Quaternion rotation formula — rotating vector v by quaternion q

3.5.8 · Physics › Guidance, Navigation & Control (GNC)

80/20 core: Ek 3D vector $v$ ko unit quaternion $q$ se rotate karne ke liye, $v$ ko ek pure quaternion ki tarah embed karo aur compute karo $v^{\prime }=qv{q}^{-1}$. Bas yahi poora game hai. Baaki sab samajhna hai ki kyun sandwich kaam karta hai aur kyun half-angle use hota hai.

The Setup — WHAT are we even doing?

The Quaternion Product — ek rule jo humein chahiye

The Rotation Formula — DERIVED, dumped nahi

Step 1: Right side par conjugate kyun (the "sandwich")?

Hum chahte hain ki output $v^{\prime }=(0,{\mathbf{v}}^{\prime })$ phir bhi ek pure quaternion ho — warna ye ek 3D vector nahi hoga. Product ka conjugate lo: $(qv{q}^{-1}{)}^{\ast }=(q^{-1}{)}^{\ast }{v}^{\ast }{q}^{\ast }.$ Unit $q$ ke liye: $(q^{-1}{)}^{\ast }=q$ aur $q^{\ast }=q^{-1}$. Aur $v^{\ast }=-v$ kyunki $v$ pure hai. Toh $(qv{q}^{-1}{)}^{\ast }=q(-v)q^{-1}=-(qv{q}^{-1})$. Jo object $-$(apne khud ke conjugate) ke barabar ho uska zero scalar part hota hai ⇒ output pure hai. ✓

Ye step kyun? Ye prove karta hai ki sandwich vectors par closed hai. Ek one-sided product $qv$ generally nonzero scalar deta — jo vector nahi hoga.

Step 2: Half-angle kyun? Maano $q=(\cos \frac{\theta }{2},\sin \frac{\theta }{2}\hat{n})$

Lo $q=\cos \frac{\theta }{2}+\sin \frac{\theta }{2}\hat{n}$ jahan unit axis $\hat{n}$ hai. $\mathbf{v}={\mathbf{v}}_{\parallel }+{\mathbf{v}}_{\perp }$ mein split karo ($\hat{n}$ ke parallel/perpendicular).

Parallel part ($\hat{n}\times {\mathbf{v}}_{\parallel }=0$, $\hat{n}\cdot {\mathbf{v}}_{\parallel }=|{\mathbf{v}}_{\parallel }|$): Algebra work karne par, $q{\mathbf{v}}_{\parallel }{q}^{-1}={\mathbf{v}}_{\parallel }$ — axis ke saath wala component unchanged rehta hai (jaisa ki $\hat{n}$ ke baare mein rotation mein hona chahiye). ✓

Perpendicular part: $\hat{n}\cdot {\mathbf{v}}_{\perp }=0$ use karke aur $c=\cos \frac{\theta }{2},s=\sin \frac{\theta }{2}$ lete hain, $q{\mathbf{v}}_{\perp }{q}^{-1}=(c^2-s^2){\mathbf{v}}_{\perp }+2cs(\hat{n}\times {\mathbf{v}}_{\perp }).$ Ab double-angle identities: $c^2-s^2=\cos \theta$ aur $2cs=\sin \theta$. Toh $q{\mathbf{v}}_{\perp }{q}^{-1}=\cos \theta {\mathbf{v}}_{\perp }+\sin \theta (\hat{n}\times {\mathbf{v}}_{\perp }).$ Ye exactly Rodrigues' rotation hai ${\mathbf{v}}_{\perp }$ ka angle $\theta$ se $\hat{n}$ ke baare mein. ✓

Half-angle kyun? Kyunki $q$ do baar appear karta hai (ek baar $q$ ke roop mein, ek baar $q^{-1}$ ke roop mein), har ek $\theta /2$ contribute karta hai. Do half-angles double-angle formulas ke through ek full $\theta$ mein combine hote hain. Ye yaad rakhne wali #1 cheez hai.

Step 3: Rodrigues mein collapse karo (sanity check)

Parallel + perpendicular combine karne par poora Rodrigues formula milta hai: ${\mathbf{v}}^{\prime }=\mathbf{v}\cos \theta +(\hat{n}\times \mathbf{v})\sin \theta +\hat{n}(\hat{n}\cdot \mathbf{v})(1-\cos \theta ).$ Toh $qv{q}^{-1}$ disguise mein Rodrigues hi hai — same rotation, sasta storage.

Worked Examples

Common Mistakes (Steel-manned)

Flashcards

Connections

• Quaternion algebra & Hamilton product
• Rodrigues' rotation formula
• Rotation matrices & orthogonality
• Euler angles & gimbal lock
• SLERP — quaternion interpolation
• Spacecraft attitude determination (GNC)
• Angular velocity & quaternion kinematics $\dot q=\tfrac12 q\omega$

Concept Map