Visual walkthrough — Quaternion kinematics — q̇ = ½ Ξ(q) ω

3.5.9 · D2 · Physics › Guidance, Navigation & Control (GNC) › Quaternion kinematics — q̇ = ½ Ξ(q) ω

Shuru karne se pehle: teen words, teen tasveerein.

Hum prerequisite tasveerein tab link karte hain jab zarurat ho: Angular velocity and the body frame, Rotation matrices and SO(3), Rodrigues rotation formula.

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Step 1 — Ek rotation ek axis aur ek angle hota hai

KYA. Rigid body ka koi bhi turn, chahe kitna bhi complicated lage, ek hi spin hai: ek seedhi line chuno (woh axis $\hat{n}$, ek unit arrow) aur uske around kisi angle $\theta$ se ghuma do. Ek rotation ki yahi poori baat hai.

KYUN. Hume "orientation" ka sabse seedha-sacha description chahiye pehle, uske baad hum pooch sakte hain ki yeh kaise change hoti hai. Euler's rotation theorem guarantee karta hai ki yeh axis–angle pair hamesha exist karta hai, isliye hum yahan se shuru karte hain, nau messy matrix entries se nahi (dekho Rotation matrices and SO(3)).

TASVEER.

Laal arrow axis $\hat{n}$ hai (length 1). Orange sweep angle $\theta$ hai. Aage jo bhi hai woh sirf ek chalak tarika hai in do cheezon ko four numbers mein store karne ka.

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Step 2 — Spin ko four numbers mein store karo (half angle kyun)

KYA. Hum axis $\hat{n}$ aur angle $\theta$ ko ek quaternion $q$ mein pack karte hain, jo four numbers ka ek stack hai:

$$q=\underset{\begin{array}{c}\text{one scalar}\\ q_0\end{array}}{\underbrace{\cos \frac{\theta }{2}}},\underset{\begin{array}{c}\text{three numbers}\\ {\mathbf{q}}_v=(q_1,q_2,q_3)\end{array}}{\underbrace{\hat{n}\sin \frac{\theta }{2}}}$$

Term by term padho: $q_0$ ek akela number hai jo track karta hai kitna ghume; ${\mathbf{q}}_v$ woh axis hai, $\sin \frac{\theta }{2}$ se chhooti karke, jo track karta hai kis taraf.

$\frac{\theta }{2}$ kyun, $\theta$ kyun nahi? Yahi mashoor $\frac{1}{2}$ ka beej hai. Half-angle store karna hi woh cheez hai jo quaternion multiplication ko rotations sahi se compose karne deti hai (yeh "double cover" hai: $q$ aur $-q$ dono ek hi physical orientation ko naam dete hain). Yeh half yaad rakho — yeh Step 6 mein wapas aayega.

TASVEER.

Jaise asli angle $\theta$ (orange) sweep karta hai, store hue four numbers $\frac{\theta }{2}$ ke cosine aur sine curves par sawaar hote hain (teal, plum). Dhyaan do ki yeh asli turn se aadhi speed pe chalte hain — yahi half-angle hai, drawn.

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Step 3 — Do rotations milkar quaternions ko multiply karte hain

KYA. "Rotation $q$ karo, phir thodi aur tiny rotation $\delta q$ karo" ek single product $q\otimes \delta q$ ke roop mein likha jaata hai. Symbol $\otimes$ Hamilton product hai:

$$p\otimes q=\left[\begin{array}{c}\underset{\text{new scalar part}}{\underbrace{p_0{q}_0-{\mathbf{p}}_v\cdot {\mathbf{q}}_v}}\\ \underset{\text{axes blended}}{\underbrace{p_0{\mathbf{q}}_v+q_0{\mathbf{p}}_v}}+\underset{\text{order matters}}{\underbrace{{\mathbf{p}}_v\times {\mathbf{q}}_v}}\end{array}\right]$$

Term by term: dot product ${\mathbf{p}}_v\cdot {\mathbf{q}}_v$ (ek single number jo alignment measure karta hai) scalar ko feed karta hai; cross product ${\mathbf{p}}_v\times {\mathbf{q}}_v$ (ek naya arrow, dono ke perpendicular) non-commutativity ka fingerprint hai.

KYUN. Rotations compose karna commute nahi karta — pehle turn phir roll, aur pehle roll phir turn alag hote hain. Sirf woh product jo $\times$ term carry kare is behaviour ko copy kar sakta hai. Yahi poora reason hai ki quaternions exist karte hain.

TASVEER.

Left: rotate-then-flip. Right: flip-then-rotate. Same do moves, alag final arrow — unke beech ka gap exactly cross-product term hai.

