Visual walkthrough — Converting between DCM, quaternions, Euler angles

3.5.10 · D2 · Physics › Guidance, Navigation & Control (GNC) › DCM, quaternions, Euler angles ke beech convert karna

Hum sirf EK fact se shuru karte hain: rotation ek vector ko ghoomata hai. Kuch aur assume nahi.

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Step 1 — Vector hota kya hai, aur "rotate it" ka matlab kya hai

KYA HAI: blue arrow $\vec{v}$ spin hokar ek naye blue arrow ${\vec{v}}^{\prime }$ mein badal jaata hai. KYUN MATTER KARTA HAI: is topic ka har representation — matrix, quaternion, Euler angles — bas ek recipe hai jo $\vec{v}$ leti hai aur ${\vec{v}}^{\prime }$ deti hai. Agar hum woh recipe ek baar, clearly likh sakein, toh teenon hamare paas aa jaate hain. PICTURE: note karo ki dono arrows ki same length hai (green circle jise dono touch karte hain). Rotation kabhi stretch nahi karta.

Recall Rotation se kya preserve hota hai?

Length preserve hoti hai ::: tip same radius ke ek sphere par rehti hai; sirf direction badlati hai.

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Step 2 — Axis: Euler's theorem kehta hai ek line kaafi hai

KYA HAI: orange line axis $\hat{e}$ hai; vector uske around angle $\Phi$ sweep karta hai. YEH TOOL KYUN: ek axis + ek angle ke saath kaam karna rotation ka sabse sasta honest description hai — baaki sab is dono objects ki re-packaging hai. Isliye parent note axis–angle ko "the anchor" kehta hai. PICTURE: axis (orange) par existing points bilkul nahi hilte — woh pivot hain. Axis se door sab kuch swing karta hai.

• $\hat{e}$ ::: fixed pivot line, length $1$.
• $\Phi$ ::: us line ke around kitna turn karte hain.

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Step 3 — Vector ko split karo: woh trick jo ise solvable banati hai

Poori derivation ek idea par tikhi hai: ek complete vector ka mushkil rotation aasaan ho jaata hai agar hum pehle vector ko do pieces mein kaat dein — ek jo spin ko ignore kare aur ek jo use puri tarah feel kare.

$\hat{e}\cdot \vec{v}$ (dot product) padhna: yeh matching components ko multiply karta hai aur add karta hai: $\hat{e}\cdot \vec{v}=e_1{v}_x+e_2{v}_y+e_3{v}_z$. Geometrically yeh measure karta hai ki $\vec{v}$ mein se kitna $\hat{e}$ ke along point karta hai — shadow ki length. Yeh tool kyun aur koi nahi? Humein ek number chahiye jo bataye "axis ke along kitna door," aur dot product exactly woh ek-direction-measuring-stick hai.

KYA HAI: blue $\vec{v}$ orange ${\vec{v}}_{\parallel }$ (axis ke upar) aur green ${\vec{v}}_{\perp }$ (flat spin-plane mein) mein split hota hai. KYUN: ${\vec{v}}_{\parallel }$ axis par hai, isliye — Step 2 ke according — rotation use bilkul chhodta nahi. Sirf ${\vec{v}}_{\perp }$ actually rotate hota hai. Humne 3-D rotation ko flat 2-D rotation mein reduce kar diya. PICTURE: green arrow shaded disc mein rehta hai jo orange axis ke perpendicular hai; woh disc wahan hai jahan sab action hota hai.

• $\hat{e}(\hat{e}\cdot \vec{v})$ ::: along-axis shadow ko ek arrow ke roop mein rebuild karo (direction $\hat{e}$, length = dot product).
• $\vec{v}-{\vec{v}}_{\parallel }$ ::: shadow subtract karo, in-plane part bachega.

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Step 4 — Flat part ko rotate karne ke liye ek doosra in-plane arrow chahiye

Disc ke andar, ${\vec{v}}_{\perp }$ clock hand ki tarah $\Phi$ se ghoomta hai. Ek plane mein koi bhi angle describe karne ke liye aapko do perpendicular reference arrows chahiye — ek "cosine" direction aur ek "sine" direction. Humara ek already hai (${\vec{v}}_{\perp }$ khud). Humein uska $90°$ companion chahiye.

