Visual walkthrough — DCM kinematics — Ċ = −[ω×]C

3.5.4 · D2 · Physics › Guidance, Navigation & Control (GNC) › [[3.5.04 DCM kinematics — Ċ = −[ω×]C|DCM kinematics — Ċ = −[ω×]C]]

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Step 0 — Teen words jo kisi bhi cheez se pehle chahiye

Koi bhi symbol likhne se pehle, hum teen plain-English pictures par agree karte hain.

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Step 1 — Stubborn fact, drawn

KYA. Teen body arrows ${\hat{b}}_1,{\hat{b}}_2,{\hat{b}}_3$ hamesha length-1 rehte hain aur hamesha mutually perpendicular rehte hain, tumble ke har instant par.

KYU. Woh ek rigid spacecraft pe painted hain. Rigid ka matlab hai painted marks ke beech ki distances aur angles kabhi nahi badal sakti. Yahi is derivation ka poora physical input hai — literally sirf yahi cheez hum assume karte hain.

PICTURE. Figure mein red body-triad dekho: do alag times par yeh naye orientation mein aa gaya hai, lekin arrows ke beech ke chhote right-angle squares identical hain. Kuch nahi stretch hua, kuch nahi shear hua.

Hum "unit length + mutually perpendicular" ko $C$ ke baare mein ek tidy statement mein pack karte hain. Kyunki $C$ ki rows wahi teen body arrows hain jo world numbers mein likhe hain, "woh perpendicular unit vectors hain" kehna exactly yeh hai: $\underset{\text{rows}=\text{body arrows}}{\underbrace{C(t)}}\underset{\text{same arrows, as columns}}{\underbrace{C(t{)}^T}}=\underset{\begin{array}{c}\text{diagonal par 1’s}=\text{unit length}\\ \text{off-diagonal par 0’s}=\text{perpendicular}\end{array}}{\underbrace{I}}$

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Step 2 — Stubborn fact ko differentiate karo

KYA. Hum $C{C}^T=I$ ke dono sides ka $\frac{d}{dt}$ lete hain.

KYU. Hum jaanna chahte hain ki $C$ kitni tezi se badalta hai ($\dot{C}$). Ab tak hamare paas sirf orthonormality equation hai. Toh hum usse differentiate karte hain aur woh rule nikaalte hain jo woh $\dot{C}$ par force karti hai. Yeh classic trick hai: us constraint ko differentiate karo jo tum tod nahi sakte.

Right side $I$ constant hai, toh uski rate $0$ hai. Left side ko product rule chahiye (product ki rate = pehle ki rate × doosra + pehla × doosre ki rate): $\underset{\begin{array}{c}\text{pehla arrow-set}\\ \text{badal raha hai}\end{array}}{\underbrace{\dot{C}{C}^T}}+\underset{\begin{array}{c}\text{doosra arrow-set}\\ \text{badal raha hai}\end{array}}{\underbrace{C{\dot{C}}^T}}=\underset{\text{kyunki }I\text{ frozen hai}}{\underbrace{0}}$

PICTURE. Figure mein arrow ${\hat{b}}_1$ ko ek tiny time $dt$ mein nudge hota dikhaya gaya hai. Uski tip move karti hai, lekin kyunki uski length nahi badal sakti, nudge (red) arrow ke khud ke perpendicular hona chahiye. Yeh perpendicular-nudge hi har cheez ka geometric seed hai: ek velocity jo hamesha us cheez ke sideways hai jise woh move karti hai, woh rotation hai.

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Step 3 — Twist ko naam do, prove karo ki yeh skew hai

KYA. Combination $\dot{C}{C}^T$ ko ek naam do, $\Omega$, aur uski shape discover karo.

KYU. $\dot{C}{C}^T$ baar baar aa raha hai; naam dene se hum iske baare mein reason kar sakte hain. Aur uski shape bahut restrictive nikli hai — woh restriction hi payoff hai.

Note karo ki Step 2 ka doosra term pehle ka sirf transpose hai: $C{\dot{C}}^T=(\dot{C}{C}^T{)}^T={\Omega }^T$. Toh Step 2 ban jaata hai: $\underset{\dot{C}{C}^T}{\underbrace{\Omega }}+\underset{\text{uska mirror}}{\underbrace{{\Omega }^T}}=0⟹\Omega =-{\Omega }^T.$

Ek matrix jo minus apne transpose ke barabar hoti hai, use skew-symmetric kehte hain. Concretely matlab:

• diagonal sab zeros hai (har diagonal entry khud ki negative ke barabar hai, toh $0$ hona chahiye),
• diagonal ke across flip karne se sign flip ho jaata hai.

