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Angular-velocity reconstruction from a unit-quaternion path: for $C^{1}$ $q : I \to S^{3}$ the body angular velocity of $ρ \circ q$ is $ω = 2 φ^{- 1} (q^{- 1} q^{'})$

Statement

Let $I \subseteq R$ be a nondegenerate open interval, let $q : I \to S^{3}$ be of class $C^{1}$ , and let

$R : I ⟶ S O (3), R (t) : = ρ (q (t)),$

be the associated rotation path, the matrix being taken in the basis $(i, j, k)$ of $Im H$ (The quaternion rotation map $ρ$ : conjugation on the imaginary quaternions, the unit sphere $S^{3}$ , and quaternion lifts). Then:

$q (t)^{- 1} q^{'} (t) = \overset{ˉ}{q} (t) q^{'} (t)$ has real part $0$ for every $t \in I$ , so it lies in $Im H$ and there is a unique vector $ω (t) \in R^{3}$ with

$q (t)^{- 1} q^{'} (t) = \frac{1}{2} φ (ω (t)),$

namely $ω (t) = 2 φ^{- 1} (q (t)^{- 1} q^{'} (t))$ , where $φ^{- 1} : Im H \to R^{3}$ is the inverse of the isomorphism in The quaternion rotation map $ρ$ : conjugation on the imaginary quaternions, the unit sphere $S^{3}$ , and quaternion lifts; $t \mapsto ω (t)$ is continuous. 2. The curve $q$ solves $q^{'} (t) = \frac{1}{2} q (t) φ (ω (t))$ on $I$ , the matrix curve $R$ is of class $C^{1}$ , and $R^{'} (t) = R (t) Ω_{ω (t)}$ for every $t \in I$ , where $Ω_{ω (t)}$ is the matrix $z \mapsto ω (t) \times z$ . 3. Consequently $ω$ is the body angular velocity of $R$ in the sense of The body angular velocity of a $C^{1}$ rotation path: $Ω = R^{T} R^{'}$ is skew, $Ω z = ω \times z$ determines a unique continuous $ω$ , and $R^{'} = R Ω$ : $R$ is a $C^{1}$ rotation path whose body angular velocity is $ω$ .

Facts & Assumptions

Given: The interval $I$ , the $C^{1}$ curve $q : I \to S^{3}$ , and the objects $φ, S^{3}, ρ$ of The quaternion rotation map $ρ$ : conjugation on the imaginary quaternions, the unit sphere $S^{3}$ , and quaternion lifts.

[F1]

$H$ is $R^{4}$ with the quaternion product; every quaternion has a unique decomposition $x = Re (x) + φ (w)$ with $Re (x) = x_{0}$ and $w \in R^{3}$ , and $φ$ is a linear isomorphism $R^{3} \to Im H$ (The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ , The quaternion rotation map $ρ$ : conjugation on the imaginary quaternions, the unit sphere $S^{3}$ , and quaternion lifts).

[F2]

For $C^{1}$ curves $p, q : I \to H$ the product $p q$ is $C^{1}$ with $(p q)^{'} = p^{'} q + p q^{'}$ ; for all $p, q \in H$ one has $N (p q) = N (p) N (q)$ , $Re (p \overset{ˉ}{q}) = ⟨ p, q ⟩_{2}$ , and $(N \circ q)^{'} = 2 Re (q^{'} \overset{ˉ}{q})$ ; also $\hat{N (q)} = q \overset{ˉ}{q} = \overset{ˉ}{q} q$ . Thus for $q \in S^{3}$ one has $q^{- 1} = \overset{ˉ}{q}$ and $q \overset{ˉ}{q} = \overset{ˉ}{q} q = 1_{H}$ (The product rule for quaternion and matrix curves, multiplicativity of the quaternion norm, and the conjugation formula $q φ (v) \overset{ˉ}{q} = φ ((a^{2} - ∥ u ∥^{2}) v + 2 ⟨ u, v ⟩ u + 2 a (u \times v))$ clauses 1-2, The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ , $H$ is a division ring that is not commutative, hence not a field: $q^{- 1} = \overset{ˉ}{q} / N (q)$ for $q \neq 0$ , while $i j = k$ and $j i = - k$ ).

[F3]

$q_{0} \in S^{3}$ implies $ρ (q_{0}) \in S O (3)$ , and $ρ$ is a homomorphism on $S^{3}$ (The quaternion rotation map $ρ$ : conjugation on the imaginary quaternions, the unit sphere $S^{3}$ , and quaternion lifts).

