Exercises — Modified Rodrigues parameters — singularity-free, compact

KYA: Definition hai $\mathbf{\sigma }=\hat{e}\tan (\Phi /4)$. KYUN: Direction rotation axis khud hai; length angle store karti hai, $\tan (\Phi /4)$ ke through squeeze hokar. $\Phi =180^{\circ }$ par: quarter-angle $=45^{\circ }$, aur $\tan 45^{\circ }=1$. Toh $\Vert \mathbf{\sigma }\Vert =1$. Figure s01 par: yeh woh single point hai jahan blue curve orange switch line se milti hai — boundary case, jahan (upar diye policy ke according) tum abhi switch nahi karte.

Singularity $\Phi =360^{\circ }$ par. Tab $q_0=\cos (\Phi /2)=\cos 180^{\circ }=-1$, jisse denominator $1+q_0=0$ ho jaata hai. Zero se divide → blow-up. Figure s01 par yeh woh red vertical wall hai jis taraf blue curve dauraata hai.

Jhooth. MRPs mein ek singularity hai, $\Phi =360^{\circ }$ par. Jo cheez inhe practical banati hai woh yeh hai ki yeh singularity shadow set ke zariye avoidable hai. Sirf quaternions aur DCM globally nonsingular hain.

Step 1 — magnitude. $\Vert \mathbf{\sigma }\Vert =\tan (90^{\circ }/4)=\tan 22.5^{\circ }\approx 0.4142$. Kyun: magnitude quarter-angle tangent hai. Step 2 — direction. Woh length axis se attach karo: $\mathbf{\sigma }=(0,0,0.4142)$. Step 3 — safety check. $0.4142<1$, toh hum figure s01 par orange line ke left mein hain — safe, koi switch nahi.

KYUN invert karo: magnitude quarter-angle batati hai. $\Vert \mathbf{\sigma }\Vert =0.2679$. KYA: $\Phi /4=\arctan (0.2679)=15^{\circ }$, toh $\Phi =60^{\circ }$. Axis hai $\hat{x}=(1,0,0)$.

Quaternion route. $q_0=\cos 60^{\circ }=0.5$; $\mathbf{q}=\hat{e}\sin 60^{\circ }=(0,0,0.8660)$. $\mathbf{\sigma }=\mathbf{q}/(1+q_0)=(0,0,0.8660)/1.5=(0,0,0.5774)$. Axis–angle route. $\tan (120^{\circ }/4)=\tan 30^{\circ }=0.5774$. ✅ Same answer — dono definitions ek hi hain.

(a) $\Vert \mathbf{\sigma }\Vert =\tan (270^{\circ }/4)=\tan 67.5^{\circ }\approx 2.4142$. Toh $\mathbf{\sigma }=(0,2.4142,0)$. (b) $2.4142>1$ → orange line ke right mein, red wall ki taraf ja rahe hain. Switch karo. (c) ${\mathbf{\sigma }}^S=-\mathbf{\sigma }/({\mathbf{\sigma }}^{\top }\mathbf{\sigma })=-\mathbf{\sigma }/(2.4142{)}^2$. Component: $-2.4142/5.8284=-0.4142$, toh ${\mathbf{\sigma }}^S=(0,-0.4142,0)$. Equivalent angle: $\Vert {\mathbf{\sigma }}^S\Vert =0.4142=\tan 22.5^{\circ }$, toh ${\Phi }^{\prime }=90^{\circ }=360^{\circ }-270^{\circ }$. ✅ Same attitude, tiny magnitude.

KYA skew matrix hai: kisi bhi 3-vector $\mathbf{a}$ ke liye, matrix $[{\mathbf{a}}^{\times }]$ aise banaya jaata hai ki $[{\mathbf{a}}^{\times }]\mathbf{b}=\mathbf{a}\times \mathbf{b}$ — yeh cross product ko plain matrix multiply mein badal deta hai. KYUN chahiye: kinematics matrix $B$ ko axes ke beech cross-product coupling chahiye, aur matrix woh cheez hai jise hum multiply aur invert kar sakte hain. $[{\mathbf{\sigma }}^{\times }]=\left(\begin{array}{ccc}0& -{\sigma }_3& {\sigma }_2\\ {\sigma }_3& 0& -{\sigma }_1\\ -{\sigma }_2& {\sigma }_1& 0\end{array}\right)$ $\mathbf{\sigma }=(0,0,0.5)$ ke liye: $[{\mathbf{\sigma }}^{\times }]=\left(\begin{array}{ccc}0& -0.5& 0\\ 0.5& 0& 0\\ 0& 0& 0\end{array}\right)$ Notice karo ki diagonal hamesha zero hota hai aur $[{\mathbf{\sigma }}^{\times }{]}^{\top }=-[{\mathbf{\sigma }}^{\times }]$ — yahi "skew-symmetric" ka matlab hai.

