Converting between DCM, quaternions, Euler angles: Question bank

These are concept traps, not calculators. Each line is a claim, an error, a "why", or an edge case. Cover the answer, commit out loud, then reveal. If your reason is "yes/no", you haven't answered — the justification is the point.

Parent: 3.5.10 · Converting between DCM, quaternions, Euler angles. Prerequisites worth re-opening: Euler's Rotation Theorem, Direction Cosine Matrix Properties, Gimbal Lock, Rotation Sequences (3-2-1 vs 3-1-3).

Notation you need before the traps

Definition The three languages and their symbols

Everything below describes one fact: how a body frame is turned relative to a reference frame. Three ways to write it:

DCM = Direction Cosine Matrix, written $C$ : a $3 \times 3$ grid of numbers. $C_{ij}$ means the entry in row $i$ , column $j$ . Applying it to a column of three numbers (a vector $v$ , an arrow in space) rotates that arrow: $v^{'} = C v$ . $I$ is the identity matrix ( $1$ s on the diagonal, $0$ s elsewhere) — the "do nothing" rotation.
Quaternion $q$ : four numbers written scalar-first as $q = (q_{0}, q)$ where $q_{0}$ is the scalar part and $q = (q_{1}, q_{2}, q_{3})$ is the vector part. So $q = (q_{0}, q_{1}, q_{2}, q_{3})$ . It rotates a vector through the sandwich product $v \mapsto q v q^{- 1}$ (multiply by $q$ on the left and its inverse $q^{- 1}$ on the right).
Euler angles $(ϕ, θ, ψ)$ : three human-readable angles — roll $ϕ$ , pitch $θ$ , yaw $ψ$ — applied in the aerospace 3-2-1 (Tait–Bryan) sequence: yaw $ψ$ about $z$ , then pitch $θ$ about the new $y$ , then roll $ϕ$ about the new $x$ . (A different order, e.g. 3-1-3, is a different convention — see Rotation Sequences (3-2-1 vs 3-1-3).)

One more tool: $[\overset{e}{^}]_{\times}$ is the cross-product matrix of the axis $\overset{e}{^}$ — the $3 \times 3$ matrix for which $[\overset{e}{^}]_{\times} v = \overset{e}{^} \times v$ . It is antisymmetric ( $[\overset{e}{^}]_{\times}^{⊤} = - [\overset{e}{^}]_{\times}$ ).

Every trap on this page assumes the body-from-inertial convention and the 3-2-1 sequence unless a line says otherwise. The picture below pins the three languages to one geometric rotation so the traps have something to point at.

Figure — Converting between DCM, quaternions, Euler angles

True or false — justify

A rotation matrix and its transpose describe the same physical orientation.

False.

C

and

C^{⊤}

are inverse rotations (body-from-inertial vs inertial-from-body). They only coincide for the identity or a

180°

turn; otherwise feeding the wrong one reverses every mapping. See Direction Cosine Matrix Properties.

The quaternions

q

and

- q

point along opposite axes, so they are opposite rotations.

False. The rotation map

v \mapsto q v q^{- 1}

is quadratic in

q

, so the two minus signs cancel and both give the identical DCM — same axis, same angle. Negating

q

Φ \to Φ + 360°

, which is no change. (See the double-cover figure.)

3 \times 3

matrix of nine real numbers can encode any rotation freely.

False. A DCM must be orthonormal with determinant

+ 1

: that's

6

constraints, leaving only

3

free numbers — the true dimension of rotation. Nine "free" numbers would allow shears and reflections.

Euler angles have a singularity because the physical rotation itself becomes undefined.

False. The rotation is perfectly fine; only the coordinate chart fails. At pitch

\pm 90°

two gimbal axes align so yaw and roll can't be separated — a flaw of the representation, not the motion. See Gimbal Lock and the gimbal figure below.

The quaternion set never has a singularity, so it can perfectly report human-readable attitude.

