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

Before we begin, one promise: we use no symbol before it is a picture. So we start with the most primitive thing there is — a heavy dot sitting somewhere on a stick.

Step 1 — Two heavy dots on a stick

WHAT. Lay a rocket down flat and look at it edge-on. Forget curves and tanks. It is really just two lumps of mass sitting at two places along one line: the dry structure (engine + payload + skin) and the propellant (the fuel). Draw a ruler along the rocket's length — call the reading on that ruler the station, written $x$ , measured in metres from the nose.

$m_{d}$ = mass of the dry part (a number of kilograms).
$x_{d}$ = the station (ruler mark) where the dry part balances by itself.
$m_{p}$ = mass of the propellant.
$x_{p}$ = the station where the propellant balances by itself.

That's the entire cast of characters.

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

WHY start here. The scary integral $\int r d m$ in the parent note is nothing more than "add up lots of tiny dots." If we can master two dots, the integral is just doing the same thing with a million dots. Master the atom first.

Step 2 — What does "balance point" even mean?

WHAT. Put a knife-edge (a pivot) under the stick and slide it until the stick doesn't tip either way. That pivot station is the centre of gravity, written $x_{cg}$ . "Doesn't tip" means the turning effort from the left lump exactly cancels the turning effort from the right lump.

Turning effort of a lump = its mass $\times$ its lever arm (how far it sits from the pivot). A lump to the left of the pivot tips one way; a lump to the right tips the other way. Balance = they cancel:

$dry mass m_{d} lever arm, signed (x_{d} - x_{cg}) + fuel mass m_{p} lever arm, signed (x_{p} - x_{cg}) = 0$

Each term is $mass \times signed distance from pivot$ — a turning effort.
The distance is signed: negative if the lump is left of the pivot, positive if right. That sign is what lets the two efforts cancel.
The whole sum being $0$ is the mathematical spelling of "it balances."

WHY this equation and not something else. We didn't define the CG by a formula and hope it balances — we demanded balance (sum of turning efforts $= 0$ ) and will now let algebra tell us where the pivot must be. That is the honest order.

Step 3 — Solve for the pivot: the CG formula appears

WHAT. Take the balance equation and just isolate $x_{cg}$ . Multiply out:

$m_{d} x_{d} - m_{d} x_{cg} + m_{p} x_{p} - m_{p} x_{cg} = 0.$

Gather the two $x_{cg}$ pieces on one side:

$m_{d} x_{d} + m_{p} x_{p} = total mass M (m_{d} + m_{p}) x_{cg} .$

Divide by the total mass $M = m_{d} + m_{p}$ :

$x_{cg} = \frac{m _{d} x _{d} + m _{p} x _{p}}{m _{d} + m _{p}}$

Top line $m_{d} x_{d} + m_{p} x_{p}$ : each station weighted by how heavy that lump is. A heavy lump drags the answer strongly toward its own station.
Bottom line $m_{d} + m_{p}$ : the total mass, so the result is a proper average (weights sum to 1).
Result $x_{cg}$ : a single station — the mass-weighted average position.

WHY it looks the way it does. It is a weighted average. If the fuel is much heavier than the dry part, the average sits almost on top of the fuel. This is the same formula as the parent note's $r_{cg} = \frac{1}{M} \sum m_{i} r_{i}$ — we just derived it for two dots by hand.

Step 4 — Watch the CG march as fuel burns

WHAT. Now let $m_{p}$ drop toward zero and re-run the formula. Use the parent note's numbers: $m_{d} = 1000 kg$ at $x_{d} = 6 m$ , full fuel $m_{p 0} = 4000 kg$ at $x_{p} = 2 m$ .

moment	fuel $m_{p}$	$x_{cg}$
liftoff	$4000$	$\frac{1000 ( 6 ) + 4000 ( 2 )}{5000} = 2.8 m$
half-burnt	$2000$	$\frac{1000 ( 6 ) + 2000 ( 2 )}{3000} = 3.33 m$
burnout	$0$	$\frac{1000 ( 6 )}{1000} = 6 m$

PICTURE. The pivot slides from $2.8 m$ (deep in the fuel tank) all the way up to $6 m$ (right on the dry structure). It chases the mass that is left behind.

