Worked examples — Euler angles — roll φ, pitch θ, yaw ψ; rotation sequence (3-2-1 convention)

3.5.2 · D3 · Physics › Guidance, Navigation & Control (GNC) › Euler angles — roll φ, pitch θ, yaw ψ; rotation sequence (3-

Shuru karne se pehle, do tools ka ek-line refresher jo hum sabse zyada use karte hain:

Neeche ke saare angles degrees mein hain jab tak formula radians maange na.

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Scenario matrix

Is topic mein jo bhi problem aa sakti hai woh inn cells mein se ek mein aati hai. Neeche har worked example par uska cell tag lagaya gaya hai.

#	Case class	Kya mushkil hai	Example
A	Single-axis, positive angle	baseline sanity check	Ex 1
B	Negative angle	$\sin$ ka sign ek column flip kar deta hai	Ex 2
C	Quadrant care in recovery (angle > 90° ya < −90°)	plain $\arctan$ fail ho jaata hai, atan2 chahiye	Ex 3
D	Teeno angles ek saath	full matrix multiply, koi zero chhupne ke liye nahi	Ex 4
E	Zero / identity input	degenerate — $I$ return karna zaroori	Ex 5
F	Gimbal lock $\theta =\pm 90^{\circ }$	recovery 0/0 ban jaati hai, DOF kho jaata hai (dono signs)	Ex 6
G	Word problem (real vehicle heading)	English → angles → vector mein translate karo	Ex 7
H	Exam twist: non-uniqueness ($\varphi ,\psi$ vs $\varphi \pm 180,\psi \pm 180$)	do angle-triples, ek orientation	Ex 8

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Ex 1 — Case A: pure pitch, positive angle

Forecast: Compute karne se pehle andaza lagao — kya naak ka positive Down component hoga, negative, ya koi nahi? (Naak upar matlab naak horizon se upar uthti hai.)

1. Sirf pitch ke saath, $R_n^b=R_y(30^{\circ })$. Yeh step kyun? $\psi =\varphi =0$ hone se $R_z$ aur $R_x$ identity ban jaate hain, toh product sirf pitch matrix pe collapse ho jaata hai.
2. Naak body frame mein $[1,0,0{]}^T$ hai; nav mein convert karo $R_b^n=(R_n^b{)}^T=R_y(30^{\circ }{)}^T$ se. Yeh step kyun? Vector jo "naak se bahar point karta hai" woh trivially $[1,0,0{]}^T$ hai body frame mein. Use nav mein dekhne ke liye body → nav jaana padega, jo transpose hai.
3. Compute karo. $R_y(\theta {)}^T[1,0,0{]}^T=[\cos \theta ,0,-\sin \theta {]}^T=[0.866,0,-0.5{]}^T$. Yeh step kyun? $R_y^T$ ka pehla column exactly $[\cos \theta ,0,-\sin \theta ]$ hota hai.

Figure s01 dekho: naak vector (amber) North–East plane se upar uth gaya hai, isliye uska Down component negative hai ($-0.5$). Naak upar = negative Down. Forecast check: haan, negative.

Verify karo: magnitude $\sqrt{0.866^2+0+0.5^2}=1.000$ (abhi bhi unit direction ✓). North component $\cos 30=0.866$ aur Down $-\sin 30=-0.5$ ek $30^{\circ }$ lift se agree karte hain.

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Ex 2 — Case B: negative yaw

Forecast: North (naak aage) rotate hokar... East ya West ki taraf jaayegi? Passive/right-hand convention mein negative yaw.

1. $R_n^b=R_z(-90^{\circ })$. $\sin (-90)=-1,\cos (-90)=0$ use karke: $R_z(-90^{\circ })=\left[\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$ Yeh step kyun? Negative angle substitute karne se har $\sin$ term ka sign flip ho jaata hai; $-\sin$ jo lower-left mein tha woh $+1$ ban jaata hai.
2. Nav mein naak $=(R_n^b{)}^T[1,0,0{]}^T$. Transpose ka pehla column $[0,-1,0{]}^T$ hai. Yeh step kyun? Same body→nav logic jo Ex 1 mein tha.
3. Result: naak $=[0,-1,0{]}^T$ = negative East = West. Yeh step kyun? East slot mein $-1$ ka matlab hai "ek unit West ki taraf."

Verify karo: $-90^{\circ }$ yaw naak ko North se West ki taraf point karna chahiye. $[0,-1,0]$ exactly West hai ✓. Compare karo parent ke Ex 1 se jahan $\psi =+90^{\circ }$ ne East diya tha — $\psi$ ka sign ne East vs West decide kiya, jaisa hona chahiye.

