Exercises — Quaternion kinematics — q̇ = ½ Ξ(q) ω

$\Xi (q)$ ek $==4\times 3==$ matrix hai. Kyun: yeh 3-vector $\omega$ (3 columns) khata hai aur 4-vector $\dot{q}$ (4 rows) produce karta hai. Ek matrix jo 3-cheez ko 4-cheez mein badalta hai woh $4\times 3$ hi honi chahiye. Upar ke formula mein entries count karke confirm karo: 4 rows, 3 columns.

Yeh ek pure quaternion hai (ise angular-velocity quaternion bhi kehte hain). Ek pure quaternion ka scalar part zero hota hai; iske teen vector slots mein $\omega =({\omega }_1,{\omega }_2,{\omega }_3)$ hota hai. Yeh unit quaternion nahi hai aur rotation represent nahi karta — yeh sirf ek convenient package hai taaki hum $\omega$ ko $q$ mein multiply kar sakein.

(b). Ek quaternion $\cos \frac{\theta }{2}$ aur $\sin \frac{\theta }{2}$ store karta hai — half-angle. Jab tum $\sin \frac{\theta }{2}$ ko time mein differentiate karte ho, chain rule se $\frac{1}{2}$ bahar aata hai: $\frac{d}{dt}\sin \frac{\theta }{2}=\frac{1}{2}\dot{\theta }\cos \frac{\theta }{2}$. Wahi $\frac{1}{2}$ poori kahaani hai — yeh double-cover ki cost hai, jahan $q$ aur $-q$ dono same physical rotation mean karte hain.

$q=(1,0,0,0)$ par: $q_0=1,q_1=q_2=q_3=0$, isliye $\Xi (q)=\left[\begin{array}{ccc}0& 0& 0\\ 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right].$ Phir $\dot{q}=\frac{1}{2}\Xi (q)\left[\begin{array}{c}0\\ 0.2\\ 0\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}0\\ 0\\ 0.2\\ 0\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0.1\\ 0\end{array}\right].$ Matlab: $q_2$ slot pehle on hota hai — yeh body $y$-axis ke baare mein pitch ka shuru hona hai. Aur ${\dot{q}}_0=0$: identity par norm abhi change nahi ho sakta ($q^{\top }\dot{q}=0$).

Yahan $q_0=0.7071,q_1=0.7071,q_2=0,q_3=0$. $\Xi (q)$ banao: $\Xi (q)=\left[\begin{array}{ccc}-0.7071& 0& 0\\ 0.7071& 0& 0\\ 0& 0.7071& -0.7071\\ 0& 0.7071& 0.7071\end{array}\right].$ Sirf ${\omega }_3=1$ nonzero hai, isliye hum $\Xi (q)$ ka teesra column lete hain aur use aadha kar dete hain: $\dot{q}=\frac{1}{2}\left[\begin{array}{c}0\\ 0\\ -0.7071\\ 0.7071\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ -0.35355\\ 0.35355\end{array}\right].$ Sanity check: $q^{\top }\dot{q}=0.7071\cdot 0+0.7071\cdot 0+0\cdot (-0.35355)+0\cdot 0.35355=0$. Sphere ke tangent. ✔

Left side ($\Xi (q)\omega$, $\frac{1}{2}$ se pehle): 2.1 se yeh $(0,0,0.2,0)$ hai. Right side ($\Omega (\omega )q$): $\omega =(0,0.2,0)$ ke saath, $\Omega (\omega )=\left[\begin{array}{cccc}0& 0& -0.2& 0\\ 0& 0& 0& -0.2\\ 0.2& 0& 0& 0\\ 0& 0.2& 0& 0\end{array}\right],\Omega (\omega )\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0.2\\ 0\end{array}\right].$ Dono $(0,0,0.2,0)$ dete hain. ✔ Same physics, do bookkeeping choices — ek $q$ factor out karta hai, doosra $\omega$.

