinated turn — acceleration hiding in plain sight

A rocket cruises with constant body velocity $u = 200$ m/s, $v = w = 0$ , while yawing at a steady $r = 0.1$ rad/s ( $p = q = 0$ ). Body speed feels constant. Mass $m = 1500$ kg. Find the sideways force $F_{y}$ needed, and the true inertial acceleration.

Forecast: with $\overset{v}{˙} = 0$ , is $F_{y}$ zero? (It is not — that is the whole lesson.)

Newton- $y$ : $F_{y} = m (\overset{v}{˙} + r u - pw)$ . Why this step? The sideways force lives here; the term $r u$ is the transport correction.
Constant body velocity → $\overset{v}{˙} = 0$ . With $p = 0$ : $F_{y} = m r u$ . Why this step? Even though the numbers $u, v, w$ don't change, the velocity arrow is being swung around by the yaw — its inertial direction changes, so there is real acceleration.
Plug in: $F_{y} = 1500 \times 0.1 \times 200 = 30000$ N. Why this step? Substitute the given numbers into $F_{y} = m r u$ to get the required force.
The inertial acceleration magnitude is $a = F_{y} / m = r u = 20$ m/s². Why this step? This is exactly centripetal acceleration $v^{2} / R$ with $R = u / r = 2000$ m: $u^{2} / R = 20 0^{2} /2000 = 20$ m/s² ✓.

Look at the figure below. The gray dashed curve is the flight arc of radius $R = 2000$ m. The blue arrow is the velocity $v$ , tangent to the arc — its length (200 m/s) never changes, but the green curved arrow (the yaw $r$ ) keeps swinging its direction. The orange arrow is the force $F_{y} = 30000$ N pointing toward the turn center: that force is the only thing bending the straight flight into a circle. So a "constant speed" turn still demands a real sideways push.

Verify: the two routes to $a$ (transport term $r u$ and centripetal $u^{2} / R$ ) agree at 20 m/s². Units of $F_{y}$ : kg·(rad/s)·(m/s) = kg·m/s² = N ✓ (radians are dimensionless).

Ex 3 — Case C: sign of a coupling term

Worked example Which way does the gyroscopic torque push?

A rocket spins with roll $p = 5$ rad/s and yaw $r = 2$ rad/s, pitch $q$ = 0 at this instant. Inertias: $I_{x} = 10$ , $I_{y} = 20$ , $I_{z} = 40$ kg·m². No applied moments ( $M = 0$ ). Find the pitch acceleration $\overset{q}{˙}$ and its sign.

Forecast: with zero pitch and zero torque, will the nose stay put? Guess the sign of $\overset{q}{˙}$ .

Euler- $y$ : $M_{y} = I_{y} \overset{q}{˙} + (I_{x} - I_{z}) r p$ . Set $M_{y} = 0$ . Why this step? We want $\overset{q}{˙}$ ; the coupling term $(I_{x} - I_{z}) r p$ decides its sign.
Solve: $\overset{q}{˙} = - \frac{( I _{x} - I _{z} )}{I _{y}} r p = - \frac{( 10 - 40 )}{20} (2) (5)$ . Why this step? Rearranging Euler- $y$ ; the sign of $(I_{x} - I_{z})$ times $r p$ is what we read.
Evaluate the sign pieces: $(I_{x} - I_{z}) = - 30$ (negative), $r p = 10$ (positive). Product of the minus in front with a negative bracket gives a positive $\overset{q}{˙}$ . Why this step? Track signs before crunching the number, so the physics (nose pitches which way) is clear rather than buried in arithmetic.
Number: $\overset{q}{˙} = - \frac{- 30}{20} (10) = + 15$ rad/s². Why this step? A body spinning about two axes at once develops a third twist for free — that is the gyroscopic coupling.

Verify: flip the yaw sign to $r = - 2$ : then $r p = - 10$ and $\overset{q}{˙} = - \frac{- 30}{20} (- 10) = - 15$ rad/s² — same size, opposite sign, as the symmetry of the formula demands ✓.

Ex 4 — Case D: symmetric spinner → steady coning

Worked example Sounding rocket coning frequency

A spin-stabilized rocket is axisymmetric: $I_{y} = I_{z} = I_{t} = 30$ kg·m², spin inertia $I_{x} = 6$ kg·m². It spins at $p_{0} = 20$ rad/s, torque-free. A small disturbance gives it transverse rates. Find the coning (nutation) frequency.

Forecast: will the transverse wobble grow, decay, or oscillate forever?

