Exercises — DCM kinematics — Ċ = −[ω×]C

KYA: vee-map apply karo. KYUN: $[\vec{w}\times ]$ ka pattern exactly batata hai ki har component kahaan chhupa hai.

• Entry $(3,2)=w_1\Rightarrow w_1=3$
• Entry $(1,3)=w_2\Rightarrow w_2=2$
• Entry $(2,1)=w_3\Rightarrow w_3=5$

Toh $\vec{\omega }=[3,2,5{]}^T$. Top row ka sanity check: $-w_3=-5$ ✔, $w_2=2$ ✔.

(b) sahi hai. KYUN: $C$ inertial vectors ko body coordinates mein le jaata hai. Jab body forward spin karti hai, to fixed inertial vectors body frame mein ulti direction mein spin karte dikhte hain — minus sign uss reversal ko record karta hai. (Choice (a) transpose $C^T:\mathcal{B}\to \mathcal{N}$ ka equation hai.)

KYA: $\dot{C}=-[\vec{\omega }\times ]C$ mein plug karo. KYUN $C=I$ help karta hai: multiply sirf $-[\vec{\omega }\times ]$ tak collapse ho jaata hai. $[\vec{\omega }\times ]=\left[\begin{array}{ccc}0& -4& 0\\ 4& 0& 0\\ 0& 0& 0\end{array}\right]\Rightarrow \dot{C}=-\left[\begin{array}{ccc}0& -4& 0\\ 4& 0& 0\\ 0& 0& 0\end{array}\right]=\left[\begin{array}{ccc}0& 4& 0\\ -4& 0& 0\\ 0& 0& 0\end{array}\right].$

KYA: $I$ mein $\dot{C}\Delta t$ add karo. $C(0.01)\approx \left[\begin{array}{ccc}1& 0.04& 0\\ -0.04& 1& 0\\ 0& 0& 1\end{array}\right].$ Orthogonality check: top-left $2\times 2$ block lo. Uska pehla column length $\sqrt{1^2+0.04^2}=\sqrt{1.0016}\approx 1.0008\ne 1$ hai. Conclusion: yeh $SO(3)$ se $\approx 8\times 10^{-4}$ drift kar gaya. Exact solution ki length exactly 1 hoti; Euler overshoot karta hai. Isliye real code renormalize karta hai — exact step ke liye Matrix exponential and rotation about a fixed axis dekho.

KYA: parent note exact exponential solution deta hai. $C(t)=\left[\begin{array}{ccc}\cos \omega t& \sin \omega t& 0\\ -\sin \omega t& \cos \omega t& 0\\ 0& 0& 1\end{array}\right].$ $\omega t=\pi /2$ par: $\cos =0,\sin =1$, toh $C=\left[\begin{array}{ccc}0& 1& 0\\ -1& 0& 0\\ 0& 0& 1\end{array}\right].$ Check: $\det =0\cdot 0-1\cdot (-1)\cdot 1=1$ ✔ (proper rotation), aur har column ek unit vector hai ✔.

KYA: differentiate karo, $C^T$ se multiply karo, vee se padho. KYUN $-\dot{C}{C}^T$: $\dot{C}=-[\omega \times ]C$ se, $C^T$ se right-multiply karo aur $C{C}^T=I$ use karo. $\dot{C}=\left[\begin{array}{ccc}-2\sin 2t& 2\cos 2t& 0\\ -2\cos 2t& -2\sin 2t& 0\\ 0& 0& 0\end{array}\right],C^T=\left[\begin{array}{ccc}\cos 2t& -\sin 2t& 0\\ \sin 2t& \cos 2t& 0\\ 0& 0& 1\end{array}\right].$ Multiply karo aur negate karo (trig ${\cos }^2+{\sin }^2=1$ se collapse ho jaata hai): $-\dot{C}{C}^T=\left[\begin{array}{ccc}0& -2& 0\\ 2& 0& 0\\ 0& 0& 0\end{array}\right]=[\vec{\omega }\times ].$ Vee-map: ${\vec{\omega }}_{\mathcal{B}}=[0,0,2{]}^T$ rad/s — body-$z$ ke baare mein $2$ rad/s ka constant spin. Aisi measured histories kahaan se aati hain, yeh Attitude propagation & determination (TRIAD, QUEST) mein dekho.

KYUN: $C$, $\mathcal{N}\to \mathcal{B}$ map karta hai. Stored solution $C(t)=e^{-[\omega \times ]t}$ mein built-in minus hai. Toh physical body spin $+\omega$, $C$ ke andar $-\omega t$ ke rotation ke roop mein dikhta hai. Jab hum $-\dot{C}{C}^T$ se invert karte hain, woh minus undo ho jaata hai aur hum true, positive body rate $+2$ recover karte hain. Bookkeeping self-consistent hai: $\dot{C}$ mein minus, recovery mein minus, dono cancel ho jaate hain.

