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

3.4.5 · Physics › Rocket Flight Mechanics

1. Setup: do frames, aur hume dono kyun chahiye

WHAT hai tension? Forces/velocities body frame mein measure karna sabse aasaan hai (ek thruster hamesha body- $x$ ke along point karta hai), lekin Newton ka law $F = m a$ acceleration ke liye sirf inertial frame mein sach hai. "Rate seen in $B$ " aur "rate seen in $I$ " ke beech ka bridge transport theorem hai.

2. Translational equations (Newton)

Scratch se derivation. Total momentum $p = m v$ (jahan $v$ = CoM velocity in $I$ ). $F_{e x t} = \frac{d p}{d t}_{I} .$ Ek constant-mass body ke liye, $F = m \dot{v} ∣_{I}$ . Lekin hum aam taur par $v$ ko body components mein store karte hain. $v$ par transport theorem apply karo:

3. Rotational equations (Euler)

Scratch se derivation. CoM ke baare mein angular momentum: $H = I ω$ . Newton ka rotational law: torque = angular momentum ka rate in $I$ : $M = \frac{d H}{d t}_{I} .$ $H$ par transport theorem apply karo:

= \mathbf I\,\dot{\vec\omega}\big|_B + \vec\omega\times(\mathbf I\vec\omega).$$ *$\dot{\mathbf I}=0$ term kyun drop hota hai?* Body axes mein inertia tensor constant hai, isliye sirf $\vec\omega$ ka body-frame derivative hota hai. > [!formula] Euler's rotational equations (principal axes) > Body axes = principal axes choose karne se $\mathbf I=\mathrm{diag}(I_x,I_y,I_z)$ ho jaata hai. Tab: > $$\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}}$$ > *Ye step kyun?* $\vec\omega\times(\mathbf I\vec\omega)$ ko component-wise compute karo: > iska $x$-component hai $q(I_z r)-r(I_y q)=(I_z-I_y)qr$. Wahi pattern cyclically. > [!mistake] Steel-man: "Torque-free matlab $\vec\omega$ = constant." > **Kyun sahi lagta hai:** analogy $\vec F=0\Rightarrow\vec v$ const se. **Trap:** $\vec M=0$ hone par bhi, > agar $I_x\neq I_y\neq I_z$ ho to coupling terms $p,q,r$ ko *energy swap* karate hain — body > tumble karti hai (jaise "tennis-racket / Dzhanibekov effect"). Jo actually conserved hai wo $\vec H$ > hai $I$ mein aur kinetic energy, **na ki** body-frame $\vec\omega$. **Fix:** $\dot{\vec\omega}=0$ > sirf tab jab $\vec\omega$ ek principal axis ke along ho (spin stabilization). ![[3.4.05-6DOF-equations-—-translational-(Newton),-rotational-(Euler's-equations).png]] --- ## 4. Worked examples > [!example] Ex 1 — Seedha climb, koi spin nahi (sanity check) > Rocket vertically climb karta hai: $\vec\omega=0$, thrust $T$ upar, gravity $mg$ neeche, body-$x$ upar, $u$=speed. > **Translational:** $F_x=m(\dot u + qw-rv)=m\dot u$. $F_x=T-mg$ ke saath: $\;\dot u=(T-mg)/m$. ✓ > *Ye step kyun?* Saare $\vec\omega$ terms vanish ho jaate hain, elementary 1D rocket motion recover hoti hai — acha hai. > **Rotational:** $M=0,\ \vec\omega=0\Rightarrow\dot{\vec\omega}=0$. Non-spinning rehta hai. ✓ > [!example] Ex 2 — Ek symmetric spinning rocket ki coning ($I_y=I_z=I_t$, spin $I_x$) > Torque-free. Euler-$x$: $M_x=I_x\dot p=0\Rightarrow p=p_0$ (spin rate constant). > Euler-$y,z$: $I_t\dot q=(I_x-I_t)rp_0$, $\;I_t\dot r=-(I_x-I_t)qp_0$. > Maano $\lambda=\frac{I_x-I_t}{I_t}p_0$. Tab $\dot q=\lambda r,\ \dot r=-\lambda q$ > $\Rightarrow \ddot q=-\lambda^2 q$: frequency $|\lambda|$ par **oscillation**. > *Ye step kyun?* Do transverse rates ek doosre mein rotate karte hain — spin axis ek cone trace karta hai. > **Interpretation:** symmetry axis ke baare mein spinning steady, predictable coning deti hai → *isliye > sounding rockets spin-stabilize karte hain.