Worked examples — Direction cosine matrix (DCM) — construction from Euler angles

3.5.3 · D3 · Physics › Guidance, Navigation & Control (GNC) › Direction cosine matrix (DCM) — construction from Euler angl

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Scenario matrix

Har DCM problem in cells mein se ek (ya blend) hoti hai. Neeche ke worked examples mein har ek kaunsa cell cover karta hai yeh announce kiya gaya hai.

Cell	Kya special hai	Trouble kahan chhupa hai	Example
A. Zero input	saare angles $=0$	identity $I$ milni chahiye — ek sanity anchor	Ex 1
B. Single axis, positive	ek angle $>0$, baaki $0$	$R_i$ ka sign convention	Ex 2
C. Single axis, negative	ek angle $<0$	kya $-\theta$ se right off-diagonal flip hoti hai?	Ex 3
D. Quadrant / large angle	angle $>90^{\circ }$ (e.g. $135^{\circ }$)	$\sin ,\cos$ signs; kaunsa quadrant	Ex 4
E. Full 3-2-1, all nonzero	yaw, pitch, roll sab active	matrix-multiply order, cross terms	Ex 5
F. Inverse problem	given $C$, angles nikalo	atan2 quadrants, $\arcsin$ range	Ex 6
G. Degenerate: gimbal lock	pitch $=\pm 90^{\circ }$	angles unique nahi rehte; $\cos \theta =0$	Ex 7
H. Word problem (real GNC)	ek sensor vector rotate karo	frame direction; forward vs inverse	Ex 8
I. Exam twist: orthonormality	prove/exploit karo $C^{-1}=C^T$	$C^T$ pe trust karo re-inverting ki jagah	Ex 9

Hum parent ki shorthand use karte hain: $c=\cos$, $s=\sin$. Angles: yaw $\psi$ axis 3 ke baare mein, pitch $\theta$ axis 2 ke baare mein, roll $\varphi$ axis 1 ke baare mein. Saari matrices inertial → body transform karti hain: ${\mathbf{v}}_B=C_{BN}{\mathbf{v}}_N$.

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Example 1 — Cell A: zero rotation

Forecast: padhne se pehle guess karo — agar body bilkul nahi mudi, toh DCM kya honi chahiye, aur vector ka kya hoga?

1. Har elementary matrix mein plug karo. Jab har angle $0$ ho: $c0=1,s0=0$, toh $R_3(0)=R_2(0)=R_1(0)=I$ (identity). Yeh step kyun? Elementary matrices identity mein reduce ho jaati hain kyunki unke $\sin$ entries vanish ho jaate hain aur $\cos$ entries $1$ ban jaate hain — yeh baseline hai jiske against baaki saare cases measure hote hain.
2. Multiply karo. $C_{BN}=I\cdot I\cdot I=I$. Yeh step kyun? Identity se multiply karne se kuch nahi badalta; "no turn" attitude ka matlab $I$ hi hona chahiye.
3. Apply karo. ${\mathbf{v}}_B=I[2,-3,5{]}^T=[2,-3,5{]}^T$.

Verify: agar frames coincide karte hain, toh same physical vector ke components dono mein same hote hain. Yahan hain bhi. ✔

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Example 2 — Cell B: single-axis positive yaw

Forecast: body ko $30^{\circ }$ left turn karne ke baad, kya old North axis body ko thoda right ($-y_B$ direction mein) dikhega ya left?

1. Sirf $R_3$ bachega. $C_{BN}=R_3(30^{\circ })=\left[\begin{array}{ccc}c30& s30& 0\\ -s30& c30& 0\\ 0& 0& 1\end{array}\right]$ jahan $c30=\frac{\sqrt{3}}{2}\approx 0.866$, $s30=0.5$. Yeh step kyun? Pitch aur roll matrices identity hain, isliye product ek hi yaw block mein collapse ho jaata hai.
2. $[1,0,0{]}^T$ pe apply karo. Row by row: $x_B=0.866$, $y_B=-0.5$, $z_B=0$. Yeh step kyun? Transform batata hai "inertial ${\hat{n}}_1$ ka kitna hissa har body axis ke along hai." Negative $y_B$ bolta hai purana North ab body $-y$ ke saath kuch angle banata hai.

Figure: Gray axes (${\hat{n}}_1,{\hat{n}}_2$) fixed inertial frame hain; blue axes (${\hat{b}}_1,{\hat{b}}_2$) body hain, $+30^{\circ }$ counter-clockwise ghuma hua. Kya notice karna hai: fixed North arrow ${\hat{n}}_1$ ka ${\hat{b}}_1$ pe ek bada positive projection hai (orange dashed, $x_B=+0.866$) aur ${\hat{b}}_2$ pe ek negative projection hai (red dashed, $y_B=-0.5$). Body ko CCW ghumane se fixed vector doosri taraf — body ke $-y$ half mein — lean karta dikhai deta hai.

