DCM kinematics — Ċ = −[ω×]C: Worked examples

3.5.4 · D3 · Physics › Guidance, Navigation & Control (GNC) › [[3.5.04 DCM kinematics — Ċ = −[ω×]C|DCM kinematics — Ċ = −[ω×]C]]

Yeh page ek drill hai. [[3.5.04 DCM kinematics — Ċ = −[ω×]C (index 3.5.4)|Parent note]] ne yeh equation prove ki thi: $\dot{C} = - [ω_{B} \times] C .$ Yahan hum isme har tarah ka input daalenge — har spin axis, ek general tilted axis, zero spin, sign-changing spin, ek limiting long-time case, time-varying spin, ek real spacecraft word problem, aur ek exam twist. Agar tum yeh sab kar sako, toh exam mein kuch bhi surprise nahi karega.

Shuru karne se pehle, ek reminder plain words mein. Symbol $[w \times]$ ("skew matrix of $w$ ") ek $3 \times 3$ box of numbers hai jo vector $w = (w_{1}, w_{2}, w_{3})$ se banta hai, aur ise kisi bhi vector $a$ se multiply karne par cross product $w \times a$ milta hai. Dekho Skew-symmetric matrices & cross-product operator. Likha hua: $[w \times] = 0 w_{3} - w_{2} - w_{3} 0 w_{1} w_{2} - w_{1} 0 .$ Dhyaan do ki diagonal hamesha zero hai aur ise diagonal ke paas flip karne par har sign flip ho jaata hai — yahi "skew-symmetric" ka matlab hai, aur yehi wajah hai ki $C$ ek rotation rehta hai.

Scenario matrix

Is topic ke har problem ka jawab in mein se kisi ek cell mein aata hai. Neeche ke worked examples mein cell ka tag laga hai.

#	Case class	Kya khaas hai	Example
A	Kisi ek body axis ( $x$ , $y$ , ya $z$ ) ke baare mein Spin	ek 2-D rotation ban jaata hai	Ex 1, Ex 2
B	Zero spin $ω = 0$	degenerate: $\dot{C} = 0$ , kuch nahi hilta	Ex 3
C	Sign flip / reverse spin ( $- ω$ )	rotation ki direction ulti ho jaati hai	Ex 4
D	Limiting / long-time behaviour ( $t \to \infty$ , periodicity)	solution periodic hai, kabhi blow up nahi hoti	Ex 5
E	General (tilted) constant axis (Rodrigues / matrix exponential)	full 3-component $ω$ , ek integration	Ex 6
F	Extract $ω$ from data (vee-map)	inverse problem	Ex 7
G	Time-varying $ω (t)$	simple exponential nahi; dhyan se integrate karo	Ex 8
H	Real-world word problem (spacecraft rate)	physical scenario se $[ω \times]$ banao	Ex 9
I	Exam twist: wrong-sign trap / frame trap	plus/minus aur N-vs-B error pakdo	Ex 10
J	Degenerate check: kya integration $det C = + 1$ preserve karti hai?	orthonormality + handedness	Ex 11

Cell A — kisi ek body axis ke baare mein spin

Worked example Ex 1 — Body

$z$ ke baare mein Constant spin, $C (t)$ nikalo Ek rigid body inertial frame ke saath aligned shuru hota hai ( $C (0) = I$ ) aur apne body $z$ -axis ke baare mein constant rate $ω = 2 rad/s$ se ghoomta hai, toh $ω_{B} = (0, 0, 2)$ . $C (t)$ nikalo aur ise $t = \frac{π}{4}$ s par evaluate karo.

Forecast: pehle guess karo — $t = \frac{π}{4} \approx 0.79$ s par body $ω t = \frac{π}{2}$ (ek quarter turn) ghoom chuki hogi. Kya top-left $2 \times 2$ block $+$ ya $-$ rotation jaisa dikhega? Equation mein minus sign yaad rakho.