[F4]

If $ω : I \to R^{3}$ is continuous and $q : I \to H$ is a $C^{1}$ solution of $q^{'} = \frac{1}{2} q φ (ω (t))$ with $q (t_{0}) \in S^{3}$ , then $q (I) \subseteq S^{3}$ , $R = ρ \circ q$ is $C^{1}$ with $R^{'} = R Ω_{ω}$ , and $ω$ is the body angular velocity of $R$ (The quaternion lift of a prescribed continuous angular velocity: $q^{'} = \frac{1}{2} q φ (ω (t))$ keeps $S^{3}$ invariant, its rotation path has body angular velocity $ω$ , and the initial value problem has a unique solution on the whole interval clause 1).

[F5]

A $C^{1}$ rotation path has exactly one body angular velocity, characterized by $R^{'} = R Ω_{ω}$ (The body angular velocity of a $C^{1}$ rotation path: $Ω = R^{T} R^{'}$ is skew, $Ω z = ω \times z$ determines a unique continuous $ω$ , and $R^{'} = R Ω$ clauses 2-3).

Proof

technique · direct; the product $q^{-1}q'$ is forced to be pure by the unit constraint, and the lift lemma of the previous item applies

1.1F1F2algebra

By [F2], $(N \circ q)^{'} = 2 Re (q^{'} \overset{ˉ}{q})$ ; since $N (q (t)) = 1$ for all $t$ , the left side vanishes, so $Re (q^{'} \overset{ˉ}{q}) = 0$ . By [F1] the scalar part of a product is $Re (x y) = x_{0} y_{0} - \sum_{j = 1}^{3} x_{j} y_{j}$ , symmetric in the two factors, and $q^{- 1} = \overset{ˉ}{q}$ on $S^{3}$ by [F2]; hence $Re (q^{- 1} q^{'}) = Re (\overset{ˉ}{q} q^{'}) = Re (q^{'} \overset{ˉ}{q}) = 0$ , so $q^{- 1} q^{'}$ is imaginary. By the unique decomposition of [F1] there is exactly one $w (t) \in R^{3}$ with $q (t)^{- 1} q^{'} (t) = φ (w (t))$ ; setting $ω (t) : = 2 w (t)$ gives $q^{- 1} q^{'} = \frac{1}{2} φ (ω)$ on $I$ . The components of $q^{- 1} q^{'}$ are finite sums of products of the continuous components of $q, q^{'}$ , so $t \mapsto ω (t)$ is continuous. This proves clause 1.

2.1step 1.1F2algebra

Multiplying $q^{- 1} q^{'} = \frac{1}{2} φ (ω (t))$ on the left by $q (t)$ and using $q q^{- 1} = 1$ gives $q^{'} = \frac{1}{2} q φ (ω (t))$ on $I$ , and $ω$ is continuous by [step 1.1].

3.1step 2.1F3F4F5∎

The curve $q$ satisfies the hypotheses of [F4] with the continuous $ω$ of [step 1.1] and with $q (t_{0}) \in S^{3}$ for any fixed $t_{0} \in I$ ; the curve $R = ρ \circ q$ takes values in $S O (3)$ by [F3] and is $C^{1}$ with $R^{'} = R Ω_{ω}$ by [F4]. By [F5] the body angular velocity of $R$ is unique, so it equals this $ω$ , which proves clauses 2 and 3.

Remarks

The formula $ω = 2 φ^{- 1} (q^{- 1} q^{'})$ is the reconstruction formula used by the examples: from a $C^{1}$ path in $S^{3}$ one reads off the angular velocity of the associated rotation path, with no additional data and no choice of sign — replacing $q$ by $- q$ multiplies $q^{- 1} q^{'}$ by $(- 1)^{- 1} (- 1) = 1$ and leaves $ω$ unchanged.
The proposition only computes the body angular velocity; the spatial angular velocity belongs to the other convention recorded in The body angular velocity of a $C^{1}$ rotation path: $Ω = R^{T} R^{'}$ is skew, $Ω z = ω \times z$ determines a unique continuous $ω$ , and $R^{'} = R Ω$ and is not asserted here.

Angular-velocity reconstruction from a unit-quaternion path: for $C^{1}$ $q : I \to S^{3}$ the body angular velocity of $ρ \circ q$ is $ω = 2 φ^{- 1} (q^{- 1} q^{'})$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Angular-velocity reconstruction from a unit-quaternion path: for C1C1 q:I→S3q:I→S3 the body angular velocity of ρ∘qρ∘q is ω=2φ−1(q−1q′)ω=2φ−1(q−1q′)

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Angular-velocity reconstruction from a unit-quaternion path: for $C^{1}$ $q : I \to S^{3}$ the body angular velocity of $ρ \circ q$ is $ω = 2 φ^{- 1} (q^{- 1} q^{'})$