Exercises — Transformation between frames — direction cosine matrices

3.4.2 · D4 · Physics › Rocket Flight Mechanics › Transformation between frames — direction cosine matrices

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Elementary matrices, yahan copy ki hain taaki scroll na karna pade:

\mathbf C_2(\theta)=\begin{pmatrix}\cos\theta & 0 & -\sin\theta\\ 0&1&0\\ \sin\theta & 0 & \cos\theta\end{pmatrix},\quad \mathbf C_1(\theta)=\begin{pmatrix}1&0&0\\ 0&\cos\theta & \sin\theta\\ 0&-\sin\theta&\cos\theta\end{pmatrix}$$ ![[deepdives/dd-physics-3.4.02-d4-s01.png]] Upar ki figure woh **passive** convention dikhati hai jo hum poore note mein use karte hain: kali arrow ek physical vector hai; grey axes frame $A$ hain; orange axes frame $B$ hain jo $+\theta$ se ghumi hain. Dhyan do arrow kabhi nahi hilti — sirf woh axes hilte hain jinke against hum ise measure karte hain. DCM ki poori kahaani yahi hai. --- ## Level 1 — Recognition Yeh check karte hain ki aap ek DCM *padh* sako: uska type, uska inverse, uska determinant identify karo, bina bhaari arithmetic ke. ### Exercise 1.1 Yeh kaunsi single-axis rotation hai, aur kis angle se? $$\mathbf M=\begin{pmatrix}1&0&0\\0&\tfrac12&\tfrac{\sqrt3}{2}\\0&-\tfrac{\sqrt3}{2}&\tfrac12\end{pmatrix}$$ > [!recall]- Solution > Top-left mein ek akela $1$ hai jiska row aur column baaki zero hai, toh **axis 1** untouched hai — yeh ek $\mathbf C_1$ rotation hai. Lower-right block ko $\mathbf C_1(\theta)$ se compare karo: $\cos\theta=\tfrac12$ aur $\sin\theta=\tfrac{\sqrt3}{2}$ (upper-right slot mein baitha hai, $\mathbf C_1$ ke $+\sin$ se match karta hai). Woh angle jiska cosine $\tfrac12$ aur sine $\tfrac{\sqrt3}{2}$ hai woh $\theta=60°$ hai. > **Answer:** $\mathbf M=\mathbf C_1(60°)$. ### Exercise 1.2 Compute kiye bina, $\mathbf C_2(40°)$ ka inverse do alag-alag tarike se likho. > [!recall]- Solution > Ek rotation table ka inverse uska transpose hota hai (rows orthonormal hain, isliye $\mathbf C\mathbf C^\top=\mathbf I$). Aur $+40°$ spin ko undo karna $-40°$ spin karne ke barabar hai. Toh: > $$\mathbf C_2(40°)^{-1}=\mathbf C_2(40°)^\top=\mathbf C_2(-40°).$$ > Dono se milta hai $\begin{pmatrix}\cos40° & 0 & \sin40°\\0&1&0\\-\sin40° & 0 & \cos40°\end{pmatrix}$. ### Exercise 1.3 Kya neeche di matrix ek valid DCM (proper rotation) hai? Ek-line ka reason do. $$\mathbf N=\begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}$$ > [!recall]- Solution > Iske rows orthonormal hain, isliye $\mathbf N\mathbf N^\top=\mathbf I$ — woh part pass hai. Lekin $\det\mathbf N=-1$ hai (do axes swap karna handedness palat deta hai). Do right-handed frames ke beech ek proper rotation ko $\det=+1$ chahiye. **Answer: nahi** — yeh ek reflection hai, rotation nahi. > [!mistake] L1 ka trap: "orthogonal kaafi hai" > **Kyun sahi lagta hai:** tumne seekha ki $\mathbf C\mathbf C^\top=\mathbf I$ DCM ki pehchaan hai, toh koi bhi matrix jo isse pass kare legal lagti hai. **Trap:** woh test sirf yeh prove karta hai ki columns orthonormal hain — ek *mirror* ($\det=-1$) bhi