Euler angles — roll φ, pitch θ, yaw ψ; rotation sequence (3-2-1 convention): Worked examples

Before we start, a one-line refresher of the two tools we lean on the most:

Definition The two tools we reuse everywhere

$atan2 (y, x)$ — reads "the angle whose sine is proportional to $y$ and cosine to $x$ ." Unlike plain $arctan (y / x)$ , it looks at the signs of both $y$ and $x$ separately, so it knows which of the four quadrants you are in. That is exactly why the recovery formulas use it.
Passive convention: $R_{n}^{b}$ takes a vector's numbers in the nav frame and re-writes them in the body frame. Its transpose $R_{b}^{n} = (R_{n}^{b})^{T}$ goes the other way (body → nav), because rotation matrices are orthogonal (see Rotation matrices & orthogonality).
DCM = Direction Cosine Matrix — just another name for the $3 \times 3$ rotation matrix $R_{n}^{b}$ ; its nine entries are the cosines of the angles between each pair of frame axes (see Direction Cosine Matrix (DCM)).
Shorthand: to keep matrices readable we write $c θ \equiv cos θ$ and $s θ \equiv sin θ$ (likewise $c ϕ, s ϕ, c ψ, s ψ$ ). This is the same shorthand the parent note used.

All angles below are in degrees unless a formula demands radians.

The scenario matrix

Every problem this topic can pose falls into one of these cells. Each worked example below is tagged with the cell(s) it covers.

#	Case class	What makes it tricky	Example
A	Single-axis, positive angle	baseline sanity	Ex 1
B	Negative angle	sign of $sin$ flips a column	Ex 2
C	Quadrant care in recovery (angle > 90° or < −90°)	plain $arctan$ fails, need atan2	Ex 3
D	All three angles combined	full matrix multiply, no zeros to hide behind	Ex 4
E	Zero / identity input	degenerate — must return $I$	Ex 5
F	Gimbal lock $θ = \pm 9 0^{\circ}$	recovery becomes 0/0, DOF lost (both signs)	Ex 6
G	Word problem (real vehicle heading)	translate English → angles → vector	Ex 7
H	Exam twist: non-uniqueness ( $ϕ, ψ$ vs $ϕ \pm 180, ψ \pm 180$ )	two angle-triples, one orientation	Ex 8

Ex 1 — Case A: pure pitch, positive angle

Forecast: Guess before computing — will the nose gain a positive Down component, a negative one, or none? (Nose up means the nose lifts above the horizon.)

With only pitch, $R_{n}^{b} = R_{y} (3 0^{\circ})$ . Why this step? $ψ = ϕ = 0$ make $R_{z}$ and $R_{x}$ into the identity, so the product collapses to just the pitch matrix.
The nose in body frame is $[1, 0, 0]^{T}$ ; convert to nav via $R_{b}^{n} = (R_{n}^{b})^{T} = R_{y} (3 0^{\circ})^{T}$ . Why this step? The vector "points out the nose" is trivially $[1, 0, 0]^{T}$ in the body frame. To see it in nav we go body → nav, which is the transpose.
Compute. $R_{y} (θ)^{T} [1, 0, 0]^{T} = [cos θ, 0, - sin θ]^{T} = [0.866, 0, - 0.5]^{T}$ . Why this step? The first column of $R_{y}^{T}$ is exactly $[cos θ, 0, - sin θ]$ .

Look at figure s01: the nose vector (amber) has lifted above the North–East plane, so its Down component is negative ( $- 0.5$ ). Nose-up = negative Down. Forecast check: yes, negative.

Figure — Euler angles — roll φ, pitch θ, yaw ψ; rotation sequence (3-2-1 convention)

Verify: magnitude $0.86 6^{2} + 0 + 0. 5^{2} = 1.000$ (still a unit direction ✓). The North component $cos 30 = 0.866$ and Down $- sin 30 = - 0.5$ agree with a $3 0^{\circ}$ lift.

Ex 2 — Case B: negative yaw

Forecast: North (nose forward) will rotate to... East or West? Negative yaw in the passive/right-hand convention.

