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

The master formulas we test against (all derived in the parent):

Level 1 — Recognition

L1.1

In the 3-2-1 convention, list the three rotations in the order they are physically applied, naming the axis, the Greek letter, and the plain-English motion for each.

Recall Solution

Yaw $ψ$ about the $z$ -axis (the "3rd" axis) — nose swings left/right (like a merry-go-round).
Pitch $θ$ about the new $y$ -axis (the "2nd" axis) — nose tips up/down.
Roll $ϕ$ about the newest $x$ -axis (the "1st" axis) — wings bank sideways.

The name 3-2-1 is literally the axis numbers in application order: $3 \to 2 \to 1$ .

L1.2

Which entry of $R_{n}^{b}$ depends on pitch alone (no roll, no yaw)? Write it.

Recall Solution

Look at the top-right entry $R_{13} = - s θ = - sin θ$ . It contains only $θ$ . That is exactly why pitch is the easiest angle to recover: no other angle contaminates it, so $θ = - arcsin (R_{13})$ needs no ratio.

L1.3

True/False: $R_{x} (ϕ) R_{y} (θ) R_{z} (ψ) = R_{z} (ψ) R_{y} (θ) R_{x} (ϕ)$ . Explain in one sentence.

Recall Solution

False. Matrix multiplication (and physical rotation) does not commute — spinning yaw-then-roll leaves you somewhere different from roll-then-yaw. Order is part of the definition; you must always state the convention.

Level 2 — Application

L2.1

Compute $R_{z} (9 0^{\circ})$ numerically, then find where the nose vector $[1, 0, 0]^{T}$ (body frame) points in nav coordinates.

Recall Solution

With $ψ = 9 0^{\circ}$ : $c ψ = 0, s ψ = 1$ , so $R_{z} (9 0^{\circ}) = 0 - 1 0 100001 .$ The nose lives in the body frame, so to see it in nav we apply $R_{b}^{n} = (R_{n}^{b})^{T}$ : $(R_{z} (9 0^{\circ}))^{T} [1, 0, 0]^{T} = 010 - 1 00 001 100 = [0, 1, 0]^{T} .$ That is $y_{n}$ = East. A $+ 9 0^{\circ}$ yaw swings the nose from North to East. ✓

L2.2

For $ϕ = 0, θ = 9 0^{\circ}, ψ = 0$ , compute the first row of $R_{n}^{b}$ (which tells us where nav-North lands in body coordinates through row·vector). Interpret.

Recall Solution

First row $= [c θ c ψ, c θ s ψ, - s θ]$ . With $θ = 9 0^{\circ}, ψ = 0$ : $c θ = 0, s θ = 1$ , so first row $= [0, 0, - 1]$ . Meaning: the body- $x$ axis (nose) is now $R_{13}$ -aligned with nav Down direction negated — the nose points straight up ( $- z_{n}$ ). A $9 0^{\circ}$ pitch-up stands the aircraft on its tail. ✓

L2.3

Given $θ = 2 0^{\circ}$ , what is $R_{13}$ ? Then invert: from $R_{13} = - 0.342$ recover $θ$ .

Recall Solution

$R_{13} = - sin 2 0^{\circ} = - 0.342$ . Inverting, $θ = - arcsin (- 0.342) = + 2 0^{\circ}$ . The minus signs cancel, confirming the recovery formula is self-consistent. ✓

Level 3 — Analysis

L3.1

Build the third row of $R_{n}^{b}$ for $ϕ = 3 0^{\circ}, θ = 2 0^{\circ}, ψ = 0$ , and verify it is a unit vector.

Recall Solution

Third row $= [c ϕ s θ c ψ + s ϕ s ψ, c ϕ s θ s ψ - s ϕ c ψ, c ϕ c θ]$ . With $ψ = 0$ ( $c ψ = 1, s ψ = 0$ ): $= [c ϕ s θ, - s ϕ, c ϕ c θ]$ . Numbers: $c 30 = 0.866, s 30 = 0.5, s 20 = 0.342, c 20 = 0.940$ : $= [0.866 \times 0.342, - 0.5, 0.866 \times 0.940] = [0.296, - 0.5, 0.814] .$ Magnitude: $0.29 6^{2} + 0. 5^{2} + 0.81 4^{2} = 0.0877 + 0.25 + 0.663 = 1.00 = 1.00$ . ✓ Interpretation: this row is the body-Down axis seen in nav — its downward component $0.814 < 1$ shows the belly is no longer pointing straight down; the craft is tilted.