Skew matrix banao. $ω_{B} = (0, 0, 2)$ ke saath, $[ω \times] = 020 - 2 00 000 .$ Yeh step kyun? Har DCM-kinematics problem $ω$ ko uske skew box mein convert karke shuru hoti hai; woh box hi woh operator hai jo $C$ ko "twist" karta hai.
ODE type pehchano. Kyunki $ω$ constant hai, $\dot{C} = - [ω \times] C$ ek linear ODE with constant coefficients hai. Iska solution matrix exponential hai: $C (t) = e^{- [ω \times] t} C (0) .$ Yeh step kyun? Jaise numbers ke liye $\overset{x}{˙} = a x \Rightarrow x = e^{a t} x_{0}$ , matrix version bhi kaam karta hai kyunki $[ω \times]$ kabhi nahi badlta.
Exponential evaluate karo. Ek $z$ -axis skew matrix ka exponential ek rotation hota hai. Minus sign ke saath yeh $z$ ke baare mein $- ω t$ angle ki rotation hai: $C (t) = cos ω t - sin ω t 0 sin ω t cos ω t 0 001 .$ Yeh step kyun? Minus $+ ω t$ ko clockwise turn mein badal deta hai — Lie-algebra picture dekho jahan skew matrix $I$ ke tangent space mein rehta hai.
$t = π /4$ , $ω = 2$ plug karo: $ω t = π /2$ , toh $cos = 0, sin = 1$ : $C (\frac{π}{4}) = 0 - 1 0 100001 .$

Verify: Check karo $C C^{T} = I$ (hota hai, columns orthonormal hain), aur $det C = + 1$ . $t = 0$ par, $ω t = 0$ deta hai $C = I$ ✔.

Figure 1 (neeche): white arrows fixed inertial axes $\overset{n}{^}_{1}, \overset{n}{^}_{2}$ hain; blue aur yellow arrows quarter turn ke baad body axes $\hat{b}_{1}, \hat{b}_{2}$ hain. Red arc dikhata hai rotation clockwise ( $- ω t$ ) hai, aur yehi wajah hai ki $- sin ω t$ $C$ ke lower-left mein hai — equation mein minus sign apne aap clockwise sweep ki tarah dikh raha hai.

Worked example Ex 2 — Body

$x$ ke baare mein Spin: kaun si entries hilti hain? Same setup ( $C (0) = I$ ) lekin $ω_{B} = (3, 0, 0)$ rad/s. $C (t)$ likho aur $t = \frac{π}{6}$ s par $C$ do.

Forecast: kaun sa $2 \times 2$ block frozen rehega, aur kaun sa rotate karega?

Skew matrix: $[ω \times] = 000003 0 - 3 0$ . Yeh step kyun? Sirf $w_{1} = 3$ nonzero hai, toh sirf $w_{1}$ entries survive karti hain — $x$ -axis untouched rehta hai.
$x$ -axis skew ka exponential (minus ke saath) $x$ ke baare mein $- ω t$ ki rotation hai: $C (t) = 100 0 cos ω t - sin ω t 0 sin ω t cos ω t .$ Yeh step kyun? Pehli row aur pehla column (index 1, yaani hamare one-based convention mein $x$ -row/column) identity ke roop mein rehte hain; $y$ – $z$ block turning karta hai.
$ω t = 3 \cdot π /6 = π /2$ plug karo: $cos = 0, sin = 1$ : $C = 100 00 - 1 010 .$

Verify: $det C = + 1$ , $C C^{T} = I$ , aur pehli row/column $(1, 0, 0)$ unhili hai — $x$ -spin ke liye sahi ✔.

Cell B — zero spin (degenerate)

Cell C — reverse spin (sign flip)

Worked example Ex 4 — Spin direction palto

Ex 1 lo lekin $ω_{B} = (0, 0, - 2)$ ke saath (dusri taraf ghoomna). $C (t)$ nikalo aur Ex 1 se $t = π /4$ par compare karo.

