Question bank — Transformation between frames — direction cosine matrices

3.4.2 · D5 · Physics › Rocket Flight Mechanics › Transformation between frames — direction cosine matrices

Shuru karne se pehle, teen words jinka hum baar baar use karenge, simple language mein define kiye gaye hain:

Related maps jo tum kaam karte waqt open rakh sakte ho: Euler angles — yaw pitch roll, Orthogonal matrices and rotation groups SO(3), Reference frames in rocketry — inertial, body, wind, Quaternions and attitude representation.

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Teen single-axis matrices (reference ke liye yahan rakhe gaye hain)

Neeche har ek question in building blocks par lean karta hai, toh chaliye inhe ek baar explicit row/column positions ke saath likh lete hain. Har ek angle $\theta$ se ek axis ke around frame ko rotate karta hai. Rows upar se neeche $1,2,3$ numbered hain; columns left se right $1,2,3$. "Slot $(i,j)$" ka matlab hai row $i$, column $j$.

\mathbf C_2(\theta)=\begin{pmatrix}\cos\theta & 0 & -\sin\theta\\ 0&1&0\\ \sin\theta & 0 & \cos\theta\end{pmatrix},\quad \mathbf C_3(\theta)=\begin{pmatrix}\cos\theta & \sin\theta & 0\\ -\sin\theta & \cos\theta & 0\\ 0&0&1\end{pmatrix}$$ ![[deepdives/dd-physics-3.4.02-d5-s01.png]] > [!intuition] $-\sin\theta$ kahan rehta hai — figure dekho > $\mathbf C_1$ aur $\mathbf C_3$ mein $-\sin\theta$ diagonal ke **neeche** baitha hai (lower off-diagonal slot mein). $\mathbf C_2$ mein woh diagonal ke **upar** jump karta hai, slot $(1,3)$ mein — upper off-diagonal — kyunki axis-2 cyclic order $3\!\to\!1$ use karta hai $1\!\to\!2$ ki jagah. Yahi woh ek trap hai jo zyaadatar logon ko pakadta hai; figure mein odd-one-out ko coral color mein highlight kiya gaya hai. > [!mnemonic] "Inner indices kiss karte hain, outer indices survive karte hain" — explain kiya gaya > Jab tum $\mathbf C^{CB}\,\mathbf C^{BA}$ multiply karte ho, superscripts ko line up karo: $\;C\underbrace{B}_{\text{inner}}\;\underbrace{B}_{\text{inner}}A$. Dono $B$ **beech mein ek doosre ke next** khade hain — ye *inner* indices hain. Ye equal hone chahiye, aur ye cancel ho jaate hain ("kiss" karte hain). Jo dono **bahari edges** par baithte hain, $C$ aur $A$, wo *outer* indices hain; ye answer $\mathbf C^{CA}$ mein survive karte hain. Toh $\mathbf C^{CB}\mathbf C^{BA}=\mathbf C^{CA}$ tabhi kaam karta hai jab inner frames match karein — warna tum ek $B$-machine ko aisa vector feed kar rahe ho jo frame $B$ mein nahi hai. --- ## Rightmost matrix pehle kyun act karta hai (step-by-step) "Pehla physical rotation rightmost jaata hai" — ye rule sirf folklore nahi hai — ye isi se nikalta hai ki matrices vectors par kaise apply hoti hain. Ek fixed vector $\mathbf v$ par 3-2-1 sequence follow karo: $$\mathbf v^{B}=\underbrace{\mathbf C_1(\phi)\,\mathbf C_2(\theta)\,\mathbf C_3(\psi)}_{\mathbf C^{BA}}\;\mathbf v^{A}$$ Matrix multiplication *inside out* apply hoti hai — vector ko touch karne wali matrix pehle act karti hai: $$\mathbf C_1(\phi)\,\mathbf C_2(\theta)\,\big[\mathbf C_3(\psi)\,\mathbf v^{A}\big]$$ 1. **KYA:** $\mathbf C_3(\psi)$ pehle $\mathbf v^{A}$ se milta hai, isse yaw ke baad express karta hai. **KYUN:** ye rightmost hai, yaani innermost factor hai. **DIKHTA HAI:** axes $z$ ke around $\psi$ se swing karte hain (figure, frame 1). 2. **KYA:** $\mathbf C_2(\theta)$ phir us intermediate result par act karta hai (pitch *naye* $y$ ke around). **KYUN:** parentheses inward-out close hote hain. **DIKHTA HAI:** figure ka frame 2. 