Step 1 — $B$ banao. ${\sigma }^2=0.25$, toh $(1-{\sigma }^2)=0.75$.

• Identity term: $0.75\mathbb{I}$.
• Skew term: $2[{\mathbf{\sigma }}^{\times }]$ L3.2 se.
• Outer-product term: $2\mathbf{\sigma }{\mathbf{\sigma }}^{\top }$ mein sirf $(3,3)$ entry nonzero hai: $2(0.5)(0.5)=0.5$. Step 2 — $\mathbf{\omega }=(0,0,0.1)$ pe apply karo. Kyunki $\mathbf{\omega }$ $\hat{z}$ ke along point karta hai aur $\mathbf{\sigma }$ bhi $\hat{z}$ ke along hai, skew term (parallel vectors ka cross product) zero deta hai. Sirf third component bachta hai: $B\mathbf{\omega }$ third entry $=(0.75+0.5)(0.1)=1.25\times 0.1=0.125$. Step 3 — quarter factor. $\dot{\mathbf{\sigma }}=\frac{1}{4}B\mathbf{\omega }=(0,0,0.03125)$ per second.

KYUN inverse form: control laws prescribe karte hain ki $\mathbf{\sigma }$ kaise move karna chahiye; hume $\mathbf{\omega }$ command back out karni padti hai. Parent note deta hai $\mathbf{\omega }=4B^{-1}\dot{\mathbf{\sigma }}$ with $B^{-1}=B^{\top }/(1+{\sigma }^2{)}^2$. Step 1. ${\sigma }^2=0.09+0.16=0.25$. Toh $(1+{\sigma }^2{)}^2=1.25^2=1.5625$. Step 2 — $B$ banao. $(1-{\sigma }^2)=0.75$. $B=0.75\mathbb{I}+2\left(\begin{array}{ccc}0& 0& -0.4\\ 0& 0& -0.3\\ 0.4& 0.3& 0\end{array}\right)+2\left(\begin{array}{ccc}0.09& -0.12& 0\\ -0.12& 0.16& 0\\ 0& 0& 0\end{array}\right)$ $B=\left(\begin{array}{ccc}0.93& -0.24& -0.8\\ -0.24& 1.07& -0.6\\ 0.8& 0.6& 0.75\end{array}\right)$ Step 3 — inverse-multiply. $\mathbf{\omega }=4\frac{B^{\top }}{1.5625}\dot{\mathbf{\sigma }}$. Kyun agla move legal hai — matrix–vector rule. Ek matrix $M$ times vector $\mathbf{v}$ ek naya vector produce karta hai jiske entries $M$ ke columns $\mathbf{v}$ ki entries se weighted hote hain: $M\mathbf{v}=v_1(\text{col }1)+v_2(\text{col }2)+v_3(\text{col }3)$. Yahan $\dot{\mathbf{\sigma }}=(0.05,0,0)$ ki sirf pehli entry nonzero hai, toh sirf $B^{\top }$ ka first column bachta hai, $0.05$ se scale hokar. Aur $B^{\top }$ ka first column by definition $B$ ki first row hai, yaani $(0.93,-0.24,-0.8)$. Isliye: $\mathbf{\omega }=\frac{4\times 0.05}{1.5625}(0.93,-0.24,-0.8)=0.128(0.93,-0.24,-0.8)$ $\mathbf{\omega }\approx (0.11904,-0.03072,-0.1024)\text{rad/s}.$

Shuru karo ${\mathbf{\sigma }}^S=-\mathbf{\sigma }/{\sigma }^2$ se jahan ${\sigma }^2=\Vert \mathbf{\sigma }{\Vert }^2$. Norm lo: $\Vert {\mathbf{\sigma }}^S\Vert =\Vert \mathbf{\sigma }\Vert /{\sigma }^2=\Vert \mathbf{\sigma }\Vert /\Vert \mathbf{\sigma }{\Vert }^2=1/\Vert \mathbf{\sigma }\Vert$. Interpretation: switch magnitude par ek reciprocal map hai. Agar $\Vert \mathbf{\sigma }\Vert =2.41$, toh shadow ka $\Vert {\mathbf{\sigma }}^S\Vert =0.414<1$. Dono magnitudes ka product exactly $1$ hai, aur $\Vert \mathbf{\sigma }\Vert =1$ woh fixed point hai jahan original aur shadow milte hain — exactly figure s01 par orange line, aur exactly isliye boundary policy us point ko akela chhod deti hai.

