Converting between DCM, quaternions, Euler angles: Exercises

3.5.10 · D4 · Physics › Guidance, Navigation & Control (GNC) › DCM, quaternions, Euler angles ke beech convert karna

Ye parent topic ki practice arena hai. Har problem L1 → L5 tak graded hai. Pehle khud try karo, phir collapsible solution kholo. Saari machinery parent note mein hai: Rodrigues Rotation Formula, Direction Cosine Matrix Properties, Quaternion Kinematics & Propagation, Gimbal Lock, Rotation Sequences (3-2-1 vs 3-1-3).

Shuru karne se pehle, ek figure saare conventions aur symbols fix kar deta hai jo hum baar baar reuse karenge.

Figure — Converting between DCM, quaternions, Euler angles

Definition Symbols jo neeche har jagah use hote hain (ek baar define kiye)

$C$ — Direction Cosine Matrix, $3 \times 3$ numbers ka grid; row $i$ , column $j$ ko $C_{ij}$ likhte hain. $C v$ multiply karne se vector $v$ rotate hota hai.
$\overset{e}{^} = (e_{1}, e_{2}, e_{3})$ — ek unit axis: length $1$ ka ek arrow jiske around rotation ghoomta hai.
$Φ$ — us axis ke baare mein turn angle.
$q = (q_{0}, q_{1}, q_{2}, q_{3})$ — quaternion: $q_{0}$ "kitna" wala scalar hai, $(q_{1}, q_{2}, q_{3})$ "kis direction mein" wala vector hai. Constraint: $q_{0}^{2} + q_{1}^{2} + q_{2}^{2} + q_{3}^{2} = 1$ .
$(ϕ, θ, ψ)$ — roll, pitch, yaw (3-2-1 Euler angles).
$atan2 (y, x)$ — angle-finder jo dono signs track karta hai (dekho Rotation Sequences (3-2-1 vs 3-1-3)). Ye jawaab deta hai "woh kaun sa angle hai jiska vertical part $y$ aur horizontal part $x$ hai?" poore circle $[- 180°, 180°]$ par, unlike plain $arctan$ jo sirf $[- 90°, 90°]$ jaanta hai.

Level 1 — Recognition

L1.1 — Ek quaternion ka axis aur angle padhna

Quaternion $q = (\frac{3}{2}, 0, 0, \frac{1}{2})$ diya gaya hai. Rotation axis $\overset{e}{^}$ aur turn angle $Φ$ kya hai?

Recall Solution — L1.1

KYA use karein: axis–angle definition $q_{0} = cos \frac{Φ}{2}$ aur $(q_{1}, q_{2}, q_{3}) = \overset{e}{^} sin \frac{Φ}{2}$ . KYU DCM nahi balki ye: axis aur angle seedha quaternion mein directly bake in hote hain; koi matrix ki zaroorat nahi.

$q_{0} = cos \frac{Φ}{2} = \frac{3}{2}$ se milta hai $\frac{Φ}{2} = 30°$ , toh $Φ = 60°$ . Vector part $(0, 0, \frac{1}{2})$ equal hona chahiye $\overset{e}{^} sin 30° = \overset{e}{^} \cdot \frac{1}{2}$ ke. $\frac{1}{2}$ se divide karke: $\overset{e}{^} = (0, 0, 1)$ .

Jawab: axis $\overset{e}{^} = (0, 0, 1)$ ( $z$ -axis), angle $Φ = 60°$ .

L1.2 — Trace shortcut ko pehchaanna

Kisi bhi rotation quaternion ke liye, uske DCM ka trace $tr (C) = C_{11} + C_{22} + C_{33}$ hota hai. $tr (C) = 2$ diya gaya ho toh $q_{0}$ nikalo (positive root lete hue).