* > [!example] Ex 3 — Thruster dwara pitch maneuver > CoM se $\ell$ aage side thruster force $F$ pitch moment $M_y=F\ell$ produce karta hai. > Agar $p=r=0$ se shuru karo: $M_y=I_y\dot q\Rightarrow \dot q=F\ell/I_y$. > *Ye step kyun?* $p=r=0$ ke saath coupling term $(I_x-I_z)rp=0$ hai, to pitch decouple ho jaata hai — ek clean > angular acceleration milti hai. Burn time mein pitch rate paane ke liye integrate karo. --- > [!recall]- Feynman: ek 12-saal ke bachche ko explain karo > Ek toy rocket imagine karo. Ye **slide** teen tareekon se kar sakta hai (aage, sideways, upar) aur **twist** teen tareekon se > (log ki tarah roll, nose upar/neeche, nose left/right). Kul chhe moves hain. Sliding predict karne ke liye hum > "push = mass × speeding-up" (Newton) use karte hain. Twisting predict karne ke liye hum ek spin-rule (Euler) use karte hain — lekin > spinning sneaky hai: ek twist par push karna ise ek *alag* tarike se twist karna shuru kar sakta hai, jaise > toss kiya hua phone ajeeb se flip karta hai. Equations mein extra "$\vec\omega\times$" bits sirf > math ka politely yaad dilana hai ki rocket spin kar raha hai jab hum use dekh rahe hain. > [!mnemonic] > **"New Fear, Euler Hears"** — **N**ewton: **F** = m(v-dot + ω×v). **E**uler: **M** = I·ω-dot + ω×(Iω). > Dono ki shape ek jaisi hai: *(rate in body) + (ω cross the momentum-thing)*. Translational > **linear** momentum $m\vec v$ use karta hai; rotational **angular** momentum $\mathbf I\vec\omega$ use karta hai. --- ## #flashcards/physics Ek rigid rocket ke 6 degrees of freedom kya hain? ::: CoM ke 3 translations (x,y,z) + body axes ke baare mein 3 rotations (roll p, pitch q, yaw r). Transport theorem state karo. ::: $\frac{d\vec A}{dt}|_I = \frac{d\vec A}{dt}|_B + \vec\omega\times\vec A$; ye ek body-frame rate ko inertial-frame rate mein convert karta hai. Translational body-axis equation mein $\vec\omega\times\vec v$ term kyun hai? ::: Kyunki ek turning body inertial frame mein accelerate karti hai jab bhi body-frame velocity components constant hon (velocity direction change hoti hai). Newton ki 6DOF translational equation body axes mein likho. ::: $\vec F = m(\dot{\vec v}|_B + \vec\omega\times\vec v)$. Euler ki equation body frame mein kyun likhi jaati hai? ::: Kyunki inertia tensor $\mathbf I$ body frame mein constant hai (body ke saath rotate karta hai), equations tractable ban jaati hain. Principal axes ke along Euler's rotational equations state karo. ::: $M_x=I_x\dot p+(I_z-I_y)qr$; $M_y=I_y\dot q+(I_x-I_z)rp$; $M_z=I_z\dot r+(I_y-I_x)pq$. $(I_z-I_y)qr$ coupling kahan se aata hai? ::: Angular momentum par apply transport theorem ke $\vec\omega\times(\mathbf I\vec\omega)$ (gyroscopic) term se. Kya $\vec M=0$ matlab $\vec\omega$ constant hai? ::: Nahi. Sirf tab jab $\vec\omega$ ek principal axis ke along ho; warna body tumble karti hai (Dzhanibekov effect). $\vec H$ (inertial frame mein) aur KE conserved hain, body-frame $\vec\omega$ nahi. Ek symmetric spinner ($I_y=I_z$) ke liye kya constant rehta hai? ::: Symmetry axis ke baare mein spin rate $p$ (kyunki $M_x=I_x\dot p=0$); transverse rates frequency $\frac{I_x-I_t}{I_t}p$ par cone karte hain. Translational equation mein mass loss ka sahi handling? ::: Thrust $\vec T=-\dot m\,\vec u_e$ ko $\vec F$ mein external force ki tarah treat karo; clean constant-mass Newton use karo, $m(t)$ alag se bookkeep karo (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 ```