Euler- $x$ : $I_{x} \overset{p}{˙} = 0 \Rightarrow p = p_{0}$ constant. Why this step? With $I_{y} = I_{z}$ the $x$ -coupling term $(I_{z} - I_{y}) q r = 0$ ; spin rate is locked.
Euler- $y, z$ : $I_{t} \overset{q}{˙} = (I_{x} - I_{t}) r p_{0}$ and $I_{t} \overset{r}{˙} = - (I_{x} - I_{t}) q p_{0}$ . Why this step? Substituting the symmetric inertias; note $q$ and $r$ feed into each other.
Define $λ = \frac{I _{x} - I _{t}}{I _{t}} p_{0} = \frac{6 - 30}{30} (20) = - 16$ rad/s. Then $\overset{q}{˙} = λ r$ , $\overset{r}{˙} = - λ q \Rightarrow \overset{q}{¨} = - λ^{2} q$ . Why this step? Two first-order coupled equations combine into one oscillator equation — the signature of circular motion of the vector $(q, r)$ .
The coning frequency is $∣ λ ∣ = 16$ rad/s (about 2.5 Hz). Why this step? $\overset{q}{¨} = - λ^{2} q$ is the equation of a spring; its frequency is $∣ λ ∣$ .

Look at the figure below. The blue and orange curves are the two transverse rates $q (t)$ and $r (t)$ . Notice they are a cosine and a (negative) sine — a quarter-cycle apart — so as one rises the other falls: the pair chases itself around in a circle. Crucially the curves neither grow nor shrink; they stay bounded between $\pm 1$ . The green dashed line marks one full period $2 π /∣ λ ∣ \approx 0.39$ s. That bounded, repeating wobble is exactly what makes spin stabilization safe.

Verify: the wobble neither grows nor decays — it is bounded oscillation, exactly why spin stabilization works. Units: (kg·m²/kg·m²)(rad/s) = rad/s ✓.

Ex 5 — Case E: fully asymmetric torque-free (intermediate axis)

Worked example Tumbling about the middle axis

A body has three different inertias $I_{x} = 2 < I_{y} = 5 < I_{z} = 9$ kg·m² (all distinct). It spins almost purely about the intermediate axis $y$ : $q = 10$ rad/s, tiny $p = r = 0.01$ rad/s. Torque-free. Find whether small $p, r$ grow or stay small.

Forecast: which axis is unstable — smallest, middle, or largest?

Euler- $x$ : $I_{x} \overset{p}{˙} = (I_{y} - I_{z}) q r$ , Euler- $z$ : $I_{z} \overset{r}{˙} = (I_{x} - I_{y}) pq$ (with $q \approx$ const). Why this step? We linearize about the big spin $q$ ; $p, r$ are the small deviations.
Differentiate Euler- $x$ and substitute Euler- $z$ : $\overset{p}{¨} = \frac{( I _{y} - I _{z} )}{I _{x}} q \overset{r}{˙} = \frac{( I _{y} - I _{z} ) ( I _{x} - I _{y} )}{I _{x} I _{z}} q^{2} p$ . Why this step? We want a single equation in $p$ to read off growth vs oscillation.
Compute the coefficient $k = \frac{( I _{y} - I _{z} ) ( I _{x} - I _{y} )}{I _{x} I _{z}} q^{2} = \frac{( 5 - 9 ) ( 2 - 5 )}{2 \cdot 9} (100) = \frac{( - 4 ) ( - 3 )}{18} (100) = \frac{12}{18} (100) = \frac{200}{3} \approx 66.7$ . Why this step? The sign of $k$ decides everything: $k > 0$ means $\overset{p}{¨} = + k p$ → exponential blow-up.
Since $k > 0$ , $p$ grows like $e^{k t}$ — the spin is unstable. The body flips. Why this step? Spinning about the middle axis always gives one negative and one positive bracket, whose product is positive → instability.

Verify: repeat with the largest axis $z$ as spin axis ( $r$ large): the two brackets become $(I_{z} - I_{x}) (I_{y} - I_{z})$ -type with opposite signs, giving $k < 0$ → oscillation → stable. This is the Dzhanibekov effect / intermediate axis theorem. Growth rate $k = 200/3 \approx 8.16$ /s ✓.

Ex 6 — Case F: degenerate inertia (perfect sphere)

Worked example A ball-shaped satellite

A satellite is a uniform sphere: $I_{x} = I_{y} = I_{z} = I = 12$ kg·m². A torque $M_{z} = 3$ N·m is applied about $z$ while it already spins $p = 4$ , $q = 5$ rad/s. Find $\overset{r}{˙}$ .

Forecast: do the $p, q$ spins produce any coupling torque about $z$ ? Guess yes/no.

