Converting between DCM, quaternions, Euler angles: Exercises

This is the practice arena for the parent topic. Every problem is graded L1 → L5. Try it first, then open the collapsible solution. All the machinery lives in the parent note: Rodrigues Rotation Formula, Direction Cosine Matrix Properties, Quaternion Kinematics & Propagation, Gimbal Lock, Rotation Sequences (3-2-1 vs 3-1-3).

Before we start, one figure fixes all the conventions and symbols we will reuse.

Figure — Converting between DCM, quaternions, Euler angles

Definition Symbols used everywhere below (defined once)

$C$ — the Direction Cosine Matrix, a $3 \times 3$ grid of numbers; row $i$ , column $j$ is written $C_{ij}$ . Multiplying $C v$ rotates the vector $v$ .
$\overset{e}{^} = (e_{1}, e_{2}, e_{3})$ — a unit axis: an arrow of length $1$ that the rotation spins around.
$Φ$ — the turn angle about that axis.
$q = (q_{0}, q_{1}, q_{2}, q_{3})$ — the quaternion: $q_{0}$ is the "how much" scalar, $(q_{1}, q_{2}, q_{3})$ is the "which way" vector. Constraint: $q_{0}^{2} + q_{1}^{2} + q_{2}^{2} + q_{3}^{2} = 1$ .
$(ϕ, θ, ψ)$ — roll, pitch, yaw (the 3-2-1 Euler angles).
$atan2 (y, x)$ — the angle-finder that keeps track of both signs (see Rotation Sequences (3-2-1 vs 3-1-3)). It answers "which angle has vertical part $y$ and horizontal part $x$ ?" over the full circle $[- 180°, 180°]$ , unlike plain $arctan$ which only knows $[- 90°, 90°]$ .

Level 1 — Recognition

L1.1 — Read a quaternion's axis and angle

The quaternion $q = (\frac{3}{2}, 0, 0, \frac{1}{2})$ is given. What is the rotation axis $\overset{e}{^}$ and the turn angle $Φ$ ?

Recall Solution — L1.1

WHAT we use: the axis–angle definition $q_{0} = cos \frac{Φ}{2}$ and $(q_{1}, q_{2}, q_{3}) = \overset{e}{^} sin \frac{Φ}{2}$ . WHY this and not the DCM: the axis and angle are directly baked into the quaternion; no matrix needed.

From $q_{0} = cos \frac{Φ}{2} = \frac{3}{2}$ we get $\frac{Φ}{2} = 30°$ , so $Φ = 60°$ . The vector part $(0, 0, \frac{1}{2})$ must equal $\overset{e}{^} sin 30° = \overset{e}{^} \cdot \frac{1}{2}$ . Dividing by $\frac{1}{2}$ : $\overset{e}{^} = (0, 0, 1)$ .

Answer: axis $\overset{e}{^} = (0, 0, 1)$ (the $z$ -axis), angle $Φ = 60°$ .

L1.2 — Recognise the trace shortcut

For any rotation quaternion, the trace of its DCM is $tr (C) = C_{11} + C_{22} + C_{33}$ . Given $tr (C) = 2$ , find $q_{0}$ (taking the positive root).

Recall Solution — L1.2

WHAT/WHY: the parent note proves $tr (C) = 4 q_{0}^{2} - 1$ . This one scalar carries $q_{0}$ , so it's the fastest handle. $4 q_{0}^{2} - 1 = 2 \Rightarrow q_{0}^{2} = \frac{3}{4} \Rightarrow q_{0} = \frac{3}{2} \approx 0.8660.$ Sanity: $q_{0} = cos \frac{Φ}{2} = \frac{3}{2} \Rightarrow Φ = 60°$ — matches L1.1's kind of rotation.

Level 2 — Application

L2.1 — Quaternion → DCM (build the matrix)

Convert $q = (\frac{2}{2}, 0, 0, \frac{2}{2})$ into its DCM $C (q)$ .

Recall Solution — L2.1

WHAT: plug into the parent's $C (q)$ formula. WHY: to rotate vectors we need matrix form. Here $q_{0} = q_{3} = \frac{2}{2}$ , so $q_{0}^{2} = q_{3}^{2} = \frac{1}{2}$ , and $q_{1} = q_{2} = 0$ .