KYA: kyunki $\vec{\omega }$ constant hai, $C(t)=e^{-[\omega \times ]t}$, axis $\hat{a}=[1,1,1{]}^T/\sqrt{3}$ ke baare mein angle $\theta =|\vec{\omega }|t=t$ (kyunki $|\vec{\omega }|=1$) se rotation, negated (kinematic equation mein minus ke kaaran). KYUN exponential: $e^{-[\omega \times ]t}$ constant-coefficient linear matrix ODE ka the solution hai; yeh exactly Matrix exponential and rotation about a fixed axis hai. Rodrigues' formula use karke angle $-t$ ke saath $\hat{a}$ ke baare mein: $C(t)=I-\sin t[\hat{a}\times ]+(1-\cos t)[\hat{a}\times {]}^2,$ jahan $-\sin t$ (nahi $+\sin t$) minus sign carry karta hai. Yahan $[\hat{a}\times ]=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}0& -1& 1\\ 1& 0& -1\\ -1& 1& 0\end{array}\right]$. $t=0$ par check: $C(0)=I$ ✔. Axis fixed hai yeh check: $C(t)\hat{a}=\hat{a}$ sabhi $t$ ke liye, kyunki $\hat{a}$ ke baare mein rotate karne se $\hat{a}$ untouched rehta hai — verify karo $[\hat{a}\times ]\hat{a}=\vec{0}$ ✔.

KYA: $D=RC$ ko differentiate karo jahan $R$ constant hai ($\dot{R}=0$). $\dot{D}=R\dot{C}=R(-[{\vec{\omega }}_{\mathcal{B}}\times ]C)=-R[{\vec{\omega }}_{\mathcal{B}}\times ]R^{T}RC.$ KYUN $R^{T}R=I$ insert karo: $D=RC$ par act karte ek skew operator expose karne ke liye. Rotation identity $R[\vec{w}\times ]R^T=[(R\vec{w})\times ]$ use karo (yeh Rotation group SO(3) and Lie algebra so(3) mein prove ki gayi hai). ${\vec{\omega }}_{\mathcal{S}}=R{\vec{\omega }}_{\mathcal{B}}$ ke saath: $\boxed{\dot{D}=-[{\vec{\omega }}_{\mathcal{S}}\times ]D.}$ Matlab: kinematic equation apna form rakhta hai — bas $\vec{\omega }$ ko sensor frame mein express karo. Woh form-invariance exactly isliye hai ki equation itni beloved hai.

KYA/KYUN: Jacobi's formula deta hai $\frac{d}{dt}\det C=\det C\cdot \mathrm{tr}(C^{-1}\dot{C})$. Yeh determinant rate ko trace mein convert karta hai — kaafi aasaan. $\dot{C}=-[\vec{\omega }\times ]C$ aur $C^{-1}=C^T$ substitute karo: $\mathrm{tr}(C^{-1}\dot{C})=\mathrm{tr}(C^T(-[\vec{\omega }\times ]C))=-\mathrm{tr}([\vec{\omega }\times ]C{C}^T)=-\mathrm{tr}[\vec{\omega }\times ].$ KYUN last trace zero hai: ek skew-symmetric matrix ka diagonal all-zero hota hai, isliye uska trace $0$ hai. Isliye $\frac{d}{dt}\det C=\det C\cdot 0=0$. $\det C$ constant hai. $■$ Yeh length-preservation proof ka determinant-level companion hai; saath mein yeh kehte hain ki $C$ hamesha ke liye ek proper rotation rehta hai. Yeh Lie-algebra fact $\mathrm{tr}(\mathfrak{so}(3))=0$ se connect karta hai jo Rotation group SO(3) and Lie algebra so(3) mein hai.

DCM side: L2.3 se, $C(t)$ ek aisa rotation hai jiska $z$ ke baare mein angle $\omega t$ hai (magnitude). Quaternion side: constant $\vec{\omega }$ deta hai $q(t)=[\cos \frac{\omega t}{2},0,0,\sin \frac{\omega t}{2}]$ (scalar-first). Ek unit quaternion $[\cos \frac{\varphi }{2},\hat{a}\sin \frac{\varphi }{2}]$, $\hat{a}$ ke baare mein angle $\varphi$ ka rotation encode karta hai. Match: yahan $\varphi =\omega t$, $\hat{a}=\hat{z}$ ke baare mein — DCM ke identical. KYUN half-angle: quaternions $SO(3)$ ko double-cover karte hain, toh woh internally $\varphi /2$ carry karte hain lekin same physical $\varphi$ represent karte hain. Numeric check $\omega t=\pi /2$ par: quaternion $=[\cos 45^{\circ },0,0,\sin 45^{\circ }]=[0.7071,0,0,0.7071]$, jo exactly L2.3 mein mila $C$ rebuild karta hai. ✔

L1 operator padho/chuno → L2 $\dot{C}$ banao aur step karo → L3 ulta karo $\omega$ paane ke liye → L4 frames badlo form rakhte hue → L5 invariants prove karo aur quaternions se bridge banao. Har rung wahi same equation $\dot{C}=-[\omega \times ]C$ hai, ek naye angle se dekhi gayi.