Exercises — Gimbal lock — problem with Euler angles at θ = ±90°

KYA: Hum defining configuration yaad karte hain. KYUN: Gimbal lock woh pitch value hai jahan roll axis, yaw axis par aa jaata hai. Middle rotation $R_y(\theta )$ last ($x$) axis ko first ($z$) axis ke relative mein tilt karta hai. Yeh tab poori tarah overlap karta hai jab tilt ek quarter turn hoti hai: $\theta =+90°$ ya $\theta =-90°$. Answer: $\theta =\pm 90°$. Yaw $\psi$ aur roll $\varphi$ redundant ho jaate hain (yeh ek effective angle mein combine ho jaate hain); pitch $\theta$ abhi bhi meaningful hai.

KYA: Parent note se kinematic matrix padhte hain. KYUN: $\dot{\varphi }$ aur $\dot{\psi }$ ko feed karne wali entries singular functions carry karti hain. Khatarnak terms hain $\tan \theta$ aur $\frac{1}{\cos \theta }$. Jaise hi $\theta \to 90°$, $\cos \theta \to 0^{+}$, toh dono $\to +\infty$. Answer: $\tan \theta \to +\infty$ aur $1/\cos \theta \to +\infty$.

KYA: Kinematic map ki yaw-rate row mein plug in karte hain. KYUN: Hum feel karna chahte hain ki lock ke paas coordinate rate kitni tezi se diverge hoti hai. Relevant row hai $\dot{\psi }=\frac{\sin \varphi }{\cos \theta }q+\frac{\cos \varphi }{\cos \theta }r$. $\varphi =0$ ke saath: $\sin \varphi =0,\cos \varphi =1$, toh $\dot{\psi }=\frac{r}{\cos \theta }.$ $89.9°$ par: $\cos (89.9°)=1.7453\times 10^{-3}$, toh $\dot{\psi }=0.01/1.7453\times 10^{-3}\approx 5.73\text{rad/s}$. $89.99°$ par: $\cos (89.99°)=1.7453\times 10^{-4}$, toh $\dot{\psi }=0.01/1.7453\times 10^{-4}\approx 57.3\text{rad/s}$. Answer: $\approx 5.73$ phir $\approx 57.3\text{rad/s}$; $90°$ se 10× zyada kareeb aane par 10× ka jump. Ek finite $r$ ek exploding $\dot{\psi }$ maangta hai.

KYA: Collapse rule use karo: "$R$ sirf $\varphi -\psi$ par depend karta hai $\theta =+90°$ par." KYUN: Koi bhi pair jiska same difference $\varphi -\psi$ ho woh same matrix dega. Yahan $\varphi -\psi =40°-10°=30°$. Koi bhi $(\psi ,\varphi )$ lo jinka difference $30°$ ho, jaise $(\psi ,\varphi )=(0°,30°)$ ya $(50°,80°)$. Answer: jaise $(\psi ,\varphi )=(0°,30°)$ — $(10°,40°)$ jaisa hi $R$.

KYA: $c_{\theta }=0,s_{\theta }=-1$ set karo aur $R_z(\psi )R_y(-90°)R_x(\varphi )$ multiply karo. KYUN: Hum doosre singular pitch ko directly test karte hain, symmetry par blindly trust karne ke bajaye. Pehle, $R_y(-90°)=\left(\begin{array}{ccc}0& 0& -1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right)$. Multiply out karne par (parent ke $+90°$ case jaisi procedure) milta hai

KAISA DIKHTA HAI: har nonzero off-column entry single combination $\varphi +\psi$ ka function hai. Answer: $\psi$ ko $\delta$ badhane aur $\varphi$ ko $\delta$ ghatane se $R$ unchanged rehta hai ⇒ effective DOF hai $\varphi +\psi$. Confirmed.

KYA: Compare karo ki $\pm 90°$ pitch ke baad roll axis kahan jaata hai. KYUN: Combination ka sign is baat se set hota hai ki roll axis yaw axis ke saath parallel land karta hai ya anti-parallel. KAISA DIKHTA HAI: Figure mein, $+90°$ pitch karne se body $x$-axis $+z$ direction mein point karne lagta hai, toh same size ka yaw aur roll same directed axis ke baare mein co-rotations ki tarah add up hote hain — lekin yaw tilt se pehle apply hota hai aur roll baad mein, aur beech ka flip ek sense reverse karta hai, result mein difference $\varphi -\psi$ bachta hai. $-90°$ pitch karne se $x$-axis $-z$ ki taraf point karta hai (anti-parallel), woh intervening sense dobara reverse hota hai, toh ab sum $\varphi +\psi$ bachta hai. Answer: Middle tilt roll axis ko $\pm z$ bheijti hai; us landing direction ka sign dono remaining rotations ke relative sense ko flip karta hai, $\varphi -\psi$ ko $\varphi +\psi$ mein badal deta hai.

