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

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1. Setup: two frames, and why we need both

WHAT is the tension? Forces/velocities are easiest to measure in the body frame (a thruster always points along body-$x$), but Newton's law $\vec F = m\vec a$ is only true for acceleration in the inertial frame. The bridge between "rate seen in $B$" and "rate seen in $I$" is the transport theorem.

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2. Translational equations (Newton)

Derivation from scratch. Total momentum $\vec p = m\vec v$ (with $\vec v$ = CoM velocity in $I$). $\vec F_{ext} = \frac{d\vec p}{dt}\Big|_I.$ For a constant-mass body, $\vec F = m\,\dot{\vec v}|_I$. But we usually store $\vec v$ in body components. Apply the transport theorem to $\vec v$:

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3. Rotational equations (Euler)

Derivation from scratch. Angular momentum about the CoM: $\vec H = \mathbf I\,\vec\omega$. Newton's rotational law: torque = rate of angular momentum in $I$: $\vec M = \frac{d\vec H}{dt}\Big|_I.$ Apply transport theorem to $\vec H$:

= \mathbf I\,\dot{\vec\omega}\big|_B + \vec\omega\times(\mathbf I\vec\omega).$$ *Why $\dot{\mathbf I}=0$ term drops?* In body axes the inertia tensor is constant, so only $\vec\omega$ has a body-frame derivative. > [!formula] Euler's rotational equations (principal axes) > Choosing body axes = principal axes makes $\mathbf I=\mathrm{diag}(I_x,I_y,I_z)$. Then: > $$\boxed{\begin{aligned} > M_x &= I_x\dot p + (I_z - I_y)\,qr\\ > M_y &= I_y\dot q + (I_x - I_z)\,rp\\ > M_z &= I_z\dot r + (I_y - I_x)\,pq > \end{aligned}}$$ > *Why this step?* Compute $\vec\omega\times(\mathbf I\vec\omega)$ component-wise: > its $x$-component is $q(I_z r)-r(I_y q)=(I_z-I_y)qr$. Same pattern cyclically. > [!mistake] Steel-man: "Torque-free means $\vec\omega$ = constant." > **Why it feels right:** analogy to $\vec F=0\Rightarrow\vec v$ const. **The trap:** even with > $\vec M=0$, if $I_x\neq I_y\neq I_z$ the coupling terms make $p,q,r$ *swap energy* — the body > tumbles (e.g. the "tennis-racket / Dzhanibekov effect"). What's actually conserved is $\vec H$ > in $I$ and kinetic energy, **not** the body-frame $\vec\omega$. **Fix:** $\dot{\vec\omega}=0$ > only if $\vec\omega$ is along a principal axis (spin stabilization). ![[3.4.05-6DOF-equations-—-translational-(Newton),-rotational-(Euler's-equations).png]] --- ## 4. Worked examples > [!example] Ex 1 — Straight climb, no spin (sanity check) > Rocket climbs vertically: $\vec\omega=0$, thrust $T$ up, gravity $mg$ down, body-$x$ up, $u$=speed. > **Translational:** $F_x=m(\dot u + qw-rv)=m\dot u$. With $F_x=T-mg$: $\;\dot u=(T-mg)/m$. ✓ > *Why this step?* All $\vec\omega$ terms vanish, recovering elementary 1D rocket motion — good. > **Rotational:** $M=0,\ \vec\omega=0\Rightarrow\dot{\vec\omega}=0$. Stays non-spinning. ✓ > [!example] Ex 2 — Coning of a symmetric spinning rocket ($I_y=I_z=I_t$, spin $I_x$) > Torque-free. Euler-$x$: $M_x=I_x\dot p=0\Rightarrow p=p_0$ (spin rate constant). > Euler-$y,z$: $I_t\dot q=(I_x-I_t)rp_0$, $\;I_t\dot r=-(I_x-I_t)qp_0$. > Let $\lambda=\frac{I_x-I_t}{I_t}p_0$. Then $\dot q=\lambda r,\ \dot r=-\lambda q$ > $\Rightarrow \ddot q=-\lambda^2 q$: **oscillation** at frequency $|\lambda|$. > *Why this step?* The two transverse rates rotate into each other — the spin axis traces a cone. > **Interpretation:** spinning about the symmetry axis gives steady, predictable coning → *this is > why sounding rockets spin-stabilize.* > [!example] Ex 3 — Pitch maneuver by a thruster > Side thruster force $F$ at distance $\ell$ ahead of CoM produces pitch moment $M_y=F\ell$. > If starting from rest with $p=r=0$: $M_y=I_y\dot q\Rightarrow \dot q=F\ell/I_y$. > *Why this step?* With $p=r=0$ the coupling term $(I_x-I_z)rp=0$, so pitch decouples — a clean > angular acceleration. Integrate to get pitch rate over burn time. --- > [!recall]- Feynman: explain to a 12-year-old > Imagine a toy rocket. It can **slide** three ways (forward, sideways, up) and **twist** three ways > (roll like a log, nose up/down, nose left/right). That's six moves total. To predict sliding we > use "push = mass × speeding-up" (Newton). To predict twisting we use a spin-rule (Euler) — but > spinning is sneaky: pushing on one twist can make it start twisting a *different* way, like how a > tossed phone flips weirdly. The extra "$\vec\omega\times$" bits in the equations are just the > math politely reminding us the rocket is spinning while we watch it. > [!mnemonic] > **"New Fear, Euler Hears"** — **N**ewton: **F** = m(v-dot + ω×v). **E**uler: **M** = I·ω-dot + ω×(Iω). > Both have the same shape: *(rate in body) + (ω cross the momentum-thing)*. Translational uses > **linear** momentum $m\vec v$; rotational uses **angular** momentum $\mathbf I\vec\omega$. --- ## #flashcards/physics What are the 6 degrees of freedom of a rigid rocket? ::: 3 translations of the CoM (x,y,z) + 3 rotations about body axes (roll p, pitch q, yaw r). State the transport theorem. ::: $\frac{d\vec A}{dt}|_I = \frac{d\vec A}{dt}|_B + \vec\omega\times\vec A$; it converts a body-frame rate into an inertial-frame rate. Why does the translational body-axis equation have an $\vec\omega\times\vec v$ term? ::: Because a turning body accelerates in the inertial frame even when body-frame velocity components are constant (velocity direction changes). Write Newton's 6DOF translational equation in body axes. ::: $\vec F = m(\dot{\vec v}|_B + \vec\omega\times\vec v)$. Why is Euler's equation written in the body frame? ::: Because the inertia tensor $\mathbf I$ is constant in the body frame (it rotates with the body), making the equations tractable. State Euler's rotational equations along principal axes. ::: $M_x=I_x\dot p+(I_z-I_y)qr$; $M_y=I_y\dot q+(I_x-I_z)rp$; $M_z=I_z\dot r+(I_y-I_x)pq$. Where does the $(I_z-I_y)qr$ coupling come from? ::: From the $\vec\omega\times(\mathbf I\vec\omega)$ (gyroscopic) term of the transport theorem applied to angular momentum. Does $\vec M=0$ imply $\vec\omega$ constant? ::: No. Only if $\vec\omega$ lies along a principal axis; otherwise the body tumbles (Dzhanibekov effect). $\vec H$ (in inertial frame) and KE are conserved, not body-frame $\vec\omega$. For a symmetric spinner ($I_y=I_z$), what stays constant? ::: The spin rate $p$ about the symmetry axis (since $M_x=I_x\dot p=0$); transverse rates cone at frequency $\frac{I_x-I_t}{I_t}p$. Correct handling of mass loss in the translational equation? ::: Treat thrust $\vec T=-\dot m\,\vec u_e$ as an external force in $\vec F$; use clean constant-mass Newton, bookkeep $m(t)$ separately (Meshchersky). --- ## Connections - [[Reference frames and rotation matrices]] - [[Transport theorem (rotating frames)]] - [[Inertia tensor and principal axes]] - [[Angular momentum conservation]] - [[Meshchersky / Tsiolkovsky variable-mass dynamics]] - [[Spin stabilization and coning motion]] - [[Dzhanibekov effect / intermediate axis theorem]] - [[Quaternion attitude kinematics]] ## 🖼️ Concept Map ```mermaid flowchart TD A[Rigid rocket in flight] -->|decouples into| B[Translation of CoM] A -->|decouples into| C[Rotation about CoM] B -->|3 DOF| D[6DOF total] C -->|3 DOF| D E[Inertial frame I] -->|Newton valid here| F[Newton 2nd law] G[Body frame B] -->|inertia tensor constant| H[Euler equations] T[Transport theorem] -->|bridges I and B| E T -->|adds omega cross A term| G F -->|apply transport to v| I[Translational eqns: F = m dv/dt + omega x v] H -->|apply transport to L| J[Rotational eqns: M = I dw/dt + omega x Iw] I -->|governs| B J -->|governs| C ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, 6DOF ka matlab hai ki ek rigid rocket sirf 6 tarike se move kar sakta hai: teen tarah se > slide (aage, side, upar) aur teen tarah se ghoom (roll, pitch, yaw). Slide wale part ke liye Newton > ka law — $\vec F = m\vec a$ — kaafi hai, aur ghoomne wale part ke liye Euler ki equations lagti hain. > Ek rigid body ki motion hamesha do saaf hisson mein toot jaati hai: center of mass ki motion + uske > around rotation. Isiliye hum ek complicated udta hua object bhi easily handle kar lete hain. > > Twist yeh hai ki Newton ka law sirf inertial frame (Earth-fixed) mein sach hai, lekin humein forces > aur velocity rocket ke apne body frame mein likhna convenient hota hai (thruster hamesha body-x pe > point karta hai). In dono ke beech pul hai **transport theorem**: jab tum body frame mein koi rate > naapte ho, inertial rate paane ke liye ek extra $\vec\omega\times$ term add karna padta hai. Yahi > term dono equations mein aata hai — translational mein $\vec\omega\times\vec v$, aur rotational mein > $\vec\omega\times(\mathbf I\vec\omega)$. > > Euler equations mein jo coupling terms hain, jaise $(I_z-I_y)qr$, woh hi asli maza hain. Inke wajah > se agar rocket ka spin axis principal axis pe nahi hai, toh $\vec M=0$ hone par bhi woh tumble kar > sakta hai (Dzhanibekov effect — space mein phone flip karne jaisa). Isiliye sounding rockets ko > symmetry axis ke around spin karwate hain — tab spin rate constant rehta hai aur motion predictable > coning banti hai. Yeh cheez samajh lo toh guidance, stability aur control sab clear ho jayega. ![[audio/3.4.05-6DOF-equations-—-translational-(Newton),-rotational-(Euler's-equations).mp3]]