Cross product kyun aur kuch nahi? Humein specifically ek aisa arrow chahiye jo (a) spin-plane mein ho aur (b) quarter-turn aage ho. Cross product woh ek operation hai jo ek perpendicular partner manufacture karta hai, isliye yeh natural sine-direction hai.

KYA HAI: disc mein, green ${\vec{v}}_{\perp }$ aur red $\hat{e}\times \vec{v}$ do arrows hain $90°$ par, same length ke. KYUN: ek $0°$ arrow aur ek $90°$ arrow ke saath, koi bhi rotated arrow pehle ka $\cos \Phi$ plus doosre ka $\sin \Phi$ hai — usi tarah jaise $(\cos \Phi ,\sin \Phi )$ ek circle trace karta hai. PICTURE: rotated tip ${\vec{v}}_{\perp }^{\prime }$ (dashed) tak pahuncha jaata hai green ke along $\cos \Phi$ chalke aur red ke along $\sin \Phi$ chalke.

${\vec{v}}_{\perp }^{\prime }=\underset{\text{shrink the }{0}^{\circ }\text{ arrow}}{\underbrace{{\vec{v}}_{\perp }\cos \Phi }}+\underset{\text{add the }90^{\circ }\text{ arrow}}{\underbrace{(\hat{e}\times \vec{v})\sin \Phi }}$

• $\cos \Phi$ ::: original in-plane arrow kitna survive karta hai.
• $\sin \Phi$ ::: perpendicular companion kitna mix in karte hain.

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Step 5 — Untouched piece wapas add karo: Rodrigues' formula saamne aata hai

Ab reassemble karo. Rotated full vector rotated flat part plus parallel part hai jo kabhi hili nahi.

${\vec{v}}^{\prime }=\underset{\text{Step 4}}{\underbrace{{\vec{v}}_{\perp }\cos \Phi }}+\underset{\text{Step 4}}{\underbrace{(\hat{e}\times \vec{v})\sin \Phi }}+\underset{\text{untouched, Step 3}}{\underbrace{\hat{e}(\hat{e}\cdot \vec{v})}}$

${\vec{v}}_{\perp }=\vec{v}-\hat{e}(\hat{e}\cdot \vec{v})$ substitute karo aur $\hat{e}(\hat{e}\cdot \vec{v})$ terms collect karo:

KYA HAI: teen coloured arrows tip-to-tail stack hoke exactly ${\vec{v}}^{\prime }$ par land karte hain. $(1-\cos \Phi )$ correction kyun: pehla term along-axis part ko bhi fade kar gaya (woh nahi hona chahiye tha — woh part kabhi nahi hilta). Teesra term exactly woh missing $(1-\cos \Phi )$ share $\hat{e}$ ke along wapas add karta hai. Check karo: axis ke along fade $\cos \Phi$ plus top-up $(1-\cos \Phi )$ milke $1$ hote hain — full restoration. Yeh Rodrigues Rotation Formula flesh mein hai. PICTURE: orange top-up arrow dekho — woh seedha axis ke upar point karta hai aur uska ek hi kaam hai: over-fade undo karna.

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Step 6 — Arrows se matrix tak: DCM padhna

Ek matrix $C$ bas ek machine hai jo $\vec{v}$ khata hai aur ${\vec{v}}^{\prime }=C\vec{v}$ rule se ${\vec{v}}^{\prime }$ output karta hai. Rodrigues ko aisi machine mein convert karne ke liye hum har term ko "(matrix) times $\vec{v}$" ke roop mein rewrite karte hain.

Yeh teen matrices kyun? Har ek bas ek Rodrigues term ko ek matrix ke roop mein re-express karta hai jo $\vec{v}$ par act kare — koi naya physics nahi, sirf naya bookkeeping taaki hum rotation ko 9 numbers ke roop mein store kar sakein.

$\boxed{C=I\cos \Phi +[\hat{e}{]}_{\times }\sin \Phi +\hat{e}{\hat{e}}^{\top }(1-\cos \Phi )}$

KYA HAI: same teen geometric pieces, ab teen $3\times 3$ blocks ke roop mein dikhaaye gaye jo $C$ mein add hote hain. KYUN: jab $C$ exist kare, toh koi bhi vector rotate karna ek multiply hai — isliye DCM vectors actually move karne ka workhorse hai. Dekho Direction Cosine Matrix Properties. PICTURE: diagonal $\cos \Phi$ se dominate hai; anti-symmetric off-diagonal $\sin \Phi$ carry karta hai (cross-product part); symmetric leftover projection carry karta hai.

