A quaternion is a rotation stored as 4 numbers . When you rotate by q 1 q_1 q 1 then by q 2 q_2 q 2 single quaternion. The Hamilton product is the rule that tells you how to multiply two rotations into one. It is the quaternion version of "stacking rotation matrices" — but with only 4 numbers and no trig.
WHY care in GNC? A spacecraft/drone attitude filter (like a MEKF or complementary filter) integrates gyro data as tiny quaternion multiplications every millisecond. Get the product rule wrong → your attitude drifts or flips. This is the arithmetic of orientation.
A quaternion is q = w + x i + y j + z k q = w + x\,i + y\,j + z\,k q = w + x i + y j + z k i 2 = j 2 = k 2 = i j k = − 1 i^2 = j^2 = k^2 = ijk = -1 i 2 = j 2 = k 2 = ij k = − 1 scalar + vector pair:
q = ( w , v ) , v = ( x , y , z ) q = (w,\ \mathbf{v}), \qquad \mathbf{v} = (x,y,z) q = ( w , v ) , v = ( x , y , z ) w w w real (scalar) part ; v \mathbf{v} v vector (imaginary) part .
From i j k = − 1 ijk=-1 ij k = − 1
i j = k , j k = i , k i = j (cyclic, right-handed) ij = k,\quad jk = i,\quad ki = j \qquad\text{(cyclic, right-handed)} ij = k , j k = i , k i = j (cyclic, right-handed) j i = − k , k j = − i , i k = − j (anti-cyclic) ji = -k,\quad kj = -i,\quad ik = -j \qquad\text{(anti-cyclic)} j i = − k , k j = − i , ik = − j (anti-cyclic)
Intuition Why do these signs exist?i , j , k i,j,k i , j , k right-handed frame under cross product: x ^ × y ^ = z ^ \hat x \times \hat y = \hat z x ^ × y ^ = z ^ i j = k ij=k ij = k also square to − 1 -1 − 1 i i i cross product (rotation-like) with a dot product (alignment/shrinking) into one operation. Non-commutativity (i j ≠ j i ij \ne ji ij = j i
Compute i j i\,j i j
p = i p=i p = i w 1 = 0 , v 1 = ( 1 , 0 , 0 ) w_1=0,\ \mathbf{v}_1=(1,0,0) w 1 = 0 , v 1 = ( 1 , 0 , 0 ) Why? i i i q = j q=j q = j w 2 = 0 , v 2 = ( 0 , 1 , 0 ) w_2=0,\ \mathbf{v}_2=(0,1,0) w 2 = 0 , v 2 = ( 0 , 1 , 0 ) scalar = 0 − ( 1 , 0 , 0 ) ⋅ ( 0 , 1 , 0 ) = 0 = 0 - (1,0,0)\!\cdot\!(0,1,0) = 0 = 0 − ( 1 , 0 , 0 ) ⋅ ( 0 , 1 , 0 ) = 0 Why? v 1 ⊥ v 2 \mathbf v_1 \perp \mathbf v_2 v 1 ⊥ v 2
vector = 0 + 0 + ( 1 , 0 , 0 ) × ( 0 , 1 , 0 ) = ( 0 , 0 , 1 ) = 0 + 0 + (1,0,0)\times(0,1,0) = (0,0,1) = 0 + 0 + ( 1 , 0 , 0 ) × ( 0 , 1 , 0 ) = ( 0 , 0 , 1 ) Why? right-hand cross gives z ^ \hat z z ^
Result = ( 0 , ( 0 , 0 , 1 ) ) = k =(0,(0,0,1)) = k = ( 0 , ( 0 , 0 , 1 )) = k i j = k ij=k ij = k