Forecast: kya Ex 1 ke compare mein off-diagonal signs swap honge?

Skew matrix sign flip karta hai: $[ω \times] = 0 - 2 0 200000$ . Yeh step kyun? $[(- ω) \times] = - [ω \times]$ — skew map linear hai.
Exponential $z$ ke baare mein $- ω t = + 2 t$ ki rotation deta hai: $C (t) = cos ω t sin ω t 0 - sin ω t cos ω t 0 001,$ $cos (- x) = cos x$ , $sin (- x) = - sin x$ use karke. Yeh step kyun? $ω$ reverse karna turn ki handedness reverse karta hai; sirf $sin$ terms ka sign badlta hai.
$t = π /4$ , $ω t = π /2$ par: $C = 010 - 1 00 001 .$

Verify: Yeh bilkul Ex 1 ke answer ka transpose hai — spin reverse karna inverse rotation deta hai ✔. $C_{Ex4} = C_{Ex1}^{T}$ , aur $C_{Ex1} C_{Ex4} = I$ .

Cell D — limiting / long-time behaviour

Worked example Ex 5 — Kya

$C$ kabhi blow up karta hai? Periodicity. Constant $ω_{B} = (0, 0, ω)$ , $ω = 2$ rad/s ke saath, $t \to \infty$ hone par $C (t)$ describe karo. Sabse chhota $T > 0$ nikalo jahan $C (T) = C (0)$ .

Forecast: unstable growth, ya hamesha bounded?

Entries hain $cos ω t, sin ω t$ — hamesha $- 1$ aur $+ 1$ ke beech bounded. Yeh step kyun? Real $a$ ke liye $e^{a t}$ ke unlike, skew matrix ka exponential purely imaginary eigenvalues rakhta hai, toh yeh barhne ki jagah oscillate karta hai. Yeh $S O (3)$ ki pahchaan hai: rotations kabhi blow up nahi hote.
Period: $C$ tab repeat karta hai jab $ω t$ mein $2 π$ ka izaafa hota hai, yaani $T = \frac{2 π}{ω} = \frac{2 π}{2} = π$ s. Yeh step kyun? Ek poora revolution frame ko apni shuru wali jagah wapas le aata hai.
Limit: $t \to \infty$ ke saath koi ek limit nahi hai; $C (t)$ $π$ s period se cycle karta hai.

Verify: $ω T = 2 \cdot π = 2 π$ , toh $cos = 1, sin = 0$ , deta hai $C (T) = I = C (0)$ ✔. Bounded aur periodic — "stable corner."

Cell E — general tilted constant axis (Rodrigues / matrix exponential)

Worked example Ex 6 — Ek tilted body axis ke baare mein Spin

Ab axis ek coordinate axis nahi hai. Maano constant body rate $ω_{B} = ω \overset{e}{^}$ hai jahan unit axis $\overset{e}{^} = \frac{1}{2} (1, 1, 0)$ aur $ω = \frac{π}{2}$ rad/s hai. $C (0) = I$ ke saath, $C (t)$ closed form mein nikalo aur $t = 1$ s par evaluate karo.

Forecast: tilted axis ke saath tum sirf ek $2 \times 2$ block nahi bhar sakte — lekin phir bhi ek clean formula hai. Guess karo kaun se trig functions aate hain.