$C_{11} = q_{0}^{2} + q_{1}^{2} - q_{2}^{2} - q_{3}^{2} = \frac{1}{2} - \frac{1}{2} = 0$
$C_{22} = q_{0}^{2} - q_{1}^{2} + q_{2}^{2} - q_{3}^{2} = \frac{1}{2} - \frac{1}{2} = 0$
$C_{33} = q_{0}^{2} - q_{1}^{2} - q_{2}^{2} + q_{3}^{2} = \frac{1}{2} + \frac{1}{2} = 1$
$C_{12} = 2 (q_{1} q_{2} + q_{0} q_{3}) = 2 \cdot \frac{1}{2} = 1$ , $C_{21} = 2 (q_{1} q_{2} - q_{0} q_{3}) = - 1$
all other off-diagonals contain $q_{1}$ or $q_{2} \Rightarrow 0$ . $C = 0 - 1 0 100001 .$ This is a pure $90°$ yaw about $z$ (see the parent's Example 1). ✓

L2.2 — Euler → DCM (3-2-1) for a single pitch

Compute $C = R_{x} (0) R_{y} (30°) R_{z} (0)$ , i.e. pure pitch $θ = 30°$ , roll $ϕ = 0$ , yaw $ψ = 0$ .

Recall Solution — L2.2

WHAT: substitute into the parent's 3-2-1 matrix with $ϕ = ψ = 0$ ( $c ϕ = c ψ = 1, s ϕ = s ψ = 0$ ). WHY: confirms you can read the general formula by killing terms. $C = c θ 0 s θ 010 - s θ 0 c θ = 0.8660 0 0.5 010 - 0.5 0 0.8660 .$ The middle row/column is untouched because we spin about $y$ : the $y$ -axis is fixed. That's the geometric "look".

L2.3 — DCM → Euler extraction

From the matrix of L2.2, recover $(ϕ, θ, ψ)$ using the parent's extraction rules.

Recall Solution — L2.3

WHAT/WHY: we read specific entries because they isolate one angle each (parent "DCM → Euler").

$θ = - arcsin (C_{13}) = - arcsin (- 0.5) = + 30°$ ✓
$ϕ = atan2 (C_{23}, C_{33}) = atan2 (0, 0.8660) = 0°$ ✓
$ψ = atan2 (C_{12}, C_{11}) = atan2 (0, 0)$ — but here $c θ \neq = 0$ , so $C_{11} = c θ c ψ = 0.866, C_{12} = c θ s ψ = 0$ , giving $atan2 (0, 0.866) = 0°$ ✓ Round trip recovered exactly. (Note: only truly undefined when $c θ = 0$ , i.e. gimbal lock — not here.)

Level 3 — Analysis

L3.1 — DCM → Quaternion, safe branch

Given $C = 0 - 1 0 100001,$ recover $q$ using the trace branch, and enforce $q_{0} \geq 0$ .

Recall Solution — L3.1

WHAT: trace first. $tr (C) = 0 + 0 + 1 = 1$ , so $q_{0} = \frac{1}{2} 1 + 1 = \frac{1}{2} 2 = \frac{2}{2} \approx 0.7071$ . WHY the trace branch is safe here: $q_{0} \approx 0.707$ is comfortably away from $0$ , so dividing by $4 q_{0}$ is fine.

$q_{1} = \frac{C _{23} - C _{32}}{4 q _{0}} = \frac{0 - 0}{4 q _{0}} = 0$
$q_{2} = \frac{C _{31} - C _{13}}{4 q _{0}} = \frac{0 - 0}{4 q _{0}} = 0$
$q_{3} = \frac{C _{12} - C _{21}}{4 q _{0}} = \frac{1 - ( - 1 )}{4 \cdot 0.7071} = \frac{2}{2.828} = 0.7071$ $q = (\frac{2}{2}, 0, 0, \frac{2}{2}) .$ This is exactly the input of L2.1 — the conversion is a perfect inverse. ✓

L3.2 — DCM → Quaternion, singular branch (near 180°)

For $C = diag (- 1, - 1, 1)$ (a $180°$ rotation about $z$ ), the trace branch fails. Recover $q$ correctly.