Recall Solution — L1.2

KYA/KYU: parent note prove karta hai $tr (C) = 4 q_{0}^{2} - 1$ . Ye ek scalar $q_{0}$ carry karta hai, isliye ye sabse fast handle hai. $4 q_{0}^{2} - 1 = 2 \Rightarrow q_{0}^{2} = \frac{3}{4} \Rightarrow q_{0} = \frac{3}{2} \approx 0.8660.$ Sanity check: $q_{0} = cos \frac{Φ}{2} = \frac{3}{2} \Rightarrow Φ = 60°$ — L1.1 wale rotation se match karta hai.

Level 2 — Application

L2.1 — Quaternion → DCM (matrix banana)

$q = (\frac{2}{2}, 0, 0, \frac{2}{2})$ ko uske DCM $C (q)$ mein convert karo.

Recall Solution — L2.1

KYA: parent ke $C (q)$ formula mein plug karo. KYU: vectors ko rotate karne ke liye matrix form chahiye. Yahan $q_{0} = q_{3} = \frac{2}{2}$ , toh $q_{0}^{2} = q_{3}^{2} = \frac{1}{2}$ , aur $q_{1} = q_{2} = 0$ .

$C_{11} = q_{0}^{2} + q_{1}^{2} - q_{2}^{2} - q_{3}^{2} = \frac{1}{2} - \frac{1}{2} = 0$
$C_{22} = q_{0}^{2} - q_{1}^{2} + q_{2}^{2} - q_{3}^{2} = \frac{1}{2} - \frac{1}{2} = 0$
$C_{33} = q_{0}^{2} - q_{1}^{2} - q_{2}^{2} + q_{3}^{2} = \frac{1}{2} + \frac{1}{2} = 1$
$C_{12} = 2 (q_{1} q_{2} + q_{0} q_{3}) = 2 \cdot \frac{1}{2} = 1$ , $C_{21} = 2 (q_{1} q_{2} - q_{0} q_{3}) = - 1$
baaki saare off-diagonals mein $q_{1}$ ya $q_{2} \Rightarrow 0$ hain. $C = 0 - 1 0 100001 .$ Ye $z$ ke baare mein pure $90°$ yaw hai (parent ka Example 1 dekho). ✓

L2.2 — Euler → DCM (3-2-1) ek single pitch ke liye

$C = R_{x} (0) R_{y} (30°) R_{z} (0)$ compute karo, yaani pure pitch $θ = 30°$ , roll $ϕ = 0$ , yaw $ψ = 0$ .

Recall Solution — L2.2

KYA: parent ke 3-2-1 matrix mein $ϕ = ψ = 0$ ke saath substitute karo ( $c ϕ = c ψ = 1, s ϕ = s ψ = 0$ ). KYU: confirm karta hai ki aap general formula se terms ko zero karke padh sakte ho. $C = c θ 0 s θ 010 - s θ 0 c θ = 0.8660 0 0.5 010 - 0.5 0 0.8660 .$ Middle row/column untouched hai kyunki hum $y$ ke around ghoom rahe hain: $y$ -axis fixed rehta hai. Ye geometric "look" hai.

L2.3 — DCM → Euler extraction

L2.2 ke matrix se, parent ke extraction rules use karke $(ϕ, θ, ψ)$ recover karo.

Recall Solution — L2.3

KYA/KYU: hum specific entries padhte hain kyunki woh ek ek angle ko isolate karte hain (parent "DCM → Euler").

$θ = - arcsin (C_{13}) = - arcsin (- 0.5) = + 30°$ ✓
$ϕ = atan2 (C_{23}, C_{33}) = atan2 (0, 0.8660) = 0°$ ✓
$ψ = atan2 (C_{12}, C_{11}) = atan2 (0, 0)$ — lekin yahan $c θ \neq = 0$ hai, toh $C_{11} = c θ c ψ = 0.866, C_{12} = c θ s ψ = 0$ , jo deta hai $atan2 (0, 0.866) = 0°$ ✓ Round trip exactly recover ho gaya. (Note: sirf tab truly undefined hota hai jab $c θ = 0$ , yaani gimbal lock — yahan nahi.)

Level 3 — Analysis

L3.1 — DCM → Quaternion, safe branch

Diya gaya $C = 0 - 1 0 100001,$ trace branch use karke $q$ recover karo, aur $q_{0} \geq 0$ enforce karo.

