Converting between DCM, quaternions, Euler angles: Worked examples

3.5.10 · D3 · Physics › Guidance, Navigation & Control (GNC) › Converting between DCM, quaternions, Euler angles

Is page mein parent topic ke conversions ko itna drill karenge ki koi bhi case surprise na kar sake. Har quadrant, har sign, har degenerate input, aur do exam traps jo sab ko pakad lete hain — sab cover karenge. Prereqs jinpe hum rely karte hain: Euler's Rotation Theorem, Rodrigues Rotation Formula, Direction Cosine Matrix Properties, Gimbal Lock, Rotation Sequences (3-2-1 vs 3-1-3), Quaternion Kinematics & Propagation.

Numbers touch karne se pehle, ek reminder plain words mein. Ek DCM ek $3 \times 3$ table of numbers hai jo ek vector ko doosre mein badalta hai; ek quaternion 4 numbers $(q_{0}, q_{1}, q_{2}, q_{3})$ hain jo ek twist ka axis aur half the angle store karte hain; Euler angles $(ϕ, θ, ψ)$ teen ordinary turn-amounts (roll, pitch, yaw) hain. Teeno ek hi same physical turn describe karte hain.

Definition Woh convention jo hum lock karte hain (koi bhi example dekhne se pehle yeh padho)

Teen choices pin down karni padengi warna har sign galat ho jayega:

Column vectors, matrix on the left. Ek vector ek column $v = v_{1} v_{2} v_{3}$ hai, aur hum left-multiplying karke rotate karte hain: $v_{new} = C v$ .
Body-from-inertial. $C$ ek fixed physical vector ke coordinates inertial frame mein jaise dikhte hain leke body frame mein unke coordinates return karta hai. Symbols mein $v_{body} = C v_{inertial}$ . Transpose $C^{⊤}$ ulta karta hai (inertial-from-body).
Sequence 3-2-1 (yaw about $z$ , phir new $y$ ke baare mein pitch, phir new $x$ ke baare mein roll), isliye $C = R_{x} (ϕ) R_{y} (θ) R_{z} (ψ)$ — sabse rightmost factor pehle act karta hai.

$C (q)$ neeche bilkul wahi matrix hai jo parent ke §2 mein hai, isi column/left convention ke liye likhi gayi.

Formula Toolbox — exact formulas jo is page par baar baar use honge (taaki kabhi peeche na jaana pade)

Notation: throughout, $c α$ means $cos α$ aur $s α$ means $sin α$ (toh $c θ = cos θ$ , $s ψ = sin ψ$ ). Yeh shorthand sirf neeche di gayi 3-2-1 matrix ke andar appear hoti hai.

Rodrigues (axis–angle → DCM) — woh anchor jisse har rotation build hoti hai, yahan reproduce ki gayi hai taaki page kabhi na chodna pade:

