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

1. Yaw $\psi$ about the $z$-axis (the "3rd" axis) — nose left/right ghoomta hai (merry-go-round ki tarah).
2. Pitch $\theta$ about the new $y$-axis (the "2nd" axis) — nose upar/neeche jhukta hai.
3. Roll $\varphi$ about the newest $x$-axis (the "1st" axis) — wings sideways bank karti hain.

Naam 3-2-1 literally application order mein axis numbers hain: $3\to 2\to 1$.

Top-right entry $R_{13}=-s\theta =-\sin \theta$ dekho. Ismein sirf $\theta$ hai. Yahi wajah hai ki pitch recover karna sabse aasaan hai: koi doosra angle isko contaminate nahin karta, isliye $\theta =-\arcsin (R_{13})$ mein koi ratio nahin chahiye.

False. Matrix multiplication (aur physical rotation) commute nahin karti — yaw-then-roll se alag jagah pahunchte ho roll-then-yaw se. Order definition ka hissa hai; convention hamesha clearly state karni chahiye.

$\psi =90^{\circ }$ ke saath: $c\psi =0,s\psi =1$, toh $R_z(90^{\circ })=\left[\begin{array}{ccc}0& 1& 0\\ -1& 0& 0\\ 0& 0& 1\end{array}\right].$ Nose body frame mein hai, toh use nav mein dekhne ke liye $R_b^n=(R_n^b{)}^T$ apply karo: $(R_z(90^{\circ }){)}^T[1,0,0{]}^T=\left[\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]=[0,1,0{]}^T.$ Yeh $y_n$ = East hai. $+90^{\circ }$ yaw nose ko North se East ki taraf swing karta hai. ✓

Pehli row $=[c\theta c\psi ,c\theta s\psi ,-s\theta ]$. $\theta =90^{\circ },\psi =0$ ke saath: $c\theta =0,s\theta =1$, toh pehli row $=[0,0,-1]$. Matlab: body-$x$ axis (nose) ab $R_{13}$ ke through nav Down direction ke negated direction mein aligned hai — nose seedha upar point kar raha hai ($-z_n$). $90^{\circ }$ pitch-up aircraft ko uske tail par khada kar deta hai. ✓

$R_{13}=-\sin 20^{\circ }=-0.342$. Ulta karte hain: $\theta =-\arcsin (-0.342)=+20^{\circ }$. Minus signs cancel ho jaate hain, jo confirm karta hai ki recovery formula self-consistent hai. ✓

Teesri row $=[c\varphi s\theta c\psi +s\varphi s\psi ,c\varphi s\theta s\psi -s\varphi c\psi ,c\varphi c\theta ]$. $\psi =0$ ke saath ($c\psi =1,s\psi =0$): $=[c\varphi s\theta ,-s\varphi ,c\varphi c\theta ]$. Numbers: $c30=0.866,s30=0.5,s20=0.342,c20=0.940$: $=[0.866\times 0.342,-0.5,0.866\times 0.940]=[0.296,-0.5,0.814].$ Magnitude: $\sqrt{0.296^2+0.5^2+0.814^2}=\sqrt{0.0877+0.25+0.663}=\sqrt{1.00}=1.00$. ✓ Interpretation: yeh row body-Down axis ko nav mein dikhata hai — downward component $0.814<1$ dikhata hai ki belly ab seedha neeche point nahin kar rahi; craft tilt hai.

• $\theta =-\arcsin (-0.5)=30^{\circ }$. Kyun: $R_{13}=-\sin \theta$ uncontaminated hai.
• $\varphi =atan2(0.433,0.75)=30^{\circ }$. Kyun: $R_{23}=s\varphi c\theta$ aur $R_{33}=c\varphi c\theta$ mein $c\theta$ common factor hai; unka ratio $\tan \varphi$ hai, aur atan2 dono signs use karke quadrant pakad leta hai.
• $\psi =atan2(0,0.866)=0^{\circ }$. Same trick $R_{12}=c\theta s\psi ,R_{11}=c\theta c\psi$ ke saath. ✓

Simplicity ke liye $c\theta =1$ lo, toh $R_{23}=\sin \varphi ,R_{33}=\cos \varphi$.

• Case $\varphi =45^{\circ }$: $R_{23}=+0.707,R_{33}=+0.707$. Ratio $=+1$, $\arctan (1)=45^{\circ }$. Sahi hai.
• Case $\varphi =225^{\circ }$ (equivalently $-135^{\circ }$): $R_{23}=-0.707,R_{33}=-0.707$. Ratio phir bhi $+1$ hai, toh $\arctan (1)=45^{\circ }$ — $180^{\circ }$ galat.
• $atan2(-0.707,-0.707)=-135^{\circ }$ — sahi hai, kyunki atan2 dekhta hai ki dono signs negative hain aur angle ko third quadrant mein rakhta hai.

Plain $\arctan$ sirf $(-90^{\circ },90^{\circ })$ mein answers deta hai kyunki $\tan$ har $180^{\circ }$ par repeat karta hai; atan2 numerator aur denominator ke alag-alag signs use karke poora $360^{\circ }$ range recover karta hai. Figure dikhata hai ki dono vectors alag quadrants mein land karte hain jabki ratio identical hai.