$\dot{q}=\frac{1}{2}\Xi (q)\omega$ se shuru karo, toh $q^{\top }\dot{q}=\frac{1}{2}(q^{\top }\Xi (q))\omega .$ Block form use karte hue (upar definition callout mein restated) $\Xi (q)=\left[\begin{array}{c}-{\mathbf{q}}_v^{\top }\\ q_{0}I+[{\mathbf{q}}_v\times ]\end{array}\right]$ aur $q=(q_0,{\mathbf{q}}_v)$, hum $q^{\top }=\left[q_0{\mathbf{q}}_v^{\top }\right]$ split karte hain aur block-by-block multiply karte hain: $q^{\top }\Xi (q)=q_0(-{\mathbf{q}}_v^{\top })+{\mathbf{q}}_v^{\top }(q_{0}I+[{\mathbf{q}}_v\times ])=-q_0{\mathbf{q}}_v^{\top }+q_0{\mathbf{q}}_v^{\top }+{\mathbf{q}}_v^{\top }[{\mathbf{q}}_v\times ].$ Last term mein $[{\mathbf{q}}_v\times {]}^{\top }=-[{\mathbf{q}}_v\times ]$ (skew) aur ${\mathbf{q}}_v\times {\mathbf{q}}_v=0$ use hota hai: ${\mathbf{q}}_v^{\top }[{\mathbf{q}}_v\times ]=-({\mathbf{q}}_v\times {\mathbf{q}}_v{)}^{\top }=0^{\top }$. Toh $q^{\top }\Xi (q)=0^{\top }$ aur isliye har $\omega$ ke liye $q^{\top }\dot{q}=0$. Geometrically (upar figure s01 dekho): kaala arrow $q$ radially sphere ki taraf point karta hai; red arrow $\dot{q}$ marked right angle par milta hai, surface ke flat saath lie karta hai. Woh right angle hai algebra $q^{\top }\dot{q}=0$ jo visible ho gayi — motion $S^3$ ke saath slide karti hai, kabhi usse nahi hatati, isliye $\Vert q\Vert$ exactly 1 rehta hai.

Pehle equation differentiate karo aur doosri substitute karo: ${\ddot{q}}_0=-\frac{1}{2}{\omega }_z{\dot{q}}_3=-\frac{1}{2}{\omega }_z\left(\frac{1}{2}{\omega }_z{q}_0\right)=-(\frac{{\omega }_z}{2}{)}^2{q}_0.$ Yeh simple harmonic motion hai angular frequency $==\frac{{\omega }_z}{2}==$ ke saath. $q_0(0)=1,q_3(0)=0$ ke saath: $q_0(t)=\cos \frac{{\omega }_{z}t}{2},q_3(t)=\sin \frac{{\omega }_{z}t}{2}.$ Kyun half: physical body angle $\theta ={\omega }_{z}t$ se ghoomta hai, lekin quaternion $\cos \frac{\theta }{2},\sin \frac{\theta }{2}$ store karta hai. Toh quaternion components body ke rotation rate ke half par oscillate karte hain. Ek full $360^{\circ }$ body turn ($\theta =2\pi$) quaternion ko half-angle mein sirf $\pi$ aage le jaata hai — aur indeed $q$ ek full turn ke baad $-q$ par return karta hai, famous double cover.

$\Xi (q)$ ke columns: $c_1=\left[\begin{array}{c}-q_1\\ q_0\\ q_3\\ -q_2\end{array}\right],c_2=\left[\begin{array}{c}-q_2\\ -q_3\\ q_0\\ q_1\end{array}\right],c_3=\left[\begin{array}{c}-q_3\\ q_2\\ -q_1\\ q_0\end{array}\right].$ $c_1\cdot c_2=(-q_1)(-q_2)+q_0(-q_3)+q_3{q}_0+(-q_2)(q_1)=q_1{q}_2-q_0{q}_3+q_0{q}_3-q_1{q}_2=0.$ Orthogonal. ✔ Aur $c_1\cdot c_1=q_1^2+q_0^2+q_3^2+q_2^2=\Vert q{\Vert }^2=1$. Toh teen columns ek orthonormal set banate hain (jab $q$ unit ho). Matlab: $\Xi (q)$ $\omega$-space se tangent plane mein ek isometry hai — $\omega$ ko stretch nahi karta, bas ise 4D tangent space ke sahi 3D slice mein rotate kar deta hai. Isliye $\Vert \dot{q}\Vert =\frac{1}{2}\Vert \omega \Vert$ exactly.