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Ex 3 — Case C: recovery jab yaw 90° se zyada ho

Forecast: $R_{11}$ aur $R_{12}$ dono negative hain. Isse $\psi$ kis quadrant mein aata hai? Kya blindly $\arctan (R_{12}/R_{11})$ karna sahi hoga?

1. Yaad karo $R_{12}=\cos \theta \sin \psi$ aur $R_{11}=\cos \theta \cos \psi$. Yeh step kyun? Parent note ke full matrix se; shared $\cos \theta$ ratio mein cancel ho jaata hai.
2. Naive route (trap): $\arctan \left(\frac{-0.5}{-0.866}\right)=\arctan (0.577)=30^{\circ }$. Yeh step kyun? Failure dikhane ke liye: $\arctan$ sirf $(-90^{\circ },90^{\circ })$ mein return kar sakta hai, isliye woh $30^{\circ }$ aur $30^{\circ }+180^{\circ }=210^{\circ }$ mein fark nahi kar sakta.
3. Sahi route: $\psi =atan2(-0.5,-0.866)$. Dono arguments negative → third quadrant → $\psi =-150^{\circ }$ (equivalently $210^{\circ }$). Yeh step kyun? $atan2$ dono signs alag padhta hai: negative sine aur negative cosine sirf third quadrant ho sakta hai.

Figure s02 mein $(R_{12},R_{11})$ ke chaar quadrant sign-patterns dikhaye hain taaki aankh se dekh sako ki tum kis yaw region mein ho.

Verify karo: $\cos (-150^{\circ })=-0.866=R_{11}$ ✓ aur $\sin (-150^{\circ })=-0.5=R_{12}$ ✓ ($\cos \theta =1$ lekar). Naive $30^{\circ }$ se $\cos 30=+0.866$ aata hai — galat sign, trap confirm ✓.

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Ex 4 — Case D: teeno angles ek saath

Forecast: Pehli row $[\cos \theta \cos \psi ,\cos \theta \sin \psi ,-\sin \theta ]$ hai. Notice karo isme $\varphi$ bilkul nahi hai — step 3 padhne se pehle kya tum samajh sakte ho kyun?

1. $R_n^b$ ki pehli row = $[c\theta c\psi ,c\theta s\psi ,-s\theta ]$ (yaad karo $c\theta \equiv \cos \theta ,s\theta \equiv \sin \theta$). Yeh step kyun? Seedha parent note ke multiplied-out matrix se.
2. Numbers plug karo. $c20=0.9397,c45=0.7071,s45=0.7071,s20=0.3420$: $[0.9397(0.7071),0.9397(0.7071),-0.3420]=[0.6645,0.6645,-0.3420]$ Yeh step kyun? Direct substitution; magnitude check ke liye 4 decimal rakhna.
3. $\varphi$ kyun nahi hai? Roll aakhri spin hai, body-$x$ axis ke baare mein. Pehli row batati hai ki body-$x$ axis nav se kaise relate karti hai — aur ek axis ke baare mein spin kabhi us axis ko nahi hilata. Isliye $\varphi$ row 1 mein invisible hai. Yeh step kyun? Yeh 3-2-1 ordering ka geometric core hai.

Verify karo: magnitude $\sqrt{0.6645^2+0.6645^2+0.3420^2}=\sqrt{0.4416+0.4416+0.1170}=\sqrt{1.0002}=1.000$ (orthogonal matrix ki unit row ✓).

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Ex 5 — Case E: degenerate zero input

Forecast: Agar body nav frame ke saath perfectly aligned hai, toh transform ek vector ke saath kya "karega"?

1. Har elementary matrix $0$ pe identity ban jaati hai kyunki $\cos 0=1,\sin 0=0$ (yaani har $c\cdot =1$ aur har $s\cdot =0$). Yeh step kyun? $0$ substitute karne se har off-diagonal sine zero ho jaata hai aur har diagonal pe $1$ aa jaata hai.
2. Teen identities ka product identity hai: $R_n^b=III=I$. Yeh step kyun? $I$ matrices ke liye multiplicative identity hai.
3. Matlab: har vector ke liye nav coordinates = body coordinates; kuch rotate nahi hota. Yeh step kyun? $I\mathbf{v}=\mathbf{v}$ saare numbers unchanged rakhta hai.

Verify karo: $\det I=1$ (proper rotation, reflection nahi ✓) aur $I^{T}I=I$ (orthogonal ✓). Yeh akela fixed point hai jahan teeno recovery formulas $0$ dete hain: $-\arcsin (0)=0$, $atan2(0,1)=0$, $atan2(0,1)=0$.

