Visual walkthrough — Quaternion rotation formula — rotating vector v by quaternion q

3.5.8 · D2 · Physics › Guidance, Navigation & Control (GNC) › Quaternion rotation formula — rotating vector v by quaternio

Yeh parent formula ka picture version hai. Hum shuru karte hain ek simple cheez se — space mein ek akela arrow — aur dheere-dheere sandwich $qv{q}^{-1}$ banate hain, ek drawing at a time. Har symbol wahin define hota hai jab pehli baar aata hai, aur har step mein WHAT / WHY / PICTURE hai.

End tak aap dekh paoge ki $q$ ke andar angle sirf half kyun hota hai.

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Step 1 — Ek arrow, aur jo turn chahiye

WHAT: Humare paas ek arrow $\mathbf{v}$ hai aur ek turn instruction hai: "angle $\theta$ se axis $\hat{n}$ ke around spin karo."

Yahan $\hat{n}$ (padho "n-hat") ek unit axis hai — ek aise line ke along exactly $1$ length ka arrow jiske around hum spin karte hain. Chhota hat $\hat{}$ hamesha matlab hai "length ek." $\theta$ (theta) kitna spin karna hai, usual tarike se measure kiya jaata hai: $90^{\circ }$ ek quarter turn hai, $180^{\circ }$ ek half turn.

WHY: 3D mein har rotation, chahe kitna bhi complicated ho, exactly ek spin hota hai kisi angle $\theta$ se kisi single axis $\hat{n}$ ke around. Yeh Euler's theorem hai. Toh agar hum arbitrary $\hat{n}$ ke around rotate kar sakte hain, toh hum koi bhi rotation kar sakte hain. Yahi poora target hai.

PICTURE: Kala arrow $\mathbf{v}$ hai; lal arrow woh jagah hai jahan hum chahte hain ki yeh land kare, ${\mathbf{v}}^{\prime }$; vertical line spin axis $\hat{n}$ hai.

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Step 2 — Arrow ko do honest pieces mein todna

WHAT: Hum $\mathbf{v}$ ko ${\mathbf{v}}_{\parallel }$ (jo $\hat{n}$ ke along point karta hai) aur ${\mathbf{v}}_{\perp }$ (jo sideways point karta hai, $\hat{n}$ se $90^{\circ }$ par) mein tod dete hain.

WHY split karna? Kyunki dono pieces spin ke under bilkul alag-alag behave karte hain:

• ${\mathbf{v}}_{\parallel }$ axis par hai — spin ise move nahi kar sakta (socho ek pencil jo ek wheel ke axle ke along rakha hai; wheel ghoomta hai, pencil nahi).
• ${\mathbf{v}}_{\perp }$ sideways nikal raha hai — yahi ek part hai jo actually swing around karta hai.

Har piece ko alag-alag solve karna poore tilted cheez ko solve karne se kaafi aasaan hai. Yahi "divide and conquer" derivation ko clean rakhta hai.

PICTURE: Kala arrow $\mathbf{v}$ apne do components mein; dhyan do chhota right-angle square jahan ${\mathbf{v}}_{\perp }$ axis se milta hai.

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Step 3 — Turn ko quaternion $q$ ke roop mein package karo

WHAT: Hum object $q$ banate hain jo turn ko carry karta hai. Dhyan do andar ka angle $\frac{\theta }{2}$ hai — half actual turn. Yeh baat yaad rakho; Step 6 mein iska fal milega.

Do pieces ko ek baar naam dete hain aur reuse karte hain: $c=\cos \frac{\theta }{2},s=\sin \frac{\theta }{2}.$ Toh $q=(c,s\hat{n})$. Kyunki $c^2+s^2=1$ aur $\hat{n}$ ki length $1$ hai, iss $q$ ki length $1$ hai — ek unit quaternion, exactly wahi jo ek rotation ko chahiye.

WHY half? Hum abhi iska justification nahi kar sakte — yeh arbitrary lagta hai. Yeh honest hai: half-angle baad mein algebra ki wajah se hampar thopa jaata hai, upfront choose nahi kiya jaata. Abhi ke liye, bas $q=(c,s\hat{n})$ ko recipe ki tarah accept karo aur dekho yeh kya karta hai.

PICTURE: $(w,\text{axis})$ plane mein unit circle. Jab $\theta$ $0$ se badhta hai, point $(\cos \frac{\theta }{2},\sin \frac{\theta }{2})$ (red mein) circle ke around sirf half speed se creep karta hai — half-angle ka visual seed.

