Question bank — DCM kinematics — Ċ = −[ω×]C

True — yeh equation $C$ mein linear hai: derivative wo $C$ hai jo ek fixed (us instant pe) skew matrix se left-multiply hoti hai. Isliye solution ek matrix exponential hai, polynomial nahi.

False — $\dot{C}$ SO(3) ke tangent space mein rehta hai, group pe nahi. Yeh (skew) $\times$ (rotation) hai, jo generally na orthogonal hota hai aur na hi determinant 1 wala.

True — zero spin se $\dot{C}=-[0\times ]C=0$ milta hai, toh $C$ freeze ho jaata hai. Physically body ghoom nahi rahi, toh frames ke beech map change nahi ho sakta.

True — yeh directly constraint $C{C}^T=I$ ko differentiate karne se aata hai, jo sab $t$ ke liye hold karta hai. Yeh akela fact hi puri skew structure ko force karta hai.

Exact ODE ke liye False — equation guarantee karta hai ki $\frac{d}{dt}(C{C}^T)=0$, toh orthonormality automatically conserve hoti hai agar shuru mein true thi. Sirf numerical integration mein hi isse dobara impose karna padta hai.

True — inverse map ke liye tumhe milta hai ${\dot{C}}^T=+[{\vec{\omega }}_{\mathcal{B}}\times ]C^T$. Sign bookkeeping ka consequence hai ki matrix kis direction mein map karta hai, koi physical fact nahi.

True — ek $3\times 3$ skew matrix ke diagonal pe zeros hote hain aur har off-diagonal ka ek sign-flipped copy, jisse 3 free entries bachti hain. Yeh vee-map ke zariye $({\omega }_1,{\omega }_2,{\omega }_3)$ se one-to-one map hoti hain.

Generally False — clean exponential tab hi kaam karta hai jab $[{\vec{\omega }}_{\mathcal{B}}\times ]$ constant ho. Agar $\vec{\omega }$ time ke saath vary kare aur alag-alag times pe skew matrices commute na karein, toh tumhe time-ordered exponential chahiye, $e^{\int (\cdot )dt}$ nahi.

Galat frame — $C:\mathcal{N}\to \mathcal{B}$ ke saath skew ko body-frame components ${\vec{\omega }}_{\mathcal{B}}=C{\vec{\omega }}_{\mathcal{N}}$ use karni chahiye, taaki output consistently body frame mein aaye. ${\vec{\omega }}_{\mathcal{N}}$ use karna dono sides ke frames ko mismatch kar deta hai.

Plus formula ek vector ko inertial frame se dekhta hai; $C$ inertial ko body mein map karta hai, toh body ke viewpoint se inertial vectors ulte spin karte dikhte hain — isliye minus. Dono formulas opposite viewpoints describe karte hain.

Continuous ODE $C{C}^T=I$ ko exactly preserve karta hai; lekin ek discrete Euler step curved surface SO(3) se seedhi line le jaata hai, toh $C$ drift karta hai. Tumhe renormalize karna hoga (Gram–Schmidt / SVD) ya quaternion propagate karna hoga.

Sign error — kyunki $[{\vec{\omega }}_{\mathcal{B}}\times ]=-\dot{C}{C}^T$, sahi extraction mein minus aata hai: ${\omega }_i=-\frac{1}{2}{\epsilon }_{ijk}(\dot{C}{C}^T{)}_{jk}$. Isse chhod dene se recovered spin direction ulti ho jaati hai.

Sirf proper rotations ke liye ($\det C=+1$). Ek reflection ($\det C=-1$) handedness flip karta hai aur ek extra minus sign introduce karta hai, jo sign derivation mein use ki gayi identity tod deta hai.

Ek $3\times 3$ skew matrix singular hoti hai (odd-dimensional skew matrices ka determinant 0 hota hai), toh uska koi inverse nahi hota. Recovery transpose trick $[\omega \times ]=-\dot{C}{C}^T$ use karti hai, $C^{-1}=C^T$ ka fayda uthati hai, skew ke kisi inverse ka nahi.

