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

3.5.5 · D3 · Physics › Guidance, Navigation & Control (GNC) › Gimbal Lock — Euler angles mein θ = ±90° par problem

Parent note ne bataya kyun gimbal lock hota hai. Yeh page tumhe fluent banata hai har us situation ko walk-through karke jo arise ho sakti hai — pitch ki har sign, zero case, exact singularity, near-singularity, ek real aircraft, aur ek exam-style trap.

Symbols se pehle: parent note ke teen dials yaad karo.

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

Hum Euler angles se yeh do maps use karte rahenge:

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

jahan $c_{α} = cos α$ , $s_{α} = sin α$ , $t_{α} = tan α$ . Matrix $E$ hai rate map: yeh real body spin ko convert karta hai is mein ki tere teen dials kitni tezi se ghoomne chahiye.

Scenario matrix

Cell	Kya test karta hai	Example
A. Generic / safe	$θ$ poles se door, saare dials independent	Ex 1
B. Zero / identity	saare angles $0$ — degenerate par harmless	Ex 2
C. Exact lock $θ = + 90°$	matrix collapse ho ke $(ϕ - ψ)$ ban jaata hai	Ex 3
D. Exact lock $θ = - 90°$	matrix collapse ho ke $(ϕ + ψ)$ ban jaata hai (opposite sign!)	Ex 4
E. Near-lock rate spike	$1/ cos θ$ tiny rates ko amplify karta hai	Ex 5
F. Non-unique inverse	kaafi $(ψ, ϕ)$ → ek orientation	Ex 6
G. Real-world word problem	aircraft loop zenith cross karta hua	Ex 7
H. Exam twist: quaternion escape	dikhao quaternion finite rehta hai jahan Euler explode karta hai	Ex 8

Yeh saath milke dono signs of the pole, zero case, ek generic case, limit jab $θ \to 90°$ , uniqueness, ek real vehicle, aur contrasting fix ko cover karte hain.

Example 1 — Cell A: ek safe, generic orientation

Worked example Statement

Ek drone $ψ = 30°$ , $θ = 20°$ , $ϕ = 10°$ ke saath body rates $p = 0.5$ , $q = 0.2$ , $r = 0.1$ rad/s hold karta hai. Kya Euler rates finite aur well-behaved hain?

Forecast: Pitch $\pm 90°$ ke kahin paas nahi hai, toh guess: sab kuch finite, koi drama nahi.

Danger term check karo. $cos θ = cos 20° \approx 0.9397$ . Yeh step kyun? Sirf woh cheez jo blow up kar sakti hai woh hai $1/ cos θ$ ; agar $cos θ$ comfortably non-zero hai, toh hum safe hain.
$\dot{θ}$ compute karo (middle row, koi division nahi): $\dot{θ} = c_{ϕ} q - s_{ϕ} r = cos 10° (0.2) - sin 10° (0.1)$ . $\approx 0.9848 (0.2) - 0.1736 (0.1) = 0.1970 - 0.01736 = 0.1796$ rad/s. Yeh step kyun? Pitch-rate kabhi $cos θ$ se divide nahi karta, isliye yeh "clean" component hai — ek accha baseline.
$\dot{ψ}$ compute karo (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. Yeh step kyun? Is row mein singular factor hai; hum confirm karte hain ki yeh ek modest number hai, blow-up nahi.

Verify: Teeno rates order $0.1$ – $0.2$ rad/s ki hain — body rates ke jaisi magnitude. Sensible hai. Units: (dimensionless) $\times$ (rad/s) = rad/s. ✓

Example 2 — Cell B: zero / identity case

Worked example Statement

Saare angles zero: $ψ = θ = ϕ = 0$ . Body rates $(p, q, r) = (0.3, 0.4, 0.5)$ . $R$ kya hai, aur $E$ kya hai?

Forecast: Zero angles = "kuch mat karo" orientation. Guess: $R$ identity hai aur $E$ identity hai, toh Euler rates body rates ke barabar hain.