L3.2

Recover all three Euler angles from $R_{13} = - 0.5, R_{23} = 0.433, R_{33} = 0.75, R_{12} = 0, R_{11} = 0.866.$

Recall Solution

$θ = - arcsin (- 0.5) = 3 0^{\circ}$ . Why: $R_{13} = - sin θ$ is uncontaminated.
$ϕ = atan2 (0.433, 0.75) = 3 0^{\circ}$ . Why: $R_{23} = s ϕ c θ$ and $R_{33} = c ϕ c θ$ share the factor $c θ$ ; their ratio is $tan ϕ$ , and atan2 uses both signs to nail the quadrant.
$ψ = atan2 (0, 0.866) = 0^{\circ}$ . Same trick with $R_{12} = c θ s ψ, R_{11} = c θ c ψ$ . ✓

L3.3

The DCM read-out uses $atan2 (R_{23}, R_{33})$ rather than $arctan (R_{23} / R_{33})$ . Give a concrete pair of $(ϕ)$ values where plain $arctan$ would give the wrong answer, and show atan2 fixes it. (See figure.)

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

Recall Solution

Take $c θ = 1$ for simplicity, so $R_{23} = sin ϕ, R_{33} = cos ϕ$ .

Case $ϕ = 4 5^{\circ}$ : $R_{23} = + 0.707, R_{33} = + 0.707$ . Ratio $= + 1$ , $arctan (1) = 4 5^{\circ}$ . Correct.
Case $ϕ = 22 5^{\circ}$ (equivalently $- 13 5^{\circ}$ ): $R_{23} = - 0.707, R_{33} = - 0.707$ . Ratio is still $+ 1$ , so $arctan (1) = 4 5^{\circ}$ — wrong by $18 0^{\circ}$ .
$atan2 (- 0.707, - 0.707) = - 13 5^{\circ}$ — correct, because atan2 sees both signs are negative and places the angle in the third quadrant.

Plain $arctan$ only ever returns answers in $(- 9 0^{\circ}, 9 0^{\circ})$ because $tan$ repeats every $18 0^{\circ}$ ; atan2 uses the two separate signs of numerator and denominator to recover the full $36 0^{\circ}$ range. The figure shows the two vectors landing in different quadrants despite an identical ratio.

Level 4 — Synthesis

L4.1

Set $θ = 9 0^{\circ}$ and show algebraically that the recovery formulas for $ϕ$ and $ψ$ both collapse to $0/0$ . Identify what physical thing has been lost. (See figure.)

Recall Solution

At $θ = 9 0^{\circ}$ : $c θ = 0, s θ = 1$ . Look at the four entries the read-out needs: $R_{11} = c θ c ψ = 0, R_{12} = c θ s ψ = 0, R_{23} = s ϕ c θ = 0, R_{33} = c ϕ c θ = 0.$ So $ϕ = atan2 (0, 0)$ and $ψ = atan2 (0, 0)$ — both undefined ( $0/0$ ). Physically, when the nose points straight up the roll axis and the yaw axis point the same way in space, so a roll and a yaw produce the same motion. Two knobs now do one job → one degree of freedom is lost. This is gimbal lock. The figure shows the roll and yaw circles becoming coincident. Practical fix: propagate attitude with quaternions or the Direction Cosine Matrix (DCM) directly, and only convert to Euler angles for human display away from $θ = \pm 9 0^{\circ}$ .

L4.2

Show that the composed matrix at gimbal lock ( $θ = 9 0^{\circ}$ ) depends only on the sum $ϕ + ψ$ , not on $ϕ$ and $ψ$ separately. Verify with $(ϕ, ψ) = (5 0^{\circ}, 1 0^{\circ})$ vs $(2 0^{\circ}, 4 0^{\circ})$ .