Verify: length preserved? $\sqrt{0.866^2+(-0.5{)}^2}=\sqrt{0.75+0.25}=1$. ✔ Signs figure se match karte hain: body $+30^{\circ }$ (CCW) turn karti hai, toh fixed North vector ke body-frame coordinates $-30^{\circ }$ rotate ho jaate hain. ✔

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Example 3 — Cell C: single-axis NEGATIVE roll

Forecast: parent ke $R_1$ mein lower block ke top-right mein $+\sin$ hai. Negative angle ke saath, kaunsa off-diagonal entry negative ho jaayega?

1. $\varphi =-45^{\circ }$ substitute karo. $\cos (-45^{\circ })=\frac{\sqrt{2}}{2}\approx 0.707$ (cosine is even), $\sin (-45^{\circ })=-0.707$ (sine is odd). Yeh step kyun? Cosine angle ki sign ignore karta hai; sine flip karta hai. Isliye $R_1$ ke do $\sin$ entries $\varphi =+45^{\circ }$ ke relative sign swap kar lete hain.
2. Matrix likho. $R_1(-45^{\circ })=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0.707& -0.707\\ 0& 0.707& 0.707\end{array}\right].$ Yeh step kyun? Row 2 ko teesre slot mein $s\varphi =-0.707$ milta hai; row 3 ko $-s\varphi =+0.707$ milta hai.
3. $[0,1,0{]}^T$ pe apply karo. $x_B=0$, $y_B=0.707$, $z_B=0.707$.

Verify: length $=\sqrt{0.707^2+0.707^2}=1$. ✔ Negative roll body ko is tarah tip karta hai ki inertial ${\hat{n}}_2$ ko $+z_B$ component milta hai — "up" direction ko apni taraf roll karne ke consistent. ✔

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Example 4 — Cell D: obtuse-angle yaw (quadrant care)

Forecast: $135^{\circ }$ second quadrant mein hai. Compute karne se pehle $\cos 135^{\circ }$ aur $\sin 135^{\circ }$ ke signs guess karo.

1. Trig evaluate karo. $\cos 135^{\circ }=-\frac{\sqrt{2}}{2}\approx -0.707$ (second quadrant: cosine negative), $\sin 135^{\circ }=+\frac{\sqrt{2}}{2}\approx 0.707$ (sine positive). Yeh step kyun? $90^{\circ }$ ke baad cosine negative ho jaata hai — ek aisa case jo single-axis-small-angle intuition miss kar sakti hai. Quadrant sahi karna is cell ka poora point hai.
2. Matrix. $R_3(135^{\circ })=\left[\begin{array}{ccc}-0.707& 0.707& 0\\ -0.707& -0.707& 0\\ 0& 0& 1\end{array}\right]$.
3. $[1,0,0{]}^T$ pe apply karo. $x_B=-0.707$, $y_B=-0.707$, $z_B=0$.

Verify: length $=\sqrt{0.5+0.5}=1$ ✔. Old North ab body ke third quadrant ($-x_B,-y_B$) mein point karta hai — exactly wahan jahan $135^{\circ }$ body turn usse bhejni chahiye. ✔

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Example 5 — Cell E: full 3-2-1, saare angles nonzero

Forecast: parent ne note kiya tha ki $C_{13}=-s\theta$ yaw aur roll se untouched hai. Compute karne se pehle $C_{13}$ predict karo.

1. Closed form se $C_{13}$ padho. Parent ke full matrix se, $C_{13}=-s\theta =-\sin 45^{\circ }\approx -0.707$. Yeh step kyun? Yeh entry sirf pitch pe depend karti hai — ek free sanity anchor jo messy cross terms se independent hai.
2. $C_{33}$ padho. $C_{33}=c\varphi c\theta =\cos 60^{\circ }\cos 45^{\circ }=0.5\times 0.707=0.354$. Yeh step kyun? $C_{33}$ sirf roll aur pitch mix karta hai (koi yaw nahi) — multiply errors pakadne ke liye ek doosra, independent anchor.
3. Multiply karke cross-check karo. Pehle $R_2(45^{\circ })R_3(30^{\circ })$ banao, phir $R_1(60^{\circ })$ se premultiply karo, aur wohi do entries padho. Yeh step kyun? Agar closed-form shortcut aur brute-force product agree karte hain, toh hamara order-of-multiplication (last rotation leftmost) sahi hai.

