Gimbal lock — problem with Euler angles at θ = ±90°: Worked examples

The parent note showed you why gimbal lock happens. This page makes you fluent by walking through every kind of situation that can arise — every sign of pitch, the zero case, the exact singularity, the near-singularity, a real aircraft, and an exam-style trap.

Before symbols: recall the three dials from the parent note.

$ψ$ (yaw) — turn left/right about the world vertical $z$ .
$θ$ (pitch) — nose up/down about the tilted $y$ .
$ϕ$ (roll) — barrel-roll about the nose $x$ .
$(p, q, r)$ — the physical body spin rates the gyroscopes actually measure (roll-rate, pitch-rate, yaw-rate), in radians per second.

We will keep using the two maps from Euler angles:

R = R_{z} (ψ) R_{y} (θ) R_{x} (ϕ), \dot{ϕ} \dot{θ} \dot{ψ} = call this E (ϕ, θ) 100 s_{ϕ} t_{θ} c_{ϕ} \frac{s _{ϕ}}{c _{θ}} c_{ϕ} t_{θ} - s_{ϕ} \frac{c _{ϕ}}{c _{θ}} p q r

where $c_{α} = cos α$ , $s_{α} = sin α$ , $t_{α} = tan α$ . The matrix $E$ is the rate map: it converts real body spin into how fast your three dials must turn.

The scenario matrix

Cell	What it tests	Hit by
A. Generic / safe	$θ$ far from poles, all dials independent	Ex 1
B. Zero / identity	all angles $0$ — degenerate but harmless	Ex 2
C. Exact lock $θ = + 90°$	matrix collapses to $(ϕ - ψ)$	Ex 3
D. Exact lock $θ = - 90°$	matrix collapses to $(ϕ + ψ)$ (opposite sign!)	Ex 4
E. Near-lock rate spike	$1/ cos θ$ amplifies tiny rates	Ex 5
F. Non-unique inverse	many $(ψ, ϕ)$ → one orientation	Ex 6
G. Real-world word problem	aircraft loop crossing zenith	Ex 7
H. Exam twist: quaternion escape	show quaternion stays finite where Euler explodes	Ex 8

Together these hit both signs of the pole, the zero case, a generic case, the limit as $θ \to 90°$ , uniqueness, a real vehicle, and the contrasting fix.

Example 1 — Cell A: a safe, generic orientation

Worked example Statement

A drone holds $ψ = 30°$ , $θ = 20°$ , $ϕ = 10°$ with body rates $p = 0.5$ , $q = 0.2$ , $r = 0.1$ rad/s. Are the Euler rates finite and well-behaved?

Forecast: Pitch is nowhere near $\pm 90°$ , so guess: everything finite, no drama.

Check the danger term. $cos θ = cos 20° \approx 0.9397$ . Why this step? The only thing that can blow up is $1/ cos θ$ ; if $cos θ$ is comfortably non-zero, we are safe.
Compute $\dot{θ}$ (middle row, no division): $\dot{θ} = c_{ϕ} q - s_{ϕ} r = cos 10° (0.2) - sin 10° (0.1)$ . $\approx 0.9848 (0.2) - 0.1736 (0.1) = 0.1970 - 0.01736 = 0.1796$ rad/s. Why this step? Pitch-rate never divides by $cos θ$ , so it is the "clean" component — a good baseline.
Compute $\dot{ψ}$ (bottom row): $\dot{ψ} = \frac{s _{ϕ}}{c _{θ}} q + \frac{c _{ϕ}}{c _{θ}} r = \frac{sin 10° ( 0.2 ) + cos 10° ( 0.1 )}{cos 20°}$ . $= \frac{0.1736 ( 0.2 ) + 0.9848 ( 0.1 )}{0.9397} = \frac{0.03473 + 0.09848}{0.9397} \approx 0.1418$ rad/s. Why this step? This row carries the singular factor; we confirm it is a modest number, not a blow-up.