Recall Solution — L3.1

KYA: pehle trace. $tr (C) = 0 + 0 + 1 = 1$ , toh $q_{0} = \frac{1}{2} 1 + 1 = \frac{1}{2} 2 = \frac{2}{2} \approx 0.7071$ . KYU trace branch yahan safe hai: $q_{0} \approx 0.707$ comfortably $0$ se door hai, toh $4 q_{0}$ se divide karna theek hai.

$q_{1} = \frac{C _{23} - C _{32}}{4 q _{0}} = \frac{0 - 0}{4 q _{0}} = 0$
$q_{2} = \frac{C _{31} - C _{13}}{4 q _{0}} = \frac{0 - 0}{4 q _{0}} = 0$
$q_{3} = \frac{C _{12} - C _{21}}{4 q _{0}} = \frac{1 - ( - 1 )}{4 \cdot 0.7071} = \frac{2}{2.828} = 0.7071$ $q = (\frac{2}{2}, 0, 0, \frac{2}{2}) .$ Ye exactly L2.1 ka input hai — conversion ek perfect inverse hai. ✓

L3.2 — DCM → Quaternion, singular branch ( $180°$ ke paas)

$C = diag (- 1, - 1, 1)$ ke liye ( $z$ ke baare mein $180°$ rotation), trace branch fail ho jaati hai. $q$ sahi se recover karo.

Recall Solution — L3.2

KYA: $tr (C) = - 1 - 1 + 1 = - 1 \Rightarrow q_{0} = \frac{1}{2} 1 + (- 1) = 0$ . $4 q_{0} = 0$ se divide karna illegal hai — isliye Shepperd kehta hai sabse bade component se divide karo. KYU: $180°$ par scalar part vanish ho jaata hai; sabse bada diagonal entry batata hai ki kaun sa vector component dominate karta hai. Diagonal entries hain $C_{11} = - 1, C_{22} = - 1, C_{33} = + 1$ ; sabse bada $C_{33}$ hai, toh $q_{3}$ solve karo: $q_{3} = \frac{1}{2} 1 + C_{33} - C_{11} - C_{22} = \frac{1}{2} 1 + 1 + 1 + 1 = \frac{1}{2} 4 = 1.$ Phir off-diagonals baaki sab dete hain $q_{0} = \frac{C _{21} - C _{12}}{4 q _{3}} = \frac{0 - 0}{4} = 0$ ke zariye, aur $q_{1} = q_{2} = 0$ . $q = (0, 0, 0, 1) .$ Check: $q_{0} = cos \frac{Φ}{2} = 0 \Rightarrow Φ = 180°$ ; axis $(0, 0, 1)$ ✓.

L3.3 — Gimbal lock detect aur handle karna

Ek DCM mein $C_{13} = - 1$ hai. Dikhao ki ye gimbal lock hai, $θ$ nikalo, aur explain karo ki kya (aur kya nahi) recoverable hai.

Recall Solution — L3.3

KYA: $θ = - arcsin (C_{13}) = - arcsin (- 1) = + 90°$ . KYU ye singularity hai: $θ = 90°$ ke saath, $c θ = 0$ , toh parent matrix mein $C_{11} = c θ c ψ = 0$ aur $C_{12} = c θ s ψ = 0$ . Tab $ψ = atan2 (0, 0)$ undefined hai — yaw aur roll axes align ho gaye hain (dekho Gimbal Lock). Kya bachta hai: off-diagonal $2 \times 2$ block collapse ho jaata hai jisse sirf sum $ϕ + ψ$ determine hota hai ( $θ = + 90°$ ke liye), har ek alag nahi. Convention: $ϕ = 0$ set karo, phir $ϕ + ψ$ ko $atan2 (C_{21}, C_{22})$ se padho — woh combined angle fully valid hai; split ek free choice hai. Ek degree of freedom reported as lost ho jaata hai.