Verify: brute-force multiply $C_{13}=-0.7071$, $C_{33}=0.3536$ deta hai — steps 1–2 se match karta hai. Poora matrix (Example 6 mein dobara use hoga) hai

$C=\left[\begin{array}{ccc}0.6124& 0.3536& -0.7071\\ 0.0474& 0.7891& 0.6124\\ 0.7891& -0.4967& 0.3536\end{array}\right].$ ✔

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Example 6 — Cell F: inverse problem (angles recover karo)

Forecast: kaunsi entry directly pitch deti hai, aur plain $\arctan$ ki jagah atan2 kyun use karna chahiye?

1. $C_{13}$ se pitch. $\theta =-\arcsin (C_{13})=-\arcsin (-0.7071)=45^{\circ }$. Yeh step kyun? $C_{13}=-s\theta$, toh $\arcsin$ usse invert karta hai. $\arcsin$ $[-90^{\circ },90^{\circ }]$ return karta hai — 3-2-1 ke liye valid pitch range, toh yahan koi quadrant ambiguity nahi.
2. $C_{11},C_{12}$ se yaw. $\psi =atan2(C_{12},C_{11})=atan2(0.3536,0.6124)=30^{\circ }$. Yeh step kyun? Plain $\arctan (C_{12}/C_{11})$ $30^{\circ }$ aur $210^{\circ }$ mein distinguish nahi kar sakta (tan har $180^{\circ }$ pe repeat karta hai). atan2 dono numerator aur denominator signs leta hai, sahi quadrant pin karta hai.
3. $C_{23},C_{33}$ se roll. $\varphi =atan2(C_{23},C_{33})=atan2(0.6124,0.3536)=60^{\circ }$. Yeh step kyun? $C_{23}=s\varphi c\theta$ aur $C_{33}=c\varphi c\theta$; unka ratio $\tan \varphi$ hai, aur shared positive $c\theta$ factor matlab hai atan2 seedha sahi quadrant padhta hai. Dono entries positive ⇒ first-quadrant roll ⇒ $60^{\circ }$.

Verify: recovered triple $(\psi ,\theta ,\varphi )=(30^{\circ },45^{\circ },60^{\circ })$ exactly Example-5 ka input hai. Round-trip closed. ✔

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Example 7 — Cell G: gimbal lock (degenerate pitch)

Forecast: $\cos 90^{\circ }=0$ ke saath, matrix mein kitne independent numbers bach sakte hain — teen angles ke barabar, ya kam?

1. $c\theta =0,s\theta =1$ ke saath matrix collapse karo. Parent ke full 3-2-1 form mein substitute karne se har woh term kill ho jaati hai jo $c\theta$ factor carry karti hai. Jo bachta hai woh yaw aur roll par sirf unke difference $\psi -\varphi$ ke through depend karta hai. (3-2-1 mein $\theta =+90^{\circ }$ ke liye surviving off-diagonal entries $\sin (\psi -\varphi )$ aur $\cos (\psi -\varphi )$ hain.) Yeh step kyun? Jab $\cos \theta =0$ hota hai toh yaw axis (3) aur roll axis (1) ek hi physical direction ke saath line up kar lete hain, isliye sirf ek combined turn observable hota hai.
2. Case $(\psi ,\varphi )=(20^{\circ },50^{\circ })$: yahan $\psi -\varphi =-30^{\circ }$. Plug in karne par milta hai $C_1=\left[\begin{array}{ccc}0& 0& -1\\ \sin (\psi -\varphi )& \cos (\psi -\varphi )& 0\\ \cos (\psi -\varphi )& -\sin (\psi -\varphi )& 0\end{array}\right]=\left[\begin{array}{ccc}0& 0& -1\\ -0.5& 0.8660& 0\\ 0.8660& 0.5& 0\end{array}\right].$ Yeh step kyun? Actual numbers likhne se hum collapse dekh sakte hain: poori first row $[0,0,-1]$ hai, sirf pitch se fix.
3. Case $({\psi }^{\prime },{\varphi }^{\prime })=(45^{\circ },25^{\circ })$: yahan ${\psi }^{\prime }-{\varphi }^{\prime }=+20^{\circ }\ne -30^{\circ }$ — alag! Isliye $({\psi }^{\prime },{\varphi }^{\prime })=(60^{\circ },90^{\circ })$ lo taaki ${\psi }^{\prime }-{\varphi }^{\prime }=-30^{\circ }$. Same difference $-30^{\circ }$ ke saath: $C_2=\left[\begin{array}{ccc}0& 0& -1\\ -0.5& 0.8660& 0\\ 0.8660& 0.5& 0\end{array}\right]=C_1.$ Yeh step kyun? Do alag angle pairs, $(20^{\circ },50^{\circ })$ aur $(60^{\circ },90^{\circ })$, ka difference $-30^{\circ }$ same hai, toh woh identical matrix produce karte hain. Ek matrix, bahut saare labels ⇒ parameterization singular hai. DCM bilkul theek hai; sirf Euler names break hote hain. Isliye GNC attitude ko quaternions ke roop mein store karta hai.