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Ex 6 — Case F: gimbal lock, dono $\theta =+90^{\circ }$ aur $\theta =-90^{\circ }$

Forecast: Parent note warn karta hai ki yahan $R_{11}=R_{12}=R_{23}=R_{33}=0$. Agar yeh sab zero hain, toh $atan2(R_{23},R_{33})=atan2(0,0)$ ka kya hoga?

Case F+: $\theta =+90^{\circ }$

1. $\theta =90^{\circ }$ pe affected entries banao ($c\theta =0,s\theta =1$): $R_{11}=c\theta c\psi =0,R_{12}=c\theta s\psi =0,R_{23}=s\varphi c\theta =0,R_{33}=c\varphi c\theta =0$. Yeh step kyun? Inme se har ek mein $\cos \theta =0$ ka factor hai.
2. Recovery collapse ho jaati hai: $\varphi =atan2(0,0)$ aur $\psi =atan2(0,0)$ undefined hain (0/0). Sirf pitch bachti hai: $\theta =-\arcsin (R_{13})=-\arcsin (-1)=90^{\circ }$. Yeh step kyun? $R_{13}=-\sin \theta =-1$ woh akela entry hai jisme $\cos \theta$ ka factor nahi hai.
3. Jo DEFINED hai: abhi bhi meaningful quantity entry $R_{21}=s\varphi s\theta c\psi -c\varphi s\psi$ hai. $\theta =90^{\circ }$ pe ($s\theta =1$) yeh $\sin \varphi \cos \psi -\cos \varphi \sin \psi =\sin (\varphi -\psi )$ ban jaata hai. Toh vehicle sirf difference $\varphi -\psi$ "jaanta" hai, alag-alag nahi — yeh lost degree of freedom hai. Yeh step kyun? $\theta =+90^{\circ }$ pe roll aur yaw axes ek hi line pe aa jaate hain, isliye sirf unka difference observable hai.

Case F−: $\theta =-90^{\circ }$

4. Wahi entries $\theta =-90^{\circ }$ pe banao ($c\theta =0,s\theta =-1$): phir se $R_{11}=R_{12}=R_{23}=R_{33}=0$, toh recovery abhi bhi $0/0$ deti hai; pitch $\theta =-\arcsin (R_{13})=-\arcsin (1)=-90^{\circ }$ hai. Yeh step kyun? Same $\cos \theta =0$ factor same chaar entries ko khatam karta hai; sirf $R_{13}=-\sin \theta =+1$ ka sign alag hai.
5. Recoverable combination SUM mein flip ho jaati hai. Ab $R_{21}=s\varphi (-1)c\psi -c\varphi s\psi =-(\sin \varphi \cos \psi +\cos \varphi \sin \psi )=-\sin (\varphi +\psi )$. Toh $\theta =-90^{\circ }$ pe observable $\varphi +\psi$ hai, difference nahi. Yeh step kyun? $s\theta$ ka sign flip hua, angle-subtraction identity ko angle-addition identity mein badal diya — dono lock cases mirror twins hain.

Figure s03 mein dikhaya gaya hai ki pitch $\pm 90^{\circ }$ ke paas aate hi roll aur yaw axes ek common line pe collapse ho jaate hain.

Verify karo: $\theta =+90^{\circ },\varphi =40^{\circ },\psi =10^{\circ }$ pe, entry $R_{21}=\sin (40^{\circ }-10^{\circ })=\sin 30^{\circ }=0.5$; $\theta =-90^{\circ }$ pe same $\varphi ,\psi$ se $R_{21}=-\sin (40^{\circ }+10^{\circ })=-\sin 50^{\circ }=-0.766$. Dono cases mein $R_{11}=R_{12}=R_{23}=R_{33}=0$ exactly. Ilaaj yeh hai ki Quaternions — avoiding gimbal lock ya raw Direction Cosine Matrix (DCM) se propagate karo aur Euler mein sirf human display ke liye convert karo.

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Ex 7 — Case G: real-world word problem

Forecast: Heading clockwise measured hoti hai North se — kya clockwise hamare convention mein positive yaw correspond karta hai (jo +ψ East ki taraf yaw karta hai)? North-East, North se East ki taraf hai.

