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

3.4.5 · D2 · Physics › Rocket Flight Mechanics › 6DOF equations — translational (Newton), rotational (Euler's

Hum the parent 6DOF note ka central result build kar rahe hain. Har woh tool jo woh borrow karta hai, hum yahan khud kamayenge.

Step 1 — Angular momentum kya hai, ek picture mein?

KYA HAI. Ek rocket kai chote chote mass ke tukdon se bana hota hai. Jab poori cheez rate $\vec{\omega }$ se spin karti hai, har tukda ek circle mein daudta hai. Mass $m_i$ ka ek tukda jo position ${\vec{r}}_i$ par baitha hai (center of mass se us tukde tak ka arrow) velocity ${\vec{v}}_i=\vec{\omega }\times {\vec{r}}_i$ se chalta hai.

$\times$ (cross product) kyun, ordinary multiplication kyun nahi? Kyunki humein ek aisi operation chahiye jo "spin axis" aur "tukda kahan hai" ko lekar return kare "tukda kis taraf ja raha hai" — aur woh moving direction dono ke perpendicular hoti hai. Cross product $\vec{a}\times \vec{b}$ define hi aise hota hai ki woh $\vec{a}$ aur $\vec{b}$ dono ke perpendicular arrow hai, jiska length $|\vec{a}||\vec{b}|\sin \theta$ ho. Yeh ek aur hai clean tool jo jawab deta hai "mujhe sideways direction do."

PICTURE. Figure mein, violet axis $\vec{\omega }$ hai. Ek tukda (magenta dot) ${\vec{r}}_i$ par baitha hai. Iska velocity (orange arrow) axis ke around wrap karta hai — $\vec{\omega }$ aur ${\vec{r}}_i$ dono ke perpendicular, exactly wahi jo $\vec{\omega }\times {\vec{r}}_i$ produce karta hai.

Ab ek tukde ka angular momentum hai ${\vec{h}}_i={\vec{r}}_i\times (m_i{\vec{v}}_i)$ — position crossed with momentum. Har tukde ko add karo:

Step 2 — Spin-momentum ko ek tensor kyun chahiye, ek number nahi

KYA HAI. $\vec{H}$ mein ${\vec{v}}_i=\vec{\omega }\times {\vec{r}}_i$ substitute karo. Sum ek aisi machine ban jaata hai jo $\vec{\omega }$ khaata hai aur $\vec{H}$ ugalta hai. Woh machine hai inertia tensor $\mathbf{I}$: $\vec{H}=\mathbf{I}\vec{\omega }.$

Tensor kyun, sirf ek mass-jaisa number kyun nahi? Sliding ke liye, resistance ek number $m$ hai (rocket ko kisi bhi direction mein push karna equally mushkil hai). Spinning ke liye, ek rocket apne lambe patle axis ke baare mein aasaani se roll karta hai lekin end-over-end tumble karna mushkil hai — resistance axis par depend karta hai. Ek akela number "alag alag directions mein alag" nahi store kar sakta; ek $3\times 3$ table (tensor) kar sakta hai. Dekho Inertia tensor and principal axes.

PICTURE. Figure mein wahi rocket do taraon se spin ho raha hai: long axis ke baare mein (chota violet $\vec{H}$, aasaan) versus end-over-end (lamba magenta $\vec{H}$, mushkil). Same $|\vec{\omega }|$, alag $|\vec{H}|$ — proof ki ek number ise describe nahi kar sakta.

Step 3 — Jis law se hum shuru karte hain: torque, $\vec{H}$ ka rate hai

KYA HAI. "Force momentum change karti hai" ka rotational version hai: torque angular momentum change karta hai. $\vec{M}={\frac{d\vec{H}}{dt}|}_I$

Chhota ${|}_I$ kyun? Symbol $\frac{d\vec{H}}{dt}$ matlab hai "har second mein $\vec{H}$ kitni tezi se change hota hai" (dot/derivative — rate of change ka hamaara tool, kyunki motion ka law exactly isi baare mein hota hai). ${|}_I$ kehta hai inertial frame $I$ mein measure kiya gaya — woh non-spinning bahari duniya jahan Newton's laws actually valid hain. Yeh "sirf $I$ mein valid" hi poori wajah hai agli do steps ke liye. Dekho Reference frames and rotation matrices.