Verify: $C_1$ ki har entry corresponding entry of $C_2$ ke barabar hai. ✔

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Example 8 — Cell H: real GNC word problem

Forecast: $90^{\circ }$ pitch nose ko bilkul upar tip kar deta hai. Compute karne se pehle guess karo ki seedha-upar wala star body coordinates mein kahan land karega.

1. $C_{BN}=R_1(0)R_2(90^{\circ })R_3(90^{\circ })$ banao. $\theta =90^{\circ }$ ke saath: $c\theta =0,s\theta =1$; $\psi =90^{\circ }$ ke saath: $c\psi =0,s\psi =1$; $\varphi =0$ ⇒ $R_1=I$. Yeh step kyun? Sensors body frame mein rehte hain; guidance star ko inertial mein deti hai. DCM translator hai, forward direction $N\to B$ mein apply hota hai.
2. $C_{BN}$ ko $[0,0,1{]}^T$ pe apply karo. $C_{BN}$ ka third column hi woh hai jo $z$-component se multiply karta hai; woh $[-s\theta ,s\varphi c\theta ,c\varphi c\theta {]}^T=[-1,0,0{]}^T$ ke barabar hai. Yeh step kyun? Kisi matrix ko $[0,0,1{]}^T$ se multiply karna sirf uska third column pick karta hai — ek shortcut yaad rakhne layak hai.
3. Result padho. ${\mathbf{v}}_B=[-1,0,0{]}^T$: star body $-x$ axis ke along hai.

Verify: length $=1$ ✔. Physically: nose ko $90^{\circ }$ upar pitch karne se body $x$-axis wahan point karne lag jaati hai jahan "upar" tha, toh up-star $-x_B$ ke along baithta hai (nose direction ke peeche). ✔

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Example 9 — Cell I: exam twist, $C^{-1}=C^T$ exploit karo

Forecast: naive move Gaussian elimination on $C$ hai. Parent ne ek shortcut diya tha — woh kya hai?

1. Property yaad karo. Kisi bhi DCM ke liye, $C^{-1}=C^T$ (orthonormality). Toh ${\mathbf{v}}_N=C_{BN}^{-1}{\mathbf{v}}_B=C_{BN}^T{\mathbf{v}}_B$. Yeh step kyun? Transpose karna free hai (rows/columns swap karo); ek general $3\times 3$ ko invert karna real kaam hai. Orthonormality pe trust karna poora exam trick hai.
2. $R_3(30^{\circ })$ transpose karo. $C^T=\left[\begin{array}{ccc}c30& -s30& 0\\ s30& c30& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}0.866& -0.5& 0\\ 0.5& 0.866& 0\\ 0& 0& 1\end{array}\right]$. Yeh step kyun? $+s$ aur $-s$ positions swap karte hain — yeh IS inverse rotation $R_3(-30^{\circ })$ hai, jo sense banata hai: $+30^{\circ }$ yaw undo karna $-30^{\circ }$ yaw hai.
3. Apply karo. ${\mathbf{v}}_N=C^T[0,1,0{]}^T=[-0.5,0.866,0{]}^T$.

Verify: confirm karo $C{\mathbf{v}}_N={\mathbf{v}}_B$: $R_3(30^{\circ })[-0.5,0.866,0{]}^T=[0,1,0{]}^T$? Row1: $0.866(-0.5)+0.5(0.866)=0$ ✔; Row2: $-0.5(-0.5)+0.866(0.866)=0.25+0.75=1$ ✔. Round-trip holds. ✔

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Recall Har cell ke liye one-line self-test

Zero input → identity ::: $C=I$, vector unchanged. Negative angle → kaunsi entries flip hoti hain? ::: sirf do $\sin$ entries (cos even hai, sin odd hai). Obtuse yaw ($135^{\circ }$) → $\cos$ ka sign? ::: negative (second quadrant). Inverse problem → atan2 kyun, arctan kyun nahi? ::: tan har $180^{\circ }$ pe repeat karta hai; atan2 dono signs use karke quadrant fix karta hai. Gimbal lock condition ::: pitch $\theta =\pm 90^{\circ }$, toh $\cos \theta =0$ aur yaw & roll ek axis pe merge ho jaate hain — sirf $\psi -\varphi$ bachta hai. DCM saste mein undo karo ::: $C^T$ use karo, kyunki $C^{-1}=C^T$.

Connections

• 3.5.03 Direction cosine matrix (DCM) — construction from Euler angles (Hinglish)
• Quaternions — avoiding gimbal lock
• Euler angles — kinematic differential equations
• Rotation group SO(3) and orthogonal matrices
• Attitude determination — TRIAD & QUEST
• Kalman filter — linearized attitude error state
• Angular velocity and skew-symmetric matrices