[\hat e]_\times=\begin{pmatrix}0&-e_3&e_2\\ e_3&0&-e_1\\ -e_2&e_1&0\end{pmatrix}$$ (Teen-term derivation ke liye [[Rodrigues Rotation Formula]] dekho.) **Quaternion → DCM** (parent §2, body-from-inertial, column convention): $$C(q)=\begin{pmatrix} q_0^2+q_1^2-q_2^2-q_3^2 & 2(q_1q_2+q_0q_3) & 2(q_1q_3-q_0q_2)\\ 2(q_1q_2-q_0q_3) & q_0^2-q_1^2+q_2^2-q_3^2 & 2(q_2q_3+q_0q_1)\\ 2(q_1q_3+q_0q_2) & 2(q_2q_3-q_0q_1) & q_0^2-q_1^2-q_2^2+q_3^2 \end{pmatrix}$$ **DCM → quaternion** (Shepperd, $q_0$-branch): $q_0=\tfrac12\sqrt{1+\text{tr}\,C}$, phir cyclic index triples $(i,j,k)=(1,2,3),(2,3,1),(3,1,2)$ ke saath: $$q_1=\frac{C_{23}-C_{32}}{4q_0},\quad q_2=\frac{C_{31}-C_{13}}{4q_0},\quad q_3=\frac{C_{12}-C_{21}}{4q_0}$$ yani $q_i=(C_{jk}-C_{kj})/(4q_0)$ jahan $(i,j,k)$, $(1,2,3)$ ka ek *cyclic* rotation hai. **DCM → quaternion (general leading component)** — jab $q_0$ bahut chhota ho, tab uske bajaye us slot se lead karo jiska *diagonal* $C_{ii}$ sabse bada ho: $$q_i=\tfrac12\sqrt{1+C_{ii}-C_{jj}-C_{kk}}\qquad (i,j,k)\ \text{cyclic},$$ phir off-diagonal pairs se baaki teen recover karo (Ex 5 mein worked out). Yeh Shepperd's method ki safe branch hai. **3-2-1 DCM** ($C=R_x(\phi)R_y(\theta)R_z(\psi)$, parent §3): row 1 hai $(c\theta c\psi,\ c\theta s\psi,\ -s\theta)$, aur $C_{21}=s\phi s\theta c\psi-c\phi s\psi$, $C_{22}=s\phi s\theta s\psi+c\phi c\psi$, $C_{23}=s\phi c\theta$, $C_{33}=c\phi c\theta$. **3-2-1 extraction:** $\theta=-\arcsin C_{13}$, $\phi=\operatorname{atan2}(C_{23},C_{33})$, $\psi=\operatorname{atan2}(C_{12},C_{11})$.

The scenario matrix

Har conversion problem jo tumhe di ja sakti hai, in mein se kisi ek cell mein aati hai. Neeche ke worked examples us cell ke saath tagged hain jo woh cover karta hai.

Cell	Isme kya tricky hai	Covered by
A. Cardinal axis, small angle	clean numbers, machinery seekho	Ex 1
B. Off-diagonal signs (which sign of $sin$ ?)	Rodrigues sign bookkeeping	Ex 2
C. Quadrant II/III/IV Euler angle	`atan2` vs `atan` — bada trap	Ex 3
D. Zero rotation (identity)	degenerate: axis undefined	Ex 4
E. $180°$ rotation ( $q_{0} \to 0$ )	Shepperd division-by-zero branch	Ex 5
F. Gimbal lock ( $θ = \pm 90°$ )	ek DOF vanishes — dono signs	Ex 6
G. Sign ambiguity $q$ vs $- q$	same attitude, do 4-tuples	Ex 7
H. Real-world word problem	ek story se rotation build karo	Ex 8
I. Exam twist: wrong convention (transpose)	reversed rotation	Ex 9
J. Composite / limiting behaviour	half-angle limit, $Φ \to 0$	Ex 10

Example 1 — Cell A: cardinal axis, small clean angle

Pehle half-angle. Kyun yeh step? Quaternion hamesha $Φ/2$ store karta hai (sandwich $q v q^{- 1}$ twist ko do baar apply karta hai; halving us doubling ko cancel karta hai). $\frac{Φ}{2} = 30°$ .
Chaar slots bharo. Kyun? $q_{0} = cos 30°$ , aur $(q_{1}, q_{2}, q_{3}) = \overset{e}{^} sin 30°$ . $\overset{e}{^} = (0, 0, 1)$ ke saath sirf $q_{3}$ bachta hai: $q_{0} = \frac{3}{2} \approx 0.8660, q_{1} = q_{2} = 0, q_{3} = \frac{1}{2} = 0.5$
$C (q)$ build karo. Toolbox $C (q)$ mein plug in kyun? Yeh neutral hub hai jiske through har representation convert hoti hai. $q_{1} = q_{2} = 0$ ke saath Toolbox matrix ki har wo term jo $q_{1}$ ya $q_{2}$ carry karti hai woh zero ho jaati hai, sirf yeh bachta hai: $C = q_{0}^{2} - q_{3}^{2} - 2 q_{0} q_{3} 0 2 q_{0} q_{3} q_{0}^{2} - q_{3}^{2} 0 00 q_{0}^{2} + q_{3}^{2} = \frac{1}{2} - \frac{3}{2} 0 \frac{3}{2} \frac{1}{2} 0 001$

Figure — Converting between DCM, quaternions, Euler angles

Verify: $q_{0}^{2} - q_{3}^{2} = 0.75 - 0.25 = 0.5 = cos 60°$ ✓ aur $2 q_{0} q_{3} = 2 (0.866) (0.5) = 0.866 = sin 60°$ ✓. Norm $0.75 + 0.25 = 1$ ✓.