Now j i j\,i j i ( 0 , 0 , − 1 ) (0,0,-1) ( 0 , 0 , − 1 ) − k -k − k Why this step? cross product is anti-commutative, so swapping order flips sign. This proves i j = − j i ij=-ji ij = − j i
Rotate 90 ∘ 90^\circ 9 0 ∘ z z z 90 ∘ 90^\circ 9 0 ∘ x x x θ \theta θ n ^ \hat n n ^ q = ( cos θ 2 , sin θ 2 n ^ ) q = (\cos\frac\theta2,\ \sin\frac\theta2\,\hat n) q = ( cos 2 θ , sin 2 θ n ^ )
q z = ( cos 45 ∘ , ( 0 , 0 , sin 45 ∘ ) ) = ( 1 2 , 0 , 0 , 1 2 ) q_z = (\cos45^\circ, (0,0,\sin45^\circ)) = (\tfrac{1}{\sqrt2}, 0,0,\tfrac{1}{\sqrt2}) q z = ( cos 4 5 ∘ , ( 0 , 0 , sin 4 5 ∘ )) = ( 2 1 , 0 , 0 , 2 1 ) q x = ( 1 2 , 1 2 , 0 , 0 ) q_x = (\tfrac{1}{\sqrt2}, \tfrac{1}{\sqrt2},0,0) q x = ( 2 1 , 2 1 , 0 , 0 )
Combined rotation ("x x x z z z q x q z q_x \, q_z q x q z Why? Hamilton convention: q total = q last q first q_{\text{total}} = q_{\text{last}}\,q_{\text{first}} q total = q last q first
w = w 1 w 2 − v 1 ⋅ v 2 = 1 2 − ( 1 2 , 0 , 0 ) ⋅ ( 0 , 0 , 1 2 ) = 1 2 − 0 = 1 2 w = w_1w_2 - \mathbf v_1\!\cdot\!\mathbf v_2 = \tfrac12 - (\tfrac{1}{\sqrt2},0,0)\!\cdot\!(0,0,\tfrac{1}{\sqrt2}) = \tfrac12 - 0 = \tfrac12 w = w 1 w 2 − v 1 ⋅ v 2 = 2 1 − ( 2 1 , 0 , 0 ) ⋅ ( 0 , 0 , 2 1 ) = 2 1 − 0 = 2 1 v = w 1 v 2 + w 2 v 1 + v 1 × v 2 \mathbf v = w_1\mathbf v_2 + w_2\mathbf v_1 + \mathbf v_1\times\mathbf v_2 v = w 1 v 2 + w 2 v 1 + v 1 × v 2 = 1 2 ( 0 , 0 , 1 2 ) + 1 2 ( 1 2 , 0 , 0 ) + ( 1 2 , 0 , 0 ) × ( 0 , 0 , 1 2 ) = \tfrac{1}{\sqrt2}(0,0,\tfrac{1}{\sqrt2}) + \tfrac{1}{\sqrt2}(\tfrac{1}{\sqrt2},0,0) + (\tfrac{1}{\sqrt2},0,0)\times(0,0,\tfrac{1}{\sqrt2}) = 2 1 ( 0 , 0 , 2 1 ) + 2 1 ( 2 1 , 0 , 0 ) + ( 2 1 , 0 , 0 ) × ( 0 , 0 , 2 1 ) = ( 0 , 0 , 1 2 ) + ( 1 2 , 0 , 0 ) + ( 0 , − 1 2 , 0 ) = ( 1 2 , − 1 2 , 1 2 ) = (0,0,\tfrac12) + (\tfrac12,0,0) + (0,-\tfrac12,0) = (\tfrac12,-\tfrac12,\tfrac12) = ( 0 , 0 , 2 1 ) + ( 2 1 , 0 , 0 ) + ( 0 , − 2 1 , 0 ) = ( 2 1 , − 2 1 , 2 1 )
So q x q z = ( 1 2 , 1 2 , − 1 2 , 1 2 ) q_xq_z = (\tfrac12, \tfrac12,-\tfrac12,\tfrac12) q x q z = ( 2 1 , 2 1 , − 2 1 , 2 1 ) Check: norm = 4 ⋅ 1 4 = 1 =\sqrt{4\cdot\tfrac14}=1 = 4 ⋅ 4 1 = 1
Definition Sandwich / conjugationTo rotate a 3D vector u \mathbf u u u = ( 0 , u ) u=(0,\mathbf u) u = ( 0 , u ) u ′ = q u q − 1 , q − 1 = q ˉ / ∣ q ∣ 2 , q ˉ = ( w , − v ) u' = q\,u\,q^{-1}, \qquad q^{-1} = \bar q/|q|^2,\ \ \bar q = (w,-\mathbf v) u ′ = q u q − 1 , q − 1 = q ˉ /∣ q ∣ 2 , q ˉ = ( w , − v ) q q q q − 1 = q ˉ q^{-1}=\bar q q − 1 = q ˉ
