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

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WHY do mass properties change at all?

WHAT is happening: Propellant is stored in tanks inside the vehicle. The engine ejects it out the nozzle. The remaining structure (dry mass) plus remaining propellant defines a new, lighter body every instant.

WHY it matters for flight:

• CG sets the moment arm for thrust misalignment, aero forces, and thrust-vector control (TVC).
• Inertia tensor sets how fast a given control torque rotates the vehicle ($\boldsymbol\tau = \mathbf{I}\dot{\boldsymbol\omega} + \dots$).
• Both move as fuel drains → a well-tuned autopilot at $t=0$ can be sluggish or unstable at $t=t_{\text{burnout}}$.

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Center of Gravity — derive it from scratch

WHY this definition? We want a single point where, if all the mass sat there, the net gravitational (and inertial) moment would be unchanged. Start from the requirement that the total moment of the weights about $\mathbf{r}_{\text{cg}}$ vanishes:

\;\Rightarrow\; \sum_i m_i \mathbf{r}_i = \mathbf{r}_{\text{cg}}\sum_i m_i.$$ Solve for $\mathbf{r}_{\text{cg}}$ → the formula above. *Why this step?* The zero-moment condition is literally the definition of "the balance point," so algebra just isolates it. ### Two-body decomposition (dry structure + propellant) A rocket is naturally two pieces: **dry mass** $m_d$ at fixed station $x_d$, and **propellant** $m_p(t)$ whose own centroid $x_p(t)$ shifts as the tank drains. $$x_{\text{cg}}(t) = \frac{m_d\,x_d + m_p(t)\,x_p(t)}{m_d + m_p(t)}$$ *Why this step?* Superposition: the CG of a composite body is the mass-weighted CG of its sub-bodies — because the defining integral is linear in $dm$. --- ## Inertia tensor — derive it from scratch > [!definition] Inertia tensor > $$\mathbf{I} = \int \big( (\mathbf{r}\!\cdot\!\mathbf{r})\,\mathbb{1} - \mathbf{r}\,\mathbf{r}^{\!\top}\big)\,dm$$ > Diagonal terms $I_{xx}=\int (y^2+z^2)\,dm$ (moments of inertia); off-diagonals $I_{xy}=-\int xy\,dm$ (products of inertia). **WHY this form?** Angular momentum is $\mathbf{L}=\int \mathbf{r}\times(\boldsymbol\omega\times\mathbf{r})\,dm$. Expand the triple product $\mathbf{r}\times(\boldsymbol\omega\times\mathbf{r}) = (\mathbf{r}\!\cdot\!\mathbf{r})\boldsymbol\omega - (\mathbf{r}\!\cdot\!\boldsymbol\omega)\mathbf{r}$. Factoring $\boldsymbol\omega$ out on the right forces exactly the tensor $\mathbf{I}$ so that $\mathbf{L}=\mathbf{I}\boldsymbol\omega$. *Why this step?* We *demand* a linear map $\boldsymbol\omega\mapsto\mathbf{L}$; the bracket is that map. > [!formula] Parallel-axis (Huygens–Steiner) theorem > To move inertia from the CG to a point offset by $\mathbf{d}$: > $$\mathbf{I}_{\text{P}} = \mathbf{I}_{\text{cg}} + M\big[(\mathbf{d}\!\cdot\!\mathbf{d})\mathbb{1} - \mathbf{d}\,\mathbf{d}^{\!\top}\big]$$ > Scalar version for one axis: $\;I_P = I_{\text{cg}} + M d^2.$ **Derivation:** Let $\mathbf{r}=\mathbf{r}' + \mathbf{d}$ with $\mathbf{r}'$ measured from CG. $$I_{xx}^P=\int\big((y'+d_y)^2+(z'+d_z)^2\big)dm.$$ Expand: the $y'^2+z'^2$ term gives $I_{xx}^{\text{cg}}$; the cross terms $2d_y\!\int y'dm$ vanish (**because $\int \mathbf{r}'\,dm=0$ by definition of CG** — *this is the key why!*); the $d_y^2+d_z^2$ term gives $M d^2$. QED. ### Composite inertia of the flying rocket Compute each sub-body's inertia **about its own CG**, then Steiner-shift each to the *vehicle* CG and add: $$\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]$$ where $D_d(t)=|x_d - x_{\text{cg}}(t)|$, $D_p(t)=|x_p(t)-x_{\text{cg}}(t)|$. *Why this step?* Inertia is additive **only about a common point**; Steiner is the tool to bring everyone to that common (moving) CG. ![[3.4.06-Mass-properties-—-CG-location,-inertia-tensor-changing-with-propellant-depletion.png]] --- ## Worked Example 1 — CG shift during burn Rocket: $m_d=1000\text{ kg}$ at $x_d=6\text{ m}$ (nose-referenced). Full propellant $m_{p0}=4000\text{ kg}$ centered at $x_p=2\text{ m}$ (assume centroid fixed for a first pass). - **At $t=0$:** $x_{\text{cg}}=\dfrac{1000(6)+4000(2)}{5000}=\dfrac{14000}{5000}=2.8\text{ m}$. *Why:* mass-weighted average, propellant dominates → CG pulled toward tank. - **Half burnt ($m_p=2000$):** $x_{\text{cg}}=\dfrac{6000+4000}{3000}=3.33\text{ m}$. - **Burnout ($m_p=0$):** $x_{\text{cg}}=x_d=6\text{ m}$. **Interpretation:** CG marched **2.8 m → 6 m**, a 3.2 m migration toward the dry structure. TVC gains must schedule against this. ## Worked Example 2 — Pitch inertia with Steiner Same numbers; treat both bodies as point-like about the pitch axis (ignore their own $I^{cg}$). At $t=0$, $x_{\text{cg}}=2.8$: $$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.$$ At burnout $x_{\text{cg}}=6$: propellant gone, $I_{yy}=1000(0)^2=0$ (point model) → real rocket keeps the dry body's own $I_d^{cg}$. *Why this step?