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Step 4 — Attitude ek 4D ball par rehta hai; velocity tangent rehni chahiye

KYA. Kyunki $\Vert q\Vert =1$ hamesha, point $q$ us surface se kabhi bahar nahi ja sakta jahan "char squares ka sum $=1$" ho. Woh surface unit 3-sphere $S^3$ hai. Hum four dimensions draw nahi kar sakte, isliye honest 2D analogy draw karte hain: circle par ek point.

KYUN. Agar ek rotation hamesha valid orientation hai, toh $q(t)$ ko is surface par slide karna hai har waqt — yeh kabhi bahar nahi bulge kar sakta ya andar nahi sink kar sakta. Yeh ek geometric demand Steps 6–7 mein $\frac{1}{2}$ aur matrix $\Xi (q)$ ko nail down kar degi.

TASVEER.

Point $q$ circle par baitha hai. Uski velocity $\dot{q}$ (green) tangent draw ki gayi hai — woh surface ke along point karti hai, $q$ ke seedha right angle par. Koi bhi velocity jisme radial (bahar ki taraf) piece ho woh $\Vert q\Vert$ grow kar deti aur forbidden hai.

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Step 5 — Thodi si si time mein ek tiny spin

KYA. Time ke ek chhote slice $\Delta t$ mein, body thode se angle $\Delta \theta =\Vert \omega \Vert \Delta t$ se current axis $\hat{\omega }=\omega /\Vert \omega \Vert$ ke around ghoomti hai. Step 2 se, woh tiny extra rotation apna khud ka quaternion hai:

$$\delta q=\left[\begin{array}{c}\cos \frac{\Delta \theta }{2}\\ \hat{\omega }\sin \frac{\Delta \theta }{2}\end{array}\right]$$

Ab $\Delta t\to 0$ shrink karo. Ek tiny angle ke liye, $\cos \frac{\Delta \theta }{2}\approx 1$ aur $\sin \frac{\Delta \theta }{2}\approx \frac{\Delta \theta }{2}$:

$$\delta q\approx \left[\begin{array}{c}1\\ \frac{1}{2}\omega \Delta t\end{array}\right]=\underset{\text{kuch nahi karo}}{\underbrace{\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]}}+\frac{\Delta t}{2}\underset{\text{woh nudge}}{\underbrace{\left[\begin{array}{c}0\\ \omega \end{array}\right]}}$$

Approximations kyun? Angle zero ke paas, cosine 1 pe flatten ho jaata hai aur sine origin se seedhi line ban jaata hai — "small-angle" facts jo tum Step 2 ki curves se seedha padh sakte ho. Yeh exact-but-messy $\delta q$ ko "identity plus a small linear nudge" mein badal deta hai, jise hum differentiate kar sakte hain.

$\frac{1}{2}$ phir kyun? Yeh $\sin \frac{\Delta \theta }{2}$ par sawaar hokar aaya — Step 2 se half-angle. Yahi literally woh jagah hai jahan mashoor factor of one-half paida hota hai.

TASVEER.

Circle par $q$ ke paas zoom in: $\sin$ ka bada orange curve uski seedhi teal tangent line se replace ho gaya. Tiny step mein, asli spin aur uska linear stand-in alag nahi dikhte — aur us line ki slope $\frac{1}{2}$ carry karta hai.

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Step 6 — "Tiny step" se asli derivative tak

KYA. Current attitude ko tiny nudge ke saath compose karo (Step 3), wahan se subtract karo jahan se shuru kiye the, aur time slice se divide karo:

$$\frac{q(t+\Delta t)-q(t)}{\Delta t}=\frac{q\otimes \delta q-q}{\Delta t}=\frac{q\otimes \left(\frac{\Delta t}{2}\left[\begin{array}{c}0\\ \omega \end{array}\right]\right)}{\Delta t}=\frac{1}{2}q\otimes \left[\begin{array}{c}0\\ \omega \end{array}\right]$$

$\Delta t\to 0$ hone do aur left side derivative $\dot{q}$ ban jaati hai — instantaneous rate, Step 4 ka green tangent arrow:

$$\boxed{\dot{q}=\frac{1}{2}q\otimes \left[\begin{array}{c}0\\ \omega \end{array}\right]}$$

Term by term: $\frac{1}{2}$ half-angle survivor hai; $q\otimes$ keh raha hai "spin body frame mein measure ki gayi, isliye right side pe multiply hoti hai"; aur $\left[\begin{array}{c}0\\ \omega \end{array}\right]$ angular-velocity quaternion hai — ek quaternion jiska zero scalar part hai, kyunki $\omega$ ek pure spin hai jiska apna koi "amount-so-far" nahi.