PICTURE. Figure ek $3\times 3$ grid ko mirror line (diagonal) par rakhta hai. Zeros mirror par baithe hain; har off-diagonal cell apne reflection ka negative hai. Free numbers count karo: sirf diagonal ke upar ke teen entries independent hain. Teen numbers. Yeh yaad rakho — ek spin ke bhi exactly teen numbers hote hain (kitni tezi se $x$ ke baare mein, $y$ ke baare mein, $z$ ke baare mein).

$\Omega =\left[\begin{array}{ccc}0& -c& b\\ c& 0& -a\\ -b& a& 0\end{array}\right]\leftarrow \text{sirf }a,b,c\text{ free hain}$

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Step 4 — Ek skew matrix ek cross product hai

KYA. Dikhao ki hamare skew $\Omega$ se ek vector ko hit karna exactly wahi karta hai jo use vector $\vec{w}=(a,b,c)$ se cross karna karta hai.

KYU. Hum suspect karte hain ki "teen free numbers" ek angular velocity hain. Confirm karne ke liye, hum skew matrix ko us operation se connect karte hain jo spinning ke liye sabko pehle se pata hai: cross product $\vec{\omega }\times \vec{r}$, jiska matlab hai "ek spinning wheel pe ek point ki velocity."

$\Omega$ ko ek test vector $\vec{r}=(x,y,z)$ se multiply karo aur cross product se compare karo: $\underset{\Omega \vec{r}}{\underbrace{\left[\begin{array}{ccc}0& -c& b\\ c& 0& -a\\ -b& a& 0\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]}}=\left[\begin{array}{c}bz-cy\\ cx-az\\ ay-bx\end{array}\right]=\underset{\vec{w}\times \vec{r}}{\underbrace{\left[\begin{array}{c}a\\ b\\ c\end{array}\right]\times \left[\begin{array}{c}x\\ y\\ z\end{array}\right]}}$

Yeh entry by entry match karte hain. Toh hum $\Omega =[\vec{w}\times ]$ likhne ki permission rakhte hain, padho "the cross-product operator of $\vec{w}$." (Yeh hat / vee map machinery hai.)

PICTURE. Figure wheel picture dikhata hai: $\vec{w}$ axle hai (red), $\vec{r}$ ek spoke hai, aur $\vec{w}\times \vec{r}$ tip ki velocity hai — dono ke perpendicular, exactly woh "sideways nudge" jo Step 2 se tha. Abstract skew matrix aur physical spinning wheel ek hi object hain.

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Step 5 — Kaunsa vector dhundho, aur minus sign kyu aata hai*

KYA. Confirm karo ki $\vec{w}$ body ki angular velocity body coordinates mein hai, ${\vec{\omega }}_{\mathcal{B}}$, aur woh minus sign carry karta hai.

KYU. "Koi vector" enough nahi hai; hume uski physical identity aur sign chahiye taaki equation usable ho. Hum ise ek arrow $\vec{p}$ ko track karke paate hain jo body par bolted hai.

Kyunki $\vec{p}$ ship pe painted hai, uske body numbers kabhi nahi badte: ${\vec{p}}_{\mathcal{B}}=C{\vec{p}}_{\mathcal{N}}$ constant hai, toh uski rate zero hai. Ise product-rule karo: $0={\dot{\vec{p}}}_{\mathcal{B}}=\underset{\text{translator badal raha hai}}{\underbrace{\dot{C}{\vec{p}}_{\mathcal{N}}}}+\underset{\text{arrow, world se dekha, spinning}}{\underbrace{C{\dot{\vec{p}}}_{\mathcal{N}}}}$

Ab do facts feed karo jo hum jaante hain:

1. World se dekha, ek body-fixed arrow spins karta hai: ${\dot{\vec{p}}}_{\mathcal{N}}={\vec{\omega }}_{\mathcal{N}}\times {\vec{p}}_{\mathcal{N}}$ (Step 4 ka wheel, world numbers mein).
2. Angular velocity kisi bhi vector ki tarah translate karti hai: ${\vec{\omega }}_{\mathcal{B}}=C{\vec{\omega }}_{\mathcal{N}}$.