False (half-true). Quaternions are singularity-free for storing and propagating, but they are not human-readable — you still convert to Euler (which is singular) to report roll/pitch/yaw.

Because

q_{0} = cos (Φ/2)

, a rotation of

0°

and a rotation of

360°

give the same quaternion.

False.

0°

gives

q_{0} = cos 0 = 1

;

360°

gives

q_{0} = cos 180° = - 1

. Same rotation, opposite quaternion — this is exactly the

q, - q

double cover.

The trace of a rotation matrix alone tells you the rotation angle.

True.

tr (C) = 1 + 2 cos Φ

, so

cos Φ = (tr (C) - 1) /2

. But it does not tell you the axis — you need the antisymmetric part for that.

Going quaternion → Euler → quaternion always returns the original quaternion.

False. Euler angles can't distinguish

q

from

- q

, and near gimbal lock they lose information entirely; so the round trip may return

- q

or a nearby but different quaternion.

Spot the error

"To get Euler angles from a DCM I use

ψ = arctan (C_{12} / C_{11})

Uses plain

arctan

, which only spans

(- 90°, 90°)

and drops both signs — a

180°

error whenever

C_{11} < 0

. Use

atan2 (C_{12}, C_{11})

, which keeps both signs and covers the full circle.

"DCM → quaternion: always

q_{0} = \frac{1}{2} 1 + tr (C)

, then

q_{1} = (C_{23} - C_{32}) / (4 q_{0})

Breaks near

180°

rotations where

tr (C) \to - 1

and

q_{0} \to 0

, so you divide by ~zero. Shepperd's fix: compute whichever of

q_{0}, q_{1}, q_{2}, q_{3}

is largest first and divide by that.

"Since

q

and

- q

are the same rotation, sign of

q

never matters anywhere."

They give the same orientation, but for attitude interpolation and error signals the sign matters: choosing the wrong sign makes a slerp take the long

> 180°

path. Enforce

q_{0} \geq 0

for the shortest arc.

"I integrated the quaternion for a while;

∥ q ∥

is now

1.02

but that's fine, close enough."

Not fine.

C (q)

built from a non-unit

q

is no longer orthonormal and will stretch/shear vectors. Renormalize

q \leftarrow q /∥ q ∥

every step. See Quaternion Kinematics & Propagation.

C = R_{x} (ϕ) R_{y} (θ) R_{z} (ψ)

and

C = R_{z} (ψ) R_{y} (θ) R_{x} (ϕ)

are the same rotation."

False — rotations don't commute, so the two orderings give different DCMs and different physical attitudes. The 3-2-1 sequence order is part of the definition. See Rotation Sequences (3-2-1 vs 3-1-3).

[\overset{e}{^}]_{\times}

in Rodrigues is symmetric because it comes from a cross product."

It's antisymmetric (

[\overset{e}{^}]_{\times}^{⊤} = - [\overset{e}{^}]_{\times}

); that's precisely why it supplies the

sin Φ

(odd, direction-flipping) term while

I

and

\overset{e}{^} \overset{e}{^}^{⊤}

supply the even terms. See Rodrigues Rotation Formula.

"To convert quaternion to Euler I'll just derive one big 12-term direct formula."

Error of practice, not algebra: the direct formula is error-prone. Route through the DCM as a neutral hub — quaternion → DCM → Euler — reusing two conversions you already trust.

"A pure

180°

rotation has

q_{0} = 1

since it's a big turn."

Backwards.

q_{0} = cos (Φ/2) = cos 90° = 0

for

180°

. The vector part

q

carries all the magnitude there — which is exactly why the

q_{0}

extraction branch fails.

Why questions

Why does the quaternion use the half angle

Φ/2

and not

Φ

The rotating action

v \mapsto q v q^{- 1}

multiplies by

q

on both sides, effectively applying the turn twice; halving the stored angle cancels that doubling so the net rotation is exactly

Φ

Why does

atan2 (y, x)

beat

arctan (y / x)

for angle extraction?