Angular velocity ko magnitude aur axis mein split karo: $ω_{B} = ω \overset{e}{^}$ , $∣ \overset{e}{^} ∣ = 1$ . Iska skew hai $[ω \times] = ω [\overset{e}{^} \times]$ . Yeh step kyun? Scalar rate ko aage nikaalne se hum unit-axis skew $[\overset{e}{^} \times]$ use kar sakte hain, jinki powers simple hain: $[\overset{e}{^} \times]^{3} = - [\overset{e}{^} \times]$ .
Rodrigues' formula se exponential likho. Kyunki $C (t) = e^{- [ω \times] t} = e^{- ϕ [\overset{e}{^} \times]}$ hai jahan turn angle $ϕ = ω t$ hai, matrix-exponential identity deta hai: $C (t) = I - sin ϕ [\overset{e}{^} \times] + (1 - cos ϕ) [\overset{e}{^} \times]^{2}, ϕ = ω t .$ Yeh step kyun? $\dot{C} = - [ω \times] C$ mein minus sign $- sin ϕ$ term ban jaata hai — $\overset{e}{^}$ ke baare mein clockwise turn. Kisi bhi constant axis ke liye yahi general answer hai; single-axis Ex 1/Ex 2 special cases hain.
$[\overset{e}{^} \times]$ aur $[\overset{e}{^} \times]^{2}$ banao. $\overset{e}{^} = \frac{1}{2} (1, 1, 0)$ ke saath: $[\overset{e}{^} \times] = \frac{1}{2} 00 - 1 001 1 - 1 0, [\overset{e}{^} \times]^{2} = \frac{1}{2} - 1 10 1 - 1 0 00 - 2 .$ Yeh step kyun? Sirf yeh do matrices Rodrigues ko chahiye hain — koi infinite series nahi.
$ϕ = ω t = \frac{π}{2} \cdot 1 = \frac{π}{2}$ plug karo: $sin ϕ = 1$ , $1 - cos ϕ = 1$ : $C (1) = I - [\overset{e}{^} \times] + [\overset{e}{^} \times]^{2} = \frac{1}{2} \frac{1}{2} \frac{1}{2} \frac{1}{2} \frac{1}{2} - \frac{1}{2} - \frac{1}{2} \frac{1}{2} 0 .$

Verify: $C C^{T} = I$ aur $det C = + 1$ (VERIFY mein check kiya gaya). Saath hi axis $\overset{e}{^}$ unhila hai: $C \overset{e}{^} = \overset{e}{^}$ , " $\overset{e}{^}$ ke baare mein rotation" ki fingerprint ✔.

Cell F — data se $ω$ extract karo (vee-map)

Worked example Ex 7 — Measured

$\dot{C} C^{T}$ se $ω$ padho Ek sensor tumhe product deta hai: $\dot{C} C^{T} = 0 - 5 4 50 - 3 - 4 30 .$ $ω_{B}$ nikalo.

Forecast: teen independent off-diagonal numbers hi $ω$ hain signs tak — compute karne se pehle guess karo.

Inverse relation use karo. Parent se, $[ω_{B} \times] = - \dot{C} C^{T}$ : $[ω \times] = 05 - 4 - 5 03 4 - 3 0 .$ Yeh step kyun? $\dot{C} = - [ω \times] C$ ko right mein $C^{T}$ se multiply karke aur $C C^{T} = I$ use karke skew matrix isolate ho jaata hai.
Vee-map apply karo $\lor$ (abhi upar define kiya gaya) skew template se components padhne ke liye: $ω_{1} = (\cdot)_{32} = 3, ω_{2} = (\cdot)_{13} = 4, ω_{3} = (\cdot)_{21} = 5.$ Yeh step kyun? Standard $[w \times]$ template se compare karo: $(3, 2)$ entry $w_{1}$ hai, $(1, 3)$ entry $w_{2}$ hai, $(2, 1)$ entry $w_{3}$ hai.
Result: $ω_{B} = (3, 4, 5)$ rad/s.

Verify: $(3, 4, 5)$ se $[ω \times]$ dobara banao — yeh step 1 se match karta hai. Magnitude $∣ ω ∣ = 9 + 16 + 25 = 50 \approx 7.071$ rad/s ✔. Yeh attitude determination ko mirror karta hai jahan tum data se rates recover karte ho.

