Intuition The big pictureA rotating rigid body (satellite, drone, aircraft) has an attitude — its orientation in space. We track it with a unit quaternion q q q ω \omega ω evolves in time . This note answers: HOW fast does q q q Ξ ( q ) \Xi(q) Ξ ( q )
The deep idea: a quaternion lives on the unit 3-sphere S 3 S^3 S 3 tangent to that sphere (so ∥ q ∥ \|q\| ∥ q ∥ 1 2 \tfrac{1}{2} 2 1 Ξ ( q ) \Xi(q) Ξ ( q )
Definition Unit quaternionA quaternion is a 4-number object
q = [ q 0 q v ] = [ q 0 q 1 q 2 q 3 ] , q 0 = cos θ 2 , q v = n ^ sin θ 2 q = \begin{bmatrix} q_0 \\ \mathbf{q}_v \end{bmatrix} = \begin{bmatrix} q_0 \\ q_1 \\ q_2 \\ q_3 \end{bmatrix}, \qquad q_0 = \cos\tfrac{\theta}{2}, \quad \mathbf{q}_v = \hat{n}\sin\tfrac{\theta}{2} q = [ q 0 q v ] = q 0 q 1 q 2 q 3 , q 0 = cos 2 θ , q v = n ^ sin 2 θ = = θ = = ==\theta== == θ == n ^ \hat{n} n ^ unit quaternion: ∥ q ∥ 2 = q 0 2 + q 1 2 + q 2 2 + q 3 2 = = = 1 = = \|q\|^2 = q_0^2 + q_1^2 + q_2^2 + q_3^2 = ==1== ∥ q ∥ 2 = q 0 2 + q 1 2 + q 2 2 + q 3 2 === 1 ==
Quaternion multiplication (Hamilton product): for p = ( p 0 , p v ) p = (p_0, \mathbf{p}_v) p = ( p 0 , p v ) q = ( q 0 , q v ) q = (q_0, \mathbf{q}_v) q = ( q 0 , q v ) p ⊗ q = [ p 0 q 0 − p v ⋅ q v p 0 q v + q 0 p v + p v × q v ] p \otimes q = \begin{bmatrix} p_0 q_0 - \mathbf{p}_v\cdot\mathbf{q}_v \\ p_0\mathbf{q}_v + q_0\mathbf{p}_v + \mathbf{p}_v\times\mathbf{q}_v \end{bmatrix} p ⊗ q = [ p 0 q 0 − p v ⋅ q v p 0 q v + q 0 p v + p v × q v ]
The WHY of the cross product term: quaternion multiplication is non-commutative , exactly like composing rotations. The p v × q v \mathbf{p}_v\times\mathbf{q}_v p v × q v
Intuition The one idea everything rests onA rotation happening now is q ( t ) q(t) q ( t ) Δ t \Delta t Δ t δ q \delta q δ q compose by multiplication . So
q ( t + Δ t ) = q ( t ) ⊗ δ q . q(t + \Delta t) = q(t) \otimes \delta q. q ( t + Δ t ) = q ( t ) ⊗ δ q . δ q \delta q δ q
A rotation of small angle Δ θ = ∥ ω ∥ Δ t \Delta\theta = \|\omega\|\,\Delta t Δ θ = ∥ ω ∥ Δ t ω ^ = ω / ∥ ω ∥ \hat\omega = \omega/\|\omega\| ω ^ = ω /∥ ω ∥ δ q = [ cos Δ θ 2 ω ^ sin Δ θ 2 ] \delta q = \begin{bmatrix} \cos\tfrac{\Delta\theta}{2} \\ \hat\omega \sin\tfrac{\Delta\theta}{2} \end{bmatrix} δ q = [ cos 2 Δ θ ω ^ sin 2 Δ θ ]
Why this step? This is just the definition of a quaternion applied to the infinitesimal rotation that occurs during Δ t \Delta t Δ t
For small Δ t \Delta t Δ t cos Δ θ 2 ≈ 1 \cos\tfrac{\Delta\theta}{2}\approx 1 cos 2 Δ θ ≈ 1 sin Δ θ 2 ≈ Δ θ 2 = ∥ ω ∥ Δ t 2 \sin\tfrac{\Delta\theta}{2}\approx \tfrac{\Delta\theta}{2} = \tfrac{\|\omega\|\Delta t}{2} sin 2 Δ θ ≈ 2 Δ θ = 2 ∥ ω ∥Δ t δ q ≈ [ 1 1 2 ω Δ t ] = [ 1 0 0 0 ] + Δ t 2 [ 0 ω ] . \delta q \approx \begin{bmatrix} 1 \\ \tfrac{1}{2}\omega\,\Delta t \end{bmatrix} = \begin{bmatrix}1\\0\\0\\0\end{bmatrix} + \frac{\Delta t}{2}\begin{bmatrix}0\\ \omega\end{bmatrix}. δ q ≈ [ 1 2 1 ω Δ t ] = 1 0 0 0 + 2 Δ t [ 0 ω ] .