Why a sandwich and not just q u qu q u A single product would mix scalar and vector parts and change length. The conjugation q ( ⋅ ) q − 1 q(\cdot)q^{-1} q ( ⋅ ) q − 1 twice-by-half , which is why the half-angle θ / 2 \theta/2 θ /2 q q q
Common mistake Multiplying in the wrong orderWrong feels right: "I did rotation A then B, so multiply q A q B q_A q_B q A q B Fix: Hamilton convention applies rotations right-to-left : q B q A q_B q_A q B q A A A A B B B i j = k ij=k ij = k j i = − k ji=-k j i = − k
Common mistake Forgetting to normalizeWrong feels right: "Products of unit quaternions stay unit, so I never renormalize."
Fix: Exactly true in math, but floating-point rounding accumulates error over thousands of gyro updates. Renormalize q ← q / ∣ q ∣ q \leftarrow q/|q| q ← q /∣ q ∣
Common mistake Treating the vector part like a plain 3-vector productWrong feels right: "v 1 × v 2 \mathbf v_1 \times \mathbf v_2 v 1 × v 2 Fix: You must add the scalar-scaling terms w 1 v 2 + w 2 v 1 w_1\mathbf v_2 + w_2\mathbf v_1 w 1 v 2 + w 2 v 1
i j k = − 1 ijk=-1 ij k = − 1 Wrong feels right: "i j k = − 1 ijk=-1 ij k = − 1 i 2 = − 1 i^2=-1 i 2 = − 1 Fix: In C \mathbb C C a b = b a ab=ba ab = ba i j ≠ j i ij\ne ji ij = j i non-commutative division algebra — that non-commutativity is precisely 3D rotation.
Recall Flashables (hide and test yourself)
What are the two parts of the Hamilton product? scalar = w 1 w 2 − v 1 ⋅ v 2 =w_1w_2-\mathbf v_1\!\cdot\!\mathbf v_2 = w 1 w 2 − v 1 ⋅ v 2 = w 1 v 2 + w 2 v 1 + v 1 × v 2 =w_1\mathbf v_2+w_2\mathbf v_1+\mathbf v_1\times\mathbf v_2 = w 1 v 2 + w 2 v 1 + v 1 × v 2
Sole source of non-commutativity? the cross product term.
i j = ? ij=? ij = ? = k \;=k = k j i = ? ji=? j i = ? = − k \;=-k = − k How to rotate a vector? q u q ˉ q\,u\,\bar q q u q ˉ q q q
Recall Feynman: explain to a 12-year-oldImagine two magic "turn cards." Each card, when you tap it, spins a toy in space. If you want one card that does both spins at once, you glue the two cards together with a special recipe. The recipe has two ingredients: one part checks how much the two spins point the same way (that's the dot part, a plain number), and the other part checks how much they twist across each other (that's the cross part, which points sideways by the right-hand rule). Because twisting-across depends on which card you tap first, tapping card A then B is not the same as B then A — just like turning a Rubik's cube.