$R$ banao. Angle $0$ par har elemental matrix identity hai, aur identity times identity identity hai: $R = I_{3}$ . Yeh step kyun? Confirm karta hai ki zero angles ek perfectly valid, non-singular orientation hai — value mein degenerate hai par broken nahi.
$E (0, 0)$ evaluate karo. $s_{ϕ} = 0, c_{ϕ} = 1, t_{θ} = 0, c_{θ} = 1$ ke saath: $E (0, 0) = 100010001 = I_{3} .$ Yeh step kyun? Origin par teen dials teen body axes ke saath exactly line up karte hain, isliye koi mixing nahi hoti.
Rates padho. $(\dot{ϕ}, \dot{θ}, \dot{ψ}) = (p, q, r) = (0.3, 0.4, 0.5)$ . Yeh step kyun? "Best case" dikhata hai: Euler rates body rates ke saath ek-ke-baad-ek equal hain.

Verify: $det R = 1$ (ek proper rotation, Rotation matrices SO(3) ka part). $cos 0 = 1 \neq = 0$ , toh koi singularity nahi. ✓

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

Worked example Statement

$θ = + 90°$ set karo, $ψ, ϕ$ symbolic rakho. Dikhao ki matrix sirf $(ϕ - ψ)$ par depend karta hai, aur $(ψ, ϕ) = (30°, 0°)$ ke saath check karo.

Forecast: Parent note ne $(ϕ - ψ)$ par collapse ka waada kiya tha. Guess: yaw aur roll ek difference mein fuse ho jaate hain.

$R_{y} (90°)$ insert karo. $c_{θ} = 0, s_{θ} = 1$ ke saath, $R_{y} (90°) = 00 - 1 010100$ . Yeh step kyun? Hum suspected singular value ko seedha test karte hain.
Multiply out karo $R_{z} (ψ) R_{y} (90°) R_{x} (ϕ)$ aur parent ka collapsed form pao: $R = 00 - 1 sin (ϕ - ψ) cos (ϕ - ψ) 0 cos (ϕ - ψ) - sin (ϕ - ψ) 0 .$ Yeh step kyun? Har free entry ek single number $ϕ - ψ$ ka function hai — wahi hai lost degree of freedom visible hua.
$(ψ, ϕ) = (30°, 0°)$ plug karo, toh $ϕ - ψ = - 30°$ . Phir top-middle entry $= sin (- 30°) = - 0.5$ , top-right $= cos (- 30°) \approx 0.8660$ . Yeh step kyun? Abstract collapse ko ek concrete number mein badalta hai jise hum machine-check kar sakte hain.

Verify: Ex 6 mein hum usi matrix par pahunchte hain ek alag $(ψ, ϕ)$ se, jo collapse prove karta hai. $det R = 1$ abhi bhi hai. ✓

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

Worked example Statement

Ab nose-down nadir ki taraf: $θ = - 90°$ . Dikhao ki collapse combination sign flip karke $(ϕ + ψ)$ ho jaata hai.

Forecast: Symmetry se guess karo ki combination $ϕ + ψ$ ban jaata hai, $ϕ - ψ$ nahi. Sign matter karta hai!

$R_{y} (- 90°)$ insert karo. Ab $c_{θ} = 0, s_{θ} = - 1$ , jo $R_{y} (- 90°) = 001010 - 1 00$ deta hai. Yeh step kyun? $- 1$ jahan pehle $+ 1$ tha woh sign flip ka seed hai.
Multiply out karo. Result hai $R = 001 - sin (ϕ + ψ) cos (ϕ + ψ) 0 - cos (ϕ + ψ) - sin (ϕ + ψ) 0 .$ Yeh step kyun? Confirm karta hai ki dono poles collapse hote hain — tum doosra pole choose karke lock se nahi bach sakte; sirf change hota hai ki kaun sa combination survive karta hai.
$(ψ, ϕ) = (20°, 10°)$ ke saath sanity check: $ϕ + ψ = 30°$ , toh top-middle $= - sin 30° = - 0.5$ , top-right $= - cos 30° \approx - 0.8660$ . Yeh step kyun? $(ϕ + ψ)$ dependence ko concrete aur checkable banata hai.