isse pass kar leta hai. **Fix:** hamesha **dono** check karo: $\mathbf C\mathbf C^\top=\mathbf I$ *aur* $\det=+1$. Determinant hi rotation ko reflection se alag karta hai. --- ## Level 2 — Application Ab tum numbers plug in karo aur crank ghuma ke chalao. ### Exercise 2.1 Frame $B$ ground frame hai jo axis 1 ke around $\phi=90°$ se rolled hai. Ek vector $\mathbf v^A=(0,\,4,\,0)$ ground-$y$ ke along point karta hai. $\mathbf v^B$ nikalo. > [!recall]- Solution > $$\mathbf C_1(90°)=\begin{pmatrix}1&0&0\\0&0&1\\0&-1&0\end{pmatrix},\qquad > \mathbf v^B=\mathbf C_1(90°)\begin{pmatrix}0\\4\\0\end{pmatrix}=\begin{pmatrix}0\\0\\-4\end{pmatrix}.$$ > **Geometrically kya hua:** humne axes ko $x$ ke around $+90°$ ghuma diya, toh $\hat b_2$ upar swing karke wahan aa gaya jahan purana $\hat a_3$ tha aur $\hat b_3$ wahan aa gaya jahan $-\hat a_2$ tha. Arrow, ab bhi purane-$y$ ke along point karta hua, ab $-\hat b_3$ ke along point karta dikhta hai. Length check: $|\mathbf v^B|=4=|\mathbf v^A|$ ✓. ### Exercise 2.2 Same frame $B=\mathbf C_3(30°)$ jaise parent Example 1, lekin ab wind $\mathbf v^A=(0,\,10,\,0)$ m/s hai (ground-$y$ ke along). $\mathbf v^B$ nikalo. > [!recall]- Solution > $$\mathbf v^B=\mathbf C_3(30°)\begin{pmatrix}0\\10\\0\end{pmatrix}=\begin{pmatrix}\sin30°\cdot10\\ \cos30°\cdot10\\0\end{pmatrix}=\begin{pmatrix}5.0\\ 8.66\\0\end{pmatrix}\text{ m/s}.$$ > **Is baar $+5$ kyun hai (Example 1 ki tarah $-5$ nahi)?** Example 1 mein vector $\hat a_1$ ke along tha; axes ko $+30°$ rotate karne se woh *negative* $\hat b_2$ ki taraf chala gaya. Yahan vector $\hat a_2$ ke along hai; usi axes rotation se woh *positive* $\hat b_1$ ki taraf jhuk jaata hai. Sign isliye flip hoti hai kyunki vector alag axis se shuru hua tha. Length: $\sqrt{5^2+8.66^2}=10$ ✓. ### Exercise 2.3 Ek body vector $\mathbf u^B=(0,\,0,\,6)$ diya gaya hai, aur $\mathbf C^{BA}=\mathbf C_2(90°)$ hai. Vector ko frame $A$ mein recover karo. > [!recall]- Solution > Hamare paas ek $B$-vector hai aur $A$ chahiye, toh hume $\mathbf C^{AB}=(\mathbf C^{BA})^\top$ chahiye: > $$\mathbf C_2(90°)=\begin{pmatrix}0&0&-1\\0&1&0\\1&0&0\end{pmatrix},\qquad > (\mathbf C_2(90°))^\top=\begin{pmatrix}0&0&1\\0&1&0\\-1&0&0\end{pmatrix}.$$ > $$\mathbf u^A=\begin{pmatrix}0&0&1\\0&1&0\\-1&0&0\end{pmatrix}\begin{pmatrix}0\\0\\6\end{pmatrix}=\begin{pmatrix}6\\0\\0\end{pmatrix}.$$ > **Answer:** $\mathbf u^A=(6,0,0)$. > [!mistake] L2 ka trap: di gayi matrix ko ulta lagaana > **Kyun sahi lagta hai:** tumhe $\mathbf C^{BA}$ *diya gaya* tha, toh lagta hai "the" matrix isi se multiply karo. **Trap:** 2.3 mein data ek $B$-vector tha. $\mathbf C^{BA}$ sirf $A$-vectors khaata hai (inner index data se match honi chahiye). $\mathbf C^{BA}\mathbf u^B$ multiply karna galat direction mein rotate karta hai. **Fix:** inner superscript ko apne data ke frame se match karo; agar match na kare, pehle transpose lo. --- ## Level 3 — Analysis Yahan tum combine, invert, aur reason karte ho ki result *kyun* aisa aata hai. ### Exercise 3.1 Yaw $\psi=90°$, pitch $\theta=0°$, roll $\phi=0°$ ke liye poora 3-2-1 attitude DCM banao. Phir words mein batao ki yeh physical orientation kya represent karta hai. > [!recall]- Solution > $\theta=\phi=0$ ke saath, $\mathbf C_2(0)=\mathbf C_1(0)=\mathbf I$, toh > $$\mathbf C^{BA}=\mathbf C_1(0)\mathbf C_2(0)\mathbf C_3(90°)=\mathbf C_3(90°)=\begin{pmatrix}0&1&0\\-1&0&0\\0&0&1\end{pmatrix}.