Verify: All three rates are of order $0.1$ – $0.2$ rad/s — same magnitude as the body rates. Sensible. Units: (dimensionless) $\times$ (rad/s) = rad/s. ✓

Example 2 — Cell B: the zero / identity case

Worked example Statement

All angles zero: $ψ = θ = ϕ = 0$ . Body rates $(p, q, r) = (0.3, 0.4, 0.5)$ . What is $R$ , and what is $E$ ?

Forecast: Zero angles = "do nothing" orientation. Guess: $R$ is the identity and $E$ is the identity, so Euler rates just equal body rates.

Build $R$ . Each elemental matrix at angle $0$ is the identity, and identity times identity is identity: $R = I_{3}$ . Why this step? Confirms zero angles are a perfectly valid, non-singular orientation — degenerate in value but not broken.
Evaluate $E (0, 0)$ . With $s_{ϕ} = 0, c_{ϕ} = 1, t_{θ} = 0, c_{θ} = 1$ : $E (0, 0) = 100010001 = I_{3} .$ Why this step? At the origin the three dials line up with the three body axes exactly, so no mixing happens.
Read the rates. $(\dot{ϕ}, \dot{θ}, \dot{ψ}) = (p, q, r) = (0.3, 0.4, 0.5)$ . Why this step? Shows the "best case": Euler rates equal body rates one-for-one.

Verify: $det R = 1$ (a proper rotation, part of Rotation matrices SO(3)). $cos 0 = 1 \neq = 0$ , so no singularity. ✓

Example 3 — Cell C: exact lock at $θ = + 90°$

Worked example Statement

Set $θ = + 90°$ , keep $ψ, ϕ$ symbolic. Show the matrix depends only on $(ϕ - ψ)$ , and check with $(ψ, ϕ) = (30°, 0°)$ .

Forecast: Parent note promised the collapse to $(ϕ - ψ)$ . Guess: yaw and roll fuse into one difference.

Insert $R_{y} (90°)$ . With $c_{θ} = 0, s_{θ} = 1$ , $R_{y} (90°) = 00 - 1 010100$ . Why this step? We test the suspected singular value head-on.
Multiply out $R_{z} (ψ) R_{y} (90°) R_{x} (ϕ)$ to get the parent's collapsed form: $R = 00 - 1 sin (ϕ - ψ) cos (ϕ - ψ) 0 cos (ϕ - ψ) - sin (ϕ - ψ) 0 .$ Why this step? Every free entry is a function of the single number $ϕ - ψ$ — that IS the lost degree of freedom made visible.
Plug $(ψ, ϕ) = (30°, 0°)$ , so $ϕ - ψ = - 30°$ . Then top-middle entry $= sin (- 30°) = - 0.5$ , top-right $= cos (- 30°) \approx 0.8660$ . Why this step? Turns the abstract collapse into a concrete number we can machine-check.

Verify: In Ex 6 we reach the same matrix from a different $(ψ, ϕ)$ , proving the collapse. $det R = 1$ still. ✓

Example 4 — Cell D: the OTHER pole, $θ = - 90°$

Worked example Statement

Now nose-down to the nadir: $θ = - 90°$ . Show the collapse combination flips sign to $(ϕ + ψ)$ .

Forecast: By symmetry guess the combination becomes $ϕ + ψ$ , not $ϕ - ψ$ . Sign matters!

Insert $R_{y} (- 90°)$ . Now $c_{θ} = 0, s_{θ} = - 1$ , giving $R_{y} (- 90°) = 001010 - 1 00$ . Why this step? The $- 1$ where before we had $+ 1$ is the seed of the sign flip.
Multiply out. The result is $R = 001 - sin (ϕ + ψ) cos (ϕ + ψ) 0 - cos (ϕ + ψ) - sin (ϕ + ψ) 0 .$ Why this step? Confirms both poles collapse — you cannot escape lock by picking the other pole; you only change which combination survives.
Sanity with $(ψ, ϕ) = (20°, 10°)$ : $ϕ + ψ = 30°$ , so top-middle $= - sin 30° = - 0.5$ , top-right $= - cos 30° \approx - 0.8660$ . Why this step? Makes the $(ϕ + ψ)$ dependence concrete and checkable.