1. English → angles mein translate karo. Heading $045^{\circ }$ (East ki taraf) $\Rightarrow \psi =+45^{\circ }$; climb $\Rightarrow \theta =+15^{\circ }$ nose-up; "wings level" $\Rightarrow \varphi =0$. Yeh step kyun? Heading Down ke baare mein ek yaw hai; climb ek pitch hai; level wings ka matlab zero roll hai.
2. Nav mein naak = $R_b^n$ ka pehla column = $R_n^b$ ki pehli row $=[c\theta c\psi ,c\theta s\psi ,-s\theta ]$. Yeh step kyun? Ex 4 mein humne establish kiya tha ki yeh row body-$x$ (naak) ki direction nav mein express karti hai.
3. Plug in karo. $c15=0.9659,c45=s45=0.7071,s15=0.2588$: $[0.9659(0.7071),0.9659(0.7071),-0.2588]=[0.683,0.683,-0.259{]}^T$ Yeh step kyun? North aur East components equal ($0.683$ dono) = exactly North-East ki taraf point kar raha hai; negative Down = climb.

Verify karo: North/East equal ⇒ ground track bearing $=atan2(0.683,0.683)=45^{\circ }$ ✓. Climb angle $=\arcsin (0.259)=15^{\circ }$ ✓. Magnitude $\sqrt{0.683^2+0.683^2+0.259^2}=1.000$ ✓.

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Ex 8 — Case H: exam twist, two-triple ambiguity

Forecast: Do bilkul alag dikhne wale angle triples, ek physical attitude. Kaunse trig identities matrices ko coincide karate hain?

1. 3-2-1 ke liye alias rule: $(\varphi ,\theta ,\psi )$ aur $(\varphi \pm 180^{\circ },180^{\circ }-\theta ,\psi \pm 180^{\circ })$ same $R_n^b$ dete hain. Yeh step kyun? Yeh standard second solution hai; yahan $180^{\circ }-20^{\circ }=160^{\circ }$.
2. Ek entry check karo, $R_{13}=-\sin \theta$. Solution 1: $-\sin 20^{\circ }=-0.342$. Solution 2: $-\sin 160^{\circ }=-\sin (180^{\circ }-20^{\circ })=-\sin 20^{\circ }=-0.342$. Same. Yeh step kyun? $\sin (180^{\circ }-x)=\sin x$ pitch entries ko agree karata hai.
3. $R_{33}=\cos \varphi \cos \theta$ check karo. Sol 1: $\cos 30\cos 20=0.866(0.9397)=0.8138$. Sol 2: $\cos 210\cos 160=(-0.866)(-0.9397)=0.8138$. Dono sign flips cancel ho jaate hain. Yeh step kyun? $180^{\circ }$ add karne se $\cos \varphi$ ka sign flip hota hai; $180^{\circ }-\theta$ lene se $\cos \theta$ ka sign flip hota hai; product unchanged rehta hai.

Verify karo: $R_{13}$ equal at $-0.342$; $R_{33}$ equal at $0.8138$. Do triples, identical matrix entries ⇒ same orientation. Exam takeaway: hamesha woh branch report karo jisme $|\theta |\le 90^{\circ }$ ho (principal solution) jab tak aur kaha na jaaye — dekho Attitude estimation (Kalman / complementary filter) kyun filters is branch ko pin karte hain.

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Active Recall

Recall Kaun sa cell logon ko sabse zyada trip karta hai, aur kyun?

Cell C (quadrant care). ::: Kyunki $\arctan (R_{12}/R_{11})$ silently sign information drop kar deta hai; $atan2$ dono signs rakhta hai aur isliye sahi quadrant identify karta hai.

Recall Gimbal lock pe kaunsi single quantity recoverable rehti hai?

Roll aur yaw ka combination ::: $\varphi -\psi$ at $\theta =+90^{\circ }$ (via $R_{21}=\sin (\varphi -\psi )$) aur $\varphi +\psi$ at $\theta =-90^{\circ }$ (via $R_{21}=-\sin (\varphi +\psi )$), plus pitch $\theta =\pm 90^{\circ }$ khud — lekin roll aur yaw alag-alag nahi.

Recall Roll

$R_n^b$ ki pehli row mein kyun kabhi appear nahi karta? Kyunki roll body-$x$ ke baare mein ek spin hai ::: aur ek rotation kabhi apni khud ki axis ko nahi hilata, isliye body-$x$ (naak) ki direction — jo exactly $R_n^b$ ki pehli row hai — $\varphi$ se bilkul independent hai.

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Connections

• Parent topic
• Rotation matrices & orthogonality
• Quaternions — avoiding gimbal lock
• Direction Cosine Matrix (DCM)
• Angular velocity & body rates p, q, r
• NED navigation frame
• Attitude estimation (Kalman / complementary filter)