$$ > **Physical meaning:** body shared vertical axis ke around $90°$ yaw hua hai. Body-$x$ (nose) ab inertial-$y$ ke along point karta hai; isliye $\mathbf C^{BA}$ ki pehli row, jo *hai* $\hat b_1$ ko $A$-coordinates mein likha, $(0,1,0)$ padhti hai. Kuch tilt nahi hota, isliye axis 3 $(0,0,1)$ rehti hai. ### Exercise 3.2 $\mathbf C^{BI}=\mathbf C_2(30°)\mathbf C_3(90°)$ diya hua (parent Example 2), direct multiplication se verify karo ki $\mathbf C^{BI}(\mathbf C^{BI})^\top=\mathbf I$ hai, sirf pehli row ko khud se aur doosri row se check karke. > [!recall]- Solution > Parent se, $\mathbf C^{BI}=\begin{pmatrix}0&0.866&-0.5\\-1&0&0\\0&0.5&0.866\end{pmatrix}$. > Row 1 hai $\mathbf r_1=(0,\,0.866,\,-0.5)$; row 2 hai $\mathbf r_2=(-1,0,0)$. > $$\mathbf r_1\cdot\mathbf r_1=0+0.866^2+0.5^2=0.75+0.25=1\ ✓$$ > $$\mathbf r_1\cdot\mathbf r_2=(0)(-1)+0.866(0)+(-0.5)(0)=0\ ✓$$ > Rows unit-length hain aur mutually perpendicular hain — exactly yeh statement hai ki $\mathbf C^{BI}(\mathbf C^{BI})^\top=\mathbf I$. **Kyun yeh *zaroor* hold karta hai:** har row ek body axis hai jo inertial coordinates mein express ki gayi hai, aur body axes construction se orthonormal hote hain. ### Exercise 3.3 Do frames $\mathbf C^{CA}=\mathbf C^{CB}\mathbf C^{BA}$ se relate hain jahan $\mathbf C^{BA}=\mathbf C_3(90°)$ aur $\mathbf C^{CB}=\mathbf C_3(-90°)$. $\mathbf C^{CA}$ compute karo aur result explain karo. > [!recall]- Solution > $$\mathbf C^{CA}=\mathbf C_3(-90°)\mathbf C_3(90°)=\mathbf C_3(-90°+90°)=\mathbf C_3(0°)=\mathbf I.$$ > (*Same* axis ke rotations apne angles add karte hain.) **$\mathbf I$ kyun:** frame $C$, $B$ se $-90°$ ghuma hua hai, jo khud $A$ se $+90°$ ghuma hua hai. Dono cancel ho jaate hain, toh $C$ aur $A$ coincide karte hain — koi re-labelling nahi chahiye, identity table. > [!mistake] L3 ka trap: *alag-alag* axes mein angles add karna > **Kyun sahi lagta hai:** 3.3 mein trick "same axis → angles add karo" khoobsurati se kaam ki, toh tum ise $\mathbf C_2(30°)\mathbf C_3(90°)$ ke liye try karte ho aur $\mathbf C(120°)$ likh dete ho. **Trap:** angle-adding tabhi kaam karta hai jab dono rotations ek axis share karte hain. Alag-alag axes commute nahi karte aur genuinely 3-D orientation produce karte hain. **Fix:** sirf $\mathbf C_k(\alpha)\mathbf C_k(\beta)=\mathbf C_k(\alpha+\beta)$ collapse karo jab subscript $k$ *identical* ho; warna matrices multiply karo. --- ## Level 4 — Synthesis Ab tum kai ideas ko ek multi-step problem mein assemble karte ho, jaise ek real attitude calculation. ### Exercise 4.1 Ek rocket ka attitude yaw $\psi=90°$, pitch $\theta=90°$, roll $\phi=0°$ hai (3-2-1 sequence). Body frame mein thrust $\mathbf F^{body}=(F,0,0)$ N nose ke along hai. Inertial frame mein thrust direction nikalo, aur batao nose kahan point karta hai. > [!recall]- Solution > $\mathbf C^{BI}=\mathbf C_1(0)\mathbf C_2(90°)\mathbf C_3(90°)$ banao. $\mathbf C_1(0)=\mathbf I$ ke saath: > $$\mathbf C_2(90°)=\begin{pmatrix}0&0&-1\\0&1&0\\1&0&0\end{pmatrix},\quad \mathbf C_3(90°)=\begin{pmatrix}0&1&0\\-1&0&0\\0&0&1\end{pmatrix}.