3. **KYA:** aakhir mein $\mathbf C_1(\phi)$ newest $x$ ke around roll apply karta hai. **DIKHTA HAI:** frame 3. ![[deepdives/dd-physics-3.4.02-d5-s02.png]] Kyunki rotations **commute nahi** karti, order swap karne se destination badal jaata hai — neeche ki figure dikhati hai ki yaw-then-pitch ek arrow ko *alag jagah* land karti hai pitch-then-yaw se, toh dono orderings genuinely alag matrices dete hain. ![[deepdives/dd-physics-3.4.02-d5-s03.png]] --- ## True or false — justify karo Ek DCM us vector ki length change kar deta hai jis par woh act karta hai. ::: **False.** Ek DCM orthogonal hota hai ($\mathbf C\mathbf C^\top=\mathbf I$), toh ye dot products aur isliye lengths preserve karta hai — ye sirf usi arrow ko naye axes mein re-express karta hai, jaise "aage" ko rename karna bina toy car ko hilaye. $\mathbf C^{BA}$ aur $\mathbf C^{AB}$ equal hain kyunki ye dono ek hi pair of frames describe karte hain. ::: **False.** Ye ek doosre ke *transposes* hain ($\mathbf C^{AB}=(\mathbf C^{BA})^\top$); ek $A\to B$ map karta hai aur doosra backwards $B\to A$ chalta hai. Ye sirf trivial case mein equal hain jahan frames coincide hote hain (identity). Har $3\times3$ matrix jiske columns unit length ke hain, ek valid DCM hai. ::: **False.** Unit columns kaafi nahi hain — columns ko ==mutually perpendicular== bhi hona chahiye *aur* determinant $+1$ hona chahiye. Warna tumhare paas ek skewed ya reflected frame hai, right-handed rotation nahi. Kisi bhi DCM ka determinant rotation angle ke hisaab se $+1$ ya $-1$ ho sakta hai. ::: **False.** Ye hamesha exactly $+1$ hota hai. Dono frames right-handed hain, aur ek right-handed→right-handed map orientation preserve karta hai; $-1$ ka matlab hoga mirror flip, jo koi physical rotation produce nahi kar sakti. Yaw phir pitch se rotate karna usi DCM deta hai jaise pitch phir yaw. ::: **False.** Rotations commute nahi karti, toh $\mathbf C_2\mathbf C_3\neq\mathbf C_3\mathbf C_2$ (upar wali non-commutativity figure dekho). Apna sir left-then-down ghoomana tumhe alag jagah point karta hai down-then-left se. Passive rotation ke liye, ek fixed vector ke numbers usi taraf rotate hote hain jis taraf axes ghoomte hain, opposite direction mein. ::: **True.** Agar axes $+\theta$ swing karte hain, toh vector unke relative *apparently* $-\theta$ lean karta hai — ye exactly wahi hai kyun $-\sin\theta$ $\mathbf C_3(\theta)$ ke lower off-diagonal mein appear karta hai. Agar do DCMs mein se har ek orthogonal hai, toh unka product bhi orthogonal hoga. ::: **True.** $(\mathbf C_2\mathbf C_1)(\mathbf C_2\mathbf C_1)^\top=\mathbf C_2\mathbf C_1\mathbf C_1^\top\mathbf C_2^\top=\mathbf I$. Do rotations chain karna phir se ek rotation hai — isliye composition kaam karta hai aur [[Orthogonal matrices and rotation groups SO(3)|SO(3)]] ke andar rehta hai. Ek DCM ka transpose aur uska inverse alag matrices hain. ::: **False.** Ek orthogonal matrix ke liye inverse *hi* transpose hota hai, toh frame change ko undo karne mein koi division nahi lagti — bas table ko ulta palat do. Ek DCM ki diagonal entry $1$ se zyada ho sakti hai. ::: **False.** Har entry kisi angle ka cosine hai, aur cosine $[-1,1]$ mein bounded hota hai. $1$ se upar ki entry ek computational error signal karti hai. $\mathbf C^{BA}$ ki rows $A$-coordinates mein likhe gaye $B$-axes hain. ::: **True.** Row $i$ mein $\hat b_i\cdot\hat a_1,\hat b_i\cdot\hat a_2,\hat b_i\cdot\hat a_3$ collect hote hain — precisely $\hat b_i$ ke components $A$ axes ke along express kiye gaye. --- ## Error dhundho "Data frame $A$ mein hai, toh main $\mathbf C^{AB}$ apply karta hoon frame $B$ paane ke liye." ::: **Wrong direction.