Har waypoint par $\Vert \mathbf{\sigma }\Vert =\tan (\Phi /4)$ compute karo:

• $\Phi =170^{\circ }$: $\tan 42.5^{\circ }=0.9163$ → safe.
• $\Phi =175^{\circ }$: $\tan 43.75^{\circ }=0.9601$ → safe.
• $\Phi =185^{\circ }$: $\tan 46.25^{\circ }=1.0416$ → 1 se exceed → yahan switch karo.
• $\Phi =190^{\circ }$: $\tan 47.5^{\circ }=1.0913$ → aur bada hota; already switch ho chuka. Trigger: switch pehle us sample par fire karta hai jahan $\Vert \mathbf{\sigma }\Vert >1$ ho (strict inequality — boundary $\Vert \mathbf{\sigma }\Vert =1$ trigger nahi karta, upar diye policy ke according), yaani jab $\Phi$ $180^{\circ }$ cross kare — $175^{\circ }$ aur $185^{\circ }$ samples ke beech. Switch ke baad stored magnitude ($185^{\circ }$ par): $\Vert {\mathbf{\sigma }}^S\Vert =1/1.0416=0.9601$, jo equals $\tan ((360-185)/4{)}^{\circ }=\tan 43.75^{\circ }$. Estimator ab ek chhota vector carry karta hai jo equivalent $175^{\circ }$ short-way rotation describe karta hai — smoothly, koi blow-up nahi.

Figure s02 aise padho: yeh same curve ke $150^{\circ }$–$200^{\circ }$ window ko zoom in karta hai. Left par do green dots safe samples hain ($170^{\circ },175^{\circ }$); right par do red dots ($185^{\circ },190^{\circ }$) dikhate hain ki raw magnitude orange $\Vert \mathbf{\sigma }\Vert =1$ line ke upar kahan badhti rehti. Gray arrows woh reciprocal hop hain jo switch perform karta hai — har red point ko neeche uske green diamond shadow par drop karta hai, jo safely $1$ ke neeche wapas hai. Woh downward hop hi singularity avoidance hai, visible bana ke.

MRP without switch: jaise $\Phi \to 360^{\circ }$, $\Vert \mathbf{\sigma }\Vert =\tan (\Phi /4)\to \tan 90^{\circ }\to \infty$. State vector ke components explode karte hain; $B$ matrix entries ($\mathbf{\sigma }$ mein quadratic) bhi explode karte hain, aur numerical integration diverge ho jaata hai. Lekin yeh ek removable failure hai — ek reciprocal switch $\Vert \mathbf{\sigma }\Vert \le 1$ hamesha ke liye rakhta hai. Euler angles: gimbal lock $\pm 90^{\circ }$ par ek structural rank loss hai — do rotation axes align ho jaate hain, ek degree of freedom gayab ho jaata hai, aur Euler angles ke andar koi coordinate relabeling ise remove nahi kar sakta. Tumhe poori parameterization badalni padti hai. Kaun sa worse hai: gimbal lock operationally worse hai, do reasons se. Pehla, $90^{\circ }$ pitch ek routine, frequently visited attitude hai, jabki ek poora $360^{\circ }$ turn ek rare extreme hai. Doosra, Euler failure representation ke andar avoidable nahi hai — singularity geometry mein baked in hai — jabki MRP failure ek single dot-product test aur ek reciprocal switch se dodgeable hai jo almost kuch cost nahi karta. MRPs kyun jeette hain overall: yeh minimum parameter count (3, Euler angles ke barabar) ko combine karte hain singularity ke saath jo $360^{\circ }$ par relocated hai (normal operation se door) aur shadow set ke zariye unreachable bana diya gaya hai. Tumhe Euler angles ki compactness milti hai gimbal lock ke bina, aur almost quaternions ki robustness milti hai redundant fourth number ya uski sign ambiguity ke bina. Woh balance — compact, $\pm 180^{\circ }$ ke upar smooth, singularity free mein avoidable — exactly isliye MRPs kaafi modern control laws ke liye preferred small attitude state hain.