PICTURE. Figure mein $\vec{H}$ time $t$ par aur thodi der baad $t+dt$ par dikhta hai. Chhota sa change $d\vec{H}$ (orange) wahi direction mein point karta hai jahan applied torque $\vec{M}$ (magenta) hai. Torque literally $\vec{H}$ arrow ki tip ko push karta hai.

Step 4 — Trap: $\mathbf{I}$ bahari duniya mein constant nahi hai

KYA HAI. Hum $\frac{d\vec{H}}{dt}{|}_I=\frac{d(\mathbf{I}\vec{\omega })}{dt}{|}_I$ compute karna chahte hain. Naively hum $\mathbf{I}\dot{\vec{\omega }}$ likhna chahenge. Lekin dhyan do: inertia tensor $\mathbf{I}$ rocket se chipka hua hai, aur rocket spin kar raha hai. Toh frame $I$ se dekha jaaye, $\mathbf{I}$ ke numbers har pal change hote rehte hain — "easy axis" ghoomti rehti hai.

Yeh kyun matter karta hai. Agar $\mathbf{I}$ aur $\vec{\omega }$ dono $I$ mein change hote hain, product rule do terms deta hai aur $\mathbf{I}$ ka term track karna bahut mushkil hai. Hum ek aisa frame chahte hain jahan $\mathbf{I}$ still rehe.

PICTURE. Figure mein body frame ke teen principal axes do instants par dikhte hain — woh rotate ho gaye hain. Inertia values unhi axes par rehti hain, toh $I$ mein woh moving targets hain (magenta = purani axes, violet = rotated axes).

Fix (preview): rate ko body frame $B$ mein compute karo, jahan $\mathbf{I}$ frozen aur diagonal hai — phir us rate ko $I$ mein convert karo. Converter hai transport theorem, Step 5.

Step 5 — Transport theorem: "body rate" ko "world rate" mein convert karna

KYA HAI. Kisi bhi arrow $\vec{A}$ ke liye jo ek spinning frame ke saath chal raha ho: ${\frac{d\vec{A}}{dt}|}_I={\frac{d\vec{A}}{dt}|}_B+\vec{\omega }\times \vec{A}$

Yeh sach kyun hai, ek picture mein. $\vec{A}$ ko body ke apne axes ${\hat{e}}_x,{\hat{e}}_y,{\hat{e}}_z$ se likho: $\vec{A}=A_x{\hat{e}}_x+A_y{\hat{e}}_y+A_z{\hat{e}}_z$. Do cheezein $\vec{A}$ ko bahar se dekhe jaane par change kar sakti hain:

1. Components $A_x,A_y,A_z$ change hote hain — yeh body ke andar ka rate hai, $\frac{d\vec{A}}{dt}{|}_B$;
2. Axes khud rotate hote hain — agar components frozen bhi hain, arrow phir bhi swing karta hai kyunki uska scaffolding spin kar raha hai. Ek fixed-length axis ${\hat{e}}_i$ jo $\vec{\omega }$ se rotate kar raha hai woh ${\dot{\hat{e}}}_i=\vec{\omega }\times {\hat{e}}_i$ se move karta hai (Step 1 wala hi "sideways" cross-product idea). $A_i(\vec{\omega }\times {\hat{e}}_i)$ sum karne par $\vec{\omega }\times \vec{A}$ milta hai.

Dono effects add karo. Bas yahi hai poora theorem — kuch nahi bas product rule, split into "components move" + "scaffolding moves." Dekho Transport theorem (rotating frames).

