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

This page assumes you know nothing. We will earn every symbol the parent note (main topic) uses, one at a time, each with a picture, before it is allowed to appear in a formula.

0. Position, and what a "vector" is here

Before mass, before twisting, we need to say where things are.

The picture: stand at the nose, point at a bolt somewhere in the rocket — that pointing arrow is $\mathbf{r}$.

Why the topic needs it: every mass element in the rocket lives somewhere. To average positions or measure distances-from-an-axis, we first need a language for "where."

Because a real rocket is long and thin, we often care about just one coordinate along its length. We call that $x$ — a plain number (a "station") measured in metres from the nose.

0b. The three axes: our reference frame

Before "one coordinate $x$" can grow into three, we must pin down which three directions we measure along.

The picture: the corner of a room — three edges meeting at right angles. The rocket sits in that corner; every crumb is located by "along, sideways, up."

Why the topic needs it: the parent writes $I_{xx}=\int (y^2+z^2)dm$. That $(y^2+z^2)$ only means something once you know $y$ and $z$ are the two directions perpendicular to the $x$-axis — the two ways a crumb can sit off the roll axis. Without a fixed frame, the symbols in the tensor are meaningless.

1. Mass and the mass element $dm$

The picture: slice the rocket into millions of tiny crumbs. Each crumb has its own little weight $dm$ and its own position $\mathbf{r}$.

Why the topic needs it: the CG and inertia are both sums over every crumb. When crumbs are countable we write $m_i$ (the $i$-th crumb); when they blur into a continuum we write $dm$ and use an integral (§4).

2. The summation sign $\sum$

The picture: a queue of boxes; $\sum$ is you walking down the queue tossing each into one bucket.

Why the topic needs it: the CG formula ${\sum }_i{m}_i{\mathbf{r}}_i$ is literally "for every crumb, multiply its mass by its position, then add all those products."

3. Multiplying a number by an arrow, and the "average position" idea

If a crumb has mass $m_i$ and position ${\mathbf{r}}_i$, then $m_i{\mathbf{r}}_i$ is that arrow stretched by the number $m_i$. Heavy crumbs get long arrows; light crumbs get short ones.

The picture: imagine each position arrow, but drawn thicker/longer the heavier its crumb. The "balance point" leans toward the fat arrows.

Now the parent's central formula is readable: ${\mathbf{r}}_{\text{cg}}=\frac{{\sum }_i{m}_i{\mathbf{r}}_i}{{\sum }_i{m}_i}$ Top = "add up (mass × position) for every crumb." Bottom = "add up all the masses" $=M$. Dividing turns the weighted sum into a weighted average position — the balance point.

4. The integral $\int \dots dm$ — a sum with infinitely many crumbs

Why this tool and not just $\sum$? A solid rocket has no natural "crumbs" — mass is smeared continuously. The integral is the exact tool for "add a quantity over a continuous smear." It answers the question "what is the total of position-weighted mass when there are infinitely many infinitely small pieces?" — which a finite sum cannot.

The picture: shrink the crumbs of §1 to dust; the queue of §2 becomes a smooth flow; $\int$ is that smooth flow into one bucket.

So the continuous CG, ${\mathbf{r}}_{\text{cg}}=\frac{1}{M}\int \mathbf{r}dm,$ is the exact same weighted-average idea, now for smooth bodies.

5. Two special multiplications of arrows

The inertia tensor is built from two operations on arrows. Both must be earned first.

Why this tool? Inertia cares about how far mass sits from the spin axis — and "far" means length, so a dot product is the natural way to extract distance-squared.

Before the next operation we need one tiny symbol: the transpose.

The picture: the same three-number list, once standing up (column $\mathbf{r}$), once lying down (row ${\mathbf{r}}^{\top }$).

The picture: $\mathbf{r}\cdot \mathbf{r}$ is one number (a length-squared); $\mathbf{r}{\mathbf{r}}^{\top }$ is a whole $3\times 3$ table of pairwise coordinate products.