Level 4 — Synthesis

L4.1 — Full chain: Euler → DCM → quaternion

$ϕ = 0, θ = 0, ψ = 90°$ lo. $C$ banao, phir $q$ extract karo, aur "via-the-hub" workflow confirm karo.

Recall Solution — L4.1

KYU via DCM: seedha Euler→quaternion formula (12 terms) yaad karna error-prone hai; do trusted maps compose karo (parent Example: DCM ko neutral hub ki tarah use karo).

Step 1 (Euler→DCM): pure yaw $90°$ . L2.2-style substitution sirf $ψ \neq = 0$ ke saath: $C = R_{z} (90°) = c ψ - s ψ 0 s ψ c ψ 0 001 = 0 - 1 0 100001 .$ Step 2 (DCM→quaternion): L3.1 ke identical → $q = (\frac{2}{2}, 0, 0, \frac{2}{2})$ . Cross-check via axis–angle: $z$ ke baare mein yaw $90°$ ⇒ $q_{0} = cos 45° = \frac{2}{2}$ , $q_{3} = sin 45° = \frac{2}{2}$ ✓. Hub route aur direct axis–angle route agree karte hain.

L4.2 — Do rotations compose karo, phir axis padho

$90°$ yaw ( $q_{A}$ L4.1 se) ke baad $x$ ke baare mein $90°$ roll ( $q_{B} = (\frac{2}{2}, \frac{2}{2}, 0, 0)$ ) apply karo. Quaternions multiply karo ( $q_{C} = q_{B} \otimes q_{A}$ ) aur combined rotation ka $Φ$ report karo.

Recall Solution — L4.2

KYA/KYU: quaternion multiplication rotations ko cheaply compose karta hai (parent: quaternions propagation ke liye best hain). Hamilton product $q = (w, x, y, z)$ ke saath: $w_{C} = w_{B} w_{A} - v_{B} \cdot v_{A}, v_{C} = w_{B} v_{A} + w_{A} v_{B} + v_{B} \times v_{A} .$ $q_{A} = (\frac{2}{2}, 0, 0, \frac{2}{2})$ , $q_{B} = (\frac{2}{2}, \frac{2}{2}, 0, 0)$ ke saath:

$w_{C} = \frac{1}{2} - 0 = \frac{1}{2}$
$v_{B} \cdot v_{A} = 0$
$w_{B} v_{A} = (0, 0, \frac{1}{2})$ , $w_{A} v_{B} = (\frac{1}{2}, 0, 0)$
$v_{B} \times v_{A} = (\frac{2}{2}, 0, 0) \times (0, 0, \frac{2}{2}) = (0 \cdot \frac{2}{2} - 0, 0 - \frac{2}{2} \cdot \frac{2}{2}, 0 - 0) = (0, - \frac{1}{2}, 0)$
$v_{C} = (\frac{1}{2}, - \frac{1}{2}, \frac{1}{2})$ $q_{C} = (\frac{1}{2}, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2}) .$ Norm check: $4 \cdot \frac{1}{4} = 1$ ✓. Angle: $q_{0} = \frac{1}{2} = cos \frac{Φ}{2} \Rightarrow \frac{Φ}{2} = 60° \Rightarrow Φ = 120°$ . Combined turn $120°$ hai axis $\overset{e}{^} = (\frac{1}{2}, - \frac{1}{2}, \frac{1}{2}) / sin 60° = (\frac{1}{3}, - \frac{1}{3}, \frac{1}{3})$ ke baare mein.

Level 5 — Mastery

L5.1 — Trace identity prove karo

Parent ke $C (q)$ se shuru karke, unit-norm constraint use karke prove karo ki $tr (C) = 4 q_{0}^{2} - 1$ .