• $C$ ka symmetric part ::: $I\cos \Phi +\hat{e}{\hat{e}}^{\top }(1-\cos \Phi )$ se aata hai.
• $C$ ka anti-symmetric part ::: $[\hat{e}{]}_{\times }\sin \Phi$ se aata hai — isliye $C_{23}-C_{32}$ baad mein axis info isolate karta hai.

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Step 7 — Edge cases: picture ko SABHON ko cover karna chahiye

Ek derivation tabhi trustworthy hai jab woh degenerate inputs mein bhi survive kare.

KYA HAI: teen miniature clocks jo same start arrow par $\Phi =0$, $\Phi =90^{\circ }$, $\Phi =180^{\circ }$ dikhate hain. KYUN: reader ko kabhi aisa spin angle nahi milna chahiye jo picture ne nahi dikhaya — "kuch mat karo" se "flip" tak. PICTURE: $\Phi =180^{\circ }$ par red $\sin$-arrow ki length $0$ hai; poora result fade + top-up se carry hota hai.

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Step 8 — Same picture, half angle: quaternion

Quaternion koi naya derivation nahi hai — yeh Step 5 hai sandwich rule ${\vec{v}}^{\prime }=q\vec{v}{q}^{-1}$ ke through dekha gaya, jo geometrically $q$ ko do baar apply karta hai. Ek hi $\Phi$ turn par land karne ke liye, har $q$ ko sirf $\Phi /2$ carry karna hoga.

KYA HAI: ek semicircle jo $\Phi /2$ (quaternion) ko exactly poore swing $\Phi$ (DCM) ka aadha dikhata hai. HALF KYUN: kyunki sandwich double karta hai, aur kyunki yeh $q$ aur $-q$ ko same physical turn describe karne deta hai — koi singularities nahi, Quaternion Kinematics & Propagation mein time-integration ke liye ideal. PICTURE: in $q_0,q_1,q_2,q_3$ ko parent note ke $C(q)$ table mein feed karo aur, $\sin \Phi =2\sin \frac{\Phi }{2}\cos \frac{\Phi }{2}$ aur $1-\cos \Phi =2{\sin }^2\frac{\Phi }{2}$ use karke, aap Step 6 ka wahi matrix regenerate kar loge. Same rotation, teen costumes.

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One-picture summary

Ek rotation → teen encodings. Axis $\hat{e}$ aur angle $\Phi$ centre mein hain. Vector split karo (dot & cross), teen arrows add karo → Rodrigues → DCM $C$ mein collect karo. Angle aadha karo → quaternion $q$. Teen planar turns chain karo → Euler angles. DCM neutral hub hai jisse sab route karte hain — kabhi bhi quaternion↔Euler directly convert mat karo (compare karo Rotation Sequences (3-2-1 vs 3-1-3) aur Gimbal Lock).

Recall Feynman: plain words mein poora walkthrough

Ek arrow lo. Woh single line dhoondo jiske around yeh spin karta hai (Euler ne promise kiya tha ek line hamesha kaafi hai) aur woh angle jitna yeh ghoomta hai. Arrow ko do mein kaato: shadow jo spin line par hai — woh part kabhi nahi hilta — aur flat leftover tumhare saamne wale disc mein. Us disc mein, ek arrow ko spin karna bas clock-hand wali baat hai: original arrow ka $\cos$ rakho, ek fresh arrow ka $\sin$ add karo jo quarter-turn aage point karta hai (cross product tumhe woh companion free mein de deta hai). Untouched shadow wapas add karo, aur — kyunki cosine ne woh shadow bhi fade kar diya tha — line ke along ek chhota top-up daalo use fix karne ke liye. Woh teen arrows Rodrigues formula hain. Har arrow-recipe ko ek chhote number-machine ke roop mein rewrite karo aur stack karo: yahi rotation matrix hai, DCM. Agar aap sirf aadha angle turn karo aur sandwich rule use karo, toh aapko quaternion milta hai — same turn, sasta aur koi singular spots nahi. Aur agar aap $z$, phir $y$, phir $x$ sequence mein spin karna prefer karo, toh aapko human roll-pitch-yaw Euler angles milte hain — readable, lekin woh pitch $90^{\circ }$ par jam jaate hain (gimbal lock). Teen languages, ek geometric fact. Matrix ke through translate karo aur aap kabhi meaning nahi khote.