Mass properties — CG location, inertia tensor changing with propellant depletion

3.4.6 · Physics › Rocket Flight Mechanics

Mass properties change kyon hote hain?

KYA ho raha hai: Propellant tanks mein vehicle ke andar store hota hai. Engine use nozzle se bahar phenk deta hai. Bacha hua structure (dry mass) aur bacha hua propellant har pal ek naya, halka body define karte hain.

WHY it matters for flight:

CG thrust misalignment, aero forces, aur thrust-vector control (TVC) ke liye moment arm set karta hai.
Inertia tensor set karta hai ki ek given control torque vehicle ko kitni tezi se rotate karta hai ( $τ = I \dot{ω} + \dots$ ).
Dono move karte hain jaise fuel drain hota hai → $t = 0$ par ek well-tuned autopilot $t = t_{burnout}$ par sluggish ya unstable ho sakta hai.

Center of Gravity — scratch se derive karo

WHY ye definition? Hum chahte hain ek aisa single point jahan, agar saari mass wahan baith jaye, to net gravitational (aur inertial) moment unchanged rahe. Shuru karo is requirement se ki $r_{cg}$ ke baare mein weights ka total moment zero ho:

\;\Rightarrow\; \sum_i m_i \mathbf{r}_i = \mathbf{r}_{\text{cg}}\sum_i m_i.$$ $\mathbf{r}_{\text{cg}}$ ke liye solve karo → upar wala formula. *Ye step kyon?* Zero-moment condition literally "balance point" ki definition hai, isliye algebra sirf use isolate karta hai. ### Do-body decomposition (dry structure + propellant) Ek rocket naturally do pieces hota hai: **dry mass** $m_d$ fixed station $x_d$ par, aur **propellant** $m_p(t)$ jiska apna centroid $x_p(t)$ shift karta hai jaise tank drain hota hai. $$x_{\text{cg}}(t) = \frac{m_d\,x_d + m_p(t)\,x_p(t)}{m_d + m_p(t)}$$ *Ye step kyon?* Superposition: ek composite body ka CG uske sub-bodies ka mass-weighted CG hota hai — kyunki defining integral $dm$ mein linear hai. --- ## Inertia tensor — scratch se derive karo > [!definition] Inertia tensor > $$\mathbf{I} = \int \big( (\mathbf{r}\!\cdot\!\mathbf{r})\,\mathbb{1} - \mathbf{r}\,\mathbf{r}^{\!\top}\big)\,dm$$ > Diagonal terms $I_{xx}=\int (y^2+z^2)\,dm$ (moments of inertia); off-diagonals $I_{xy}=-\int xy\,dm$ (products of inertia). **WHY ye form?** Angular momentum hai $\mathbf{L}=\int \mathbf{r}\times(\boldsymbol\omega\times\mathbf{r})\,dm$. Triple product expand karo $\mathbf{r}\times(\boldsymbol\omega\times\mathbf{r}) = (\mathbf{r}\!\cdot\!\mathbf{r})\boldsymbol\omega - (\mathbf{r}\!\cdot\!\boldsymbol\omega)\mathbf{r}$. Right side par $\boldsymbol\omega$ factor karne se exactly tensor $\mathbf{I}$ force hota hai taaki $\mathbf{L}=\mathbf{I}\boldsymbol\omega$. *Ye step kyon?* Hum *demand* karte hain ek linear map $\boldsymbol\omega\mapsto\mathbf{L}$; bracket wahi map hai. > [!formula] Parallel-axis (Huygens–Steiner) theorem > Inertia ko CG se ek aisi point par move karne ke liye jo $\mathbf{d}$ se offset hai: > $$\mathbf{I}_{\text{P}} = \mathbf{I}_{\text{cg}} + M\big[(\mathbf{d}\!\cdot\!\mathbf{d})\mathbb{1} - \mathbf{d}\,\mathbf{d}^{\!\top}\big]$$ > Ek axis ke liye scalar version: $\;I_P = I_{\text{cg}} + M d^2.$ **Derivation:** Let $\mathbf{r}=\mathbf{r}' + \mathbf{d}$ jahan $\mathbf{r}'$ CG se measure kiya gaya ho. $$I_{xx}^P=\int\big((y'+d_y)^2+(z'+d_z)^2\big)dm.$$ Expand karo: $y'^2+z'^2$ term se $I_{xx}^{\text{cg}}$ milta hai; cross terms $2d_y\!