Recall Solution — L5.1

KYA: teen diagonal entries add karo. $tr (C) = (q_{0}^{2} + q_{1}^{2} - q_{2}^{2} - q_{3}^{2}) + (q_{0}^{2} - q_{1}^{2} + q_{2}^{2} - q_{3}^{2}) + (q_{0}^{2} - q_{1}^{2} - q_{2}^{2} + q_{3}^{2}) .$ $q_{1}^{2}$ terms: $+ 1 - 1 - 1 = - 1$ ; similarly $q_{2}^{2}$ aur $q_{3}^{2}$ har ek net $- 1$ deta hai; $q_{0}^{2}$ net $+ 3$ deta hai. Toh $tr (C) = 3 q_{0}^{2} - q_{1}^{2} - q_{2}^{2} - q_{3}^{2} .$ KYU constraint enter hoti hai: $q_{1}^{2} + q_{2}^{2} + q_{3}^{2} = 1 - q_{0}^{2}$ use karo: $tr (C) = 3 q_{0}^{2} - (1 - q_{0}^{2}) = 4 q_{0}^{2} - 1. ■$

L5.2 — $180°$ par round-trip robustness

Dikhao ki $C = diag (1, - 1, - 1)$ ke liye (parent Example 2, $x$ ke baare mein $180°$ ), recovered quaternion se $C$ rebuild karne par original matrix wapas milti hai — verify karta hai ki singular branch sirf ek hack nahi hai.

Recall Solution — L5.2

Step 1 — extract (sabse bada diagonal $C_{11} = 1$ hai): $q_{1} = \frac{1}{2} 1 + C_{11} - C_{22} - C_{33} = \frac{1}{2} 1 + 1 + 1 + 1 = 1,$ aur $q_{0} = \frac{C _{32} - C _{23}}{4 q _{1}} = 0$ , $q_{2} = q_{3} = 0$ , milta hai $q = (0, 1, 0, 0)$ . Step 2 — rebuild $C (q)$ ke zariye $q_{1} = 1$ , baaki $0$ ke saath:

$C_{11} = q_{1}^{2} = 1$
$C_{22} = - q_{1}^{2} = - 1$ , $C_{33} = - q_{1}^{2} = - 1$
saare off-diagonals mein $q_{0}, q_{2}, q_{3} \Rightarrow 0$ hain. $C_{rebuilt} = diag (1, - 1, - 1) = C_{original} . ✓$ Singular branch exact hai — matrix ko bit-for-bit reproduce karta hai.

L5.3 — Sign ambiguity physical hai, bug nahi

Dikhao ki $q = (0, 1, 0, 0)$ aur $q = (0, - 1, 0, 0)$ same DCM produce karte hain, aur woh convention state karo jo ek unique representative choose karta hai.

Recall Solution — L5.3

KYU: parent note karta hai $C (q)$ $q$ mein quadratic hai; har entry do components ka product hai, toh saare signs flip karne se har product unchanged rehta hai. Sirf nonzero-driving term check karo: $C_{22} = - q_{1}^{2}$ . Chahe $q_{1} = + 1$ ho ya $q_{1} = - 1$ , $q_{1}^{2} = 1$ hai, toh $C_{22} = - 1$ identically; $C_{11}, C_{33}$ ke liye bhi same. Off-diagonals $0$ rehte hain. Dono quaternions $diag (1, - 1, - 1)$ dete hain. Convention: $q_{0} \geq 0$ enforce karo (aur agar $q_{0} = 0$ ho, toh pehle nonzero vector component ka sign fix karo). Ye shortest-arc representative select karta hai aur double cover ambiguity remove karta hai.

Recall Jaane se pehle ek-line self-test

$tr (C)$ deta hai ::: $4 q_{0}^{2} - 1$ , hence $q_{0} = \frac{1}{2} 1 + tr C$ (L1.2, L3.1, L5.1). Rotation $A$ phir $B$ apply karne ke liye aap multiply karte ho ::: $q_{B} \otimes q_{A}$ — doosra rotation left pe (L4.2). Gimbal lock pitch par hota hai ::: $θ = \pm 90°$ , jahan sirf $ϕ \pm ψ$ recoverable hota hai (L3.3). $180°$ ke paas aap divide karte ho ::: sabse bade quaternion component se, $q_{0}$ se nahi (L3.2, L5.2).