Recall Solution — L3.2

WHAT: $tr (C) = - 1 - 1 + 1 = - 1 \Rightarrow q_{0} = \frac{1}{2} 1 + (- 1) = 0$ . Dividing by $4 q_{0} = 0$ is illegal — this is why Shepperd says divide by the largest component. WHY: at $180°$ the scalar part vanishes; the biggest diagonal tells us which vector component dominates. The diagonal entries are $C_{11} = - 1, C_{22} = - 1, C_{33} = + 1$ ; the largest is $C_{33}$ , so solve for $q_{3}$ : $q_{3} = \frac{1}{2} 1 + C_{33} - C_{11} - C_{22} = \frac{1}{2} 1 + 1 + 1 + 1 = \frac{1}{2} 4 = 1.$ Then the off-diagonals give the rest via $q_{0} = \frac{C _{21} - C _{12}}{4 q _{3}} = \frac{0 - 0}{4} = 0$ , and $q_{1} = q_{2} = 0$ . $q = (0, 0, 0, 1) .$ Check: $q_{0} = cos \frac{Φ}{2} = 0 \Rightarrow Φ = 180°$ ; axis $(0, 0, 1)$ ✓.

L3.3 — Detect and handle gimbal lock

A DCM has $C_{13} = - 1$ . Show that this is gimbal lock, find $θ$ , and explain what is (and isn't) recoverable.

Recall Solution — L3.3

WHAT: $θ = - arcsin (C_{13}) = - arcsin (- 1) = + 90°$ . WHY it's a singularity: with $θ = 90°$ , $c θ = 0$ , so in the parent matrix $C_{11} = c θ c ψ = 0$ and $C_{12} = c θ s ψ = 0$ . Then $ψ = atan2 (0, 0)$ is undefined — the yaw and roll axes have aligned (see Gimbal Lock). What survives: the off-diagonal $2 \times 2$ block collapses so that only the sum $ϕ + ψ$ is determined (for $θ = + 90°$ ), not each separately. Convention: set $ϕ = 0$ , then $ϕ + ψ$ is read from $atan2 (C_{21}, C_{22})$ — that combined angle is fully valid; the split is a free choice. One degree of freedom is reported as lost.

Level 4 — Synthesis

L4.1 — Full chain: Euler → DCM → quaternion

Take $ϕ = 0, θ = 0, ψ = 90°$ . Build $C$ , then extract $q$ , and confirm the "via-the-hub" workflow.

Recall Solution — L4.1

WHY via DCM: memorising a direct Euler→quaternion formula (12 terms) is error-prone; compose two trusted maps (parent Example: use the DCM as neutral hub).

Step 1 (Euler→DCM): pure yaw $90°$ . Using L2.2-style substitution with only $ψ \neq = 0$ : $C = R_{z} (90°) = c ψ - s ψ 0 s ψ c ψ 0 001 = 0 - 1 0 100001 .$ Step 2 (DCM→quaternion): identical to L3.1 → $q = (\frac{2}{2}, 0, 0, \frac{2}{2})$ . Cross-check via axis–angle: yaw $90°$ about $z$ ⇒ $q_{0} = cos 45° = \frac{2}{2}$ , $q_{3} = sin 45° = \frac{2}{2}$ ✓. The hub route and the direct axis–angle route agree.

L4.2 — Compose two rotations, then read the axis

Apply a $90°$ yaw ( $q_{A}$ from L4.1) followed by a $90°$ roll about $x$ ( $q_{B} = (\frac{2}{2}, \frac{2}{2}, 0, 0)$ ). Multiply the quaternions ( $q_{C} = q_{B} \otimes q_{A}$ ) and report $Φ$ of the combined rotation.

Recall Solution — L4.2

WHAT/WHY: quaternion multiplication composes rotations cheaply (parent: quaternions are best for propagation). Hamilton product with $q = (w, x, y, z)$ : $w_{C} = w_{B} w_{A} - v_{B} \cdot v_{A}, v_{C} = w_{B} v_{A} + w_{A} v_{B} + v_{B} \times v_{A} .$ With $q_{A} = (\frac{2}{2}, 0, 0, \frac{2}{2})$ , $q_{B} = (\frac{2}{2}, \frac{2}{2}, 0, 0)$ :

$w_{C} = \frac{1}{2} - 0 = \frac{1}{2}$
$v_{B} \cdot v_{A} = 0$
$w_{B} v_{A} = (0, 0, \frac{1}{2})$ , $w_{A} v_{B} = (\frac{1}{2}, 0, 0)$
$v_{B} \times v_{A} = (\frac{2}{2}, 0, 0) \times (0, 0, \frac{2}{2}) = (0 \cdot \frac{2}{2} - 0, 0 - \frac{2}{2} \cdot \frac{2}{2}, 0 - 0) = (0, - \frac{1}{2}, 0)$
$v_{C} = (\frac{1}{2}, - \frac{1}{2}, \frac{1}{2})$ $q_{C} = (\frac{1}{2}, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2}) .$ Norm check: $4 \cdot \frac{1}{4} = 1$ ✓. Angle: $q_{0} = \frac{1}{2} = cos \frac{Φ}{2} \Rightarrow \frac{Φ}{2} = 60° \Rightarrow Φ = 120°$ . The combined turn is $120°$ about axis $\overset{e}{^} = (\frac{1}{2}, - \frac{1}{2}, \frac{1}{2}) / sin 60° = (\frac{1}{3}, - \frac{1}{3}, \frac{1}{3})$ .