Euler- $z$ : $M_{z} = I_{z} \overset{r}{˙} + (I_{y} - I_{x}) pq$ . Why this step? All three inertias appear; watch the coupling bracket.
Since $I_{x} = I_{y}$ , the bracket $(I_{y} - I_{x}) = 0$ ; the coupling term vanishes entirely. Why this step? A sphere has no "hard" or "easy" spin axis, so no gyroscopic coupling ever.
Thus $M_{z} = I \overset{r}{˙} \Rightarrow \overset{r}{˙} = M_{z} / I = 3/12 = 0.25$ rad/s². Why this step? Rotation now behaves exactly like translation: torque = inertia × angular accel.

Verify: every Euler equation for a sphere reduces to $M_{i} = I \overset{ω}{˙}_{i}$ — three independent scalar laws, no cross-talk. Units: N·m / kg·m² = 1/s² = rad/s² ✓.

Ex 7 — Case G: decoupled single-axis maneuver

Worked example Clean pitch-up with a nose thruster

A rocket at rest in rotation ( $p = q = r = 0$ ) fires a side thruster of force $F = 800$ N at lever arm $ℓ = 3$ m ahead of the CoM, producing a pure pitch moment. $I_{y} = 500$ kg·m². Find the pitch acceleration $\overset{q}{˙}$ and the pitch rate after a $t = 2$ s burn.

Forecast: does the roll/yaw state matter here? (It won't — that is why we chose $p = r = 0$ .)

Moment about pitch axis: $M_{y} = F ℓ = 800 \times 3 = 2400$ N·m. Why this step? Force times lever arm = torque; this is the drive term.
Euler- $y$ : $M_{y} = I_{y} \overset{q}{˙} + (I_{x} - I_{z}) r p$ . With $r = p = 0$ the coupling dies. Why this step? Starting non-spinning means the maneuver decouples into one clean axis.
$\overset{q}{˙} = M_{y} / I_{y} = 2400/500 = 4.8$ rad/s². Why this step? Divide torque by inertia — the final rearrangement of $M_{y} = I_{y} \overset{q}{˙}$ .
After the burn: $q = \overset{q}{˙} t = 4.8 \times 2 = 9.6$ rad/s. Why this step? Because the thruster force, lever arm and inertia are all constant during the burn, $\overset{q}{˙}$ is constant, and a constant rate of change integrates simply to $q = \overset{q}{˙} t$ (rate × time), exactly like distance = speed × time.

Verify: units N·m / kg·m² = rad/s² ✓ for $\overset{q}{˙}$ , then rad/s² · s = rad/s ✓ for $q = 9.6$ rad/s. Sanity check: if the burn lasted only $t = 0$ s the pitch rate would be $0$ , and it grows linearly with burn time — as expected for constant acceleration. If the rocket had been rolling ( $p \neq = 0$ ), Case C shows a cross term would sneak in — hence the "start from rest" trick.

Ex 8 — Case H: full word problem (force + gravity + moment)

Worked example Gravity turn with an off-axis thruster

A rocket ( $m = 1000$ kg, $I_{z} = 400$ kg·m²) flies horizontally: body- $x$ forward at $u = 150$ m/s, $v = w = 0$ , and it is yawing at $r = 0.2$ rad/s ( $p = q = 0$ ). A gimballed engine gives thrust $T = 12, 000$ N along body- $x$ , plus a small yaw-control side force $F_{s} = 500$ N at lever arm $ℓ = 4$ m behind the CoM. Gravity acts down (body- $z$ points down here) at $m g$ , $g = 9.8$ . Find (a) forward accel $\overset{u}{˙}$ , (b) sideways accel $\overset{v}{˙}$ , (c) yaw accel $\overset{r}{˙}$ .

Forecast: which of the three answers is not simply force/mass? (Guess before step 1.)

Forward: $F_{x} = T = 12000$ N. Newton- $x$ : $F_{x} = m (\overset{u}{˙} + q w - r v)$ , and $q = v = 0$ so $\overset{u}{˙} = F_{x} / m = 12000/1000 = 12$ m/s². Why this step? No transport term survives on $x$ here (needs $q$ or $v$ ).
Sideways: the side force $F_{y} = F_{s} = 500$ N. Newton- $y$ : $F_{y} = m (\overset{v}{˙} + r u - pw)$ , with $p = 0$ : $\overset{v}{˙} = F_{y} / m - r u = 500/1000 - (0.2) (150) = 0.5 - 30 = - 29.5$ m/s². Why this step? This is the non-trivial one — the yaw swings the velocity arrow, adding a big transport term $- r u$ that dwarfs the thruster's direct contribution. We isolate $\overset{v}{˙}$ by moving that transport term to the right-hand side.
Yaw: moment $M_{z} = F_{s} ℓ = 500 \times 4 = 2000$ N·m. Euler- $z$ : $M_{z} = I_{z} \overset{r}{˙} + (I_{y} - I_{x}) pq$ , with $p = q = 0$ : $\overset{r}{˙} = M_{z} / I_{z} = 2000/400 = 5$ rad/s². Why this step? Rotation decouples because $p = q = 0$ ; only the drive term $M_{z}$ survives, so the promised yaw acceleration comes straight from moment over inertia.

