Visual walkthrough — Converting between DCM, quaternions, Euler angles

We start from ONE fact only: a rotation turns a vector. Nothing else assumed.

---

Step 1 — What a vector even is, and what "rotate it" means

WHAT: the blue arrow $\vec{v}$ gets spun to a new blue arrow ${\vec{v}}^{\prime }$. WHY it matters: every representation on this topic — matrix, quaternion, Euler angles — is just a recipe that takes $\vec{v}$ and hands you ${\vec{v}}^{\prime }$. If we can write that recipe once, cleanly, we own all three. PICTURE: notice the two arrows have the same length (the green circle they both touch). Rotation never stretches.

Recall What is preserved by a rotation?

Length is preserved ::: the tip stays on a sphere of the same radius; only direction changes.

---

Step 2 — The axis: Euler's theorem says one line is enough

WHAT: the orange line is the axis $\hat{e}$; the vector sweeps through angle $\Phi$ around it. WHY this tool: working with one axis + one angle is the cheapest honest description of a rotation — everything else is a re-packaging of these two objects. That is why the parent note calls axis–angle "the anchor." PICTURE: points on the axis (orange) do not move at all — they are the pivot. Everything off-axis swings.

• $\hat{e}$ ::: the fixed pivot line, length $1$.
• $\Phi$ ::: how far around that line we turn.

---

Step 3 — Split the vector: the trick that makes it solvable

The whole derivation rests on one idea: a hard rotation of a whole vector becomes easy if we first cut the vector into two pieces — one that ignores the spin and one that fully feels it.

Reading $\hat{e}\cdot \vec{v}$ (the dot product): it multiplies matching components and adds them: $\hat{e}\cdot \vec{v}=e_1{v}_x+e_2{v}_y+e_3{v}_z$. Geometrically it measures how much of $\vec{v}$ points along $\hat{e}$ — the length of the shadow. Why this tool and not another? We need a number that says "how far along the axis," and the dot product is exactly the along-a-direction measuring stick.

WHAT: blue $\vec{v}$ splits into orange ${\vec{v}}_{\parallel }$ (up the axis) and green ${\vec{v}}_{\perp }$ (in the flat spin-plane). WHY: ${\vec{v}}_{\parallel }$ lies on the axis, so — by Step 2 — the rotation leaves it completely alone. Only ${\vec{v}}_{\perp }$ actually rotates. We reduced a 3-D rotation to a flat 2-D one. PICTURE: the green arrow lives in the shaded disc perpendicular to the orange axis; that disc is where all the action happens.

• $\hat{e}(\hat{e}\cdot \vec{v})$ ::: rebuild the along-axis shadow as an arrow (direction $\hat{e}$, length = the dot product).
• $\vec{v}-{\vec{v}}_{\parallel }$ ::: subtract the shadow off, leaving the in-plane part.

---

Step 4 — Rotating the flat part needs a second in-plane arrow

Inside the disc, ${\vec{v}}_{\perp }$ turns by $\Phi$ like a clock hand. To describe any angle in a plane you need two perpendicular reference arrows — a "cosine" direction and a "sine" direction. We already have one (${\vec{v}}_{\perp }$ itself). We need its $90°$ companion.

Why the cross product and not something else? We specifically need an arrow that is (a) in the spin-plane and (b) a quarter-turn ahead. The cross product is the one operation that manufactures a perpendicular partner, so it is the natural sine-direction.

WHAT: in the disc, green ${\vec{v}}_{\perp }$ and red $\hat{e}\times \vec{v}$ are two arrows at $90°$, same length. WHY: with a $0°$ arrow and a $90°$ arrow, any rotated arrow is $\cos \Phi$ of the first plus $\sin \Phi$ of the second — the same way $(\cos \Phi ,\sin \Phi )$ traces a circle. PICTURE: the rotated tip ${\vec{v}}_{\perp }^{\prime }$ (dashed) is reached by walking $\cos \Phi$ along green and $\sin \Phi$ along red.

${\vec{v}}_{\perp }^{\prime }=\underset{\text{shrink the }{0}^{\circ }\text{ arrow}}{\underbrace{{\vec{v}}_{\perp }\cos \Phi }}+\underset{\text{add the }90^{\circ }\text{ arrow}}{\underbrace{(\hat{e}\times \vec{v})\sin \Phi }}$

• $\cos \Phi$ ::: how much of the original in-plane arrow survives.
• $\sin \Phi$ ::: how much of the perpendicular companion we mix in.

---

Step 5 — Add the untouched piece back: Rodrigues' formula appears

Now reassemble. The rotated full vector is the rotated flat part plus the parallel part that never moved.

