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

We are building the central result of the parent 6DOF note. Every tool it borrows, we earn here.

Step 1 — What is angular momentum, as a picture?

WHAT. A rocket is made of many little chunks of mass. When the whole thing spins at rate $\vec{\omega }$, each chunk races around a circle. A chunk of mass $m_i$ sitting at position ${\vec{r}}_i$ (an arrow from the center of mass to that chunk) moves with velocity ${\vec{v}}_i=\vec{\omega }\times {\vec{r}}_i$.

WHY the $\times$ (cross product) and not ordinary multiplication? Because we need an operation that takes "the spin axis" and "where the chunk is" and returns "which way the chunk is moving" — and that moving direction is perpendicular to both. The cross product $\vec{a}\times \vec{b}$ is defined to be the arrow perpendicular to both $\vec{a}$ and $\vec{b}$, with length $|\vec{a}||\vec{b}|\sin \theta$. It is the only clean tool that answers "give me the sideways direction."

PICTURE. In the figure, the violet axis is $\vec{\omega }$. A chunk (magenta dot) sits at ${\vec{r}}_i$. Its velocity (orange arrow) wraps around the axis — perpendicular to both $\vec{\omega }$ and ${\vec{r}}_i$, exactly what $\vec{\omega }\times {\vec{r}}_i$ produces.

Now angular momentum of one chunk is ${\vec{h}}_i={\vec{r}}_i\times (m_i{\vec{v}}_i)$ — position crossed with momentum. Add up every chunk:

Step 2 — Why spin-momentum needs a tensor, not a number

WHAT. Substitute ${\vec{v}}_i=\vec{\omega }\times {\vec{r}}_i$ into $\vec{H}$. The sum turns into a machine that eats $\vec{\omega }$ and spits out $\vec{H}$. That machine is the inertia tensor $\mathbf{I}$: $\vec{H}=\mathbf{I}\vec{\omega }.$

WHY a tensor and not just a mass-like number? For sliding, resistance is one number $m$ (a rocket is equally hard to push in every direction). For spinning, a rocket is easy to roll about its long thin axis but hard to tumble end-over-end — resistance depends on the axis. A single number cannot store "different in different directions"; a $3\times 3$ table (a tensor) can. See Inertia tensor and principal axes.

PICTURE. The figure shows the same rocket spun two ways: about the long axis (small violet $\vec{H}$, easy) versus end-over-end (long magenta $\vec{H}$, hard). Same $|\vec{\omega }|$, different $|\vec{H}|$ — proof one number can't describe it.

Step 3 — The law we start from: torque is the rate of $\vec{H}$

WHAT. The rotational version of "force changes momentum" is: torque changes angular momentum. $\vec{M}={\frac{d\vec{H}}{dt}|}_I$

WHY the little ${|}_I$? The symbol $\frac{d\vec{H}}{dt}$ means "how fast $\vec{H}$ changes each second" (the dot/derivative — our tool for rate of change, because that is exactly what a law of motion is about). The ${|}_I$ says measured in the inertial frame $I$ — the non-spinning outside world where Newton's laws are actually valid. This "valid only in $I$" is the whole reason the next two steps exist. See Reference frames and rotation matrices.

PICTURE. The figure shows $\vec{H}$ at time $t$ and a moment later at $t+dt$. The tiny change $d\vec{H}$ (orange) points the same way as the applied torque $\vec{M}$ (magenta). Torque literally pushes the tip of the $\vec{H}$ arrow.

Step 4 — The trap: $\mathbf{I}$ is not constant in the outside world

WHAT. We want to compute $\frac{d\vec{H}}{dt}{|}_I=\frac{d(\mathbf{I}\vec{\omega })}{dt}{|}_I$. Naively we'd like to write $\mathbf{I}\dot{\vec{\omega }}$. But watch out: the inertia tensor $\mathbf{I}$ is glued to the rocket, and the rocket is spinning. So from the outside frame $I$, the numbers in $\mathbf{I}$ keep changing every instant — the "easy axis" keeps swinging around.

WHY this matters. If both $\mathbf{I}$ and $\vec{\omega }$ change in $I$, the product rule gives two terms and $\mathbf{I}$'s term is horrible to track. We want a frame where $\mathbf{I}$ holds still.

PICTURE. The figure shows the body frame's three principal axes at two instants — they have rotated. The inertia values live on those axes, so in $I$ they are moving targets (magenta = old axes, violet = rotated axes).

The fix (previewed): compute the rate in the body frame $B$, where $\mathbf{I}$ is frozen and diagonal — then convert that rate back to $I$. The converter is the transport theorem, Step 5.