Cell G — time-varying angular velocity

Worked example Ex 8 — Fixed body axis ke baare mein Spin-up,

$ω (t)$ ramp karta hai Body apne $z$ -axis ke baare mein ghoomti hai lekin rate linearly ramp karta hai: $ω_{B} (t) = (0, 0, α t)$ jahan $α = 2 rad/s^{2}$ , aur $C (0) = I$ . $C (t)$ nikalo aur $t = 1$ s par evaluate karo.

Forecast: shayad $e^{- [ω \times] t}$ likhne ka mann kare — lekin $ω$ ab constant nahi hai. Kya naive exponential yahan bhi kaam karta hai?

Check karo ki $[ω (t) \times]$ alag-alag times par khud se commute karta hai ya nahi. Yahan saare skew matrices $[ω (t) \times]$ ek hi same axis $z$ ke baare mein hain, bas $α t$ se scale kiye gaye hain. Matrices jo ek fixed matrix ke scalar multiples hain woh hamesha commute karte hain. Yeh step kyun? $\dot{C} = - [ω (t) \times] C$ ka general solution ek time-ordered exponential ki zaroorat hai; yeh ordinary exponential of the integral mein tab collapse hota hai jab alag-alag times par skew matrices commute karein. Yahan karte hain (single fixed axis), toh hum angle integrate kar sakte hain.
Rate ko turn angle mein integrate karo. Define karo $ϕ (t) = \int_{0}^{t} ω (s) d s = \int_{0}^{t} α s d s = \frac{1}{2} α t^{2}$ . Yeh step kyun? Total angle ghuma hua rate curve ke area under se aata hai, rate $\times$ time se nahi, kyunki rate badhta rehta hai.
$C (t)$ ko $- ϕ (t)$ se $z$ -rotation ke roop mein likho: $C (t) = cos ϕ (t) - sin ϕ (t) 0 sin ϕ (t) cos ϕ (t) 0 001, ϕ (t) = \frac{1}{2} α t^{2} .$ Yeh step kyun? Ex 1 jaisi $z$ -block structure, lekin angle ab integrated $ϕ (t)$ hai, $ω t$ nahi.
$t = 1$ , $α = 2$ plug karo: $ϕ = \frac{1}{2} \cdot 2 \cdot 1^{2} = 1$ rad, toh $cos 1 \approx 0.5403$ , $sin 1 \approx 0.8415$ : $C (1) = cos 1 - sin 1 0 sin 1 cos 1 0 001 \approx 0.5403 - 0.8415 0 0.8415 0.5403 0 001 .$ Yeh maanga hua numerical matrix $C (1)$ hai.

Verify: $C C^{T} = I$ aur $det C = + 1$ har $t$ ke liye (VERIFY mein check kiya gaya). Warning: agar $ω (t)$ ka axis khud tumble karta, toh skew matrices commute nahi karte aur yeh shortcut fail ho jaata — tumhe numerically ODE integrate karni padti (aur yehi wajah hai ki practice mein quaternions use kiye jaate hain).

Cell H — real-world word problem

Worked example Ex 9 — Spacecraft nadir spin

Ek imaging satellite apne body $y$ -axis (pitch) ke baare mein $ω = 0.10$ rad/s se rotate karta hai taaki uska camera ground ki taraf point karta rahe. $t = 0$ par uska DCM $C (0) = I$ hai. Ek body vector $\hat{b} = (0, 0, 1)$ (camera boresight, shuru mein inertial $z$ ke saath) $t = 5$ s ke baad inertial frame mein express karke kahan point karta hai?

Forecast: boresight $x$ – $z$ plane mein sweep karta hai; predict karo kis taraf.