Verify: Ex 3 ( $ϕ - ψ$ use karta hai) aur is example ( $ϕ + ψ$ use karta hai) ko compare karo. Dono singular hain, opposite combinations. ✓

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

Worked example Statement

Body rate $r = 0.01$ rad/s (bahut halka), $ϕ = 0$ , $p = q = 0$ . $θ = 85°$ , $89°$ , $89.9°$ , $89.99°$ par $\dot{ψ}$ compute karo. Use diverge hote dekho.

Forecast: Jaise $θ \to 90°$ , $cos θ \to 0$ , toh $\dot{ψ} = r / cos θ$ upar ki taraf chilla-na chahiye. Guess: roughly $\times 10$ har baar jab $θ$ $90°$ ke $10 \times$ zyada paas ho jaata hai.

Formula isolate karo. $ϕ = 0$ ke saath: $\dot{ψ} = \frac{c _{ϕ}}{c _{θ}} r = \frac{r}{cos θ}$ . Yeh step kyun? Singular division ke siwa sab kuch hat jaata hai taaki blow-up bilkul nanga ho.
Tabulate karo:
- $θ = 85°$ : $cos 85° = 0.08716 \Rightarrow \dot{ψ} = 0.01/0.08716 \approx 0.1147$ 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. Yeh step kyun? Numbers prove karte hain ki yeh ek region hai, ek point nahi — $85°$ par bhi rate amplified hai, aur yeh unbounded grow karta hai.
Limit padho. Jaise $θ \to 90°$ , kisi bhi $r \neq = 0$ ke liye $\dot{ψ} \to \infty$ . Yeh step kyun? Math demand karta hai infinite dial-speed ek finite, gentle physical turn represent karne ke liye — coordinate chart singular hai (Singularities of coordinate charts).

Verify: $90°$ ke har tenfold approach par $\dot{ψ}$ $\approx 10$ se multiply hota hai — exactly $1/ cos θ$ behaviour. Physical rate $r$ poore time $0.01$ raha; sirf representation explode hua. ✓

Example 6 — Cell F: inverse map unique nahi hai

Worked example Statement

$θ = + 90°$ par, $(ψ, ϕ) = (50°, 20°)$ lo. Dikhao ki yeh Ex 3 ke $(30°, 0°)$ jaisa SAME $R$ deta hai.

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

Combination compute karo. $ϕ - ψ = 20° - 50° = - 30°$ . Yeh step kyun? Sirf yahi difference collapsed matrix ko feed karta hai, toh equal differences equal matrices dene chahiye.
Entries fill karo. $sin (- 30°) = - 0.5$ , $cos (- 30°) \approx 0.8660$ — Ex 3 ke identical. Yeh step kyun? Confirm karta hai ki do alag dial settings ek orientation par map hote hain.
Interpret karo. Lock par orientation $\to$ Euler map many-to-one hai, toh koi bhi attitude estimator (Attitude determination and control) ek unique $(ψ, ϕ)$ recover nahi kar sakta. Yeh step kyun? Yeh estimation catastrophe hai, sirf cosmetic nahi.

Verify: Ex 3 aur Ex 6 ke matrices element-wise equal hain. ✓

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

Worked example Statement

Ek jet vertical loop fly karta hai. Climb ke top par uska nose straight-up ( $θ = 90°$ ) se guzarta hai jabki $q = 1$ rad/s (pitch-rate) par roll ho raha hai aur ek crosswind gust se $r = 0.02$ rad/s yaw ho raha hai, $ϕ = 0$ ke saath. Onboard Euler-angle filter $θ = 89.95°$ par kaun sa Euler pitch-rate aur yaw-rate report karta hai?