Why both appear: twisting resistance depends on distance and on the direction of the offset. The dot product supplies the distance-squared part; the outer product isolates the piece along the axis (which does not resist a twist about that axis). We are about to subtract one from the other — that is where the tensor, and its minus signs, are born.

6. Building the inertia tensor $\mathbf{I}$

The picture: hold a hammer by the head vs. by the end of the handle. Same hammer, but far-off mass makes it feel "heavier to swing." That $d^2$ is why the parent's $I=\int (y^2+z^2)dm$ has squares in it.

Why we need a full tensor and not one number: a rocket can be twisted about pitch, yaw, or roll — three different axes — and a lopsided (asymmetric) body can even spill a pitch-twist into a roll response. One number can't capture that; a $3\times 3$ grid can. That grid is the rotational version of mass, and we write it $\mathbf{I}$.

Assembling the grid, term by term

We now combine §5's two products. First we need $1$:

Now form, for one crumb $dm$ at position $\mathbf{r}=(x,y,z)$, the quantity $[(\mathbf{r}\cdot \mathbf{r})1-\mathbf{r}{\mathbf{r}}^{\top }]dm.$

WHAT this is: "distance-squared on every diagonal slot, minus the grid of coordinate-products from §5." Written out, the first piece is $(x^2+y^2+z^2)$ on each diagonal; subtracting the outer-product grid gives:

$(\mathbf{r}\cdot \mathbf{r})1-\mathbf{r}{\mathbf{r}}^{\top }=\left(\begin{array}{ccc}{y}^2+z^2& -xy& -xz\\ -xy& x^2+z^2& -yz\\ -xz& -yz& x^2+y^2\end{array}\right).$

WHY the diagonal loses one square (top-left is $y^2+z^2$, not $x^2+y^2+z^2$): on the $x$-axis diagonal, the dot product delivered all three squares, but the outer product's own diagonal entry $x^2$ gets subtracted off. That is deliberate — mass sitting along the $x$-axis is at zero distance from that axis, so it must not count toward $x$-twisting-laziness. The subtraction is exactly what removes the along-axis part, leaving only the two perpendicular distances $y^2+z^2$. This is the whole reason both products were introduced.

WHY the off-diagonals carry a minus sign: the diagonal (dot-product) part contributes nothing off-diagonal — $1$ is zero off the diagonal. So every off-diagonal entry comes only from the subtracted outer product, and the outer product's $(x,y)$ entry is $+xy$. Subtracting it leaves $-xy$. The minus is therefore not a convention we choose — it is forced by the "$(\cdot )-(\cdot )$" structure. Summing over all crumbs:

7. Distance-from-CG and the offset $\mathbf{d}$

The parallel-axis theorem needs one more symbol.

The picture: the CG is a dot; the axis you actually spin about is another dot; $\mathbf{d}$ is the little arrow bridging them, $D$ its length.

Why the topic needs it: in a flying rocket each part (dry structure, propellant) knows its own inertia about its own centre, but we must add everything about the vehicle's CG. The offset $\mathbf{d}$ (and the $M{D}^2$ correction) is the bridge — and because the CG moves as fuel burns, $\mathbf{d}$ and $D$ change every instant. That is the beating heart of the whole topic.

8. Time-dependence: the little $(t)$

Why the topic needs it: this single piece of notation is the reason the note exists. A textbook static body has no $(t)$; a rocket has $(t)$ on almost everything, which is why autopilot gains and thrust vectoring must be re-tuned mid-flight.

Prerequisite map

Each box is a tool you now own; the arrows show how they stack into the parent topic's live-in-flight mass properties. Related downstream ideas — Tsiolkovsky Rocket Equation, Propellant Slosh Dynamics — reuse this same foundation.

Equipment checklist

Cover the right side and test yourself.