$$ > $$\mathbf C^{BI}=\mathbf C_2(90°)\mathbf C_3(90°)=\begin{pmatrix}0&0&-1\\-1&0&0\\0&1&0\end{pmatrix}.$$ > Hamare paas ek body vector hai, inertial chahiye, toh transpose: > $$\mathbf F^I=(\mathbf C^{BI})^\top\begin{pmatrix}F\\0\\0\end{pmatrix} > =\begin{pmatrix}0&-1&0\\0&0&1\\-1&0&0\end{pmatrix}\begin{pmatrix}F\\0\\0\end{pmatrix} > =\begin{pmatrix}0\\0\\-F\end{pmatrix}.$$ > **Answer:** $\mathbf F^I=(0,0,-F)$ — thrust seedha *neeche* inertial $-z$ ke along point karta hai. Equivalently nose (body-$x$) inertial $-z$ ke along point karta hai: rocket $90°$ yaw hua phir nose $90°$ pitch hokar ground ki taraf aa gaya. Magnitude $|\mathbf F^I|=F$ ✓, kyunki koi bhi rotation length preserve karta hai. ### Exercise 4.2 Usi $\mathbf C^{BI}$ ko use karte hue jo 4.1 se mili, ek inertial gust $\mathbf g^I=(0,\,G,\,0)$ N se push karti hai (inertial-$y$ ke along). Gust ko body frame mein express karo, aur batao yeh kaunsa body axis load karta hai. > [!recall]- Solution > Data inertial hai, hume body chahiye, aur $\mathbf C^{BI}$ inertial vectors khaata hai — perfect, koi transpose nahi: > $$\mathbf g^{body}=\mathbf C^{BI}\begin{pmatrix}0\\G\\0\end{pmatrix} > =\begin{pmatrix}0&0&-1\\-1&0&0\\0&1&0\end{pmatrix}\begin{pmatrix}0\\G\\0\end{pmatrix} > =\begin{pmatrix}0\\0\\G\end{pmatrix}.$$ > **Answer:** $\mathbf g^{body}=(0,0,G)$ — gust body axis 3 load karta hai (ek pure side/normal load, koi axial component nahi). Yeh woh number hai jo ek structural engineer ko chahiye, aur yeh seedha milta hai uss matrix choose karne se jiska inner index data se match karta tha. > [!mistake] L4 ka trap: tab transpose karna jab nahi karna chahiye > **Kyun sahi lagta hai:** 4.1 mein tumhe *transpose karna hi pada* (body data, inertial chahiye tha), toh transpose karna "safe default" lagta hai. **Trap:** 4.2 mein data already inertial tha aur $\mathbf C^{BI}$ already inertial input accept karta hai — transpose karna gust ko galat direction mein rotate kar deta. **Fix:** aadat se transpose mat karo. Har baar *inner* superscript ko apne data frame se check karo; transpose sirf **tabhi** karo jab woh agree na karte hon. --- ## Level 5 — Mastery Ek problem jo orthogonality, composition, determinant, aur geometry sab ko ek saath baandhti hai — aur ek degenerate case cover karti hai. ### Exercise 5.1 Ek DCM $\mathbf C^{BA}=\mathbf C_3(\psi)\mathbf C_2(\theta)\mathbf C_3(-\psi)$ ke roop mein banaya gaya hai arbitrary $\psi$ ke liye $\theta=180°$ ke saath. (a) Dikhao ki $\det\mathbf C^{BA}=+1$ hai $\psi$ se independent. (b) $\theta=180°$ ki degenerate value ke liye $\mathbf C^{BA}$ explicitly compute karo aur interpret karo. > [!recall]- Solution > **(a)** Product ka determinant determinants ka product hota hai, aur har elementary rotation ka $\det=+1$ hota hai: > $$\det\mathbf C^{BA}=\det\mathbf C_3(\psi)\cdot\det\mathbf C_2(180°)\cdot\det\mathbf C_3(-\psi)=1\cdot1\cdot1=1.