DCM kinematics — Ċ = −[ω×]C

3.5.4 · Physics › Guidance, Navigation & Control (GNC)

WHAT is $C$?

Kyunki entries axes ke beech ke angles ke cosines hain, isliye $C_{ij}={\hat{b}}_i\cdot {\hat{n}}_j$.

WHY does $C$ change, aur HOW fast?

Step 1 — Key constraint: orthonormality hamesha preserved rehti hai

Basis vectors unit-length aur mutually perpendicular rehte hain, chahe body kaise bhi spin kare. Isliye har time par: $C(t)C(t{)}^T=I$

Yeh step kyun? Yeh ek akela fact sab kuch hai. Yeh derivative ko ek special (skew) structure lene par majboor karta hai.

Step 2 — Constraint ko differentiate karo

$C{C}^T=I$ ko product rule se differentiate karo: $\dot{C}{C}^T+C{\dot{C}}^T=0$

Notice karo ki $C{\dot{C}}^T=(\dot{C}{C}^T{)}^T$. Maano $\Omega \equiv \dot{C}{C}^T$. Tab: $\Omega +{\Omega }^T=0\Rightarrow \Omega =-{\Omega }^T$

Yeh step kyun? Humne abhi prove kiya ki $\Omega =\dot{C}{C}^T$ skew-symmetric hai. Ek $3\times 3$ skew matrix mein sirf 3 independent numbers hote hain — bilkul angular velocity vector ke components ki tarah.

Step 3 — $\Omega$ ko physically identify karo

Koi bhi skew-symmetric $3\times 3$ matrix kisi vector $\vec{w}$ ke cross-product operator ke roop mein likha ja sakta hai: $[\vec{w}\times ]=\left[\begin{array}{ccc}0& -w_3& w_2\\ w_3& 0& -w_1\\ -w_2& w_1& 0\end{array}\right],[\vec{w}\times ]\vec{a}=\vec{w}\times \vec{a}$

$\vec{w}$ kaun sa vector hai? Ek body-fixed vector $\vec{p}$ lo (body frame mein constant). Inertial coordinates mein iska rate of change hai ${\dot{\vec{p}}}_{\mathcal{N}}={\vec{\omega }}_{\mathcal{N}}\times {\vec{p}}_{\mathcal{N}}$ (rotating vector). $C$ ke through transport karne par (neeche box dekho) pata chalta hai ki $\vec{w}$ angular velocity hai jo body frame mein express hai, ${\vec{\omega }}_{\mathcal{B}}$, aur saath mein ek minus sign bhi hai.

Minus sign kyun? $C$, $\mathcal{N}\to \mathcal{B}$ map karta hai. Jab body $\vec{\omega }$ se aage spin karta hai, toh inertial vectors body frame mein peeche spin karte dikhai dete hain — isliye $-[\omega \times ]$. (Agar aap $C_{\mathcal{N}/\mathcal{B}}=C^T$ track karein, toh ${\dot{C}}^T=[\omega \times ]C^T$ milta hai, bina minus ke. Yeh sign direction ke bookkeeping choice ki wajah se hai.)

Sign cleanly derive karna (Feynman-style transport)

Maano $\vec{p}$ body mein fixed hai (${\vec{p}}_{\mathcal{B}}=$ const). Tab ${\vec{p}}_{\mathcal{B}}=C{\vec{p}}_{\mathcal{N}}$, isliye $0={\dot{\vec{p}}}_{\mathcal{B}}=\dot{C}{\vec{p}}_{\mathcal{N}}+C{\dot{\vec{p}}}_{\mathcal{N}}.$ ${\dot{\vec{p}}}_{\mathcal{N}}={\vec{\omega }}_{\mathcal{N}}\times {\vec{p}}_{\mathcal{N}}$ aur ${\vec{\omega }}_{\mathcal{B}}=C{\vec{\omega }}_{\mathcal{N}}$ use karke: $\dot{C}{\vec{p}}_{\mathcal{N}}=-C({\vec{\omega }}_{\mathcal{N}}\times {\vec{p}}_{\mathcal{N}})=-[{\vec{\omega }}_{\mathcal{B}}\times ]C{\vec{p}}_{\mathcal{N}}.$ Sabhi ${\vec{p}}_{\mathcal{N}}$ ke liye sach hai ⟹ $\dot{C}=-[{\vec{\omega }}_{\mathcal{B}}\times ]C$. Kyun valid hai? Humne identity $C(\vec{a}\times \vec{b})=(C\vec{a})\times (C\vec{b})$ use ki jo proper rotations ke liye sahi hai.

![[3.5.04-DCM-kinematics-—-Ċ-=-−[ω×]C.png]]

WHY it matters (woh 20% jo 80% deta hai)

• Yeh attitude propagation ODE hai: $\dot{C}=-[\omega \times ]C$ integrate karo taaki pata chale spacecraft kahan point kar raha hai.
• Yeh construction se hi orthonormality guarantee karta hai — derivative rotation group $SO(3)$ ke tangent space mein rehta hai, isliye exact solutions kabhi orthogonality "khoते" nahi (numerically aapko renormalize karna padega).
• Yeh quaternion/Euler-angle kinematics ka parent hai: sab usi ek equation ke re-parametrizations hain.

Worked Examples

Common Mistakes

Flashcards

Connections

• Skew-symmetric matrices & cross-product operator
• Rotation group SO(3) and Lie algebra so(3)
• Quaternion kinematics — q̇ = ½ Ω(ω) q
• Euler angle kinematics & gimbal lock
• Matrix exponential and rotation about a fixed axis
• Attitude propagation & determination (TRIAD, QUEST)
• Poisson's equation for rotating frames

![[3.5.04-DCM-kinematics-—-Ċ-=-−[ω×]C.png]]

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