** $\mathbf C^{AB}$ ek $B$-vector ko consume karta hai aur $A$ output karta hai. Superscripts ko *output←input* padho; $A\to B$ bhejna hai toh $\mathbf C^{BA}$ chahiye, jiska inner index $A$ tumhare data ke frame se kiss kare. "3-2-1 sequence ke liye roll matrix vector ko pehle multiply karta hai: $\mathbf C^{BA}=\mathbf C_3\mathbf C_2\mathbf C_1(\phi)$." ::: **Order reversed.** *Pehla* physical rotation *rightmost* hona chahiye (woh pehle vector se milta hai, jaise step figure dikhata hai). Sahi hai $\mathbf C^{BA}=\mathbf C_1(\phi)\mathbf C_2(\theta)\mathbf C_3(\psi)$ — yaw $\psi$ pehle act karta hai. "$\mathbf C_2(\theta)$ mein $-\sin\theta$ uske lower-left slot $(3,1)$ mein hai, jaise $\mathbf C_1$ aur $\mathbf C_3$ mein." ::: **Wrong slot.** Reference matrix dekho: $\mathbf C_2$ ka minus sign ==upper== off-diagonal mein baitha hai, slot $(1,3)$ mein, kyunki axis-2 cyclic order $3\!\to\!1$ use karta hai. Slot $(3,1)$ mein actually $+\sin\theta$ hota hai. Hamesha $C_{ij}=\hat b_i\cdot\hat a_j$ se derive karo instead of koi pattern copy karne ke. "$\mathbf C^{CB}$ aur $\mathbf C^{BA}$ combine karne ke liye, unhe add karo: $\mathbf C^{CA}=\mathbf C^{CB}+\mathbf C^{BA}$." ::: **Wrong operation.** Rotations ka composition *multiplication* hai, addition nahi: $\mathbf C^{CA}=\mathbf C^{CB}\mathbf C^{BA}$. Inner $B$ indices kiss karte hain aur cancel ho jaate hain, outer $C\leftarrow A$ bachta hai. "DCM symmetric hai, toh transpose karte waqt multiplication order se koi fark nahi padta." ::: **False premise.** Ek general DCM *symmetric nahi* hota ($\mathbf C\neq\mathbf C^\top$ unless ye identity hai ya $180°$ rotation). Iska transpose frame direction reverse karta hai aur generally ek alag matrix deta hai. "$\det\mathbf C=-1$, toh ye phir bhi ek valid attitude hai, bas ek odd wala." ::: **Rotation nahi hai.** $\det=-1$ ek reflection hai — ye handedness flip karta hai. Swapped column, sign slip, ya left-handed axis definition ke liye recheck karo; ek physical attitude hamesha $\det=+1$ deta hai. --- ## Why questions Entry $C^{BA}_{ij}$ specifically *cosine* kyun hoti hai, angle ki koi aur function kyun nahi? ::: Kyunki ek component ek *projection* hai, aur $\hat a_j$ ko $\hat b_i$ par project karna dot product hai $\hat b_i\cdot\hat a_j=\lVert\hat b_i\rVert\lVert\hat a_j\rVert\cos\theta_{ij}=\cos\theta_{ij}$ (dono unit vectors hain). Cosine exactly "ek direction ka kitna hissa doosre ke along hai" ka measure hai. Vector ko $\hat b_i$ se dot karna sirf $i$-th $B$-component *extract* kyun karta hai? ::: Kyunki $B$-frame orthonormal hai: $\hat b_i\cdot\hat b_i=1$ aur $\hat b_i\cdot\hat b_k=0$ for $k\neq i$. Toh dot product har term ko maar deta hai siwa us ek ke jo $\hat b_i$ ke along hai, woh ek number bachta hai. Hum ek DCM ko full matrix inversion chalaye bina transpose karke invert kyun kar sakte hain? ::: Kyunki iske rows orthonormal hain, jo $\mathbf C\mathbf C^\top=\mathbf I$ deta hai, jo *define* karta hai $\mathbf C^\top$ ko inverse ke roop mein. Ek pure rotation ko undo karna matlab sirf axes ko wapas ghooma dena, aur transpose pehle se hi woh reversed table encode karta hai. Determinant $+1$ hi kyun hona chahiye, koi aur value kyun nahi? ::: Ek DCM ek right-handed frame ko right-handed frame mein map karta hai, orientation aur volume preserve karta hai; orthogonality force karta hai $|\det|=1$, aur