KYUN. Identity terms $q\otimes \left[\begin{array}{c}1\\ 0\end{array}\right]=q$ exactly cancel ho jaate hain, sirf nudge bachta hai. Derivative ki definition (difference quotient ki limit) "har tick ek small rotation compose karo" ko ek smooth flow mein badal deti hai. Yahi poora trick hai — dekho Numerical integration RK4 ki baad mein hum is flow ko computer par kaise step karte hain.

TASVEER.

$q(t)$ se $q(t+\Delta t)$ tak secant line (dotted) tangent (solid green) par pivot ho jaati hai jaise $\Delta t$ shrink hota hai — limit lene ki tasveer.

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Step 7 — Product ko matrix $\Xi (q)$ ke roop mein package karo

KYA. Hamilton product $q\otimes \left[\begin{array}{c}0\\ \omega \end{array}\right]$ $\omega$ mein linear hai — $\omega$ double karo toh output double ho jaata hai, koi leftover terms nahi. Jo bhi linear ho use matrix times $\omega$ likha ja sakta hai. Step 3 ka formula $p=q$ aur $q_{\text{here}}=(0,\omega )$ ke saath grind karo:

$$q\otimes \left[\begin{array}{c}0\\ \omega \end{array}\right]=\left[\begin{array}{c}-{\mathbf{q}}_v\cdot \omega \\ q_0\omega +{\mathbf{q}}_v\times \omega \end{array}\right]=\Xi (q)\omega$$

toh

$$\dot{q}=\frac{1}{2}\Xi (q)\omega ,\Xi (q)=\left[\begin{array}{ccc}-q_1& -q_2& -q_3\\ q_0& -q_3& q_2\\ q_3& q_0& -q_1\\ -q_2& q_1& q_0\end{array}\right]$$

Term by term: top row $-{\mathbf{q}}_v^{\top }$ dot product se aaya (axis aur spin ka alignment). Bottom $3\times 3$ block $q_{0}I+[{\mathbf{q}}_v\times ]$ hai — scalar $q_0$ jo $\omega$ ko scale karta hai plus skew matrix jo ${\mathbf{q}}_v\times \omega$ perform karta hai.

KYUN. GNC estimators jaise Extended Kalman Filter for attitude estimation ko ek clean linear map $\dot{q}=A\omega$ chahiye jo matrix code mein daala ja sake. $\Xi (q)$ wahi map hai — Hamilton product rewrite kiya gaya taaki computer multiply kar sake.

TASVEER.

$\Xi (q)$ ka $4\times 3$ grid jisme top row (dot-product) plum shade mein hai aur lower block (scale + cross) teal mein — exactly dikhate hue ki kaunsi entries Step 3 ke kis piece se aayi.

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Step 8 — Tasveer ko sanity check karo: norm drift nahi kar sakta

KYA. Humne Step 4 mein claim kiya tha ki $\dot{q}$ tangent honi chahiye, yaani $\dot{q}\perp q$, taaki $\Vert q\Vert$ 1 rahe. Check karo:

$$\frac{d}{dt}\Vert q{\Vert }^2=2q^{\top }\dot{q}=2q^{\top }\left(\frac{1}{2}\Xi (q)\omega \right)=(q^{\top }\Xi (q))\omega$$

aur key fact $q^{\top }\Xi (q)=0^{\top }$:

$$q^{\top }\Xi (q)=\underset{\text{scalar times top row}}{\underbrace{q_0(-{\mathbf{q}}_v^{\top })}}+\underset{\text{vector part times block}}{\underbrace{{\mathbf{q}}_v^{\top }(q_{0}I+[{\mathbf{q}}_v\times ])}}=-q_0{\mathbf{q}}_v^{\top }+q_0{\mathbf{q}}_v^{\top }+{\mathbf{q}}_v^{\top }[{\mathbf{q}}_v\times ]=0^{\top }$$

Do $q_0{\mathbf{q}}_v^{\top }$ pieces cancel ho jaate hain, aur ${\mathbf{q}}_v^{\top }[{\mathbf{q}}_v\times ]=-({\mathbf{q}}_v\times {\mathbf{q}}_v{)}^{\top }=0$ kyunki koi bhi vector khud se cross karo toh zero milta hai.

KYUN. Yeh Step 4 ka poora promise hai, algebra mein cash kiya gaya: chahe koi bhi $\omega$ feed karo, rate $q$ ke perpendicular hai. $\Vert q\Vert$ kabhi nahi badalti. Equation attitude ko ball se bahar nahi bhej sakti.

TASVEER.