Ulta hai — $C$ ke rows body-frame basis vectors hain jo inertial coordinates mein expressed hain. Entry $C_{ij}={\hat{b}}_i\cdot {\hat{n}}_j$ yeh clearly dikhata hai.

Kyunki $C$ curved surface $C{C}^T=I$ pe locked hai; sirf wohi velocities jo surface ko surface pe rakhti hain, surface ki tangent directions ki taraf point karti hain, jo exactly {skew matrix} $\times C$ hain. Koi bhi aur form $C$ ko rotation group se bahar le jaata.

Ek rotation matrix, 9 entries ke bawajood, sirf 3 true degrees of freedom rakhta hai (orthonormality 6 hata deti hai). Toh $\vec{\omega }$ mein 3 numbers exactly $C$ ke 3 tarike se change hone se match karte hain — Lie-algebra dimension se ek perfect count, dekho Rotation group SO(3) and Lie algebra so(3).

Physical spin fixed hai; sign sirf record karta hai ki tumne map $\mathcal{N}\to \mathcal{B}$ (minus) likha ya $\mathcal{B}\to \mathcal{N}$ (plus). Dono identical physical motion predict karte hain — tumhe sirf declare karna hai ki tum konsa convention use kar rahe ho.

Quaternions aur Euler angles sirf alag coordinates hain usi rotation ke liye; unke rate equations (Quaternion kinematics — q̇ = ½ Ω(ω) q, Euler angle kinematics & gimbal lock) is ek DCM equation ke re-parametrizations hain. Physics same hai, bookkeeping alag.

Kyunki Poisson's equation for rotating frames yeh general statement hai ki ek body-fixed quantity ki inertial rate $\vec{\omega }\times (\cdot )$ ke barabar hoti hai; DCM ke har column/basis vector pe apply karne se exactly $\dot{C}=-[\omega \times ]C$ milta hai.

$\vec{\omega }$ ek rate hai, jo encode hota hai is mein ki $C$ kaise change ho raha hai, uski instantaneous value mein nahi. Tumhe $\dot{C}$ chahiye (ya $C$ ke do nearby samples) $-\dot{C}{C}^T$ banane aur spin padhne ke liye.

$\dot{C}=0$: map momentarily frozen hai. Lekin $C$ khud koi bhi rotation ho sakta hai — ek non-spinning body phir bhi kisi bhi fixed orientation mein point kar sakti hai.

Yeh us fixed axis ke around ek planar rotation hogi, matrix exponential $e^{-[\omega \times ]t}C(0)$ se di gayi. Spin axis ek eigenvector hai eigenvalue 1 ke saath, toh us axis ki direction preserve hoti hai.

Simple $e^{-[\omega \times ]t}$ formula fail ho jaata hai kyunki alag-alag times pe skew matrices commute nahi karti; tumhe ODE step by step integrate karna hoga (ya time-ordered product use karna hoga). Physically isliye real tumbling attitude ka koi closed form nahi hota.

Nahi — DCM equation har jagah perfectly well-behaved hai; sirf Euler-angle re-parametrization singular ho jaati hai (rates blow up), dekho Euler angle kinematics & gimbal lock. Yahi robustness ek key reason hai DCM ya quaternion propagate karne ka.

Exact equation mein kuch bhi saturate nahi hota — $C$ bas tez rotate karta hai, SO(3) pe rehta hai. Practical limit numerical hai: bada $\omega$ times finite step size bada drift karaata hai, chhote steps ya renormalization force karta hai.

Nahi — equation sirf $C{C}^T$ conserve karti hai; restore nahi karti. Agar $C(0)C(0{)}^T\ne I$ ho, toh woh error unchanged aage carry hoti hai (uski change ki rate zero hai, lekin correction bhi zero hai). Clean shuru karo.

Nahi — flow continuously $\det C=+1$ wale connected component pe rehta hai. Kyunki $\det$ continuous hai aur $+1$ se shuru hota hai, aur ODE kabhi $SO(3)$ nahi chhodta, yeh reflection mein jump nahi kar sakta.