\int y'dm$ vanish karte hain (**kyunki $\int \mathbf{r}'\,dm=0$ CG ki definition se** — *ye key why hai!*); $d_y^2+d_z^2$ term se $M d^2$ milta hai. QED. ### Flying rocket ki composite inertia Har sub-body ka inertia **uske apne CG ke baare mein** compute karo, phir har ek ko Steiner-shift karo *vehicle* CG tak aur add karo: $$\mathbf{I}_{\text{veh}}(t) = \Big[\mathbf{I}_d^{cg} + m_d D_d^2(t)\Big] + \Big[\mathbf{I}_p^{cg}(t) + m_p(t) D_p^2(t)\Big]$$ jahan $D_d(t)=|x_d - x_{\text{cg}}(t)|$, $D_p(t)=|x_p(t)-x_{\text{cg}}(t)|$. *Ye step kyon?* Inertia additive hai **sirf ek common point ke baare mein**; Steiner woh tool hai jo sabko us common (moving) CG par laata hai. ![[3.4.06-Mass-properties-—-CG-location,-inertia-tensor-changing-with-propellant-depletion.png]] --- ## Worked Example 1 — Burn ke dauran CG shift Rocket: $m_d=1000\text{ kg}$ at $x_d=6\text{ m}$ (nose-referenced). Full propellant $m_{p0}=4000\text{ kg}$ centered at $x_p=2\text{ m}$ (pehli pass ke liye centroid fixed assume karo). - **$t=0$ par:** $x_{\text{cg}}=\dfrac{1000(6)+4000(2)}{5000}=\dfrac{14000}{5000}=2.8\text{ m}$. *Kyon:* mass-weighted average, propellant dominate karta hai → CG tank ki taraf khicha jaata hai. - **Half burnt ($m_p=2000$):** $x_{\text{cg}}=\dfrac{6000+4000}{3000}=3.33\text{ m}$. - **Burnout ($m_p=0$):** $x_{\text{cg}}=x_d=6\text{ m}$. **Interpretation:** CG **2.8 m → 6 m** march kiya, dry structure ki taraf 3.2 m ka migration. TVC gains ko iske against schedule karna padega. ## Worked Example 2 — Steiner ke saath Pitch inertia Same numbers; dono bodies ko pitch axis ke baare mein point-like treat karo (unke apne $I^{cg}$ ignore karo). $t=0$ par, $x_{\text{cg}}=2.8$: $$I_{yy}=m_d D_d^2 + m_p D_p^2 = 1000(6-2.8)^2 + 4000(2-2.8)^2$$ $$=1000(10.24)+4000(0.64)=10240+2560=12800\text{ kg·m}^2.$$ Burnout par $x_{\text{cg}}=6$: propellant chala gaya, $I_{yy}=1000(0)^2=0$ (point model) → real rocket dry body ka apna $I_d^{cg}$ rakhta hai. *Ye step kyon?* Dikhata hai ki inertia **collapse** karta hai jaise fuel jaata hai — ab same torque vehicle ko bahut tezi se spin karta hai → autopilot ko gain reduce karni chahiye. --- > [!mistake] Common errors ko steel-man karo > **"CG nozzle ki taraf move karta hai jaise fuel jalta hai."** *Kyon sahi lagta hai:* mass peeche se nikal raha hai, isliye > surely CG exit ka peecha karega. **Fix:** CG **jo bacha hua hai** uski taraf move karta hai — usually dry > structure (engine/payload), jo aksar tanks ke *aage* hoti hai. Track karo ki *remaining* mass kahan baith rahi hai. > > **"Bas burnt mass ko $I$ se subtract kar do."** *Kyon sahi lagta hai:* kam mass, kam inertia. **Fix:** tumhe > (a) moving CG bhi recompute karni hai aur (b) *naye* CG ke baare mein har sub-body ke liye Steiner re-apply karna hai. Inertia yahan simply mass ke proportional nahi hai. > > **"Products of inertia hamesha zero hote hain."