orientation-preservation sign $+1$ pick karta hai. Ye wahi "special" hai [[Orthogonal matrices and rotation groups SO(3)|SO(3)]] mein. Rocket engineers attitude ko DCM (ya [[Quaternions and attitude representation|quaternion]]) ke roop mein kyun store karte hain, sirf teen Euler angles ki bajaye? ::: Kyunki Euler angles *gimbal lock* suffer karte hain — $\pm90°$ ke pitch par do axes align ho jaate hain aur ek angle undefined ho jaata hai. DCM ke nau numbers kabhi ek degree of freedom nahi khoote, toh [[Angular velocity and the kinematic equations|attitude propagation]] har jagah well-defined rehti hai. Aik hi physical thrust arrow ke body aur inertial frames mein alag components kyun hote hain? ::: Kyunki components *axes ke against measurements* hain, aur dono frames ke axes alag direction mein point karte hain. Arrow ([[Thrust vectoring and force resolution|thrust]]) unchanged hai; sirf coordinate labels alag hain, $\mathbf F^I=\mathbf C^{IB}\mathbf F^{body}$ ke through related. Product $\mathbf C^{CB}\mathbf C^{BA}$ mein, dono inner $B$ kyun match karne chahiye? ::: Kyunki right matrix ek $B$-vector output karta hai aur left matrix ek $B$-vector input ke roop mein expect karta hai. Agar wo inner frames agree nahi karte, tum ek machine ko wrong-frame numbers feed kar rahe ho — "inner indices kiss" rule precisely yahi compatibility check hai. --- ## Edge cases Zero rotation (dono frames coincide) ke liye kaunsa DCM correspond karta hai? ::: Identity matrix $\mathbf I$. Har axis apne partner ke saath aligned hai, toh har diagonal cosine $\cos 0°=1$ hai aur har off-diagonal $\cos 90°=0$ hai. $\mathbf C_3(180°)$ kaisa dikhta hai, aur kya ye symmetric hai? ::: $\theta=180°$ set karne par $\cos180°=-1$ aur $\sin180°=0$ milta hai, toh $\mathbf C_3(180°)=\begin{pmatrix}-1&0&0\\0&-1&0\\0&0&1\end{pmatrix}$. Kyunki dono off-diagonals vanish ho jaate hain, ye special case *symmetric hai* — ek rare instance jahan $\mathbf C=\mathbf C^\top$. $\theta=90°$ ke pitch par DCM ka kya hota hai aur Euler angles ke liye ye dangerous kyun hai? ::: DCM khud bilkul valid rehta hai (phir bhi orthogonal, $\det=+1$). Lekin *Euler-angle recovery* break ho jaati hai: yaw aur roll axes align ho jaate hain, toh wo angles trade off karte hain aur ambiguous ho jaate hain — classic [[Euler angles — yaw pitch roll|gimbal lock]]. Agar tumne accidentally ek left-handed frame use kiya, toh matrix se immediately kya bata deta hai? ::: Determinant $+1$ ki jagah $-1$ aata hai. Woh flip flag karta hai ki tumhari axis definition mein ek reflection chupi hui hai — frame right-handed nahi hai. Ek full $360°$ se $B$ ko $A$ rotate karne par $\mathbf C^{BA}$ kya hai? ::: Phir se identity. Ek pura chakkar har axis ko apne aap par wapas laata hai, toh sab direction cosines apne $0°$/$90°$ values par wapas aa jaate hain — matrix "yaad nahi rakh sakta" ki ek rotation hua. Ek vector jo exactly rotation axis ke along lie karta hai, us single-axis DCM ke under uske components kaise change hote hain? ::: Woh bilkul change nahi hote. Rotation axis pointwise fixed hota hai, toh us par ek vector (e.g., $\mathbf C_3$ ke under $(0,0,v_3)$) dono frames mein identical components rakhta hai. > [!recall]- Ek-line summary yaad rakhne ke liye > Ek DCM ek lossless *directions ka translation table* hai: ye ek fixed arrow ko re-label karta hai, kabhi resize nahi karta. Inner indices kiss karte hain (match karne chahiye, cancel hote hain), outer survive karte hain; inverse = transpose; determinant $= +1$ hamesha.