Forecast: Pitch-rate clean middle row use karta hai (theek hai); yaw-rate $1/ cos θ$ use karta hai (spike). Guess: $\dot{θ} \approx 1$ , $\dot{ψ}$ bahut bada.

Middle row (pitch-rate). $\dot{θ} = c_{ϕ} q - s_{ϕ} r = cos 0 (1) - sin 0 (0.02) = 1$ rad/s. Yeh step kyun? Confirm karta hai ki physically dominant rotation faithfully report hoti hai — filter har jagah galat nahi hai.
Bottom row (yaw-rate). $cos 89.95° = 0.0008727$ , toh $\dot{ψ} = \frac{c _{ϕ}}{c _{θ}} r = \frac{0.02}{0.0008727} \approx 22.92$ rad/s. Yeh step kyun? $0.02$ rad/s ka ek gust $22.9$ rad/s reported yaw-rate ban jaata hai — ek aircraft ke liye bilkul unphysical ( $> 1300°$ /s).
Engineering consequence. Integrator heading wildly overshoot kar deta hai; Apollo Guidance Computer ne famously design se aisi lock ke aas-paas kaam kiya. Yeh step kyun? Abstract spike ko ek mission-critical failure mode se jodhta hai.

Verify: $\dot{θ} = 1$ rad/s finite aur sane hai; $\dot{ψ} \approx 22.9$ rad/s $0.02$ rad/s cause se $\approx 1146 \times = 1/ cos 89.95°$ se bada hai. ✓

Example 8 — Cell H: exam twist — quaternion finite rehta hai

Worked example Statement

Ex 7 jaisa hi physical state ( $θ = 89.95°$ , tiny gust). Ek quaternion filter $\overset{q}{˙} = \frac{1}{2} q \otimes ω$ propagate karta hai $ω = (p, q, r) = (0, 1, 0.02)$ ke saath. Kya quaternion rate finite hai jahan Euler rate explode hua?

Forecast: Quaternions mein kahin bhi $1/ cos θ$ nahi hai. Guess: rate body rate ke order ka rehta hai — completely finite.

Rate bound karo. Quaternion derivative ki magnitude $∣ \overset{q}{˙} ∣ = \frac{1}{2} ∣ q ∣∣ ω ∣$ hai, aur ek unit quaternion mein $∣ q ∣ = 1$ hai. Yeh step kyun? Kisi bhi coordinate term se division nahi dikhta — yeh $E$ se structural difference hai.
Plug in karo. $∣ ω ∣ = 0^{2} + 1^{2} + 0.0 2^{2} = 1.0004 \approx 1.0002$ rad/s, toh $∣ \overset{q}{˙} ∣ \approx \frac{1}{2} (1) (1.0002) \approx 0.5001$ per second. Yeh step kyun? Same physical state, par quaternion rate ek tame $\approx 0.5$ hai — jabki Ex 7 ka $\dot{ψ}$ $\approx 22.9$ tha.
Conclude karo. Physics ( $ω$ ) hamesha finite tha; sirf Euler chart singular tha. Charts ko quaternions mein change karo aur "explosion" gayab ho jaata hai. Yeh step kyun? Yeh chapter ka poora moral hai, numerically side-by-side prove hua.

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

Recall Har example kaun sa cell cover karta hai?

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 — woh representation jin par yeh examples stress-test karte hain
Quaternions — Ex 8 mein finite-rate escape
Rotation matrices SO(3) — woh group jiska $det = 1$ humne check kiya
Angular velocity kinematics — rate map $E$ jo har example use karta hai
Attitude determination and control — jahan non-unique inverse (Ex 6) bite karta hai
Apollo Guidance Computer — Ex 7 mein referenced historical work-around
Singularities of coordinate charts — deeper reason ki chart fail hota hai, physics nahi