PICTURE. Left: components change hote hain, axes still hain. Right: components frozen hain, axes spin karti hain — arrow phir bhi move karta hai. Real motion dono ka sum hai.

Step 6 — Euler's equation assemble karo

KYA HAI. Transport theorem (Step 5) ko $\vec{A}=\vec{H}$ ke saath apply karo, aur Step 3 ka law use karo: $\vec{M}={\frac{d\vec{H}}{dt}|}_I={\frac{d\vec{H}}{dt}|}_B+\vec{\omega }\times \vec{H}.$ Ab body frame choose karne ka fayda milta hai: wahan $\vec{H}=\mathbf{I}\vec{\omega }$ hai jahan $\mathbf{I}$ frozen aur diagonal hai, toh uska body-rate simply $\mathbf{I}\dot{\vec{\omega }}$ hai ($\dot{\mathbf{I}}$ term zero hai — Step 4 ka poora point yahi tha). $\vec{H}=\mathbf{I}\vec{\omega }$ substitute karo:

Cross-term real kyun hai aur math artifact nahi. Yeh transport theorem ke scaffolding-spins piece se aaya. Yahi woh physics hai jo ek spinning top ko precess karaati hai aur ek phenke gaye phone ko flip karaati hai. Parent ka Angular momentum conservation note dekho: even when $\vec{M}=0$, yeh term $\vec{\omega }$ ko dance karaata rehta hai.

PICTURE. Figure teen arrows stack karta hai: applied torque $\vec{M}$ = "obvious" spin-up arrow plus gyroscopic side-arrow $\vec{\omega }\times (\mathbf{I}\vec{\omega })$.

Step 7 — Scalars mein tod do (principal-axis components)

KYA HAI. $\mathbf{I}=\mathrm{diag}(I_x,I_y,I_z)$ aur $\vec{\omega }=(p,q,r)$ (roll, pitch, yaw rates) ke saath, $\vec{H}=(I_{x}p,I_{y}q,I_{z}r)$ hai. Cross-product recipe se $\vec{\omega }\times \vec{H}$ compute karo — uska $x$-component hai $q(I_{z}r)-r(I_{y}q)=(I_z-I_y)qr$:

• $I_x\dot{p}$ ::: roll axis roll torque se spin up hoti hai.
• $(I_z-I_y)qr$ ::: pitch aur yaw rates roll mein feed karte hain agar do transverse inertias alag hain. Agar $I_z=I_y$ toh yeh zero ho jaata hai.

Cyclic pattern $x\to y\to z\to x$ kyun? Cross product teen axes ko symmetrically treat karta hai; labels $(x,y,z)\to (y,z,x)$ rotate karne par har line reproduce hoti hai. Ek yaad karo, teen mil jaate hain.

PICTURE. Figure teen boxes $p,q,r$ ki ek ring hai jisme arrows dikhate hain ki har pair teesre mein "leak" karta hai — coupling visually dikhaya gaya.

Step 8 — Har case: coupling term kya karta hai

Hum kabhi bhi reader ko aise scenario mein nahi chodenge joh humne dikhaya nahi. Yahan coupling term $(I_z-I_y)qr$ (aur uske cousins) ke saare regimes hain:

PICTURE. Char mini-panels: A steady zero, B steady spin, C gentle cone, D violent flip — ek term se govern hone wala poora behaviour spectrum.

Ek-picture summary

Yeh single diagram poore page ko compress karta hai: torque = duniya mein $\vec{H}$ ka rate se shuru karo (Step 3), notice karo ki $\mathbf{I}$ duniya mein still nahi baithega (Step 4), transport theorem ke zariye body frame mein jump karo (Step 5) jahan $\mathbf{I}$ frozen hai, aur $\vec{M}=\mathbf{I}\dot{\vec{\omega }}+\vec{\omega }\times (\mathbf{I}\vec{\omega })$ par land karo (Step 6) — doosra term woh keemat hai jo frames jump karte waqt spin karte rehne ki wajah se chukani padti hai.

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