Example 2 — Cell B: off-diagonal signs, tilted axis

Rodrigues ke pieces. Kyun? Toolbox formula $C = I cos Φ + [\overset{e}{^}]_{\times} sin Φ + \overset{e}{^} \overset{e}{^}^{⊤} (1 - cos Φ)$ ko $cos 120° = - \frac{1}{2}$ , $sin 120° = \frac{3}{2}$ , $1 - cos 120° = \frac{3}{2}$ chahiye.
Outer product term. Kyun? $\overset{e}{^} \overset{e}{^}^{⊤}$ ki har entry $\frac{1}{3}$ hai, $\frac{3}{2}$ se multiply karne par har jagah $\frac{1}{2}$ milta hai.
Cross-product term $[\overset{e}{^}]_{\times} sin 120°$ . Sign kyun yahan hai? Antisymmetric $[\overset{e}{^}]_{\times}$ with $\overset{e}{^} = \frac{1}{3} (1, 1, 1)$ ko $\frac{3}{2}$ se scale karne par off-diagonal $\pm \frac{1}{2}$ milte hain.
Jodo. Diagonal: $- \frac{1}{2} + \frac{1}{2} = 0$ . Hamare body-from-inertial, column convention mein teeno terms mil ke woh cyclic permutation deti hain jo $z \to y \to x \to z$ bheji hai: $C = 010001100$

Verify: $det C = 1$ , $C^{⊤} C = I$ (orthonormal), aur $tr C = 0 = 1 + 2 cos 120°$ ✓ (trace-angle identity $tr C = 1 + 2 cos Φ$ ).

Example 3 — Cell C: quadrant II/III/IV Euler extraction

Pehle pitch. Kyun? Toolbox se, $θ = - arcsin C_{13}$ (minus, $C_{13}$ mein baithe $- sin θ$ ko undo karta hai). Yahan $θ = - arcsin 0 = 0$ — koi gimbal lock nahi, aage badhna safe hai.
Yaw atan2 se. atan kyun nahi? Toolbox deta hai $ψ = atan2 (C_{12}, C_{11})$ . Plain $arctan (\frac{C _{12}}{C _{11}}) = arctan (\frac{- 0.5}{- 0.866}) = arctan (0.577) = 30°$ — galat quadrant (ratio ke andar dono minus signs cancel ho gaye). atan2 dono signs alag alag rakhta hai: $ψ = atan2 (- 0.5, - 0.866) = - 150° = 210°$ Dono arguments negative ⇒ third quadrant, jo $arctan$ (range $\pm 90°$ ) kabhi reach nahi kar sakta.
Roll. Kyun? $ϕ = atan2 (C_{23}, C_{33}) = atan2 (0, 1) = 0$ .

Verify: $cos 210° = - 0.866 = C_{11}$ ✓, $sin 210° = - 0.5 = C_{12}$ ✓ ( $θ = 0$ ke liye, row 1 hai $(cos ψ, sin ψ, 0)$ ).

Example 4 — Cell D: identity / zero rotation (degenerate axis)

Trace branch. Kyun? $tr I = 3 \Rightarrow q_{0} = \frac{1}{2} 1 + 3 = 1$ .
Vector part. Zero kyun? Off-diagonal differences $C_{23} - C_{32} = 0$ , isliye $q_{1} = q_{2} = q_{3} = 0$ . Toh $q = (1, 0, 0, 0)$ .
Axis undefined hai — aur yeh theek hai. Kyun? $Φ = 0$ se, $sin \frac{Φ}{2} = 0$ , toh $\overset{e}{^} sin \frac{Φ}{2} = 0$ chahe $\overset{e}{^}$ kuch bhi ho. Euler's theorem kehta hai "axis exist karta hai" sirf tab jab koi turn ho; $0°$ turn ka koi unique axis nahi hota. Quaternion isko sidestep karta hai — woh vector part mein $0$ store karta hai aur kabhi nahi puchta ki axis kaun sa tha.
Euler. $θ = - arcsin 0 = 0$ , $ϕ = atan2 (0, 1) = 0$ , $ψ = atan2 (0, 1) = 0$ .

