Quaternions — definition q = (q₀, q₁, q₂, q₃), unit quaternion constraint: Exercises

3.5.6 · D4 · Physics › Guidance, Navigation & Control (GNC) › Quaternions — definition q = (q₀, q₁, q₂, q₃), unit quaterni

Shuru karne se pehle, ek picture jo poori game tumhare dimag mein fix kar de. Ek unit quaternion ek aisa point hai jo 4-dimensional space mein centre se bilkul ek unit door baithta hai — yeh ek aisi sphere par rehta hai jise hum poori tarah draw nahi kar sakte, isliye hum uska ek slice draw karte hain.

Figure — Quaternions — definition q = (q₀, q₁, q₂, q₃), unit quaternion constraint

Blue circle dekho: yeh unit 3-sphere $S^{3}$ ki 2D shadow hai. Yellow dot ek valid rotation quaternion hai (yeh circle par baitha hai). Red dot drift ho gaya hai — uske chaar squares ab $1$ mein add nahi hote, isliye yeh rotation nahi hai jab tak hum ise renormalize nahi karte (red arrow use snap back karta hai).

L1 — Recognition

Yeh test karta hai ki kya tum ek quaternion ko padh sakte ho aur woh ek law check kar sakte ho jo use obey karna zaroori hai.

Recall Solution 1.1

Test hai unit constraint: $q_{0}^{2} + q_{1}^{2} + q_{2}^{2} + q_{3}^{2} = 1$ . Har ek ke squares ka sum compute karo.

(a) $1 + 0 + 0 + 0 = 1$ → valid ✔ (yeh identity hai, "kuch mat karo").
(b) $0.25 \times 4 = 1$ → valid ✔.
(c) $0.36 + 0.64 = 1$ → valid ✔.
(d) $1 + 1 = 2 \neq = 1$ → valid nahi ✗ (norm $2$ hai, pehle normalize karna padega).

Recall Solution 1.2

Scalar part = $q_{0} = 0.6$ (yeh "kitna twist" wala piece hai).
Vector part $q_{v} = (q_{1}, q_{2}, q_{3}) = (0, 0.8, 0)$ (yeh "kaunsi axis" wala piece hai).
Conjugate sirf vector part ko flip karta hai: $q^{*} = (0.6, 0, - 0.8, 0)$ .

L2 — Application

Ab tum angle aur axis se quaternions build karte ho, half-angle rule use karke.

Recall Solution 2.1

Formula: $q = (cos \frac{θ}{2}, \overset{n}{^} sin \frac{θ}{2})$ . Half-angle isliye use hota hai kyunki vector ko rotate karna $q$ ko do baar sandwich $q v q^{- 1}$ mein apply karta hai. Half-angle: $\frac{6 0 ^{\circ}}{2} = 3 0^{\circ}$ , toh $cos 3 0^{\circ} = \frac{3}{2} \approx 0.8660$ , $sin 3 0^{\circ} = 0.5$ . $q = (\frac{3}{2}, 0, 0.5, 0) \approx (0.8660, 0, 0.5, 0) .$ Norm check: $(\frac{3}{2})^{2} + 0. 5^{2} = 0.75 + 0.25 = 1$ ✔.

Recall Solution 2.2

$q = (cos \frac{θ}{2}, \overset{n}{^} sin \frac{θ}{2})$ se match karo.

Scalar: $cos \frac{θ}{2} = 0.7071 = \frac{2}{2}$ ⇒ $\frac{θ}{2} = 4 5^{\circ}$ ⇒ $θ = 9 0^{\circ}$ .
Vector part $= (0.7071, 0, 0)$ , aur uski length $sin \frac{θ}{2} = sin 4 5^{\circ} = 0.7071$ ke barabar honi chahiye. Hai bhi. Toh unit axis hai $\overset{n}{^} = (1, 0, 0)$ , $x$ -axis. Yeh $x$ ke baare mein $9 0^{\circ}$ rotation hai. (General mein reverse conversion ke liye Axis-Angle Representation dekho.)