** *Kyon sahi lagta hai:* nice symmetric rockets. **Fix:** slosh, > asymmetric drain, aur canted engines off-diagonal $I_{xy}$ terms create karte hain → cross-coupled roll/pitch. --- > [!recall]- Feynman: ek 12-saal ke bacche ko samjhao > Socho ek broom spin karna. Agar tum handle ke along ek heavy ring slide karo, to "balance spot" (CG) move karta hai, aur > twist karna aasaan ya mushkil ho jaata hai. Ek rocket ek badi tank of fuel carry karta hai bilkul usi heavy ring ki tarah. Jaise-jaise woh > fuel jalaata hai, balance spot empty end ki taraf slide karta hai, aur rocket spin karna aasaan ho jaata hai. Isliye pilot-computer ko baar-baar seekhna padta hai "abhi mera balance point *kahan* hai, aur main *abhi* kitna twisty hoon?" > [!mnemonic] Yaad rakho > **"CG LEFTOVER ka peecha karta hai; Inertia SHRINKS lekin STEINER rehta hai."** > — CG remaining mass ki taraf jaata hai; $I$ fuel jaane par drop karta hai, phir bhi tum *hamesha* parallel-axis shift re-do karte ho. --- ## Active recall > [!recall] Quick self-test > 1. Propellant jalte waqt CG kis taraf move karta hai, aur kyon? > 2. *CG se* shift karte waqt Steiner cross-term vanish kyon ho jaata hai? > 3. "$I(t)$ find karne mein less fuel" ke saath kaun se do updates hone chahiye? #flashcards/physics CG definition (continuous body) ::: $\mathbf{r}_{cg}=\frac{1}{M}\int \mathbf{r}\,dm$ — mass-weighted average position. Two-body rocket CG formula ::: $x_{cg}=\dfrac{m_d x_d + m_p(t)x_p(t)}{m_d+m_p(t)}$. Zero-moment condition CG kyon define karta hai ::: Ye woh point hai jahan $\sum m_i(\mathbf r_i-\mathbf r_{cg})=0$, matlab balance point. Inertia tensor definition ::: $\mathbf I=\int((\mathbf r\!\cdot\!\mathbf r)\mathbb 1-\mathbf r\mathbf r^\top)dm$; $\mathbf L=\mathbf I\boldsymbol\omega$ se aaya. Parallel-axis (scalar) ::: $I_P=I_{cg}+Md^2$. Steiner cross-term kyon khatam hota hai ::: Kyunki $\int\mathbf r'\,dm=0$ CG se measure kiya jaaye to. CG migration ki direction ::: Remaining (dry) mass ki taraf, usually forward, nozzle ki taraf nahi. Burnout par inertia ka effect ::: $I$ shrinks → same torque se bada $\dot\omega$ milta hai → autopilot gain drop karni chahiye. Mass loss ke alawa I(t) ke liye do updates ::: Moving CG recompute karo, phir har sub-body ke liye Steiner re-apply karo. Nonzero products of inertia ka source ::: Asymmetric drain, slosh, canted engines → roll/pitch cross-coupling. --- ## Connections - [[Thrust Vector Control]] — live CG chahiye moment arms ke liye. - [[Tsiolkovsky Rocket Equation]] — same mass-depletion story, translational side. - [[Rigid Body Rotational Dynamics]] — $\boldsymbol\tau=\mathbf I\dot{\boldsymbol\omega}+\boldsymbol\omega\times\mathbf I\boldsymbol\omega$. - [[Parallel-Axis Theorem]] — core tool jo yahan use hua. - [[Propellant Slosh Dynamics]] — fixed-centroid assumption tod deta hai. - [[Gain Scheduling in Autopilots]] — isliye time-varying $I$ aur CG scheduled control force karte hain. ## 🖼️ Concept Map ```mermaid flowchart TD PD[Propellant depletion] CG[Center of gravity] IT[Inertia tensor I] CGdef[Mass-weighted average position] ZM[Zero-moment balance condition] TB[Dry mass plus propellant decomposition] ITdef[Integral of r-r 1 minus r rT] AM[Angular momentum L] TAU[Control torque tau equals I omega-dot] AP[Autopilot tuning] PD -->|shifts| CG PD -->|shrinks| IT ZM -->|defines| CGdef CGdef -->|formula for| CG CG -->|via superposition| TB PD -->|drains tank centroid| TB AM -->|expand triple product| ITdef ITdef -->|defines| IT IT -->|sets| TAU CG -->|sets moment arm for TVC| TAU TAU -->|must be retuned in flight| AP ```