Why the 1 2 \tfrac12 2 1 the half-angle in sin θ 2 \sin\tfrac{\theta}{2} sin 2 θ This is the origin of the famous 1 2 \tfrac{1}{2} 2 1
q ( t + Δ t ) − q ( t ) = q ⊗ δ q − q = q ⊗ ( δ q − q identity ) = q ⊗ Δ t 2 [ 0 ω ] q(t+\Delta t) - q(t) = q \otimes \delta q - q = q\otimes(\delta q - q_{\text{identity}}) = q \otimes \frac{\Delta t}{2}\begin{bmatrix}0\\\omega\end{bmatrix} q ( t + Δ t ) − q ( t ) = q ⊗ δ q − q = q ⊗ ( δ q − q identity ) = q ⊗ 2 Δ t [ 0 ω ]
where q identity = ( 1 , 0 ) q_{\text{identity}}=(1,\mathbf{0}) q identity = ( 1 , 0 ) Δ t \Delta t Δ t Δ t → 0 \Delta t\to 0 Δ t → 0
q ˙ = 1 2 q ⊗ [ 0 ω ] \boxed{\;\dot q = \frac{1}{2}\, q \otimes \begin{bmatrix} 0 \\ \omega \end{bmatrix}\;} q ˙ = 2 1 q ⊗ [ 0 ω ]
Why this step? We turned "compose a tiny rotation" into a genuine time-derivative. The pure quaternion ( 0 , ω ) (0,\omega) ( 0 , ω ) angular-velocity quaternion .
Quaternion multiplication is linear , so q ⊗ ( 0 , ω ) q\otimes(0,\omega) q ⊗ ( 0 , ω ) 4 × 3 4\times 3 4 × 3 ω \omega ω q ⊗ ( 0 , ω ) q\otimes(0,\omega) q ⊗ ( 0 , ω )
q ⊗ [ 0 ω ] = [ − q v ⋅ ω q 0 ω + q v × ω ] = Ξ ( q ) ω q\otimes\begin{bmatrix}0\\\omega\end{bmatrix} = \begin{bmatrix} -\mathbf{q}_v\cdot\omega \\ q_0\omega + \mathbf{q}_v\times\omega\end{bmatrix} = \Xi(q)\,\omega q ⊗ [ 0 ω ] = [ − q v ⋅ ω q 0 ω + q v × ω ] = Ξ ( q ) ω
with
Why this step? GNC filters (EKF, etc.) need a linear relation q ˙ = A ω \dot q = A\,\omega q ˙ = A ω Ξ ( q ) \Xi(q) Ξ ( q )
The rate of change of ∥ q ∥ 2 \|q\|^2 ∥ q ∥ 2 zero — otherwise q q q
d d t ∥ q ∥ 2 = 2 q ⊤ q ˙ = 2 q ⊤ ( 1 2 Ξ ( q ) ω ) = ( q ⊤ Ξ ( q ) ) ω . \frac{d}{dt}\|q\|^2 = 2\,q^{\!\top}\dot q = 2\,q^{\!\top}\!\left(\tfrac12 \Xi(q)\omega\right) = \big(q^{\!\top}\Xi(q)\big)\,\omega. d t d ∥ q ∥ 2 = 2 q ⊤ q ˙ = 2 q ⊤ ( 2 1 Ξ ( q ) ω ) = ( q ⊤ Ξ ( q ) ) ω .