$y$ -spin ke liye Skew: $ω_{B} = (0, 0.10, 0)$ deta hai: $[ω \times] = 00 - 0.10 000 0.10 00 .$ Yeh step kyun? Sirf $w_{2} = 0.10$ nonzero hai; $y$ -axis spin axis hai.
$C (t)$ = $y$ ke baare mein $- ω t$ ki rotation: $C (t) = cos ω t 0 sin ω t 010 - sin ω t 0 cos ω t .$ Yeh step kyun? Cell A ka standard body-axis result, $y$ -block ke liye adapted.
Inertial coords mein boresight hai $b_{N} = C^{T} \hat{b}$ (kyunki $C$ N→B map karta hai, iska transpose B→N map karta hai). $ω t = 0.10 \cdot 5 = 0.5$ rad ke saath: $b_{N} = C^{T} (0, 0, 1)^{T} = (- sin 0.5, 0, cos 0.5) = (- 0.4794, 0, 0.8776) .$ Yeh step kyun? Hum boresight ground/inertial frame se dekhna chahte hain, toh body se inertial mein transform karte hain.

Verify: $∣ b_{N} ∣ = 0.479 4^{2} + 0.877 6^{2} = 1.000$ (unit length preserved) ✔. Camera $0.5$ rad ( $\approx 28. 6^{\circ}$ ) $- x$ ki taraf tilt hua hai — ek sensible pitch sweep.

Cell I — exam twist (sign & frame traps)

Worked example Ex 10 — Do galtiyan pakdo

Ek student likhta hai: " $C : N \to B$ aur $ω_{N} = (0, 0, ω)$ ke liye, kinematics hai $\dot{C} = + [ω_{N} \times] C$ ." Dono galtiyan nikalo aur theek karo, phir sahi equation do.

Forecast: padhne se pehle galtiyan count karo — bilkul do hain.

Galti 1 — sign. N→B map karne wale matrix ke liye, sahi sign minus hai: $\dot{C} = - [\cdot] C$ . Plus $\dot{C}^{T} = [ω \times] C^{T}$ (B→N map) se belong karta hai. Kyun: forward body rotation ki wajah se inertial vectors body frame mein backward rotate hote hue dikhte hain.
Galti 2 — frame. Skew ko body components $ω_{B}$ use karne chahiye, $ω_{N}$ nahi, kyunki output side ( $\dot{C}$ ka left index) body frame mein rehta hai.
Corrected equation: $\dot{C} = - [ω_{B} \times] C, ω_{B} = C ω_{N} .$ Yeh step kyun? Sign convention aur $ω$ ka frame dono ko $C$ ki mapping direction se match karna chahiye.

Verify: Agar hum (galat tarike se) plus rakhte, toh $\frac{d}{d t} (C C^{T})$ nonzero hoti, orthonormality violate hoti — ek contradiction jo sign error flag karta hai. Sahi minus ke saath, $\frac{d}{d t} (C C^{T}) = 0$ ✔. Quaternion kinematics — q̇ = ½ Ω(ω) q aur Euler angle kinematics & gimbal lock mein analogous convention care se compare karo.

Cell J — degenerate handedness check

Worked example Ex 11 — Kya flow

$det C = + 1$ rakhta hai? Ex 1 ke constant $z$ -spin solution use karke, confirm karo ki $det C (t) = + 1$ har $t$ ke liye hai (naki $- 1$ , jo ek reflection hoti).

Forecast: rotations ko handedness preserve karni chahiye — lekin prove karo, assume mat karo.

Determinant compute karo Ex 1 se $C (t)$ ka. Last row ke baad expand karte hue: $det C = 1 \cdot (cos^{2} ω t + sin^{2} ω t) = cos^{2} ω t + sin^{2} ω t .$ Yeh step kyun? $z$ -row $(0, 0, 1)$ hai, toh determinant $2 \times 2$ block mein reduce ho jaata hai.
Pythagorean identity: $cos^{2} + sin^{2} = 1$ , toh $det C = + 1$ har $t$ ke liye. Yeh step kyun? Yeh "proper rotation, no reflection" ki algebraic fingerprint hai.
General reason: $\frac{d}{d t} det C = det C \cdot tr (- [ω \times]) = 0$ kyunki har skew matrix ka trace zero hota hai. Toh $det C$ apni starting value $+ 1$ se kabhi nahi badlta. Yeh step kyun? Jacobi's formula aur $tr [ω \times] = 0$ (zero diagonal) determinant ko lock kar deta hai.