Recall Solution

At $θ = 9 0^{\circ}$ , plug $c θ = 0, s θ = 1$ into the lower-left $2 \times 2$ block of $R_{n}^{b}$ : $R_{21} = s ϕ c ψ - c ϕ s ψ = sin (ϕ - ψ), R_{22} = s ϕ s ψ + c ϕ c ψ = cos (ϕ - ψ),$ $R_{31} = c ϕ c ψ + s ϕ s ψ = cos (ϕ - ψ), R_{32} = c ϕ s ψ - s ϕ c ψ = - sin (ϕ - ψ) .$ Every non-trivial entry is a function of $ϕ - ψ$ only (the sign convention here makes the difference the free combination; the exact combination depends on convention, but the point stands: one combined angle, not two). Check: $(5 0^{\circ}, 1 0^{\circ}) \Rightarrow ϕ - ψ = 4 0^{\circ}$ ; $(2 0^{\circ}, 4 0^{\circ}) \Rightarrow ϕ - ψ = - 2 0^{\circ}$ — these give different matrices, but any other pair with the same difference (e.g. $(6 0^{\circ}, 2 0^{\circ})$ , difference $4 0^{\circ}$ ) gives the same matrix as $(5 0^{\circ}, 1 0^{\circ})$ . That collapse from two independent numbers to one is exactly the lost degree of freedom. ✓

Level 5 — Mastery

L5.1

A UAV holds $ϕ = 0, ψ = 0$ and pitches slowly from $θ = 8 0^{\circ}$ up through $9 0^{\circ}$ to $10 0^{\circ}$ . Explain what the recovered Euler yaw $ψ$ does numerically as it crosses $9 0^{\circ}$ , and why an attitude estimator running on Euler angles would misbehave here. Then state the design fix.

Recall Solution

As $θ \to 9 0^{\circ}$ , the entries $R_{11}, R_{12}$ shrink toward $0$ while any tiny sensor noise remains. So $ψ = atan2 (R_{12}, R_{11})$ divides small noisy numbers by small noisy numbers — the recovered yaw becomes wildly sensitive and can jump by up to $18 0^{\circ}$ in a single step as $θ$ passes through $9 0^{\circ}$ (the frame effectively "flips" which way it is looking). A filter integrating $\dot{ψ}$ would see an enormous spurious rate and inject a false correction. Why this happens: the map from orientation to $(ϕ, θ, ψ)$ has a coordinate singularity at $θ = \pm 9 0^{\circ}$ (the Jacobian of the Euler-rate relation blows up because it contains a $1/ cos θ$ term — the same $c θ$ that cancelled harmlessly in L3 now sits in a denominator for the rates). Near the singularity, small state changes demand infinite angle changes. Design fix: run the estimator's internal state as a quaternion or a DCM (both singularity-free — see Quaternions — avoiding gimbal lock, Direction Cosine Matrix (DCM)), propagate using body rates $p, q, r$ from Angular velocity & body rates p, q, r, and only convert to $(ϕ, θ, ψ)$ for the pilot's display, avoiding the display update exactly at $θ = \pm 9 0^{\circ}$ .

L5.2

Prove that $R_{n}^{b}$ is orthogonal (i.e. $(R_{n}^{b})^{T} R_{n}^{b} = I$ ) using the fact that each elementary rotation is orthogonal — without multiplying out the full $3 \times 3$ .

Recall Solution

Each elementary matrix satisfies $R_{a}^{T} R_{a} = I$ (its columns are perpendicular unit vectors — the definition of a rotation, see Rotation matrices & orthogonality). For the product $R_{n}^{b} = R_{x} R_{y} R_{z}$ : $(R_{x} R_{y} R_{z})^{T} (R_{x} R_{y} R_{z}) = R_{z}^{T} R_{y}^{T} R_{x}^{T} R_{x} R_{y} R_{z} .$ Now peel from the inside: $R_{x}^{T} R_{x} = I$ , leaving $R_{z}^{T} R_{y}^{T} R_{y} R_{z}$ ; then $R_{y}^{T} R_{y} = I$ , leaving $R_{z}^{T} R_{z} = I$ . Hence $(R_{n}^{b})^{T} R_{n}^{b} = I$ : the product of orthogonal matrices is orthogonal. Why it matters: this is why the inverse is just the transpose, which we used freely in L2.1 to map body → nav. ✓

Connections

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)