Verify: $q = (1, 0, 0, 0)$ ke saath $C (q)$ $diag (1, 1, 1) = I$ deta hai ✓. Norm $= 1$ ✓.

Example 5 — Cell E: $180°$ rotation, Shepperd branch

Khatre ko dekho. Kyun? $q_{0} = \frac{1}{2} 1 + tr C = \frac{1}{2} 1 - 1 = 0$ . Toolbox ka $q_{i} = (C_{j k} - C_{k j}) / (4 q_{0})$ zero se divide karta hai.
Sabse bada diagonal chuno — general leading-component formula use karke. Kyun? Shepperd ka rule: us $q_{i}$ se lead karo jo provably sabse bada ho, taaki denominator zero se door rahe. Sabse bada diagonal entry $C_{33} = + 1$ hai, jo $q_{3}$ ki taraf point karta hai. Toolbox formula $q_{i} = \frac{1}{2} 1 + C_{ii} - C_{j j} - C_{k k}$ apply karo $(i, j, k) = (3, 1, 2)$ ke saath: $q_{3} = \frac{1}{2} 1 + C_{33} - C_{11} - C_{22} = \frac{1}{2} 1 + 1 + 1 + 1 = 1$
Baaki bharo — index order spell out karke. Kyun? Jab hum $q_{3}$ se lead karte hain, baaki slots cyclic triple $(i, j, k) = (3, 1, 2)$ ko $4 q_{3}$ se divide karke aate hain. Concretely: $q_{0} = (C_{12} - C_{21}) / (4 q_{3})$ , $q_{1} = (C_{31} + C_{13}) / (4 q_{3})$ , $q_{2} = (C_{23} + C_{32}) / (4 q_{3})$ (plus signs isliye kyunki $q_{3}$ -led rows symmetric off-diagonal pair use karti hain). Inke saare numerators yahan $0$ hain, isliye $q_{0} = q_{1} = q_{2} = 0$ . Toh $q = (0, 0, 0, 1)$ .

Verify: $q = (0, 0, 0, 1)$ ke saath $C (q)$ $diag (q_{0}^{2} + q_{1}^{2} - q_{2}^{2} - q_{3}^{2}, \dots) = diag (- 1, - 1, 1)$ deta hai ✓.

Example 6 — Cell F: gimbal lock DONO $θ = + 90°$ aur $θ = - 90°$ par

Case (a): $θ = + 90°$ .

Pitch. Kyun? $θ = - arcsin (C_{13}) = - arcsin (- 1) = + 90°$ . Pitch seedha upar hai.
Lock detect karo. Undefined kyun? $θ = 90°$ par Toolbox row 1 hai $(c θ c ψ, c θ s ψ, - s θ) = (0, 0, - 1)$ , isliye $C_{11} = C_{12} = 0$ aur $ψ = atan2 (0, 0)$ undefined hai. Yaw aur roll axes parallel ho gayi hain — Gimbal Lock.
Determined combination par collapse karo. Yahan $ϕ - ψ$ kyun bachta hai? $θ = 90°$ ( $s θ = 1$ ) ko $C_{21} = s ϕ s θ c ψ - c ϕ s ψ = s ϕ c ψ - c ϕ s ψ = sin (ϕ - ψ)$ mein substitute karo, aur $C_{22} = s ϕ s ψ + c ϕ c ψ = cos (ϕ - ψ)$ . Toh $θ = + 90°$ par matrix sirf difference $ϕ - ψ$ par depend karta hai. Convention se $ϕ = 0$ set karo, phir: $- ψ = atan2 (C_{21}, C_{22}) = atan2 (- 0.5, 0.866) = - 30° \Rightarrow ψ = 30°, ϕ - ψ = - 30°.$

Case (b): $θ = - 90°$ .