Verify: Compare Ex 3 (uses $ϕ - ψ$ ) with this (uses $ϕ + ψ$ ). Both are singular, opposite combinations. ✓

Example 5 — Cell E: the near-lock rate spike (the limit)

Worked example Statement

Body rate $r = 0.01$ rad/s (a whisper), $ϕ = 0$ , $p = q = 0$ . Compute $\dot{ψ}$ at $θ = 85°$ , $89°$ , $89.9°$ , $89.99°$ . Watch it diverge.

Forecast: As $θ \to 90°$ , $cos θ \to 0$ , so $\dot{ψ} = r / cos θ$ should scream upward. Guess: it roughly $\times 10$ each time $θ$ gets $10 \times$ closer to $90°$ .

Isolate the formula. With $ϕ = 0$ : $\dot{ψ} = \frac{c _{ϕ}}{c _{θ}} r = \frac{r}{cos θ}$ . Why this step? Strips away everything but the singular division so the blow-up is naked.
Tabulate:
- $θ = 85°$ : $cos 85° = 0.08716 \Rightarrow \dot{ψ} = 0.01/0.08716 \approx 0.1147$ rad/s.
- $θ = 89°$ : $cos 89° = 0.01745 \Rightarrow \dot{ψ} \approx 0.5730$ rad/s.
- $θ = 89.9°$ : $cos 89.9° = 0.001745 \Rightarrow \dot{ψ} \approx 5.730$ rad/s.
- $θ = 89.99°$ : $cos 89.99° = 0.0001745 \Rightarrow \dot{ψ} \approx 57.30$ rad/s. Why this step? Numbers prove it is a region, not a point — already at $85°$ the rate is amplified, and it grows without bound.
Read the limit. As $θ \to 90°$ , $\dot{ψ} \to \infty$ for any $r \neq = 0$ . Why this step? The math demands infinite dial-speed to represent a finite, gentle physical turn — the coordinate chart is singular (Singularities of coordinate charts).

Verify: Each tenfold approach to $90°$ multiplies $\dot{ψ}$ by $\approx 10$ — exactly $1/ cos θ$ behaviour. Physical rate $r$ stayed $0.01$ throughout; only the representation exploded. ✓

Example 6 — Cell F: the inverse map is not unique

Worked example Statement

At $θ = + 90°$ , take $(ψ, ϕ) = (50°, 20°)$ . Show it yields the SAME $R$ as Ex 3's $(30°, 0°)$ .

Forecast: $ϕ - ψ = 20 - 50 = - 30°$ , identical to Ex 3. Guess: exact same matrix.

Compute the combination. $ϕ - ψ = 20° - 50° = - 30°$ . Why this step? Only this difference feeds the collapsed matrix, so equal differences must give equal matrices.
Fill the entries. $sin (- 30°) = - 0.5$ , $cos (- 30°) \approx 0.8660$ — identical to Ex 3. Why this step? Confirms two different dial settings map to one orientation.
Interpret. The map orientation $\to$ Euler is many-to-one at lock, so no attitude estimator (Attitude determination and control) can recover a unique $(ψ, ϕ)$ . Why this step? This is the estimation catastrophe, not just cosmetic.

Verify: Matrices from Ex 3 and Ex 6 are element-wise equal. ✓

Example 7 — Cell G: real-world word problem (aircraft loop)

Worked example Statement

A jet flies a vertical loop. At the top of the climb its nose passes through straight-up ( $θ = 90°$ ) while rolling at $q = 1$ rad/s (pitch-rate) and yawing at $r = 0.02$ rad/s from a crosswind gust, with $ϕ = 0$ . What Euler pitch-rate and yaw-rate does the onboard Euler-angle filter report at $θ = 89.95°$ ?