Key algebraic fact: q ⊤ Ξ ( q ) = 0 ⊤ q^{\!\top}\Xi(q) = \mathbf{0}^{\!\top} q ⊤ Ξ ( q ) = 0 ⊤ 1 × 3 1\times3 1 × 3 q ⊤ Ξ ( q ) = q 0 ( − q v ⊤ ) + q v ⊤ ( q 0 I + [ q v × ] ) = − q 0 q v ⊤ + q 0 q v ⊤ + q v ⊤ [ q v × ] = 0 ⊤ q^{\!\top}\Xi(q) = q_0(-\mathbf{q}_v^{\!\top}) + \mathbf{q}_v^{\!\top}\big(q_0 I + [\mathbf q_v\times]\big) = -q_0\mathbf{q}_v^{\!\top} + q_0\mathbf{q}_v^{\!\top} + \mathbf{q}_v^{\!\top}[\mathbf q_v\times] = \mathbf 0^{\!\top} q ⊤ Ξ ( q ) = q 0 ( − q v ⊤ ) + q v ⊤ ( q 0 I + [ q v × ] ) = − q 0 q v ⊤ + q 0 q v ⊤ + q v ⊤ [ q v × ] = 0 ⊤ q v ⊤ [ q v × ] = − ( q v × q v ) ⊤ = 0 \mathbf{q}_v^{\!\top}[\mathbf q_v\times] = -(\mathbf q_v\times\mathbf q_v)^{\!\top}=\mathbf 0 q v ⊤ [ q v × ] = − ( q v × q v ) ⊤ = 0 d d t ∥ q ∥ 2 = 0 \frac{d}{dt}\|q\|^2 = 0 d t d ∥ q ∥ 2 = 0
Intuition Geometric meaningq ˙ \dot q q ˙ orthogonal to q q q along the sphere, never off it. This is why the equation is guaranteed self-consistent.
If ω \omega ω body frame (usual case, gyros measure body rates): use right multiplication q ˙ = 1 2 q ⊗ ( 0 , ω ) \dot q = \tfrac12 q\otimes(0,\omega) q ˙ = 2 1 q ⊗ ( 0 , ω ) Ξ ( q ) \Xi(q) Ξ ( q )
If ω \omega ω inertial frame : use left multiplication q ˙ = 1 2 ( 0 , ω ) ⊗ q \dot q = \tfrac12 (0,\omega)\otimes q q ˙ = 2 1 ( 0 , ω ) ⊗ q Ω ( ω ) \Omega(\omega) Ω ( ω )
Worked example Example 1 — Pure spin about bodyz z z ω = ( 0 , 0 , ω z ) \omega = (0,0,\omega_z) ω = ( 0 , 0 , ω z ) q ( 0 ) = ( 1 , 0 , 0 , 0 ) q(0)=(1,0,0,0) q ( 0 ) = ( 1 , 0 , 0 , 0 ) q ˙ = 1 2 q ⊗ ( 0 , 0 , 0 , ω z ) . \dot q = \tfrac12\,q\otimes(0,0,0,\omega_z). q ˙ = 2 1 q ⊗ ( 0 , 0 , 0 , ω z ) . Why: we expect q q q q 0 = cos ω z t 2 , q 3 = sin ω z t 2 q_0=\cos\tfrac{\omega_z t}{2},\ q_3=\sin\tfrac{\omega_z t}{2} q 0 = cos 2 ω z t , q 3 = sin 2 ω z t z z z q ˙ 0 = − 1 2 ω z q 3 \dot q_0 = -\tfrac12\omega_z q_3 q ˙ 0 = − 2 1 ω z q 3 q ˙ 3 = 1 2 ω z q 0 \dot q_3 = \tfrac12\omega_z q_0 q ˙ 3 = 2 1 ω z q 0 ω z 2 \tfrac{\omega_z}{2} 2 ω z q 0 = cos ω z t 2 , q 3 = sin ω z t 2 q_0=\cos\tfrac{\omega_z t}{2},\ q_3 = \sin\tfrac{\omega_z t}{2} q 0 = cos 2 ω z t , q 3 = sin 2 ω z t
Worked example Example 2 — Numerical single stepLet q = ( 1 , 0 , 0 , 0 ) q = (1,0,0,0) q = ( 1 , 0 , 0 , 0 ) ω = ( 0.1 , 0 , 0 ) \omega = (0.1,\,0,\,0) ω = ( 0.1 , 0 , 0 ) q ˙ \dot q q ˙ q ˙ = 1 2 Ξ ( q ) ω , Ξ ( q ) = [ 0 0 0 1 0 0 0 1 0 0 0 1 ] , q ˙ = 1 2 [ 0 0.1 0 0 ] = [ 0 0.05 0 0 ] . \dot q = \tfrac12\Xi(q)\omega,\quad \Xi(q)=\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\\0&0&1\end{bmatrix}\!,\ \ \dot q = \tfrac12\begin{bmatrix}0\\0.1\\0\\0\end{bmatrix}=\begin{bmatrix}0\\0.05\\0\\0\end{bmatrix}. q ˙ = 2 1 Ξ ( q ) ω , Ξ ( q ) = 0 1 0 0 0 0 1 0 0 0 0 1 , q ˙ = 2 1 0 0.1 0 0 = 0 0.05 0 0 . Why this step? q 1 q_1 q 1 x x x x x x q ˙ 0 = 0 \dot q_0 = 0 q ˙ 0 = 0 q ⊤ q ˙ = 0 q^\top\dot q = 0 q ⊤ q ˙ = 0