$q$ (blue radial) aur $\dot{q}$ (green tangent) ek perfect right angle par milte hain; corner par woh chhota box $q^{\top }\dot{q}=0$ hai jo humne abhi prove kiya.

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Step 9 — Edge aur degenerate cases (kuch bhi uncovered nahi)

KYA & KYUN — har corner dekho:

1. Zero spin, $\omega =0$. Toh $\Xi (q)0=0$, isliye $\dot{q}=0$. Point ball par still baithta hai — non-rotating body apna attitude rakhta hai. ✔
2. Identity par, $q=(1,0,0,0)$. Top row $-{\mathbf{q}}_v^{\top }=0^{\top }$, isliye ${\dot{q}}_0=0$; sirf vector part on hota hai. Pehli motion purely $q_1,q_2,q_3$ mein jaati hai — exactly tangent direction. ✔
3. Stored axis ke along spin (jaise $\omega \parallel \hat{n}$): cross term ${\mathbf{q}}_v\times \omega =0$ vanish ho jaata hai, isliye body simply usi axis ke around ghoomti rehti hai — koi wobble nahi, clean single-axis spin. ✔
4. $q$ ki jagah $-q$ (double cover): $q$ mein har sign flip karo aur $\Xi (q)$ bhi sign flip kar leta hai, $-\dot{q}$ deta hai — same physical rotation mirror-labelled point se trace hoti hai. Koi contradiction nahi; dono ek orientation ko naam dete hain.
5. Bada $\omega$ finite step par. $\frac{1}{2}$ ki derivation continuous time mein exact hai, lekin ek discrete integrator (jaise Numerical integration RK4) thodi-si ball se drift kar jaata hai — isliye real flight code har tick $q\leftarrow q/\Vert q\Vert$ renormalize karta hai. Compare karo Euler angles and gimbal lock se, jahan koi bhi renormalization singularity se nahi bachata.

TASVEER.

Char mini-circles: (a) $\omega =0$ frozen dot; (b) identity with tangent-only motion; (c) axis-aligned spin, no wobble; (d) discrete steps jo bahar drift kar rahe hain, ek plum arrow unhe surface par wapas snap karta hua (renormalization).

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Ek-tasveer summary

Ek frame poori derivation carry karta hai: sphere par point $q$; tiny orange nudge $\delta q\approx 1+\frac{\Delta t}{2}(0,\omega )$ use aage kheeenchta hua; green tangent $\dot{q}=\frac{1}{2}\Xi (q)\omega$ jo result hoti hai; aur right angle jo prove karta hai ki $\Vert q\Vert$ 1 rehta hai. Half-angle in, tangent out.

Recall Feynman retelling — seedhe shabdon mein bolo

Ek ghoomti cheez ka ek "main kis taraf point kar raha hun" state hota hai, aur hum ise chalakiyon se char numbers mein store karte hain jinke squares ek mein add hote hain — toh state ek round ball ki surface par stuck hai. Thoda aur ghoomna matlab hai multiply karna ek small turn-quaternion se, aur kyunki humne half-angle store kiya, woh small turn hai "kuch mat karo plus spin ka ek-haalf". Do-nothing subtract karo, time se divide karo, aur rate milti hai: ek-haalf, times current state, times spin. Multiply ko ek tidy matrix $\Xi (q)$ ke roop mein rewrite karo taaki computer use kar sake. Aakhir mein, woh khoobsurat check: rate hamesha ball ke along sideways point karti hai, kabhi bahar nahi, isliye state kabhi fall off nahi kar sakti — jo exactly matlab hai "abhi bhi ek valid rotation hai". Poora formula sirf hai "ball par half-speed se slide karo."

Recall Quick self-test

Factor $\frac{1}{2}$ kahan se aata hai? ::: Quaternion mein half-angle $\sin \frac{\theta }{2}$ store karne se (Step 2); zero par uski slope $\frac{1}{2}$ hai (Step 5). $\dot{q}\perp q$ kyun honi chahiye? ::: Kyunki $\Vert q\Vert =1$ hamesha rehta hai, isliye velocity sirf sphere ke along slide kar sakti hai, kabhi radially bahar nahi (Step 4 / Step 8). $\Xi (q)$ ek matrix kyun hai, nonlinear map kyun nahi? ::: Kyunki Hamilton product $q\otimes (0,\omega )$ $\omega$ mein linear hai (Step 7) — $\omega$ double karo toh $\dot{q}$ double ho jaata hai. Jab $\omega =0$ toh $\dot{q}$ kya hai? ::: Zero — non-spinning body apna attitude pakde rehta hai (Step 9, case 1).

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