* Shows inertia **collapses** as fuel leaves — same torque now spins the vehicle far faster → autopilot must reduce gain. --- > [!mistake] Steel-man the common errors > **"CG moves toward the nozzle as fuel burns."** *Why it feels right:* mass is leaving out the back, so > surely the CG chases the exit. **Fix:** the CG moves toward whatever is **left behind** — usually the > dry structure (engine/payload), which is often *forward* of the tanks. Track where the *remaining* mass sits. > > **"Just subtract burnt mass from $I$."** *Why it feels right:* less mass, less inertia. **Fix:** you must > also (a) recompute the moving CG and (b) re-apply Steiner for every sub-body about the *new* CG. Inertia > is not simply proportional to mass here. > > **"Products of inertia are always zero."** *Why it feels right:* nice symmetric rockets. **Fix:** slosh, > asymmetric drain, and canted engines create off-diagonal $I_{xy}$ terms → cross-coupled roll/pitch. --- > [!recall]- Feynman: explain to a 12-year-old > Imagine spinning a broom. If you slide a heavy ring along the handle, the "balance spot" (CG) moves, and > it gets easier or harder to twist. A rocket carries a huge tank of fuel like that heavy ring. As it burns > the fuel away, the balance spot slides toward the empty end, and the rocket becomes easier to spin. So the > pilot-computer has to keep re-learning "where is my balance point *right now*, and how twisty am I *right now*?" > [!mnemonic] Remember it > **"CG chases the LEFTOVER; Inertia SHRINKS but STEINER stays."** > — CG heads to remaining mass; $I$ drops as fuel goes, yet you *always* re-do the parallel-axis shift. --- ## Active recall > [!recall] Quick self-test > 1. Which way does CG move as propellant burns, and why? > 2. Why does the Steiner cross-term vanish when shifting *from* the CG? > 3. What two updates must accompany "less fuel" when finding $I(t)$? #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 ::: $x_{cg}=\dfrac{m_d x_d + m_p(t)x_p(t)}{m_d+m_p(t)}$. Why the zero-moment condition defines CG ::: It's the point where $\sum m_i(\mathbf r_i-\mathbf r_{cg})=0$, i.e. the balance point. Inertia tensor definition ::: $\mathbf I=\int((\mathbf r\!\cdot\!\mathbf r)\mathbb 1-\mathbf r\mathbf r^\top)dm$; came from $\mathbf L=\mathbf I\boldsymbol\omega$. Parallel-axis (scalar) ::: $I_P=I_{cg}+Md^2$. Why Steiner cross-term dies ::: Because $\int\mathbf r'\,dm=0$ measured from the CG. Direction of CG migration ::: Toward the remaining (dry) mass, usually forward, not the nozzle. Effect of burnout on inertia ::: $I$ shrinks → same torque gives larger $\dot\omega$ → autopilot gain must drop. Two updates needed for I(t) besides mass loss ::: Recompute moving CG, then re-apply Steiner for each sub-body. Source of nonzero products of inertia ::: Asymmetric drain, slosh, canted engines → roll/pitch cross-coupling. --- ## Connections - [[Thrust Vector Control]] — needs live CG for moment arms. - [[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 used here. - [[Propellant Slosh Dynamics]] — breaks the fixed-centroid assumption. - [[Gain Scheduling in Autopilots]] — why time-varying $I$ and CG force scheduled control. ## 🖼️ 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 ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, rocket ka fundamental funda ye hai ki wo apna hi zyada-tar weight fuel ke roop mein > bahar phenk deta hai. Jaise-jaise propellant burn hota hai, rocket ka **CG (center of gravity)** > shift karta jaata hai aur **inertia tensor** chhota hota jaata hai. Isse control par direct asar > padta hai: thrust ka moment arm CG se naapa jaata hai, aur kitni jaldi rocket ghoomega wo inertia > pe depend karta hai. Matlab liftoff wala rocket aur burnout wala rocket practically alag machine > hain. > > CG ka formula simple mass-weighted average hai: $x_{cg}=\frac{m_d x_d+m_p x_p}{m_d+m_p}$. Yaad > rakho — CG us taraf jaata hai jahan **bacha hua mass** hai, nozzle ki taraf nahi (ye common galti > hai). Humare example mein CG 2.8 m se 6 m tak khisak gaya. Inertia ke liye **parallel-axis > (Steiner) theorem** use karo: har sub-body ka apna inertia lo, phir naye CG tak shift karke add karo. > Yaad rakho ki Steiner ka cross-term isliye zero hota hai kyunki CG se naapne par $\int \mathbf r'\,dm=0$. > > Practical baat: sirf "mass kam ho gaya to inertia kam" mat socho. Tumhe (1) naya CG nikalna hai > aur (2) har piece ke liye Steiner dobara lagana hai — har instant. Isiliye autopilot mein **gain > scheduling** hota hai: time ke saath gains change karte hain kyunki I aur CG dono badal rahe hote > hain. Agar ye nahi karoge to rocket burnout ke time over-sensitive ho ke unstable ho sakta hai. ![[audio/3.4.06-Mass-properties-—-CG-location,-inertia-tensor-changing-with-propellant-depletion.mp3]]