Doosre pole par pitch. Sign flip kyun? $θ = - arcsin (+ 1) = - 90°$ , isliye $s θ = - 1$ . Row-1 zeros ( $C_{11} = C_{12} = 0$ ) abhi bhi hold karte hain — yeh phir se lock hai.
Fused combination ab $ϕ + ψ$ hai. Difference nahi, sum kyun? $s θ = - 1$ ko same entries mein daalo: $C_{21} = - s ϕ c ψ - c ϕ s ψ = - sin (ϕ + ψ)$ aur $C_{22} = - s ϕ s ψ + c ϕ c ψ = cos (ϕ + ψ)$ . Toh $θ = - 90°$ par sirf $ϕ + ψ$ determined hai. $ϕ = 0$ set karo: $atan2 (- C_{21}, C_{22}) = atan2 (0.5, 0.866) = 30° \Rightarrow ϕ + ψ = 30° \Rightarrow ψ = 30°.$
Loss report karo (dono poles). Zabar se kyun kehna padega? Sirf ek combined number real hai: $ϕ - ψ$ at $+ 90°$ , $ϕ + ψ$ at $- 90°$ . Ise ek specific $ϕ$ aur $ψ$ mein todna ek arbitrary choice hai. Representation ka ek DOF kissi bhi pole par chala jaata hai; physical attitude bilkul theek hai.

Verify: (a) $θ = 90°, ϕ = 0, ψ = 30°$ : $C_{21} = sin (ϕ - ψ) = sin (- 30°) = - 0.5$ ✓, $C_{22} = cos (- 30°) = 0.866$ ✓. (b) $θ = - 90°, ϕ = 0, ψ = 30°$ : $C_{21} = - sin (ϕ + ψ) = - sin (30°) = - 0.5$ ✓, $C_{22} = cos (30°) = 0.866$ ✓.

Example 7 — Cell G: $q$ vs $- q$ same attitude hain

Dono build karo. Kyun? Toolbox $C (q)$ ki har entry do $q$ -components ka product hai (quadratic). Har sign flip karne par $(- 1) \times (- 1) = + 1$ hota hai — koi change nahi.
Concretely $2 q_{0} q_{3}$ : $q$ se yeh $2 (0.866) (0.5) = 0.866$ hai; $- q$ se $2 (- 0.866) (- 0.5) = 0.866$ hai. Identical.
Unique representative. $q_{0} \geq 0$ enforce kyun karein? "Shortest-arc" twist pick karne ke liye aur ek database mein ek attitude ke do naam avoid karne ke liye.

Verify: $C (q) = C (- q)$ entrywise; dono Example 1 ki $60°$ -about- $z$ matrix ke barabar hain ✓.

Example 8 — Cell H: real-world word problem

Do elementary quaternions. Compose karne ke liye quaternions kyun? Quaternion multiplication rotations ko cleanly compose karta hai ( $180°$ ambiguity nahi). Yaw $ψ = 90°$ about $z$ : $q_{ψ} = (cos 45°, 0, 0, sin 45°) = (0.7071, 0, 0, 0.7071)$ . Pitch $θ = 30°$ about $y$ : $q_{θ} = (cos 15°, 0, sin 15°, 0) = (0.9659, 0, 0.2588, 0)$ .
Sahi order mein multiply karo. $q_{ψ} \otimes q_{θ}$ kyun? Ek 3-2-1 chain ke liye total $C = R_{y} R_{z}$ (pitch yaw ke baad apply hoti hai); matching quaternion product $q = q_{ψ} \otimes q_{θ}$ hai Hamilton product $(a_{0} b_{0} - a \cdot b, a_{0} b + b_{0} a + a \times b)$ use karke.
Compute karo. Scalar: $0.7071 (0.9659) - 0 = 0.6830$ . Vector: $0.7071 (0, 0.2588, 0) + 0.9659 (0, 0, 0.7071) + (0, 0, 0.7071) \times (0, 0.2588, 0)$ . Cross term $(0, 0, 0.7071) \times (0, 0.2588, 0) = (0 (0) - 0.7071 (0.2588), 0.7071 (0) - 0 (0), 0 (0.2588) - 0 (0)) = (- 0.1830, 0, 0)$ . Sum vector $= (- 0.1830, 0.1830, 0.6830)$ . $q \approx (0.6830, - 0.1830, 0.1830, 0.6830)$

