Modified Rodrigues parameters — singularity-free, compact

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

MRPs ki zaroorat kyun hai?

Humare paas pehle se orientation store karne ke kai tarike hain. Har ek mein koi na koi kami hai:

• Direction Cosine Matrix (DCM): 9 numbers, 6 constraints → wasteful, integrate karna awkward.
• Euler angles: 3 numbers, lekin gimbal lock — $\pm 90^{\circ }$ par ek singularity jahan do axes align ho jaate hain aur ek degree of freedom kho jaata hai.
• Quaternions: 4 numbers, kabhi bhi singularity nahi, lekin ek redundant constraint $\Vert q\Vert =1$ aur ek sign ambiguity ($q$ aur $-q$ ek hi rotation hain) carry karte hain.

MRPs kya hain? — Axis–angle se derivation

Har rotation = angle $\Phi$ se unit axis $\hat{e}$ ke baare mein rotate karo. Euler symmetric parameters (quaternion) se shuru karte hain:

$q_0=\cos \frac{\Phi }{2},\mathbf{q}=\hat{e}\sin \frac{\Phi }{2}$

Classical Rodrigues parameters (Gibbs vector) hain $\mathbf{g}=\hat{e}\tan (\Phi /2)=\mathbf{q}/q_0$. Yeh $\Phi =180^{\circ }$ par blow up ho jaate hain (jahan $q_0=0$ hota hai). Bahut jaldi.

Trick: half-angle ko dobara use karo. MRPs ko $q_0$ ki jagah $(1+q_0)$ se divide karke define karo:

$\boxed{\mathbf{\sigma }=\frac{\mathbf{q}}{1+q_0}}$

$\mathbf{q}/(1+q_0)$ aakhir $\hat{e}\tan (\Phi /4)$ kyun hota hai? Quaternion substitute karo: $\frac{\mathbf{q}}{1+q_0}=\frac{\hat{e}\sin (\Phi /2)}{1+\cos (\Phi /2)}=\hat{e}\tan \frac{\Phi }{4}$ Yeh step kyun? Humne identity $\frac{\sin \theta }{1+\cos \theta }=\tan \frac{\theta }{2}$ use ki, jahan $\theta =\Phi /2$ hai. Half-angle ka half-angle = quarter-angle. Yahi extra halving hai jo singularity ko $180^{\circ }$ (Gibbs) se badhaakar $360^{\circ }$ tak push karti hai.

Shadow set — aakhri singularity ko kaise khatam karein

Kyunki $q$ aur $-q$ ek hi physical rotation hain, isliye har attitude ke liye do MRP vectors hote hain:

$\mathbf{\sigma }=\frac{\mathbf{q}}{1+q_0},{\mathbf{\sigma }}^S=\frac{-\mathbf{q}}{1-q_0}=-\frac{\mathbf{\sigma }}{\Vert \mathbf{\sigma }{\Vert }^2}$

Yeh kyun kaam karta hai: $\Vert \mathbf{\sigma }\Vert =\tan (\Phi /4)$. $\Phi =180^{\circ }$ par, $\Vert \mathbf{\sigma }\Vert =1$; $\Phi \to 360^{\circ }$ par, $\Vert \mathbf{\sigma }\Vert \to \infty$. Shadow set $\Phi \mapsto 360^{\circ }-\Phi$ map karta hai, isliye jab original 1 se zyada hota hai tab $\Vert {\mathbf{\sigma }}^S\Vert <1$ hamesha hota hai. $\Vert \mathbf{\sigma }\Vert \le 1$ rakhne se guarantee milti hai ki tum singularity ke paas kabhi nahi pahunchoge.

Kinematic differential equation (propagate kaise karein)

Body angular velocity $\mathbf{\omega }$ diya hua ho, to MRPs is tarah evolve karte hain:

$\dot{\mathbf{\sigma }}=\frac{1}{4}B(\mathbf{\sigma })\mathbf{\omega }$

$B(\mathbf{\sigma })=(1-{\sigma }^2){\mathbb{I}}_3+2[{\mathbf{\sigma }}^{\times }]+2\mathbf{\sigma }{\mathbf{\sigma }}^{\top },{\sigma }^2={\mathbf{\sigma }}^{\top }\mathbf{\sigma }$

jahan $[{\mathbf{\sigma }}^{\times }]$ cross-product (skew-symmetric) matrix hai.

Ek useful property (Steel-man): $B(\mathbf{\sigma })$ almost orthogonal hai: $B^{\top }B=(1+{\sigma }^2{)}^2{\mathbb{I}}_3\Rightarrow B^{-1}=\frac{B^{\top }}{(1+{\sigma }^2{)}^2}.$ Isse kinematics ko invert karna (yaani $\mathbf{\omega }$ solve karna) trivial ho jaata hai — control law design ke liye practically bahut bada fayda.

Worked examples

Common mistakes

Flashcards

Connections

• Quaternions (Euler symmetric parameters) — parent representation; $\mathbf{\sigma }=\mathbf{q}/(1+q_0)$
• Classical Rodrigues parameters (Gibbs vector) — sibling, $\tan (\Phi /2)$, $180^{\circ }$ par singular
• Euler angles and Gimbal Lock — woh singularity problem jise MRPs improve karte hain
• Direction Cosine Matrix (DCM) — 9-number global standard
• Attitude kinematics and $\boldsymbol\omega$ — jahan $\dot{\mathbf{\sigma }}=\frac{1}{4}B\mathbf{\omega }$ rehta hai
• Spacecraft attitude control laws — MRP feedback controllers $B^{-1}$ ki simplicity exploit karte hain

Concept Map