$R_{n}^{b} = R_{z} (- 9 0^{\circ})$ . Using $sin (- 90) = - 1, cos (- 90) = 0$ : $R_{z} (- 9 0^{\circ}) = 010 - 1 00 001$ Why this step? Substituting a negative angle simply flips the sign of every $sin$ term; the $- sin$ that sat in the lower-left becomes $+ 1$ .
Nose in nav $= (R_{n}^{b})^{T} [1, 0, 0]^{T}$ . The transpose's first column is $[0, - 1, 0]^{T}$ . Why this step? Same body→nav logic as Ex 1.
Result: nose $= [0, - 1, 0]^{T}$ = negative East = West. Why this step? $- 1$ in the East slot means "one unit toward West."

Verify: A $- 9 0^{\circ}$ yaw should point the nose from North to West. $[0, - 1, 0]$ is exactly West ✓. Contrast with the parent's Ex 1 where $ψ = + 9 0^{\circ}$ gave East — the sign of $ψ$ decided East vs West, as it must.

Ex 3 — Case C: recovery when yaw exceeds 90°

Forecast: Both $R_{11}$ and $R_{12}$ are negative. In which quadrant does that put $ψ$ ? Would blindly doing $arctan (R_{12} / R_{11})$ get it right?

Recall $R_{12} = cos θ sin ψ$ and $R_{11} = cos θ cos ψ$ . Why this step? From the full matrix in the parent note; the shared $cos θ$ cancels in the ratio.
Naive route (the trap): $arctan (\frac{- 0.5}{- 0.866}) = arctan (0.577) = 3 0^{\circ}$ . Why this step? To expose the failure: $arctan$ can only return angles in $(- 9 0^{\circ}, 9 0^{\circ})$ , so it cannot tell $3 0^{\circ}$ apart from $3 0^{\circ} + 18 0^{\circ} = 21 0^{\circ}$ .
Correct route: $ψ = atan2 (- 0.5, - 0.866)$ . Both arguments negative → third quadrant → $ψ = - 15 0^{\circ}$ (equivalently $21 0^{\circ}$ ). Why this step? $atan2$ reads the two signs separately: negative sine and negative cosine can only be the third quadrant.

Figure s02 shows the four quadrant sign-patterns of $(R_{12}, R_{11})$ so you can eyeball which yaw region you are in.

Verify: $cos (- 15 0^{\circ}) = - 0.866 = R_{11}$ ✓ and $sin (- 15 0^{\circ}) = - 0.5 = R_{12}$ ✓ (taking $cos θ = 1$ ). The naive $3 0^{\circ}$ gives $cos 30 = + 0.866$ — wrong sign, confirming the trap.

Ex 4 — Case D: all three angles at once

Forecast: The first row is $[cos θ cos ψ, cos θ sin ψ, - sin θ]$ . Notice it does not contain $ϕ$ at all — can you see why before reading step 3?

First row of $R_{n}^{b}$ = $[c θ c ψ, c θ s ψ, - s θ]$ (recall $c θ \equiv cos θ, s θ \equiv sin θ$ ). Why this step? Straight from the multiplied-out matrix in the parent note.
Plug numbers. $c 20 = 0.9397, c 45 = 0.7071, s 45 = 0.7071, s 20 = 0.3420$ : $[0.9397 (0.7071), 0.9397 (0.7071), - 0.3420] = [0.6645, 0.6645, - 0.3420]$ Why this step? Direct substitution; keep 4 decimals to survive the magnitude check.
Why no $ϕ$ ? Roll is the last spin, about the body- $x$ axis. The first row describes how the body- $x$ axis relates to nav — and a spin about an axis never moves that axis. So $ϕ$ is invisible in row 1. Why this step? This is the geometric heart of the 3-2-1 ordering.

Verify: magnitude $0.664 5^{2} + 0.664 5^{2} + 0.342 0^{2} = 0.4416 + 0.4416 + 0.1170 = 1.0002 = 1.000$ (unit row of an orthogonal matrix ✓).

Ex 5 — Case E: the degenerate zero input

Forecast: If the body is perfectly aligned with the nav frame, what should the transform "do" to a vector?

Each elementary matrix at $0$ becomes the identity because $cos 0 = 1, sin 0 = 0$ (i.e. every $c \cdot = 1$ and every $s \cdot = 0$ ). Why this step? Substituting $0$ zeroes every off-diagonal sine and puts $1$ on every diagonal.
Product of three identities is the identity: $R_{n}^{b} = I I I = I$ . Why this step? $I$ is the multiplicative identity for matrices.
Meaning: nav coordinates = body coordinates for every vector; nothing is rotated. Why this step? $I v = v$ leaves all numbers unchanged.