Worked example Example 3 — Forecast-then-verifyForecast: if ω \omega ω q ˙ \dot q q ˙ q q q q ˙ \dot q q ˙ linear in ω \omega ω double .
Verify: 1 2 Ξ ( q ) ( 2 ω ) = 2 ⋅ 1 2 Ξ ( q ) ω = 2 q ˙ \tfrac12\Xi(q)(2\omega)=2\cdot\tfrac12\Xi(q)\omega = 2\dot q 2 1 Ξ ( q ) ( 2 ω ) = 2 ⋅ 2 1 Ξ ( q ) ω = 2 q ˙ Ξ \Xi Ξ
Common mistake "The factor should be 1, not ½."Why it feels right: in 3D vector kinematics R ˙ = [ ω × ] R \dot R = [\omega\times]R R ˙ = [ ω × ] R no ½. So people expect quaternions to match.
The fix: quaternions store the half-angle (sin θ 2 \sin\tfrac\theta2 sin 2 θ 1 2 \tfrac12 2 1 q q q − q -q − q
Common mistake "Order doesn't matter — I'll write( 0 , ω ) ⊗ q (0,\omega)\otimes q ( 0 , ω ) ⊗ q Why it feels right: multiplication order seems cosmetic.
The fix: quaternion product is non-commutative. Body-frame ω \omega ω must post-multiply (q ⊗ q\otimes q ⊗
Common mistake "I don't need to renormalize during integration."Why it feels right: the math proves ∥ q ∥ = 1 \|q\|=1 ∥ q ∥ = 1 The fix: that's only exact in continuous time. Discrete numerical steps (Euler/RK) leak norm. Real GNC code renormalizes q ← q / ∥ q ∥ q \leftarrow q/\|q\| q ← q /∥ q ∥
Recall Feynman: explain to a 12-year-oldImagine you're spinning on an office chair. To describe which way you face , you use a special 4-number "compass" called a quaternion. Now, if I tell you how fast you're spinning, you want to know how fast your compass numbers change. It turns out the compass numbers change at half the spin speed — because the compass secretly measures half-turns (a quaint quirk that makes the math turn out beautifully). There's also a little bookkeeping table (the Ξ \Xi Ξ
"Half the spin, times XI, right-multiply the fin(body)."
Half → the 1 2 \tfrac12 2 1 XI → Ξ ( q ) \Xi(q) Ξ ( q ) ω \omega ω right-multiply body → q ⊗ ( 0 , ω ) q\otimes(0,\omega) q ⊗ ( 0 , ω )
Why does the ½ appear in q ˙ = 1 2 Ξ ( q ) ω \dot q = \tfrac12\Xi(q)\omega q ˙ = 2 1 Ξ ( q ) ω Because quaternions encode the
half-angle (
sin θ 2 , cos θ 2 \sin\tfrac\theta2,\cos\tfrac\theta2 sin 2 θ , cos 2 θ ); differentiating the half-angle gives ½ by the chain rule.