Parent note ke Example 2 se, $\dot{q}=\frac{1}{2}\Xi (q)\omega =(0,0.05,0,0)$. Euler step: $q^{\text{new}}=(1,0,0,0)+0.5(0,0.05,0,0)=(1,0.025,0,0)$. Norm: $\Vert q^{\text{new}}\Vert =\sqrt{1^2+0.025^2}=\sqrt{1.000625}=1.0003125\dots$ — 1 se thoda bada. Renormalize: norm se divide karo: $q^{\text{norm}}=\frac{1}{1.0003125}(1,0.025,0,0)=(0.99968762,0.02499219,0,0).$ Kyun zaroori: tangency proof (3.1) exact hai sirf continuous time mein. Ek finite Euler step tangent line ke saath straight move karta hai, jo curved sphere se nikal jaati hai, isliye $\Vert q\Vert$ 1 se upar creep karta hai. Renormalize karna ise $S^3$ par snap back karta hai. Higher-order integrators jaise RK4 (ek step mein chaar rate-samples) kam drift karte hain lekin phir bhi periodic renormalization ki zaroorat hoti hai.

Exact: $q_0=\cos \frac{{\omega }_{z}t}{2},q_3=\sin \frac{{\omega }_{z}t}{2}$. ${\omega }_{z}t=\pi \cdot 1=\pi$ ke saath: $q_0=\cos \frac{\pi }{2}=0,q_3=\sin \frac{\pi }{2}=1\Rightarrow q(1)=(0,0,0,1).$ Physical angle: $q_0=\cos \frac{\theta }{2}=0\Rightarrow \frac{\theta }{2}=\frac{\pi }{2}\Rightarrow \theta =\pi =180^{\circ }$ $z$-axis ke baare mein. Consistent: rate $\pi$ rad/s for 1 s body ko $\pi$ radians ghoomata hai. ✔

$q$ ek vector jo body frame mein express hai use inertial frame mein map karta hai. Body par gyros se measured ek tiny extra spin naturally body coordinates mein describe hoti hai, toh yeh pehle act karti hai, body frame mein — aur ek map ke liye "pehle" hota hai rightmost factor: $q\otimes \delta q$. Isliye body rates post-multiply karte hain. Agar instead $\omega$ fixed inertial frame mein di gayi hai, toh extra spin $q$ ka kaam ho jaane ke baad act karti hai, inertial coordinates mein — leftmost factor: $\delta q\otimes q$. Isliye inertial rates pre-multiply karte hain. Kyunki $\otimes$ non-commutative hai, inhe silently swap karna galat attitude integrate karta hai — ek classic flight-software bug. Yeh Rotation matrices and SO(3) mein composition order se link karta hai.

Input Jacobian $\partial \dot{q}/\partial \omega$. Likho $\dot{q}=\frac{1}{2}\Xi (q)\omega$. Yeh $\omega$ mein linear hai ($\dot{q}$ ka har component ${\omega }_1,{\omega }_2,{\omega }_3$ ka fixed linear combination hai), aur ek linear map $M\omega$ ka $\omega$ ke respect mein derivative sirf $M$ hai. Isliye $\partial \dot{q}/\partial \omega =\frac{1}{2}\Xi (q)$ — ek $4\times 3$ matrix. Explicit entry: $\dot{q}$ ke Jacobian ki row 3, column 2 hai $\frac{1}{2}$ times $\Xi (q)$ ki entry $(3,2)$. $q=(1,0,0,0)$ par woh entry $q_0=1$ hai, isliye $\frac{\partial {\dot{q}}_3}{\partial {\omega }_2}=\frac{1}{2}\cdot 1=\frac{1}{2}=\mathrm{0.5.}$ State Jacobian $\partial \dot{q}/\partial q$. Doosra form use karo $\dot{q}=\frac{1}{2}\Omega (\omega )q$, jo $q$ mein linear hai, toh $\partial \dot{q}/\partial q=\frac{1}{2}\Omega (\omega )$ — ek $4\times 4$ matrix. Explicit entry: row 3, column 1. $\Omega (\omega )$ ki entry $(3,1)$ hai ${\omega }_2$; $\omega =(0,0.2,0)$ ke saath woh $0.2$ hai, isliye $\frac{\partial {\dot{q}}_3}{\partial q_0}=\frac{1}{2}\cdot 0.2=\mathrm{0.1.}$ Exercise 2.1 se cross-check: wahan $q=(1,0,0,0)$ mein sirf $q_0\ne 0$ hai, aur ${\dot{q}}_3=0.1$. Woh poori value single Jacobian path $\frac{\partial {\dot{q}}_3}{\partial q_0}\cdot q_0=0.1\cdot 1=0.1$ se flow karti hai. ✔ Dono Jacobians direct computation se agree karte hain, aur exactly isliye quaternion kinematics ek EKF ke propagation aur covariance-update steps mein cleanly slot ho jaata hai.