Step 5 — The transport theorem: converting "body rate" to "world rate"

WHAT. For any arrow $\vec{A}$ carried by a spinning frame: ${\frac{d\vec{A}}{dt}|}_I={\frac{d\vec{A}}{dt}|}_B+\vec{\omega }\times \vec{A}$

WHY it's true, in one picture. Write $\vec{A}$ using the body's own axes ${\hat{e}}_x,{\hat{e}}_y,{\hat{e}}_z$: $\vec{A}=A_x{\hat{e}}_x+A_y{\hat{e}}_y+A_z{\hat{e}}_z$. Two things can make $\vec{A}$ change as seen from outside:

1. the components $A_x,A_y,A_z$ change — that is the rate inside the body, $\frac{d\vec{A}}{dt}{|}_B$;
2. the axes themselves rotate — even if the components are frozen, the arrow swings because its scaffolding spins. A fixed-length axis ${\hat{e}}_i$ rotating at $\vec{\omega }$ moves at ${\dot{\hat{e}}}_i=\vec{\omega }\times {\hat{e}}_i$ (same "sideways" cross-product idea from Step 1). Summing $A_i(\vec{\omega }\times {\hat{e}}_i)$ gives $\vec{\omega }\times \vec{A}$.

Add the two effects. That's the whole theorem — nothing but the product rule, split into "components move" + "scaffolding moves." See Transport theorem (rotating frames).

PICTURE. Left: components change, axes still. Right: components frozen, axes spin — the arrow still moves. Real motion is the sum.

Step 6 — Assemble Euler's equation

WHAT. Apply the transport theorem (Step 5) with $\vec{A}=\vec{H}$, and use the law from Step 3: $\vec{M}={\frac{d\vec{H}}{dt}|}_I={\frac{d\vec{H}}{dt}|}_B+\vec{\omega }\times \vec{H}.$ Now the payoff of choosing the body frame: there $\vec{H}=\mathbf{I}\vec{\omega }$ with $\mathbf{I}$ frozen and diagonal, so its body-rate is simply $\mathbf{I}\dot{\vec{\omega }}$ (the $\dot{\mathbf{I}}$ term is zero — Step 4's whole point). Substituting $\vec{H}=\mathbf{I}\vec{\omega }$:

WHY the cross-term is real and not a math artifact. It came from the scaffolding-spins piece of the transport theorem. It is the same physics that makes a spinning top precess and a thrown phone flip. Look at the parent's Angular momentum conservation note: even when $\vec{M}=0$, this term keeps $\vec{\omega }$ dancing.

PICTURE. The figure stacks the three arrows: applied torque $\vec{M}$ = the "obvious" spin-up arrow plus the gyroscopic side-arrow $\vec{\omega }\times (\mathbf{I}\vec{\omega })$.

Step 7 — Break into scalars (principal-axis components)

WHAT. With $\mathbf{I}=\mathrm{diag}(I_x,I_y,I_z)$ and $\vec{\omega }=(p,q,r)$ (roll, pitch, yaw rates), $\vec{H}=(I_{x}p,I_{y}q,I_{z}r)$. Compute $\vec{\omega }\times \vec{H}$ by the cross-product recipe — its $x$-component is $q(I_{z}r)-r(I_{y}q)=(I_z-I_y)qr$:

• $I_x\dot{p}$ ::: roll axis spun up by roll torque.
• $(I_z-I_y)qr$ ::: pitch and yaw rates feeding into roll if the two transverse inertias differ. If $I_z=I_y$ this vanishes.

WHY the cyclic pattern $x\to y\to z\to x$? The cross product treats the three axes symmetrically; rotating the labels $(x,y,z)\to (y,z,x)$ reproduces each line. Memorize one, get three.

PICTURE. The figure is a ring of three boxes $p,q,r$ with arrows showing each pair "leaking" into the third — the coupling made visual.

Step 8 — Every case: what the coupling term does

We must never leave the reader in a scenario we didn't show. Here are all the regimes of the coupling term $(I_z-I_y)qr$ (and its cousins):

PICTURE. Four mini-panels: A steady zero, B steady spin, C gentle cone, D violent flip — the full spectrum of behaviours governed by one term.

The one-picture summary

This single diagram compresses the whole page: start with torque = rate of $\vec{H}$ in the world (Step 3), notice $\mathbf{I}$ won't sit still in the world (Step 4), jump into the body frame via the transport theorem (Step 5) where $\mathbf{I}$ is frozen, and land on $\vec{M}=\mathbf{I}\dot{\vec{\omega }}+\vec{\omega }\times (\mathbf{I}\vec{\omega })$ (Step 6) — with the second term being the price of having jumped frames while spinning.

Flashcards