Write the continuous-time quaternion kinematic equation (body rates). q ˙ = 1 2 q ⊗ ( 0 , ω ) = 1 2 Ξ ( q ) ω \dot q = \tfrac12\,q\otimes(0,\omega) = \tfrac12\,\Xi(q)\,\omega q ˙ = 2 1 q ⊗ ( 0 , ω ) = 2 1 Ξ ( q ) ω .
What guarantees ∥ q ∥ \|q\| ∥ q ∥ q ⊤ Ξ ( q ) = 0 ⊤ q^{\top}\Xi(q)=\mathbf 0^{\top} q ⊤ Ξ ( q ) = 0 ⊤ , so
d d t ∥ q ∥ 2 = q ⊤ q ˙ ⋅ 2 = 0 \tfrac{d}{dt}\|q\|^2 = q^{\top}\dot q\cdot 2 = 0 d t d ∥ q ∥ 2 = q ⊤ q ˙ ⋅ 2 = 0 ;
q ˙ ⊥ q \dot q \perp q q ˙ ⊥ q .
Structure of Ξ ( q ) \Xi(q) Ξ ( q ) 4 × 3 4\times3 4 × 3 : top row
− q v ⊤ -\mathbf q_v^\top − q v ⊤ ; bottom
3 × 3 3\times3 3 × 3 block
q 0 I + [ q v × ] q_0 I + [\mathbf q_v\times] q 0 I + [ q v × ] .
Body-frame ω \omega ω Post-multiply,
q ⊗ ( 0 , ω ) q\otimes(0,\omega) q ⊗ ( 0 , ω ) . Inertial-frame
ω \omega ω pre-multiplies.
What is the equivalent Ω ( ω ) \Omega(\omega) Ω ( ω ) q ˙ = 1 2 Ω ( ω ) q \dot q = \tfrac12\Omega(\omega)q q ˙ = 2 1 Ω ( ω ) q , factoring out
q q q instead of
ω \omega ω ;
Ω ( ω ) \Omega(\omega) Ω ( ω ) is
4 × 4 4\times4 4 × 4 skew.
Why renormalize in code if norm is provably preserved? Preservation is exact only in continuous time; discrete integrators leak norm, so renormalize each step.
For pure spin ω z \omega_z ω z q 3 ( t ) q_3(t) q 3 ( t ) q 3 ( t ) = sin ω z t 2 q_3(t)=\sin\tfrac{\omega_z t}{2} q 3 ( t ) = sin 2 ω z t (half-angle), with
q 0 = cos ω z t 2 q_0=\cos\tfrac{\omega_z t}{2} q 0 = cos 2 ω z t .
q at t plus dt = q times delta-q
small angle = norm omega times dt
approx identity plus half omega dt
difference quotient to derivative
Rotation theta about axis n-hat
Rotations compose by multiplication
Hamilton product with cross term
Intuition Hinglish mein samjhoDekho, jab koi satellite ya drone space me ghoomta hai, to uska orientation (kis taraf muh hai) hum ek unit quaternion q q q ω \omega ω q ˙ = 1 2 Ξ ( q ) ω \dot q = \tfrac12\,\Xi(q)\,\omega q ˙ = 2 1 Ξ ( q ) ω
Sabse important cheez woh 1 2 \tfrac12 2 1 half-angle store karta hai (cos θ 2 \cos\tfrac\theta2 cos 2 θ sin θ 2 \sin\tfrac\theta2 sin 2 θ 1 2 \tfrac12 2 1 Ξ ( q ) \Xi(q) Ξ ( q ) ω \omega ω q q q q ⊗ ( 0 , ω ) q\otimes(0,\omega) q ⊗ ( 0 , ω ) q q q left pe rakho.
Ek bahut khoobsurat property ye hai: ∥ q ∥ \|q\| ∥ q ∥ q ⊤ Ξ ( q ) = 0 q^\top\Xi(q)=0 q ⊤ Ξ ( q ) = 0 q ˙ \dot q q ˙ q q q
Ek practical warning: paper pe norm perfectly conserve hota hai, lekin computer me jab tum discrete steps (Euler/RK4) se integrate karte ho, thoda-thoda norm leak hota hai. Isliye real flight software har step ke baad q ← q / ∥ q ∥ q \leftarrow q/\|q\| q ← q /∥ q ∥