Verify: $∥ q ∥^{2} = 0.683 0^{2} + 0.183 0^{2} + 0.183 0^{2} + 0.683 0^{2} \approx 1.000$ ✓ (unit quaternions ka composite unit hota hai).

Example 9 — Cell I: exam twist, transpose trap

Convention mismatch pehchano. Kyun care karna? Hamara locked convention (definition box dekho) body-from-inertial hai $v_{body} = C v_{inertial}$ ke saath. Ek inertial-from-body matrix uska transpose $C^{⊤}$ hota hai. Galat wala feed karne par har rotation reverse ho jaati hai.
Transpose karo. Kyun? $C_{correct} = C^{⊤} = 010 - 1 00 001$ .
$q$ nikalo. Toolbox $q_{0}$ -branch use karke: $tr = 1 \Rightarrow q_{0} = \frac{1}{2} 2 = 0.7071$ ; $q_{3} = (C_{12} - C_{21}) /4 q_{0} = (- 1 - 1) / (2.828) = - 0.7071$ . Toh $q = (0.7071, 0, 0, - 0.7071)$ — ek $- 90°$ yaw, woh reverse jo un-transposed matrix se milta.

Verify: un-transposed ne $ψ = + 90°$ diya tha (Ex 1-style); transposed ne $ψ = - 90°$ diya. Angles exact negatives hain ✓; $∥ q ∥ = 1$ ✓.

Example 10 — Cell J: limiting behaviour, half-angle jab $Φ \to 0$

Exact quaternion. $q_{0} = cos (0.01) = 0.99995$ , $q_{2} = sin (0.01) = 0.0099998$ .
Small-angle limit. Taylor-expand kyun? $Φ \to 0$ ke liye, $cos \frac{Φ}{2} \to 1$ aur $sin \frac{Φ}{2} \to \frac{Φ}{2}$ . Toh $q \to (1, 0, \frac{Φ}{2}, 0) = (1, 0, 0.01, 0)$ . Yeh linearised form hai jo Quaternion Kinematics & Propagation mein use hoti hai.
Limit ki sanity. Propagation ke liye yeh kyun matter karta hai? Ek integration step pe update lagbhag $q \approx (1, \frac{1}{2} ω d t)$ hoti hai; use match karna propagator ko first-order accurate rakhta hai aur dikhata hai ki renormalization kyun zaroori hai (exact $q_{0} = 0.99995 \neq = 1$ norm ko drift karta hai).

Verify: $sin (0.01) = 0.00999983$ vs approximation $0.01$ — relative error $\approx 1.7 \times 1 0^{- 5}$ , tiny as claimed ✓. $cos (0.01)^{2} + sin (0.01)^{2} = 1$ ✓.

Recall Self-test: cell ka naam bolo

Sirf $tr C = - 1$ diya ho toh kaunsa extraction branch use karoge? ::: Shepperd's largest-diagonal branch (Cell E) — $q_{0} = 0$ , sabse bade $q_{i}$ se lead karo $q_{i} = \frac{1}{2} 1 + C_{ii} - C_{j j} - C_{k k}$ ke zariye. Tumne yaw padha aur dono atan2 arguments negative hain — kaunsa quadrant? ::: Third (Cell C); plain atan $180°$ -galat first-quadrant answer deta. $θ = + 90°$ vs $θ = - 90°$ par, kaun sa Euler combination bachta hai? ::: $ϕ - ψ$ at $+ 90°$ , $ϕ + ψ$ at $- 90°$ (Cell F, gimbal lock — dono poles). Kya $q$ aur $- q$ kabhi alag DCMs dete hain? ::: Nahi — $C (q)$ mein $q$ quadratic hai (Cell G).