Rearrange karo aur rotation identity $C(\vec{a}\times \vec{b})=(C\vec{a})\times (C\vec{b})$ use karo: $\dot{C}{\vec{p}}_{\mathcal{N}}=-C({\vec{\omega }}_{\mathcal{N}}\times {\vec{p}}_{\mathcal{N}})=-(C{\vec{\omega }}_{\mathcal{N}})\times (C{\vec{p}}_{\mathcal{N}})=-[{\vec{\omega }}_{\mathcal{B}}\times ]C{\vec{p}}_{\mathcal{N}}$

Yeh har possible ${\vec{p}}_{\mathcal{N}}$ ke liye hold karta hai, toh aage ke matrices equal hone chahiye: $\boxed{\dot{C}=-[{\vec{\omega }}_{\mathcal{B}}\times ]C}$

MINUS KYU, ek sentence mein, picture ke saath: $C$ world ko cockpit se dekha hua report karta hai. Jab tum (body) apna sir left ghumate ho, poori duniya right ko swing karti dikhti hai. Forward body-spin $\Rightarrow$ backward apparent world-spin $\Rightarrow$ minus.

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Step 6 — Edge & degenerate cases (koi scenario kabhi unshown mat chhodna)

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Ek-picture summary

Poori derivation ek arrow hai: rigid axes orthonormal rehte hain ($C{C}^T=I$) $\to$ differentiate karo ($\dot{C}{C}^T+C{\dot{C}}^T=0$) $\to$ twist skew hai ($\Omega =-{\Omega }^T$, 3 free numbers) $\to$ skew = cross product ($\Omega =[\vec{w}\times ]$) $\to$ woh vector $-{\vec{\omega }}_{\mathcal{B}}$ hai $\Rightarrow \dot{C}=-[{\vec{\omega }}_{\mathcal{B}}\times ]C$.

Recall Poori walkthrough ki Feynman retelling

Tum ek tumbling spaceship mein bandhe ho. Diwaaron pe painted hain teen arrows right angles par — tumhara body frame. Bahar, sitare still baithe hain — world frame. DCM $C$ tumhara translator hai: isko do "Sun kahan hai, star-coordinates mein," yeh batayega "Sun kahan hai, wall-coordinates mein." Jaise-jaise tum tumble karte ho, translator badalta rehta hai. Lekin tumhare painted arrows kabhi stretch ya bend nahi ho sakte — woh length-1 aur ek-doosre ke square rehte hain. Yeh vaada $C{C}^T=I$ ki tarah likho aur poochho "yeh per second kitna badalta hai?" Differentiate karo: translator ki change, uske transpose se multiply, plus uska mirror, zero ke barabar. Matlab: woh change-object minus apne khud ke mirror ke barabar hai. Aise objects ke sirf teen loose numbers hote hain. Teen numbers... kitni tezi se tum apne teen arrows mein se har ek ke baare mein spin karte ho! Aur kisi vector ko aise object se multiply karna exactly usse apne spin axle se cross karna hai — wheel-and-spoke picture. Toh $C$ ki change tumhara spin hai, crossed in. Last twist: kyunki $C$ world ko tumhare cockpit se dekha hua report karta hai, apna sir left ghuma ne se duniya right ko swing karti hai — ek minus sign. Sab milaao: $\dot{C}=-[{\vec{\omega }}_{\mathcal{B}}\times ]C$. Poori cheez ek stubborn fact se nikli: rigid axes square rehte hain.

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Recall

Orthonormality constraint se final equation tak derivation chain batao. ::: $C{C}^T=I\Rightarrow \dot{C}{C}^T+C{\dot{C}}^T=0\Rightarrow \Omega =\dot{C}{C}^T$ skew hai $\Rightarrow \Omega =[\vec{w}\times ]\Rightarrow$ ek body-fixed vector track karne par $\vec{w}=-{\vec{\omega }}_{\mathcal{B}}$ milta hai, toh $\dot{C}=-[{\vec{\omega }}_{\mathcal{B}}\times ]C$. Ek $3\times 3$ skew matrix exactly angular velocity ki information kyu carry karta hai? ::: Iske sirf 3 independent entries hote hain (diagonal par zeros, off-diagonals sign se paired), jo 3 spin components se match karte hain; aur $\Omega \vec{r}=\vec{w}\times \vec{r}$. Minus sign kahan se aata hai, ek sentence mein? ::: $C$ world→body map karta hai (cockpit view), toh forward body spin se world backward spin karti dikhti hai.