Verify: $det I = 1$ (proper rotation, not a reflection ✓) and $I^{T} I = I$ (orthogonal ✓). This is the only fixed point where all three recovery formulas give $0$ : $- arcsin (0) = 0$ , $atan2 (0, 1) = 0$ , $atan2 (0, 1) = 0$ .

Ex 6 — Case F: gimbal lock, BOTH $θ = + 9 0^{\circ}$ and $θ = - 9 0^{\circ}$

Forecast: The parent note warns $R_{11} = R_{12} = R_{23} = R_{33} = 0$ here. If those are all zero, what happens to $atan2 (R_{23}, R_{33}) = atan2 (0, 0)$ ?

Case F+: $θ = + 9 0^{\circ}$

Build the affected entries at $θ = 9 0^{\circ}$ ( $c θ = 0, s θ = 1$ ): $R_{11} = c θ c ψ = 0, R_{12} = c θ s ψ = 0, R_{23} = s ϕ c θ = 0, R_{33} = c ϕ c θ = 0$ . Why this step? Every one of them carries a factor $cos θ = 0$ .
Recovery collapses: $ϕ = atan2 (0, 0)$ and $ψ = atan2 (0, 0)$ are undefined (0/0). Only pitch survives: $θ = - arcsin (R_{13}) = - arcsin (- 1) = 9 0^{\circ}$ . Why this step? $R_{13} = - sin θ = - 1$ is the one entry with no $cos θ$ factor.
What IS defined: the still-meaningful quantity is the entry $R_{21} = s ϕ s θ c ψ - c ϕ s ψ$ . At $θ = 9 0^{\circ}$ ( $s θ = 1$ ) this becomes $sin ϕ cos ψ - cos ϕ sin ψ = sin (ϕ - ψ)$ . So the vehicle only "knows" the difference $ϕ - ψ$ , not each separately — that is the lost degree of freedom. Why this step? At $θ = + 9 0^{\circ}$ the roll and yaw axes point along the same line, so only their difference is observable.

Case F−: $θ = - 9 0^{\circ}$

Build the same entries at $θ = - 9 0^{\circ}$ ( $c θ = 0, s θ = - 1$ ): again $R_{11} = R_{12} = R_{23} = R_{33} = 0$ , so recovery still gives $0/0$ ; pitch is $θ = - arcsin (R_{13}) = - arcsin (1) = - 9 0^{\circ}$ . Why this step? Same $cos θ = 0$ factor kills the same four entries; only the sign of $R_{13} = - sin θ = + 1$ differs.
Recoverable combination flips to a SUM. Now $R_{21} = s ϕ (- 1) c ψ - c ϕ s ψ = - (sin ϕ cos ψ + cos ϕ sin ψ) = - sin (ϕ + ψ)$ . So at $θ = - 9 0^{\circ}$ the observable is $ϕ + ψ$ , not the difference. Why this step? $s θ$ flipped sign, turning the angle-subtraction identity into an angle-addition one — the two lock cases are mirror twins.

Figure s03 shows the roll and yaw axes collapsing onto a common line as pitch approaches $\pm 9 0^{\circ}$ .

Verify: at $θ = + 9 0^{\circ}, ϕ = 4 0^{\circ}, ψ = 1 0^{\circ}$ , the entry $R_{21} = sin (4 0^{\circ} - 1 0^{\circ}) = sin 3 0^{\circ} = 0.5$ ; at $θ = - 9 0^{\circ}$ the same $ϕ, ψ$ give $R_{21} = - sin (4 0^{\circ} + 1 0^{\circ}) = - sin 5 0^{\circ} = - 0.766$ . In both cases $R_{11} = R_{12} = R_{23} = R_{33} = 0$ exactly. The cure is to propagate with Quaternions — avoiding gimbal lock or the raw Direction Cosine Matrix (DCM) and only convert to Euler for a human display.

Ex 7 — Case G: real-world word problem

Forecast: Heading measured clockwise from North — does clockwise correspond to a positive yaw in our convention (which yaws +ψ toward East)? North-East is East-ward of North.