Exercise 3.3 se, unit $q$ ke liye $\Xi (q)$ ke columns orthonormal hain, toh $\Xi (q)$ length preserve karta hai: $\Vert \Xi (q)\omega \Vert =\Vert \omega \Vert$. Isliye $\Vert \dot{q}\Vert =\Vert \frac{1}{2}\Xi (q)\omega \Vert =\frac{1}{2}\Vert \Xi (q)\omega \Vert =\frac{1}{2}\Vert \omega \Vert .$ Number: $\Vert \dot{q}\Vert =\frac{1}{2}(0.05)=0.025$ (quaternion-units per second). Interpretation: point $q$ $S^3$ ke saath body ki angular speed ke half par slide karta hai — half-angle ka geometric echo. Unit sphere par arc-length angle ke barabar hai, toh $q$ har second $0.025$ rad of great-circle arc sweep karta hai, aur $t=\frac{2\pi }{0.05}=125.66$ s ke baad (ek full physical turn, $\Vert \omega \Vert t=2\pi$) usne $\pi$ arc trace kiya hai — $-q$ par land karta hai, phir double cover. (Dekho figure s02.)

Constant $\omega$ ke liye, $\dot{q}=\frac{1}{2}\Omega (\omega )q$ ek linear ODE with constant coefficients hai, jiska exact solution matrix exponential hai $q(t+h)=\exp (\frac{1}{2}\Omega (\omega )h)q(t)$. Kyunki $\Omega (\omega )$ skew-symmetric hai (${\Omega }^{\top }=-\Omega$), uska exponential orthogonal hota hai, toh yeh norm exactly preserve karta hai: $\Vert q(t+h)\Vert =\Vert q(t)\Vert$ — koi drift nahi, koi renormalization ki zaroorat nahi. Closed form hai $\exp (\frac{1}{2}\Omega (\omega )h)=\cos \frac{\Vert \omega \Vert h}{2}I+\frac{\sin \frac{\Vert \omega \Vert h}{2}}{\Vert \omega \Vert }\Omega (\omega ),$ jo precisely angle $\Vert \omega \Vert h$ ka $\hat{\omega }$ ke baare mein ek rotation quaternion hai — Rodrigues rotation formula $R=I+\sin \theta [\hat{n}\times ]+(1-\cos \theta )[\hat{n}\times {]}^2$ ka quaternion analogue. Dono "ek axis ke baare mein ek angle se rotate karo" ko ek $\cos$ (stay) plus ek $\sin$ (turn) part mein split karte hain. Exponential map rate description ($\Omega ,\Xi$) aur finite rotation description (Rodrigues, $R\in SO(3)$) ke beech bridge hai.

Half aata hai ::: quaternion mein stored half-angle $\sin \frac{\theta }{2}$ se (chain rule). $\dot{q}$ $q$ se orthogonal hai kyunki ::: $q^{\top }\Xi (q)=0^{\top }$, motion ko $S^3$ ke tangent rakhta hai. $\Vert \dot{q}\Vert$ equals ::: $\frac{1}{2}\Vert \omega \Vert$ ($\Xi (q)$ ke columns orthonormal hain).