Forecast: Pitch-rate uses the clean middle row (fine); yaw-rate uses $1/ cos θ$ (spike). Guess: $\dot{θ} \approx 1$ , $\dot{ψ}$ huge.

Middle row (pitch-rate). $\dot{θ} = c_{ϕ} q - s_{ϕ} r = cos 0 (1) - sin 0 (0.02) = 1$ rad/s. Why this step? Confirms the physically dominant rotation is reported faithfully — the filter isn't wrong everywhere.
Bottom row (yaw-rate). $cos 89.95° = 0.0008727$ , so $\dot{ψ} = \frac{c _{ϕ}}{c _{θ}} r = \frac{0.02}{0.0008727} \approx 22.92$ rad/s. Why this step? A $0.02$ rad/s gust becomes a $22.9$ rad/s reported yaw-rate — utterly unphysical for an aircraft (that's $> 1300°$ /s).
Engineering consequence. The integrator overshoots heading wildly; the Apollo Guidance Computer famously worked around such lock by design. Why this step? Ties the abstract spike to a mission-critical failure mode.

Verify: $\dot{θ} = 1$ rad/s is finite and sane; $\dot{ψ} \approx 22.9$ rad/s dwarfs the $0.02$ rad/s cause by $\approx 1146 \times = 1/ cos 89.95°$ . ✓

Example 8 — Cell H: exam twist — the quaternion stays finite

Worked example Statement

Same physical state as Ex 7 ( $θ = 89.95°$ , tiny gust). A quaternion filter propagates $\overset{q}{˙} = \frac{1}{2} q \otimes ω$ with $ω = (p, q, r) = (0, 1, 0.02)$ . Is the quaternion rate finite where the Euler rate exploded?

Forecast: Quaternions have no $1/ cos θ$ anywhere. Guess: rate stays of order the body rate — totally finite.

Bound the rate. The magnitude of the quaternion derivative is $∣ \overset{q}{˙} ∣ = \frac{1}{2} ∣ q ∣∣ ω ∣$ , and a unit quaternion has $∣ q ∣ = 1$ . Why this step? No division by any coordinate term appears — that's the structural difference from $E$ .
Plug in. $∣ ω ∣ = 0^{2} + 1^{2} + 0.0 2^{2} = 1.0004 \approx 1.0002$ rad/s, so $∣ \overset{q}{˙} ∣ \approx \frac{1}{2} (1) (1.0002) \approx 0.5001$ per second. Why this step? Same physical state, but the quaternion rate is a tame $\approx 0.5$ — while Ex 7's $\dot{ψ}$ was $\approx 22.9$ .
Conclude. The physics ( $ω$ ) was always finite; only the Euler chart was singular. Change charts to quaternions and the "explosion" disappears. Why this step? This is the whole moral of the chapter, proven numerically side-by-side.

Verify: $∣ \overset{q}{˙} ∣ \approx 0.5$ per second (finite) vs Euler $\dot{ψ} \approx 22.9$ rad/s (spiking) for the identical body rate. ✓

Recall Which cell does each example cover?

Generic safe case ::: Ex 1 (Cell A) Zero / identity ::: Ex 2 (Cell B) Exact lock $+ 90°$ collapse to $(ϕ - ψ)$ ::: Ex 3 (Cell C) Exact lock $- 90°$ collapse to $(ϕ + ψ)$ ::: Ex 4 (Cell D) Near-lock rate spike / limit ::: Ex 5 (Cell E) Non-unique inverse ::: Ex 6 (Cell F) Real aircraft loop ::: Ex 7 (Cell G) Quaternion escape ::: Ex 8 (Cell H)

Connections

Hinglish version →
Euler angles — the representation these examples stress-test
Quaternions — the finite-rate escape in Ex 8
Rotation matrices SO(3) — the group whose $det = 1$ we checked
Angular velocity kinematics — the rate map $E$ every example uses
Attitude determination and control — where the non-unique inverse (Ex 6) bites
Apollo Guidance Computer — historical work-around referenced in Ex 7
Singularities of coordinate charts — the deeper reason the chart, not the physics, fails