KYA: Map ki singularity ko determinant ke vanishing se relate karo. KYUN (kaun sa tool aur kyun yeh): Ek linear map invertibility kho deta hai ⇔ uske matrix ka zero determinant hota hai. Yeh the algebraic detector hai "ek direction kuch mein squash ho gayi" — exactly woh DOF loss jo hum dhundh rahe hain. Hum determinant use karte hain kyunki yeh ek single scalar hai jiske zeros sare singular configurations ko ek saath pin down karte hain. Kyunki $\det (E)=1/\det (E^{-1})=1/\cos \theta$, forward map $E$ precisely wahan blow up karta hai jahan $\det (E^{-1})=\cos \theta =0$. Pitch range $\theta \in (-90°,+90°]$ par jo ek chart ke liye use hoti hai, $\cos \theta =0$ sirf $\theta =\pm 90°$ par. KAISA DIKHTA HAI: determinant ek smooth cosine bump hai, $\theta =0$ par $=1$ (perfectly invertible) aur dono poles par $0$ tak pinch karta hai. Answer: Forward Euler-rate map ki singularities exactly $\theta =\pm 90°$ hain; map sirf un poles se door well-conditioned hai, unke paas worst.

KYA: Woh single entry padho jo pitch encode karti hai. KYUN: General ZYX matrix mein $(3,1)$ entry equal hoti hai $-\sin \theta$ ke, doosre do angles se pitch alag karti hai. Lock hai $\theta =\pm 90°$ ⇔ $\sin \theta =\pm 1$ ⇔ $R_{31}=\mp 1$. Toh lock flag karo jab $|R_{31}|$ ek tolerance se $1$ ke andar ho, jaise $|R_{31}|>0.9998$ ($\theta$ ek pole ke $\approx 1.1°$ andar hai). KYUN yeh toota hai: usual formulas $\psi =atan2(R_{21},R_{11})$ aur $\varphi =atan2(R_{32},R_{33})$ woh entries read karte hain jo lock par zero ho gayi hain (parent ki $\theta =90°$ matrix dekho: $R_{11}=R_{21}=R_{32}=R_{33}=0$). Tum $atan2(0,0)$ paate ho — undefined. Sirf combination $\varphi \mp \psi$ recoverable hai, surviving entries $\sin (\varphi \mp \psi ),\cos (\varphi \mp \psi )$ se. Answer: Lock detect karo $|R_{31}|\to 1$ se; lock ke paas, ambiguity ko convention se split karo (jaise $\psi =0$ set karo aur $\varphi =atan2(R_{12},R_{22})$ solve karo), kyunki sirf $\varphi \mp \psi$ uniquely defined hai.

KYA: Quaternion→$SO(3)$ map ki algebraic form ko Euler map se contrast karo. KYUN (kaun sa tool aur kyun): Ek coordinate map ki singularities vanishing quantity se division ya rank-dropping Jacobian ki tarah appear hoti hain. Toh hum map ke formula mein koi bhi aisa denominator inspect karte hain jo zero hit kar sake. Ek unit quaternion se rotation, jaise pehli row ke liye, $R=\left(\begin{array}{ccc}1-2(y^2+z^2)& 2(xy-wz)& 2(xz+wy)\\ \cdots & \cdots & \cdots \end{array}\right)$ — pure polynomials, koi $1/\cos \theta$ nahi, koi denominator nahi. Unit sphere par $(w,x,y,z)$ ki koi value nahi hai jo koi entry blow up kare. Redundancy: $q$ aur $-q$ identical $R$ dete hain (dono rows mein $wz$, $wy$, $wx$ aur squares jaise products hain, jo sign flip karne par invariant rehte hain). Yeh ek two-to-one cover hai $S^3\to SO(3)$ — lekin yeh kabhi ek direction nahi khoता: map ka Jacobian $S^3$ par har jagah full rank rehta hai. KAISA DIKHTA HAI: $S^3$ ko smoothly aur doubly rotation group par bina kisi pinch points ke wrap hota imagine karo — Euler chart ke mukable, jo dono poles par pinch ho jaata hai. Answer: Koi singular orientation nahi (koi lost DOF nahi); ek hi quirk hai woh harmless sign ambiguity $q\equiv -q$. Quaternions aur Singularities of coordinate charts dekho.

KYA: "Hairy sphere"-style obstruction apply karo: $SO(3)$ globally 3 parameters se bina singularity ke chartable nahi hai. KYUN: $SO(3)$ ka ek curved global topology hai (yeh ${\mathbb{RP}}^3$ hai); koi bhi single 3-angle chart gol Dharti ke flat map ki tarah hai — yeh zaroor kahin tear ya pinch karega. Tum singularity ko move kar sakte ho (alag axis order lock ko alag pitch par rakhta hai) par kabhi delete nahi kar sakte. Escape: 3-parameter chart par insist mat karo. Ek 4-parameter, one-constraint representation use karo (quaternions: 4 numbers, $|q|=1$) ya $9$-number rotation matrix ($SO(3)$ with 6 constraints). Extra parameters tumhe ek smooth global description dete hain redundancy ki cost par — lekin redundancy $\ne$ lost DOF. Answer: Nahi — har 3-angle set mein kahin na kahin lock hogi; fix yeh hai ki internal computation ke liye "three-angle" family se poori tarah bahar niklo (quaternions / rotation matrices), Euler angles sirf human display ke liye convert karo, exactly jaisa parent note ka engineer's fix prescribe karta hai. Rotation matrices SO(3) aur Attitude determination and control dekho.