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Step 4 — Sandwich, aur dono taraf bread kyun

WHAT: Hum arrow ko left side se $q$ se aur right side se $q^{-1}$ se multiply karte hain — ek "sandwich," $v$ filling ki tarah.

WHY sirf $qv$ nahi (ek roti)? Kyunki akela product $qv$ generally nonzero scalar part produce karta hai — answer ab pure quaternion nahi rahega, yaani ab ek valid 3D arrow nahi. Right-hand $q^{-1}$ wahi hai jo us scalar leak ko cancel karta hai aur guarantee karta hai ki output dobara $(0,{\mathbf{v}}^{\prime })$ ho — ek clean arrow. (Parent note yeh closure $(qv{q}^{-1}{)}^{\ast }=-(qv{q}^{-1})$ ke zariye prove karta hai.)

Ise wrap → spin → unwrap ki tarah socho: $q$ arrow ko wrap karta hai, multiplication ise spin karta hai, $q^{-1}$ ise ordinary arrows ki duniya mein wapas unwrap karta hai.

PICTURE: Teen-panel strip: bare arrow, arrow jo $q$ se "wrapped" hai (red tint = abhi pure vector nahi), aur arrow jo $q^{-1}$ se wapas clean red vector ${\mathbf{v}}^{\prime }$ mein "unwrapped" hai.

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Step 5 — Parallel part kabhi nahi hilta

WHAT: On-axis piece ${\mathbf{v}}_{\parallel }$ ko sandwich mein daalo. Result hai $q{\mathbf{v}}_{\parallel }{q}^{-1}={\mathbf{v}}_{\parallel }.$ Unchanged. Bilkul waisa hi bahar aata hai jaisa andar gaya tha.

WHY: ${\mathbf{v}}_{\parallel }$ $\hat{n}$ ke along point karta hai, aur $q$ ka vector part bhi $\hat{n}$ ke along point karta hai. Jab do quaternions same axis direction share karte hain, unka cross product $\hat{n}\times {\mathbf{v}}_{\parallel }=0$ vanish ho jaata hai — koi "sideways" ingredient nahi jo swing kare. Phir scalar bookkeeping ${\mathbf{v}}_{\parallel }$ ko unchanged chhodh deta hai. Physically: $\hat{n}$ ke around spin kuch bhi nahi hila sakta jo already $\hat{n}$ ke along hai.

PICTURE: Axis jis par ${\mathbf{v}}_{\parallel }$ hai; ek curved arrow spin ko sweep karta dikhata hai, aur ${\mathbf{v}}_{\parallel }$ (red mein) axle par bilkul still baitha hai.

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Step 6 — Perpendicular part swing karta hai — aur half-angle double ho jaata hai

Yeh sab ka dil hai. Sideways piece ${\mathbf{v}}_{\perp }$ ke liye sandwich work out karne par milta hai $q{\mathbf{v}}_{\perp }{q}^{-1}=\underset{\text{stretch along }{\mathbf{v}}_{\perp }}{\underbrace{(c^2-s^2)}}{\mathbf{v}}_{\perp }+\underset{\text{stretch sideways}}{\underbrace{2cs}}(\hat{n}\times {\mathbf{v}}_{\perp }).$

Term-by-term:

• ${\mathbf{v}}_{\perp }$ — original sideways arrow.
• $\hat{n}\times {\mathbf{v}}_{\perp }$ — ek naya arrow, abhi bhi sideways, lekin ${\mathbf{v}}_{\perp }$ se spin direction mein $90^{\circ }$ aage rotated (cross product "sideways-ahead" direction manufacture karta hai jisme swing karna hai).
• $c^2-s^2$ aur $2cs$ — hum kitna mix karte hain dono ka.

Substitute karne par: $q{\mathbf{v}}_{\perp }{q}^{-1}=\cos \theta {\mathbf{v}}_{\perp }+\sin \theta (\hat{n}\times {\mathbf{v}}_{\perp }).$

Ise geometrically padho: yeh ${\mathbf{v}}_{\perp }$ hai jo apne flat sideways plane mein angle $\theta$ se turn kiya gaya hai — $\cos \theta$ "stay" ka plus $\sin \theta$ "$90^{\circ }$ aage swing" ka. Bilkul ordinary 2D rotation.

PICTURE: $\hat{n}$ ke perpendicular flat plane. Do kale basis arrows ${\mathbf{v}}_{\perp }$ aur $\hat{n}\times {\mathbf{v}}_{\perp }$ $90^{\circ }$ par; red result arrow angle $\theta$ par baitha hai, components $\cos \theta$ aur $\sin \theta$ label kiye hain.