$\theta =90^{\circ }$ par: $c\theta =0,s\theta =1$. Read-out ko jo char entries chahiye unhe dekho: $R_{11}=c\theta c\psi =0,R_{12}=c\theta s\psi =0,R_{23}=s\varphi c\theta =0,R_{33}=c\varphi c\theta =0.$ Toh $\varphi =atan2(0,0)$ aur $\psi =atan2(0,0)$ — dono undefined ($0/0$). Physically, jab nose seedha upar point karta hai toh roll axis aur yaw axis space mein ek hi direction mein point karte hain, isliye ek roll aur ek yaw ek hi motion produce karte hain. Do knobs ab ek kaam karte hain → ek degree of freedom kho gaya. Yahi hai gimbal lock. Figure dikhata hai ki roll aur yaw circles coincident ho jaate hain. Practical fix: quaternions ya Direction Cosine Matrix (DCM) se seedha attitude propagate karo, aur Euler angles mein sirf human display ke liye convert karo jab $\theta =\pm 90^{\circ }$ se door ho.

$\theta =90^{\circ }$ par, $c\theta =0,s\theta =1$ ko $R_n^b$ ke lower-left $2\times 2$ block mein plug karo: $R_{21}=s\varphi c\psi -c\varphi s\psi =\sin (\varphi -\psi ),R_{22}=s\varphi s\psi +c\varphi c\psi =\cos (\varphi -\psi ),$ $R_{31}=c\varphi c\psi +s\varphi s\psi =\cos (\varphi -\psi ),R_{32}=c\varphi s\psi -s\varphi c\psi =-\sin (\varphi -\psi ).$ Har non-trivial entry sirf $\varphi -\psi$ ka function hai (yahan sign convention ki wajah se difference free combination hai; exact combination convention par depend karta hai, lekin point yeh hai: ek combined angle, do nahin). Check: $(50^{\circ },10^{\circ })\Rightarrow \varphi -\psi =40^{\circ }$; $(20^{\circ },40^{\circ })\Rightarrow \varphi -\psi =-20^{\circ }$ — yeh alag matrices dete hain, lekin koi bhi doosra pair jo same difference rakhta ho (jaise $(60^{\circ },20^{\circ })$, difference $40^{\circ }$) same matrix dega jaise $(50^{\circ },10^{\circ })$. Do independent numbers se ek par yeh collapse exactly woh lost degree of freedom hai. ✓

Jab $\theta \to 90^{\circ }$, entries $R_{11},R_{12}$ $0$ ki taraf shrink hoti hain jabki koi bhi chhota sensor noise rehta hai. Toh $\psi =atan2(R_{12},R_{11})$ chhote noisy numbers ko chhote noisy numbers se divide karta hai — recovered yaw wildly sensitive ho jaata hai aur ek hi step mein $180^{\circ }$ tak jump kar sakta hai jab $\theta$ $90^{\circ }$ se guzarta hai (frame effectively "flip" kar leta hai ki woh kis taraf dekh raha hai). Ek filter jo $\dot{\psi }$ integrate kar raha ho usse ek bada spurious rate dikhega aur woh ek false correction inject kar dega. Kyun hota hai: orientation se $(\varphi ,\theta ,\psi )$ tak ka map $\theta =\pm 90^{\circ }$ par coordinate singularity rakhta hai (Euler-rate relation ka Jacobian blow up karta hai kyunki ismein $1/\cos \theta$ term hai — wahi $c\theta$ jo L3 mein harmlessly cancel hua tha ab rates ke liye denominator mein baitha hai). Singularity ke paas, chhoti state changes ke liye infinite angle changes chahiye. Design fix: estimator ki internal state ko quaternion ya DCM ke roop mein chalao (dono singularity-free hain — dekho Quaternions — avoiding gimbal lock, Direction Cosine Matrix (DCM)), Angular velocity & body rates p, q, r se body rates $p,q,r$ use karke propagate karo, aur $(\varphi ,\theta ,\psi )$ mein sirf pilot ke display ke liye convert karo, exactly $\theta =\pm 90^{\circ }$ par display update avoid karke.

Har elementary matrix satisfy karta hai $R_a^T{R}_a=I$ (iske columns perpendicular unit vectors hain — rotation ki definition, dekho Rotation matrices & orthogonality). Product $R_n^b=R_x{R}_y{R}_z$ ke liye: $(R_x{R}_y{R}_z{)}^T(R_x{R}_y{R}_z)=R_z^T{R}_y^T{R}_x^T{R}_x{R}_y{R}_z.$ Ab andar se chhilao: $R_x^T{R}_x=I$, bachta hai $R_z^T{R}_y^T{R}_y{R}_z$; phir $R_y^T{R}_y=I$, bachta hai $R_z^T{R}_z=I$. Isliye $(R_n^b{)}^T{R}_n^b=I$: orthogonal matrices ka product orthogonal hota hai. Kyun matter karta hai: yahi wajah hai ki inverse sirf transpose hai, jise humne L2.1 mein body → nav map karne ke liye freely use kiya tha. ✓