Verify: (a) $\overset{u}{˙} = 12$ m/s², (b) $\overset{v}{˙} = - 29.5$ m/s², (c) $\overset{r}{˙} = 5$ rad/s². The sign of $\overset{v}{˙}$ is negative: even though the thruster pushes $+ y$ , the turn's centripetal demand pulls the body sideways faster the other way — a real coordinated-turn effect. Units all m/s² and rad/s² ✓.

Ex 9 — Case I: limiting / exam twist (variable mass)

Worked example Why the naive

$\frac{d ( m v )}{d t}$ blows up A rocket burns propellant: $\overset{m}{˙} = - 20$ kg/s, current $m = 800$ kg, exhaust speed relative to the rocket $u_{e} = 2500$ m/s. It flies straight, $ω = 0$ , no gravity, $u = 400$ m/s. A student writes $F = \frac{d ( m u )}{d t} = m \overset{u}{˙} + \overset{m}{˙} u$ and sets the only force to zero. Find the correct $\overset{u}{˙}$ and expose the error.

Forecast: the student concludes $\overset{u}{˙} = - \overset{m}{˙} u / m$ . Is that the real acceleration?

Correct physics (Meshchersky): $F_{e x t} + \overset{m}{˙} u_{e, rel} = m \overset{u}{˙}$ . Treat thrust $T = - \overset{m}{˙} u_{e}$ as an external force. Why this step? The ejected gas carries momentum; you cannot lump $\overset{m}{˙} u$ into $\frac{d ( m v )}{d t}$ and call it Newton — the reference momentum is the gas's, not the rocket's.
Thrust: $T = - \overset{m}{˙} u_{e} = - (- 20) (2500) = 50, 000$ N (forward). Why this step? Mass leaves ( $\overset{m}{˙} < 0$ ) rearward, so thrust is forward and positive.
No other external force, so $m \overset{u}{˙} = T \Rightarrow \overset{u}{˙} = 50000/800 = 62.5$ m/s². Why this step? The clean constant-mass Newton form with thrust as an external force.
The student's wrong answer: $\overset{u}{˙}_{wrong} = - \overset{m}{˙} u / m = - (- 20) (400) /800 = 10$ m/s² — off by more than a factor of six, and it uses flight speed $u$ instead of exhaust speed $u_{e}$ . Why this step? The trap replaces the exhaust momentum with the vehicle momentum.

Verify: correct $\overset{u}{˙} = 62.5$ m/s² vs naive $10$ m/s². The ratio of the missing physics is $u_{e} / u = 2500/400 = 6.25$ , exactly the discrepancy — proof the error is the swapped velocity ✓. Units: N/kg = m/s² ✓.

Recall One-line summary of every case

Case A ::: no spin → plain $F = ma$ . Case B ::: constant body speed but turning → real accel from the $ω \times v$ term. Case C ::: sign of $(I_{x} - I_{z}) r p$ sets which way the gyroscopic torque points. Case D ::: symmetric spinner → bounded coning at frequency $∣ λ ∣ = \frac{∣ I _{x} - I _{t} ∣}{I _{t}} p_{0}$ . Case E ::: intermediate-axis spin → coefficient $k > 0$ → exponential flip. Case F ::: sphere → all coupling zero → three independent $M_{i} = I \overset{ω}{˙}_{i}$ . Case G ::: start from rest → single-axis maneuver decouples cleanly. Case H ::: word problem → build $F, M$ , watch the transport term dominate $\overset{v}{˙}$ . Case I ::: variable mass → thrust is an external force $- \overset{m}{˙} u_{e}$ , never $\frac{d ( m v )}{d t}$ .

Flashcards

In Ex 2, why is

F_{y}

nonzero even though

\overset{v}{˙} = 0

The yaw rate

r

swings the velocity arrow, so the transport term

r u

gives real inertial (centripetal) acceleration

r u = u^{2} / R

Which spin axis is unstable for a fully asymmetric body?

The intermediate (middle) axis — the coupling coefficient

k

becomes positive, giving exponential growth (Dzhanibekov effect).

For a perfect sphere, what happens to Euler's coupling terms?

They all vanish since

I_{x} = I_{y} = I_{z}

, leaving three independent scalar laws

M_{i} = I \overset{ω}{˙}_{i}

Why is the naive

\frac{d ( m v )}{d t}

wrong for a burning rocket?

It replaces the exhaust momentum with vehicle momentum; the correct thrust is the external force

- \overset{m}{˙} u_{e}

(Meshchersky).