"Scalar Says: Multiply-minus-Dot. Vector Vibes: two Weights plus a Cross."
w 1 w 2 − v 1 ⋅ v 2 w_1w_2 - \mathbf v_1\!\cdot\!\mathbf v_2 w 1 w 2 − v 1 ⋅ v 2 w 1 v 2 + w 2 v 1 + v 1 × v 2 w_1\mathbf v_2 + w_2\mathbf v_1 + \mathbf v_1\times\mathbf v_2 w 1 v 2 + w 2 v 1 + v 1 × v 2
Rotation Matrices — SO(3) — the Hamilton product ≡ matrix multiplication of rotations.Axis-Angle & Euler Rodrigues — where the half-angle θ / 2 \theta/2 θ /2 q q q Quaternion Kinematics — $\dot q = \tfrac12 q\,\omega$ — gyro integration uses this product.Multiplicative EKF (MEKF) — attitude filters chain Hamilton products.Gimbal Lock & Euler Angles — the problem quaternions solve.Cross Product & Right-Hand Rule — the geometric root of i j = k ij=k ij = k
Hamilton product scalar part w 1 w 2 − v 1 ⋅ v 2 w_1w_2 - \mathbf v_1\cdot\mathbf v_2 w 1 w 2 − v 1 ⋅ v 2 Hamilton product vector part w 1 v 2 + w 2 v 1 + v 1 × v 2 w_1\mathbf v_2 + w_2\mathbf v_1 + \mathbf v_1\times\mathbf v_2 w 1 v 2 + w 2 v 1 + v 1 × v 2 Which term causes non-commutativity the cross product
v 1 × v 2 \mathbf v_1\times\mathbf v_2 v 1 × v 2 Value of i j ij ij Value of j i ji j i Fundamental relation defining quaternions i 2 = j 2 = k 2 = i j k = − 1 i^2=j^2=k^2=ijk=-1 i 2 = j 2 = k 2 = ij k = − 1 Order convention: rotation A then B combined quaternion q B q A q_B q_A q B q A (right-to-left)
How to rotate vector u by unit quaternion q u ′ = q u q ˉ u' = q\,u\,\bar q u ′ = q u q ˉ Inverse of a unit quaternion its conjugate
q ˉ = ( w , − v ) \bar q=(w,-\mathbf v) q ˉ = ( w , − v ) Is the product of two unit quaternions a unit quaternion Yes,
∣ p q ∣ = ∣ p ∣ ∣ q ∣ |pq|=|p||q| ∣ pq ∣ = ∣ p ∣∣ q ∣ Quaternion w plus xi yj zk
Hamilton rules i2 j2 k2 ijk eq -1
Multiplication table ij eq k etc
Scalar part w1w2 minus v1 dot v2
Vector part w1v2 plus w2v1 plus cross
Attitude filter MEKF gyro integration
Intuition Hinglish mein samjhoDekho, quaternion ek chhota sa 4-number ka packet hai jo ek rotation ko store karta hai: ek scalar part w w w ( x , y , z ) (x,y,z) ( x , y , z ) combine karne ka rule hi Hamilton product hai. Yeh basically rotation matrices ko multiply karne jaisa hai, but sirf 4 numbers ke saath, koi bhaari trigonometry nahi.
Formula ka core simple hai agar aap scalar+vector notation mein soch lo: scalar part ban-ta hai w 1 w 2 − v 1 ⋅ v 2 w_1w_2 - \mathbf v_1\cdot\mathbf v_2 w 1 w 2 − v 1 ⋅ v 2 w 1 v 2 + w 2 v 1 + v 1 × v 2 w_1\mathbf v_2 + w_2\mathbf v_1 + \mathbf v_1\times\mathbf v_2 w 1 v 2 + w 2 v 1 + v 1 × v 2 i , j , k i,j,k i , j , k i j = k ij=k ij = k j i = − k ji=-k j i = − k i j k = − 1 ijk=-1 ij k = − 1 cross product term hi non-symmetric hai, isiliye p q ≠ q p pq \ne qp pq = q p
GNC mein iska matlab bahut bada hai: drone ya satellite ka attitude filter har millisecond gyro ka data lekar ek tiny quaternion multiply karta hai. Agar aap order ulta kar do ya normalize karna bhool jao, to orientation drift ya flip ho jaayega. Convention: q total = q last q first q_{\text{total}} = q_{\text{last}}\, q_{\text{first}} q total = q last q first q u q ˉ q\,u\,\bar q q u q ˉ θ / 2 \theta/2 θ /2