$$ > Yeh **har** $\psi$ ke liye hold karta hai — orientation-preserving maps compose hoke bhi orientation preserve karti hain. > > **(b)** $\theta=180°$ par: $\cos180°=-1,\ \sin180°=0$, toh > $$\mathbf C_2(180°)=\begin{pmatrix}-1&0&0\\0&1&0\\0&0&-1\end{pmatrix}.$$ > Ab ise sandwich karo. $c=\cos\psi,\ s=\sin\psi$ likho: > $$\mathbf C_3(\psi)=\begin{pmatrix}c&s&0\\-s&c&0\\0&0&1\end{pmatrix},\quad > \mathbf C_3(-\psi)=\begin{pmatrix}c&-s&0\\s&c&0\\0&0&1\end{pmatrix}.$$ > Pehle $\mathbf C_2(180°)\mathbf C_3(-\psi)=\begin{pmatrix}-c&s&0\\s&c&0\\0&0&-1\end{pmatrix}$. Phir left side se $\mathbf C_3(\psi)$ multiply karo: > $$\mathbf C^{BA}=\begin{pmatrix}c&s&0\\-s&c&0\\0&0&1\end{pmatrix}\begin{pmatrix}-c&s&0\\s&c&0\\0&0&-1\end{pmatrix} > =\begin{pmatrix}-c^2+s^2 & 2sc & 0\\ 2sc & s^2-c^2 & 0\\ 0&0&-1\end{pmatrix}.$$ > $c^2-s^2=\cos2\psi$ aur $2sc=\sin2\psi$ use karke: > $$\mathbf C^{BA}=\begin{pmatrix}-\cos2\psi & \sin2\psi & 0\\ \sin2\psi & \cos2\psi & 0\\ 0&0&-1\end{pmatrix}.$$ > **Interpretation:** $180°$ pitch $z$-axis flip karta hai (bottom-right $=-1$) aur sandwiching yaws top-left block mein ek *reflection-like* pattern produce karte hain — phir bhi $\det=+1$ hai, toh yeh ab bhi ek proper rotation hai ($xy$-plane mein angle $\psi$ par ek axis ke around $180°$ turn). $\psi=0$ ke liye yeh $\mathrm{diag}(-1,1,-1)=\mathbf C_2(180°)$ reduce ho jaata hai, hamaari starting flip, consistency confirm karta hua. Quick check: $\psi=45°$ lene se $2\psi=90°$ milta hai, toh $\mathbf C^{BA}=\begin{pmatrix}0&1&0\\1&0&0\\0&0&-1\end{pmatrix}$, jiska determinant $(0)(0)(-1)-\dots = +1$ hai ✓. > [!mistake] L5 ka trap: diagonal par $-1$ dekhkar reflection assume karna > **Kyun sahi lagta hai:** tumne seekha ki $\det=-1$ reflection hai, aur yahan tum diagonal par $-1$s *dekhte* ho, toh "zaroor" improper hoga. **Trap:** individual diagonal $-1$s handedness ke baare mein kuch nahi kehte — sirf **poora determinant** kehta hai. Yahan do sign-flips multiply hokar $+1$ bana dete hain, ise ek genuine rotation banaye rakhte hain. **Fix:** properness ko poori matrix ke $\det$ se judge karo, kisi stray $-1$ ko dekhkar nahi. --- ## Wrap-up recall > [!recall]- Poori ladder ke one-line answers > - L1: **dono** check karo — $\mathbf C\mathbf C^\top=\mathbf I$ aur $\det=+1$. > - L2: inner superscript ko apne data frame se match karo; sirf mismatch par transpose karo. > - L3: angles sirf *same* axis ke liye add karo. > - L4: transpose default nahi hai — har baar superscripts se decide karo. > - L5: properness *poore* determinant ki property hai, kisi ek entry ki nahi. Yeh bhi dekho: [[Euler angles — yaw pitch roll]], [[Reference frames in rocketry — inertial, body, wind]], [[Orthogonal matrices and rotation groups SO(3)]], [[Thrust vectoring and force resolution]].