Translate English → angles. Heading $04 5^{\circ}$ (toward East side) $\Rightarrow ψ = + 4 5^{\circ}$ ; climbing $\Rightarrow θ = + 1 5^{\circ}$ nose-up; "wings level" $\Rightarrow ϕ = 0$ . Why this step? Heading is a yaw about Down; a climb is a pitch; level wings mean zero roll.
Nose in nav = first column of $R_{b}^{n} =$ first row of $R_{n}^{b}$ $= [c θ c ψ, c θ s ψ, - s θ]$ . Why this step? We established in Ex 4 that this row is the body- $x$ (nose) direction expressed in nav.
Plug in. $c 15 = 0.9659, c 45 = s 45 = 0.7071, s 15 = 0.2588$ : $[0.9659 (0.7071), 0.9659 (0.7071), - 0.2588] = [0.683, 0.683, - 0.259]^{T}$ Why this step? Equal North and East components ( $0.683$ each) = pointing exactly North-East; negative Down = climbing.

Verify: North/East equal ⇒ ground track bearing $= atan2 (0.683, 0.683) = 4 5^{\circ}$ ✓. Climb angle $= arcsin (0.259) = 1 5^{\circ}$ ✓. Magnitude $0.68 3^{2} + 0.68 3^{2} + 0.25 9^{2} = 1.000$ ✓.

Ex 8 — Case H: exam twist, the two-triple ambiguity

Forecast: Two totally different-looking angle triples, one physical attitude. Which trig identities make the matrices coincide?

The alias rule for 3-2-1: $(ϕ, θ, ψ)$ and $(ϕ \pm 18 0^{\circ}, 18 0^{\circ} - θ, ψ \pm 18 0^{\circ})$ give the same $R_{n}^{b}$ . Why this step? This is the standard second solution; here $18 0^{\circ} - 2 0^{\circ} = 16 0^{\circ}$ .
Check one entry, $R_{13} = - sin θ$ . Solution 1: $- sin 2 0^{\circ} = - 0.342$ . Solution 2: $- sin 16 0^{\circ} = - sin (18 0^{\circ} - 2 0^{\circ}) = - sin 2 0^{\circ} = - 0.342$ . Same. Why this step? $sin (18 0^{\circ} - x) = sin x$ makes pitch entries agree.
Check $R_{33} = cos ϕ cos θ$ . Sol 1: $cos 30 cos 20 = 0.866 (0.9397) = 0.8138$ . Sol 2: $cos 210 cos 160 = (- 0.866) (- 0.9397) = 0.8138$ . The two sign flips cancel. Why this step? Adding $18 0^{\circ}$ flips $cos ϕ$ 's sign; taking $18 0^{\circ} - θ$ flips $cos θ$ 's sign; product unchanged.

Verify: $R_{13}$ equal at $- 0.342$ ; $R_{33}$ equal at $0.8138$ . Two triples, identical matrix entries ⇒ same orientation. Exam takeaway: always report the branch with $∣ θ ∣ \leq 9 0^{\circ}$ (the principal solution) unless told otherwise — see Attitude estimation (Kalman / complementary filter) for why filters pin this branch.

Active Recall

Recall Which cell trips people up most, and why?

Cell C (quadrant care). ::: Because $arctan (R_{12} / R_{11})$ silently drops the sign information; $atan2$ keeps both signs and so identifies the correct quadrant.

Recall At gimbal lock, what single quantity stays recoverable?

The combination of roll and yaw ::: $ϕ - ψ$ at $θ = + 9 0^{\circ}$ (via $R_{21} = sin (ϕ - ψ)$ ) and $ϕ + ψ$ at $θ = - 9 0^{\circ}$ (via $R_{21} = - sin (ϕ + ψ)$ ), plus the pitch $θ = \pm 9 0^{\circ}$ itself — but not roll and yaw separately.

Recall Why does roll never appear in the first row of

$R_{n}^{b}$ ? Because roll is a spin about body- $x$ ::: and a rotation never moves its own axis, so the body- $x$ (nose) direction — which is exactly the first row of $R_{n}^{b}$ — is completely independent of $ϕ$ .

Connections

Parent topic
Rotation matrices & orthogonality
Quaternions — avoiding gimbal lock
Direction Cosine Matrix (DCM)
Angular velocity & body rates p, q, r
NED navigation frame
Attitude estimation (Kalman / complementary filter)