Verify: $t = π /4$ par ( $ω t = π /2$ ): $cos^{2} + sin^{2} = 0 + 1 = 1$ ✔. Flow $S O (3)$ mein rehta hai — koi reflection nahi, kabhi bhi. Yeh Poisson's equation structure guarantee karta hai.

Recall checklist

Recall Kya har cell cover hua?

Single-axis (A) ::: Ex 1 ( $z$ ), Ex 2 ( $x$ ), Ex 9 ( $y$ ). Zero spin (B) ::: Ex 3 — $\dot{C} = 0$ , $C$ frozen. Sign flip (C) ::: Ex 4 — reverse spin transpose deta hai. Limiting/periodic (D) ::: Ex 5 — bounded, period $2 π / ω$ . General tilted axis (E) ::: Ex 6 — Rodrigues' formula, axis $\frac{1}{2} (1, 1, 0)$ . Extract $ω$ , vee-map (F) ::: Ex 7 — $ω = (3, 4, 5)$ . Time-varying $ω (t)$ (G) ::: Ex 8 — rate ko angle $ϕ = \frac{1}{2} α t^{2}$ mein integrate karo. Word problem (H) ::: Ex 9 — 5 s ke baad boresight. Exam twist (I) ::: Ex 10 — sign + frame traps. Handedness (J) ::: Ex 11 — $det C = + 1$ hamesha.

Flashcards

ω_{B} = (0, 0, ω)

aur

C (0) = I

ke liye,

C (t)

kya hai?

z

ke baare mein

- ω t

ki rotation: top-left block

[[cos ω t, sin ω t], [- sin ω t, cos ω t]]

C_{33} = 1

Jab

ω = 0

toh

\dot{C}

ka kya hota hai?

[ω \times] = 0

, toh

\dot{C} = 0

aur

C (t) = C (0)

hamesha ke liye (degenerate case).

Spin reverse karna

ω \to - ω

C (t)

ke saath kya karta hai?

Inverse/transpose rotation deta hai; sirf

sin

terms ka sign flip hota hai.

Constant-spin

C (t)

kabhi blow up kyun nahi karta?

Skew matrices ke imaginary eigenvalues hote hain, toh

e^{- [ω \times] t}

oscillate karta hai (period

2 π / ω

) — bounded,

S O (3)

mein.

General constant axis

\overset{e}{^}

ke liye, kaun sa closed form

C (t)

deta hai?

Rodrigues:

C = I - sin ϕ [\overset{e}{^} \times] + (1 - cos ϕ) [\overset{e}{^} \times]^{2}

jahan

ϕ = ω t

Jab

ω (t)

vary karta hai lekin fixed axis rakhta hai, toh turn angle kaise nikalte hain?

Rate integrate karo:

ϕ (t) = \int_{0}^{t} ω (s) d s

(jaise linear ramp ke liye

\frac{1}{2} α t^{2}

Time-varying

ω (t)

ke liye

e^{- [ω \times] t}

hamesha kyun use nahi kar sakte?

Integral ka exponential valid hai sirf tab jab alag-alag times ke skew matrices commute karein (fixed axis); warna time-ordered exponential ya numerical integration chahiye.

- \dot{C} C^{T}

ω_{B}

kaise padhte hain?

Vee-map:

ω_{1} = (\cdot)_{32}

ω_{2} = (\cdot)_{13}

ω_{3} = (\cdot)_{21}

DCM flow ke under

det C

hamesha

+ 1

kyun rehta hai?

\frac{d}{d t} det C = det C \cdot tr (- [ω \times]) = 0

kyunki har skew matrix ka trace zero hota hai.