Level 5 — Mastery

L5.1 — Prove the trace identity

Starting from the parent's $C (q)$ , prove that $tr (C) = 4 q_{0}^{2} - 1$ using the unit-norm constraint.

Recall Solution — L5.1

WHAT: add the three diagonal entries. $tr (C) = (q_{0}^{2} + q_{1}^{2} - q_{2}^{2} - q_{3}^{2}) + (q_{0}^{2} - q_{1}^{2} + q_{2}^{2} - q_{3}^{2}) + (q_{0}^{2} - q_{1}^{2} - q_{2}^{2} + q_{3}^{2}) .$ The $q_{1}^{2}$ terms: $+ 1 - 1 - 1 = - 1$ ; likewise $q_{2}^{2}$ and $q_{3}^{2}$ each net $- 1$ ; $q_{0}^{2}$ nets $+ 3$ . So $tr (C) = 3 q_{0}^{2} - q_{1}^{2} - q_{2}^{2} - q_{3}^{2} .$ WHY the constraint enters: use $q_{1}^{2} + q_{2}^{2} + q_{3}^{2} = 1 - q_{0}^{2}$ : $tr (C) = 3 q_{0}^{2} - (1 - q_{0}^{2}) = 4 q_{0}^{2} - 1. ■$

L5.2 — Round-trip robustness at 180°

Show that for $C = diag (1, - 1, - 1)$ (parent Example 2, $180°$ about $x$ ), rebuilding $C$ from the recovered quaternion returns the original matrix — verifying the singular branch is not just a hack.

Recall Solution — L5.2

Step 1 — extract (largest diagonal is $C_{11} = 1$ ): $q_{1} = \frac{1}{2} 1 + C_{11} - C_{22} - C_{33} = \frac{1}{2} 1 + 1 + 1 + 1 = 1,$ and $q_{0} = \frac{C _{32} - C _{23}}{4 q _{1}} = 0$ , $q_{2} = q_{3} = 0$ , giving $q = (0, 1, 0, 0)$ . Step 2 — rebuild via $C (q)$ with $q_{1} = 1$ , rest $0$ :

$C_{11} = q_{1}^{2} = 1$
$C_{22} = - q_{1}^{2} = - 1$ , $C_{33} = - q_{1}^{2} = - 1$
all off-diagonals contain $q_{0}, q_{2}, q_{3} \Rightarrow 0$ . $C_{rebuilt} = diag (1, - 1, - 1) = C_{original} . ✓$ The singular branch is exact — it reproduces the matrix bit-for-bit.

L5.3 — Sign ambiguity is physical, not a bug

Show that $q = (0, 1, 0, 0)$ and $q = (0, - 1, 0, 0)$ produce the same DCM, and state the convention that picks a unique representative.

Recall Solution — L5.3

WHY: the parent notes $C (q)$ is quadratic in $q$ ; every entry is a product of two components, so flipping all signs leaves every product unchanged. Check the only nonzero-driving term: $C_{22} = - q_{1}^{2}$ . Whether $q_{1} = + 1$ or $q_{1} = - 1$ , $q_{1}^{2} = 1$ , so $C_{22} = - 1$ identically; same for $C_{11}, C_{33}$ . Off-diagonals stay $0$ . Both quaternions give $diag (1, - 1, - 1)$ . Convention: enforce $q_{0} \geq 0$ (and if $q_{0} = 0$ , fix the sign of the first nonzero vector component). This selects the shortest-arc representative and removes the double cover ambiguity.

Recall One-line self-test before you leave

$tr (C)$ gives ::: $4 q_{0}^{2} - 1$ , hence $q_{0} = \frac{1}{2} 1 + tr C$ (L1.2, L3.1, L5.1). To apply rotation $A$ then $B$ you multiply ::: $q_{B} \otimes q_{A}$ — second rotation on the left (L4.2). Gimbal lock happens at pitch ::: $θ = \pm 90°$ , where only $ϕ \pm ψ$ is recoverable (L3.3). Near $180°$ you divide by ::: the largest quaternion component, not $q_{0}$ (L3.2, L5.2).