Dividing

y / x

throws away the individual signs, collapsing four quadrants into two.

atan2

keeps both signs, resolving the full

[- π, π]

range with no ambiguity.

Why is the DCM called the "hub" for conversions?

Every representation has a clean, well-tested formula to and from the DCM, so composing two of those beats memorizing a dozen fragile direct formulas — fewer chances to flip a sign.

Why do we isolate the quaternion vector part from the antisymmetric part of

C

The symmetric part of

C

mixes squares like

q_{0}^{2} - q_{1}^{2}

, but the differences

C_{23} - C_{32} = 4 q_{0} q_{1}

cleanly pick out cross terms — the vector components — free of the diagonal clutter.

Why does Euler's rotation theorem let us derive everything from axis–angle?

It guarantees any orientation is a single turn

Φ

about one axis

\overset{e}{^}

, so

(\overset{e}{^}, Φ)

is a complete minimal description — DCM and quaternion are just two repackagings of it. See Euler's Rotation Theorem.

Why does the near-

180°

case specifically break the trace-based extraction?

Φ \to 180°

tr (C) \to - 1

q_{0} = \frac{1}{2} 1 + tr (C) \to 0

; every other component is then computed by dividing by that vanishing

q_{0}

, amplifying rounding error catastrophically.

Why does gimbal lock only cost you one degree of freedom, not two?

The pitch

θ

is still fully determined; only yaw and roll fuse into the single combination

ϕ \pm ψ

. So you lose one independent angle, not the whole attitude.

Why must you always state whether

C

is body-from-inertial or inertial-from-body?

Because the two conventions are transposes/inverses of each other; an unstated convention silently reverses every rotation you apply and every attitude you report.

Edge cases

The identity rotation (

Φ = 0

): what is its quaternion and DCM?

q = (1, 0, 0, 0)

and

C = I

. The axis

\overset{e}{^}

is undefined (nothing turns), which is harmless — any axis works because

sin (0) = 0

kills the vector part.

Rotation of exactly

180°

about

x

, i.e.

C = diag (1, - 1, - 1)

: which extraction branch do you use?

tr (C) = - 1

q_{0} = 0

; switch to the largest diagonal,

C_{11}

, giving

q_{1} = \frac{1}{2} 1 + C_{11} - C_{22} - C_{33} = 1

, so

q = (0, 1, 0, 0)

Pitch

θ = + 90°

(so

C_{13} = - 1

): what happens to yaw and roll extraction?

C_{11} = C_{12} = 0

, so

atan2 (0, 0)

is undefined. By convention set

ϕ = 0

and solve for the determined combination

ϕ + ψ

; report the lost DOF. See Gimbal Lock.

A degenerate "axis"

\overset{e}{^} = (0, 0, 0)

handed to the axis–angle formula: valid?

No —

\overset{e}{^}

must be a unit vector. A zero axis has no direction; the only rotation with no meaningful axis is the identity, which you should represent directly as

Φ = 0

Very small rotation

Φ \approx 0

with a well-defined axis: is anything numerically fragile?

The rotation is fine, but recovering the axis from the tiny antisymmetric part of

C

is noisy since those entries

\sim sin Φ \approx Φ

are near zero — a small-angle regime where quaternions integrate more stably. See Quaternion Kinematics & Propagation.

Feeding a slightly non-orthonormal

C

(measurement noise) into DCM→quaternion: what breaks?

The trace/off-diagonal identities no longer hold exactly, so

∥ q ∥ \neq = 1

and

q_{0} = \frac{1}{2} 1 + tr (C)

may even take a square root of a negative number — orthonormalize

C

first (e.g. via QUEST-style normalization).

Two different Euler triples reported for the same orientation — bug or expected?

Expected. Euler angles are periodic and non-unique (e.g.

ϕ

and

ϕ \pm 360°

, or the gimbal-lock fusion), so distinct triples can name one attitude. The DCM or normalized quaternion is the canonical arbiter.