L3 — Analysis

Quaternions ko todho: conjugates, inverses, aur non-commuting product.

Recall Solution 3.1

Kyunki $q$ ek unit quaternion hai, $∥ q ∥ = 1$ , isliye $q^{- 1} = q^{*} = (q_{0}, - q_{v})$ — bas vector signs flip karo: $q^{- 1} = (0.8660, 0, - 0.5, 0) .$ Kyun yeh conjugate hona chahiye: $q \otimes q^{*} = (q_{0}^{2} + ∣ q_{v} ∣^{2}, 0) = (∥ q ∥^{2}, 0) = (1, 0)$ exactly tab jab norm $1$ ho. $+ y$ ke baare mein $6 0^{\circ}$ rotation ko undo karna $- y$ ke baare mein $6 0^{\circ}$ rotation hai, jo $y$ -component ko negate karne se yahi keh raha hai. ✔

Recall Solution 3.2

Hamilton product: $p \otimes q = (p_{0} q_{0} - p \cdot q, p_{0} q + q_{0} p + p \times q)$ . Yahan $p_{0} = q_{0} = 0$ , $p = (1, 0, 0)$ , $q = (0, 1, 0)$ .

Scalar: $0 - (p \cdot q) = - (0) = 0$ .
Vector: $0 + 0 + p \times q = (1, 0, 0) \times (0, 1, 0) = (0, 0, 1)$ . $p \otimes q = (0, 0, 0, 1) = k .$ Ab swap karo. Cross product sign flip karta hai: $q \times p = (0, 0, - 1)$ . $q \otimes p = (0, 0, 0, - 1) = - k .$ Yeh sign se alag hain — $ij = k$ lekin $j i = - k$ ki seedhi picture. $p \times q$ term exactly isliye hai kyun rotations bhi commute nahi karte.

L4 — Synthesis

Kayi ideas ek saath combine karo: rotations compose karna, double cover, aur ek chain mein renormalization.

Recall Solution 4.1

Step 1 — har factor banao (half-angle, $4 5^{\circ}$ , $cos = sin = \frac{2}{2} \approx 0.7071$ ): $q_{1} = (\frac{2}{2}, 0, 0, \frac{2}{2}) (z -axis), q_{2} = (\frac{2}{2}, \frac{2}{2}, 0, 0) (x -axis) .$ Step 2 — Hamilton product $p = q_{2}$ , $q = q_{1}$ ke saath. Maano $c = \frac{2}{2}$ , toh $c^{2} = \frac{1}{2}$ . $p = (c, 0, 0)$ , $q = (0, 0, c)$ .

Scalar: $p_{0} q_{0} - p \cdot q = c \cdot c - 0 = c^{2} = \frac{1}{2}$ .
$p_{0} q = (0, 0, c^{2}) = (0, 0, \frac{1}{2})$ .
$q_{0} p = (c^{2}, 0, 0) = (\frac{1}{2}, 0, 0)$ .
$p \times q = (c, 0, 0) \times (0, 0, c) = (0, - c^{2}, 0) = (0, - \frac{1}{2}, 0)$ . Vector parts ka sum: $(\frac{1}{2}, - \frac{1}{2}, \frac{1}{2})$ . $q_{2} \otimes q_{1} = (\frac{1}{2}, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2}) .$ Step 3 — sanity check norm: $4 \times (\frac{1}{2})^{2} = 4 \times \frac{1}{4} = 1$ ✔. Composed object ab bhi ek valid unit quaternion hai, isliye yeh ek legal single rotation hai — Euler angles chain karne ke mukable yeh ek bada advantage hai (Euler Angles and Gimbal Lock).