WHY it moves toward the nose, not the tail. People expect the CG to chase the exhaust out the back. But the formula only knows about mass that is still on board. As $m_{p} \to 0$ , the top line loses its $m_{p} x_{p}$ term and the whole average collapses onto $x_{d}$ . The leftover mass wins. This kills the classic mistake from the parent note.

Step 5 — From balancing to twisting: why we need a second number

WHAT. Balancing tells you where to push. It does not tell you how hard the rocket resists being spun. Two rockets can share the same CG yet one whips around while the other barely budges. The number that measures "resistance to spin" is the moment of inertia, written $I$ (for the pitch axis, $I_{y y}$ ).

For a single dot of mass $m$ sitting a distance $d$ from the spin axis:

$I = m d^{2} .$

$m$ : the mass of the dot — more mass, harder to spin.
$d$ : the distance from the axis you spin about. This is the pivot we just found, $x_{cg}$ .
$d^{2}$ : squared! Distance matters far more than mass. Double the arm and inertia quadruples.

WHY the square, and why this tool. When the rocket rotates by a tiny angle, a dot far from the axis sweeps a long arc and picks up lots of speed; a dot near the axis barely moves. Kinetic energy of spin adds up like $\frac{1}{2} m v^{2}$ with $v = d ω$ , so the $d$ comes in twice — once for speed, once for the arc. That squared distance is exactly why moving fuel far from the axis makes a rocket sluggish, and why draining it makes the rocket suddenly nimble. (Full tensor form and the $L = I ω$ story live in Rigid Body Rotational Dynamics.)

Step 6 — Add the dots, but only about a common pivot

WHAT. A rocket has two lumps, so add two $m d^{2}$ terms — each distance measured from the same pivot $x_{cg}$ :

$I_{y y} = dry lump about CG m_{d} (x_{d} - x_{cg})^{2} + fuel lump about CG m_{p} (x_{p} - x_{cg})^{2}$

Each bracket $(x - x_{cg})$ is that lump's arm from the balance point.
Squared, so both terms are positive — spin resistance never subtracts.
They add because inertia about one shared axis is simply additive.

WHY "same pivot" is non-negotiable. You cannot add one lump's resistance measured about the nose to another's measured about the tail — that's adding apples to oranges. The rule that lets you move an inertia from a lump's own centre to the shared pivot is the parallel-axis (Steiner) theorem, $I_{about pivot} = I_{own centre} + m d^{2}$ . For point-like lumps the "own centre" part is $0$ , leaving just $m d^{2}$ . (Deeper: Parallel-Axis Theorem.)

Plug in liftoff numbers ( $x_{cg} = 2.8$ ):

$I_{y y} = 1000 (6 - 2.8)^{2} + 4000 (2 - 2.8)^{2} = 1000 (10.24) + 4000 (0.64) = 12800 kg\cdotm^{2} .$

Step 7 — Watch inertia collapse, and why the autopilot must react

WHAT. Re-run $I_{y y}$ as the fuel drains, each time using the new $x_{cg}$ from Step 4.

moment	$x_{cg}$	$I_{y y}$ (point model)
liftoff	$2.8$	$12800 kg\cdotm^{2}$
half-burnt	$3.33$	$1000 (6 - 3.33)^{2} + 2000 (2 - 3.33)^{2} \approx 10667 kg\cdotm^{2}$
burnout	$6$	$1000 (0)^{2} = 0$ (point model)

PICTURE. The inertia bar shrinks dramatically. In the point model it hits $0$ at burnout because we pretended the dry structure was itself a dot on the axis — a real rocket keeps its own dry-body inertia $I_{d}^{c g}$ , so the true curve flattens to a small positive floor, not zero.