${\vec{v}}^{\prime }=\underset{\text{Step 4}}{\underbrace{{\vec{v}}_{\perp }\cos \Phi }}+\underset{\text{Step 4}}{\underbrace{(\hat{e}\times \vec{v})\sin \Phi }}+\underset{\text{untouched, Step 3}}{\underbrace{\hat{e}(\hat{e}\cdot \vec{v})}}$

Substitute ${\vec{v}}_{\perp }=\vec{v}-\hat{e}(\hat{e}\cdot \vec{v})$ and collect the $\hat{e}(\hat{e}\cdot \vec{v})$ terms:

WHAT: the three coloured arrows stack tip-to-tail to land exactly on ${\vec{v}}^{\prime }$. WHY the $(1-\cos \Phi )$ correction: the first term faded the along-axis part too (it shouldn't have — that part never moves). The third term adds back exactly the missing $(1-\cos \Phi )$ share along $\hat{e}$. Check: along the axis the fade $\cos \Phi$ plus the top-up $(1-\cos \Phi )$ sum to $1$ — full restoration. This is Rodrigues Rotation Formula in the flesh. PICTURE: watch the orange top-up arrow — it points straight up the axis and its only job is to undo the over-fade.

---

Step 6 — From arrows to a matrix: reading off the DCM

A matrix $C$ is just a machine that eats $\vec{v}$ and outputs ${\vec{v}}^{\prime }$ by the rule ${\vec{v}}^{\prime }=C\vec{v}$. To turn Rodrigues into such a machine we rewrite each term as "(matrix) times $\vec{v}$."

Why these three matrices? Each just re-expresses one Rodrigues term as a matrix acting on $\vec{v}$ — no new physics, only new bookkeeping so we can store the rotation as 9 numbers.

$\boxed{C=I\cos \Phi +[\hat{e}{]}_{\times }\sin \Phi +\hat{e}{\hat{e}}^{\top }(1-\cos \Phi )}$

WHAT: the same three geometric pieces, now shown as three $3\times 3$ blocks that sum to $C$. WHY: once $C$ exists, rotating any vector is one multiply — that is why the DCM is the workhorse for actually moving vectors. See Direction Cosine Matrix Properties. PICTURE: the diagonal is dominated by $\cos \Phi$; the anti-symmetric off-diagonal carries $\sin \Phi$ (the cross-product part); the symmetric leftover carries the projection.

• The symmetric part of $C$ ::: comes from $I\cos \Phi +\hat{e}{\hat{e}}^{\top }(1-\cos \Phi )$.
• The anti-symmetric part of $C$ ::: comes from $[\hat{e}{]}_{\times }\sin \Phi$ — this is why $C_{23}-C_{32}$ later isolates axis info.

---

Step 7 — Edge cases: the picture must cover ALL of them

A derivation is only trustworthy if it survives the degenerate inputs.

WHAT: three miniature clocks showing $\Phi =0$, $\Phi =90^{\circ }$, $\Phi =180^{\circ }$ acting on the same start arrow. WHY: the reader should never meet a spin angle the picture didn't show — from "do nothing" to "flip." PICTURE: at $\Phi =180^{\circ }$ the red $\sin$-arrow has length $0$; the whole result is carried by fade + top-up.

---

Step 8 — Same picture, half the angle: the quaternion

The quaternion is not a new derivation — it is Step 5 seen through the sandwich rule ${\vec{v}}^{\prime }=q\vec{v}{q}^{-1}$, which geometrically applies $q$ twice. To land on one turn of $\Phi$, each $q$ must carry only $\Phi /2$.

WHAT: a semicircle showing $\Phi /2$ (quaternion) as exactly half the full swing $\Phi$ (DCM). WHY the half: because the sandwich doubles, and because it makes $q$ and $-q$ describe the same physical turn — no singularities, ideal for time-integration in Quaternion Kinematics & Propagation. PICTURE: feed these $q_0,q_1,q_2,q_3$ into the $C(q)$ table of the parent note and, using $\sin \Phi =2\sin \frac{\Phi }{2}\cos \frac{\Phi }{2}$ and $1-\cos \Phi =2{\sin }^2\frac{\Phi }{2}$, you regenerate the very matrix of Step 6. Same rotation, three costumes.

---

The one-picture summary

One rotation → three encodings. Axis $\hat{e}$ and angle $\Phi$ sit at the centre. Split the vector (dot & cross), add three arrows → Rodrigues → collect into DCM $C$. Halve the angle → quaternion $q$. Chain three planar turns → Euler angles. The DCM is the neutral hub everyone routes through — never convert quaternion↔Euler directly (compare Rotation Sequences (3-2-1 vs 3-1-3) and Gimbal Lock).

Recall Feynman: the whole walkthrough in plain words

Take an arrow. Pick the single line it spins around (Euler promised one line is always enough) and the angle it turns. Cut the arrow into two: the shadow lying on the spin line — that part never moves — and the flat leftover in the disc facing you. In that disc, spinning an arrow is just clock-hand stuff: keep $\cos$ of the original arrow, add $\sin$ of a fresh arrow that points a quarter-turn ahead (the cross product hands you that companion for free). Add the untouched shadow back, and — since the cosine faded that shadow too — throw in a small top-up along the line to fix it. Those three arrows are the Rodrigues formula. Rewrite each arrow-recipe as a little number-machine and stack them: that's the rotation matrix, the DCM. If instead you turn only half the angle and use the sandwich rule, you get the quaternion — same turn, cheaper and no singular spots. And if you'd rather spin about $z$, then $y$, then $x$ in sequence, you get the human roll-pitch-yaw Euler angles — readable, but they jam at pitch $90^{\circ }$ (gimbal lock). Three languages, one geometric fact. Translate through the matrix and you never lose the meaning.