Recall Solution 4.2

Dono ka $\sum q_{i}^{2} = 4 \times 0.25 = 1$ ✔, toh dono unit quaternions hain. $q$ ke liye: $cos \frac{θ}{2} = 0.5 \Rightarrow \frac{θ}{2} = 6 0^{\circ} \Rightarrow θ = 12 0^{\circ}$ axis $\overset{n}{^} \propto (0.5, - 0.5, 0.5)$ ke baare mein, yaani $\overset{n}{^} = \frac{1}{3} (1, - 1, 1)$ . $- q$ ke liye: $cos \frac{θ}{2} = - 0.5 \Rightarrow \frac{θ}{2} = 12 0^{\circ} \Rightarrow θ = 24 0^{\circ}$ axis $\propto (- 0.5, 0.5, - 0.5) = - \frac{1}{3} (1, - 1, 1)$ ke baare mein. $- \overset{n}{^}$ ke baare mein $24 0^{\circ}$ ka turn wahi hai jo $+ \overset{n}{^}$ ke baare mein $12 0^{\circ}$ ka turn hai (ulti direction mein lambe raaste se jaana tumhe usi jagah pahunchata hai). Sandwich mein, do minus signs cancel ho jaate hain: $(- q) v (- q)^{- 1} = q v q^{- 1}$ . Same physical rotation. Yahi $S^{3} \to S O (3)$ double cover hai.

L5 — Mastery

Exactly wahi karo jo ek flight computer har timestep karta hai: integrate karo, drift karo, aur repair karo.

Recall Solution 5.1

(a) Squares ka sum: $0.994 5^{2} + 0.052 3^{2} + 0.069 8^{2} + 0.034 9^{2}$ $= 0.989030 + 0.002735 + 0.004872 + 0.001218 = 0.997855$ . $∥ q ∥ = 0.997855 \approx 0.998927$ . (b) Har component ko norm se divide karo: $q \leftarrow \frac{q}{∥ q ∥} \approx (0.99557, 0.05236, 0.06987, 0.03494) .$ Naya squares ka sum $= 1$ by construction ✔. (c) Discrete integration rounding error accumulate karta hai, isliye $q$ dheere dheere unit sphere se bahar jaata hai; norm se divide karna ise seedha $S^{3}$ par wapas project karta hai, aur ise ek valid rotation rakhta hai agले Rotation Matrices SO(3) conversion ke liye jo guidance loop ko chahiye.

Recall Solution 5.2

(a) Half-angle $\frac{18 0 ^{\circ}}{2} = 9 0^{\circ}$ : $cos 9 0^{\circ} = 0$ , $sin 9 0^{\circ} = 1$ . $q = (0, \overset{n}{^} \cdot 1) = (0, \frac{1}{2}, \frac{1}{2}, 0) \approx (0, 0.7071, 0.7071, 0) .$ Norm: $0 + \frac{1}{2} + \frac{1}{2} + 0 = 1$ ✔ (ek "pure" quaternion, scalar part $0$ , kyunki $18 0^{\circ}$ turns mein hamesha $cos 9 0^{\circ} = 0$ hota hai). (b) Unit ⇒ $q^{- 1} = q^{*} = (0, - \frac{1}{2}, - \frac{1}{2}, 0)$ . (c) $q \otimes q^{- 1} = (q_{0}^{2} + ∣ q_{v} ∣^{2}, 0) = (0 + 1, 0) = (1, 0, 0, 0)$ = identity ✔. $18 0^{\circ}$ ki ek curiosity note karo: yahan $q = (0, q_{v})$ aur $- q = (0, - q_{v}) = q^{*} = q^{- 1}$ , toh yeh rotation apna khud ka inverse hai — $18 0^{\circ}$ do baar ghumane se tum wapas aa jaate ho.

Active Recall

Recall Ek line mein: tum kaise test karoge ki

$(a, b, c, d)$ ek rotation quaternion hai? Check karo $a^{2} + b^{2} + c^{2} + d^{2} = 1$ .

Recall Ek line mein:

$θ$ aur $\overset{n}{^}$ diye hain, $q$ likho. $q = (cos \frac{θ}{2}, \overset{n}{^} sin \frac{θ}{2})$ — aadha angle.

Recall Ek line mein: unit quaternion ka inverse aur kyun.

$q^{- 1} = q^{*}$ (vector part flip karo), kyunki $q \otimes q^{*} = ∥ q ∥^{2} = 1$ .