WHY this is the whole point of the chapter. Rotational law says torque $τ = I \overset{ω}{˙}$ . If the same control torque $τ$ meets an inertia $I$ that has collapsed, the angular acceleration $\overset{ω}{˙} = τ / I$ blows up — the rocket spins far faster for the same nudge. An autopilot tuned for the heavy liftoff rocket would now over-steer and oscillate. So the controller schedules its gain against $I (t)$ . That is exactly Gain Scheduling in Autopilots, driven straight by this figure, and it sets the authority budget for Thrust Vector Control.

Step 8 — Two edge cases you must be ready for

WHAT — degenerate case A: no fuel at all ( $m_{p} = 0$ ). The CG formula becomes $x_{cg} = \frac{m _{d} x _{d}}{m _{d}} = x_{d}$ , and the fuel inertia term vanishes. Everything reduces to the bare dry body. Good — the formula degrades gracefully.

WHAT — degenerate case B: fuel exactly at the dry station ( $x_{p} = x_{d}$ ). Then both lumps sit at the same station, the CG never moves as fuel burns, and $(x_{p} - x_{cg}) = (x_{d} - x_{cg})$ so inertia just scales with mass. A rocket built this way would need no CG scheduling — real rockets are never so lucky because tanks and engines live at different ends.

WHY cover these. If your formula gave a divide-by-zero or a nonsense negative inertia in a limiting case, it would be wrong. Testing the extremes is how you trust the general result. (Off-axis effects like sloshing fuel breaking the neat symmetry are handled in Propellant Slosh Dynamics; the mass budget that ties burn to velocity gain is the Tsiolkovsky Rocket Equation.)

The one-picture summary

Everything on this page, compressed: the top track shows the CG dot sliding nose-ward as the fuel bar drains; the bottom track shows the inertia bar shrinking in lock-step. Read it left-to-right as "time during the burn."

Recall Feynman: the whole walkthrough in plain words

Lay the rocket flat: it's two heavy dots on a ruler — the dry body and the fuel. Slide a knife under it until it balances; that spot is the CG, and it lands closer to whichever dot is heavier. Since the fuel is the heavy one, the CG starts near the tank. Now burn the fuel: the heavy dot fades away, so the balance point slides toward what's left — toward the dry body up front, not toward the exhaust. Second question: how hard is it to spin? Each dot fights spinning by mass times its distance-from-pivot squared — far dots fight much harder. Add both dots about the same balance point. As the far, heavy fuel disappears, that spin-resistance collapses, so the same steering nudge now flips the rocket much faster. The flight computer watches both numbers every instant and turns its steering gain down as the rocket gets light and twitchy. Balance point chases the leftover; inertia shrinks — but you always re-measure both about the new pivot.

Active recall

Recall Quick self-test

Which lump does the CG sit closer to, and why? ::: The heavier lump — it's a mass-weighted average, so a big mass drags the average onto itself. As fuel drains to zero, where does the CG end up? ::: On the dry station $x_{d}$ — the leftover mass wins; it moves nose-ward, not toward the nozzle. Why is distance squared in $I = m d^{2}$ ? ::: A far dot sweeps a long arc AND moves fast ( $v = d ω$ ), so $d$ enters twice. Why must every lump's inertia be measured about the same pivot? ::: Inertia is only additive about one common axis; Steiner ( $I = I_{c g} + m d^{2}$ ) moves each lump to that shared CG. Why does the autopilot lower its gain near burnout? ::: $\overset{ω}{˙} = τ / I$ ; as $I$ collapses the same torque over-rotates the light rocket, so gain must drop. What two updates does "less fuel" demand for $I (t)$ ? ::: Recompute the moved CG, then redo every $m d^{2}$ about that new CG.