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Step 7 — Reassemble: yahi Rodrigues' formula hai

WHAT: Fixed parallel part (Step 5) ko swung perpendicular part (Step 6) mein add karo: ${\mathbf{v}}^{\prime }={\mathbf{v}}_{\parallel }+[\cos \theta {\mathbf{v}}_{\perp }+\sin \theta (\hat{n}\times {\mathbf{v}}_{\perp })].$ ${\mathbf{v}}_{\parallel }=\hat{n}(\hat{n}\cdot \mathbf{v})$ aur ${\mathbf{v}}_{\perp }=\mathbf{v}-{\mathbf{v}}_{\parallel }$ rewrite karne par classic form milta hai: $\boxed{{\mathbf{v}}^{\prime }=\mathbf{v}\cos \theta +(\hat{n}\times \mathbf{v})\sin \theta +\hat{n}(\hat{n}\cdot \mathbf{v})(1-\cos \theta )}$

Term-by-term:

• $\mathbf{v}\cos \theta$ — poore arrow ko $\cos \theta$ se shrink karo.
• $(\hat{n}\times \mathbf{v})\sin \theta$ — sideways swing add karo.
• $\hat{n}(\hat{n}\cdot \mathbf{v})(1-\cos \theta )$ — on-axis part ko wapas full length tak top up karo ($\cos \theta$ ne ise zyada shrink kar diya tha; yahan $\hat{n}\cdot \mathbf{v}$ axis-shadow ki length hai).

WHY it matters: Yeh exactly Rodrigues' rotation formula hai. Toh quaternion sandwich koi naya rotation nahi — yeh wahi honest rotation hai, bas 9-number matrix ki jagah 4 numbers mein stored hai. Isse Rotation matrices & orthogonality se cross-check karo aur tum wahi ${\mathbf{v}}^{\prime }$ paoge.

PICTURE: Do pieces snap karke ek saath: fixed ${\mathbf{v}}_{\parallel }$ aur swung ${\mathbf{v}}_{\perp }$ milke poora red ${\mathbf{v}}^{\prime }$ reconstruct karte hain, $\hat{n}$ ke around rotation cone par baitha hua.

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Step 8 — Edge & degenerate cases (reader ko kabhi akela mat chhodho)

PICTURE: Chaar degenerate cases ka 2×2 grid, har ek mein input arrow (kala) aur output arrow (lal) dikhaya gaya hai taaki reader dekhe "haan, yahi hota hai."

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Ek-picture summary

Sab kuch ek canvas par: arrow $\mathbf{v}$ ${\mathbf{v}}_{\parallel }$ (frozen) aur ${\mathbf{v}}_{\perp }$ (swung) mein split, axis $\hat{n}$, quaternion $q$ jo $\frac{\theta }{2}$ carry karta hai jo do baar khaya jaake $\theta$ ban jaata hai, aur final red ${\mathbf{v}}^{\prime }$ rotation cone trace karta hua.

Recall Poore walkthrough ki Feynman retelling

Tumhare paas ek arrow hai aur tum ise kisi axle ke around spin karna chahte ho. Pehle tum notice karte ho ki arrow ke do parts hain: woh bit jo axle ke along hai (jise spin kabhi nahi hila sakta) aur woh bit jo sideways nikal raha hai (sirf yahi part actually swing karta hai). Tum apni spin instruction ek chhote 4-number gadget mein likhte ho jise quaternion kehte hain, aur — ajeeb lagta hai — tum usme sirf half turn amount daalte ho. Phir tum sandwich karte ho: arrow ko gadget se wrap karo, spin karo, uske mirror-twin se unwrap karo. Kyunki gadget do baar use hota hai — ek baar wrapping mein, ek baar unwrapping mein — do half-turns milke poore turn ban jaate hain jo tum actually chahte the. Sideways part exactly $\theta$ se swing karta hai (woh $\cos \theta$ aur $\sin \theta$ mixing hai), along-axle part jagah par reh jaata hai, aur dono ko wapas jodna exactly Rodrigues' formula hai. Corner cases check karo — koi turn nahi, half turn, on-axis arrow, zero arrow — aur har ek behave karta hai. Matrix jaisa hi rotation, nau ki jagah chaar numbers.

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Dekho bhi: Quaternion algebra & Hamilton product · Euler angles & gimbal lock · Spacecraft attitude determination (GNC) · Angular velocity & quaternion kinematics $\dot q=\tfrac12 q\omega$