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A time-varying rate about a fixed axis: $q (t) = q_{0} (\cos ((θ (t) - θ (0)) / 2) + φ (\hat{n}) \sin ((θ (t) - θ (0)) / 2))$ solves the lift equation with $ω (t) = θ^{'} (t) \hat{n}$

Example

Let $I \subseteq R$ be a nondegenerate open interval with $0 \in I$ , let $θ : I \to R$ be of class $C^{1}$ , let $\hat{n} \in R^{3}$ be a unit vector, let $q_{0} \in S^{3}$ , and define the normalized phase $α (t) : = θ (t) - θ (0)$ and

$e (t) : = \cos (α (t) / 2) + φ (\hat{n}) \sin (α (t) / 2), q (t) : = q_{0} e (t),$

with $φ, S^{3}, ρ$ as in The quaternion rotation map $ρ$ : conjugation on the imaginary quaternions, the unit sphere $S^{3}$ , and quaternion lifts. Then:

$e$ and $q$ are of class $C^{1}$ , they take values in $S^{3}$ , and $q (0) = q_{0}$ .
The curve $q$ solves $q^{'} (t) = \frac{1}{2} q (t) φ (ω (t))$ with $ω (t) : = θ^{'} (t) \hat{n}$ , on all of $I$ .
Consequently $R : = ρ \circ q$ is a $C^{1}$ rotation path whose body angular velocity is $ω (t) = θ^{'} (t) \hat{n}$ , and $R (t) = ρ (q_{0}) ρ (e (t))$ , where $ρ (e (t))$ is the rotation about $\hat{n}$ through the angle $α (t) = θ (t) - θ (0)$ .

Facts & Assumptions

Given: The interval $I ∋ 0$ , the curve $θ$ , the unit $\hat{n}$ , the unit quaternion $q_{0}$ , and the curves $e, q$ above.

[L1]

$α$ is $C^{1}$ , $α (0) = 0$ , and $α^{'} = θ^{'}$ ; the sine and cosine derivative rules, Pythagorean identity, chain rule, and class- $C^{1}$ closure of composites apply (The derivatives of sine and cosine are cosine and minus sine, Parity and the Pythagorean identity for sine and cosine, The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , $C^{k}$ Euclidean maps and diffeomorphisms).

[L2]

For $q = s + φ (u)$ , $q^{'} = t + φ (w)$ one has $q q^{'} = (s t - ⟨ u, w ⟩_{2}) + φ (s w + t u + u \times w)$ , hence $φ (\hat{n})^{2} = - 1$ for $∥ \hat{n} ∥_{2} = 1$ , and $N (s + φ (u)) = s^{2} + ∥ u ∥_{2}^{2}$ (The product rule for quaternion and matrix curves, multiplicativity of the quaternion norm, and the conjugation formula $q φ (v) \overset{ˉ}{q} = φ ((a^{2} - ∥ u ∥^{2}) v + 2 ⟨ u, v ⟩ u + 2 a (u \times v))$ clause 1, The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ , The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[L3]

$ρ (q) (φ (v)) = q φ (v) \overset{ˉ}{q}$ for $q \in S^{3}$ , and for $q = \cos (θ / 2) + φ (\hat{n}) \sin (θ / 2)$ this equals $φ (\cos θ v + (1 - \cos θ) ⟨ \hat{n}, v ⟩_{2} \hat{n} + \sin θ (\hat{n} \times v))$ ; moreover $ρ (p q) = ρ (p) ρ (q)$ for $p, q \in S^{3}$ (The product rule for quaternion and matrix curves, multiplicativity of the quaternion norm, and the conjugation formula $q φ (v) \overset{ˉ}{q} = φ ((a^{2} - ∥ u ∥^{2}) v + 2 ⟨ u, v ⟩ u + 2 a (u \times v))$ clause 4, The quaternion rotation map $ρ$ : conjugation on the imaginary quaternions, the unit sphere $S^{3}$ , and quaternion lifts Definition and Remark).

[L4]

The lift equation with continuous $ω$ and initial data in $S^{3}$ has a unique solution on the whole interval, and its associated rotation path has body angular velocity $ω$ (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 clauses 1-2, 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 $ω = 2 φ^{- 1} (q^{- 1} q^{'})$ clause 3).

Verification

technique · direct; verify the explicit curve solves the lift equation and read off the rotation

1.1L1L2algebra

Since $α (0) = 0$ , [L2] gives $N (e (t)) = \cos^{2} (α (t) / 2) + ∥ \hat{n} ∥_{2}^{2} \sin^{2} (α (t) / 2) = 1$ and $e (0) = 1$ ; multiplying by the unit quaternion $q_{0}$ keeps $N (q (t)) = N (q_{0}) N (e (t)) = 1$ . Both $e$ and $q$ are $C^{1}$ on $I$ by [L1], and $q (0) = q_{0} e (0) = q_{0}$ . This proves clause 1.

2.1step 1.1L1L2algebra

Differentiating $e$ with [L1] gives $e^{'} (t) = \frac{θ^{'} (t)}{2} (- \sin (α (t) / 2) + φ (\hat{n}) \cos (α (t) / 2))$ . By [L2], $φ (\hat{n}) e (t) = e (t) φ (\hat{n})$ because $e (t) = \cos (α / 2) \cdot 1 + \sin (α / 2) φ (\hat{n})$ is a real combination of $1$ and $φ (\hat{n})$ , which commute, and $φ (\hat{n})^{2} = - 1$ . Hence $\frac{1}{2} φ (θ^{'} (t) \hat{n}) e (t) = \frac{θ^{'} (t)}{2} (\cos (α / 2) φ (\hat{n}) + \sin (α / 2) φ (\hat{n})^{2}) = e^{'} (t)$ , and therefore $q^{'} = q_{0} e^{'} = q_{0} \frac{1}{2} φ (θ^{'} (t) \hat{n}) e = \frac{1}{2} q_{0} e φ (θ^{'} (t) \hat{n}) = \frac{1}{2} q φ (θ^{'} (t) \hat{n})$ , which is the displayed equation with $ω (t) = θ^{'} (t) \hat{n}$ ; $ω$ is continuous because $θ$ is $C^{1}$ . This proves clause 2.

3.1step 2.1L3L4∎

Since $q (0) = q_{0} \in S^{3}$ and $ω$ is continuous, [L4] identifies $R = ρ \circ q$ as a $C^{1}$ rotation path whose body angular velocity is $ω (t) = θ^{'} (t) \hat{n}$ ; the homomorphism property and the axis-angle form of [L3] give $R (t) = ρ (q_{0}) ρ (e (t))$ with $ρ (e (t))$ the rotation through $α (t) = θ (t) - θ (0)$ about $\hat{n}$ . This proves clause 3.

Remarks

The verification is a finite computation: it checks that the half-angle curve solves the lift equation and then invokes the lift-and-reconstruction items; no claim is made about the existence or uniqueness of $q$ beyond what those items supply, and none about the physical meaning of $θ$ or $t$ .
The phase is normalized by subtracting $θ (0)$ , so every allowed $C^{1}$ function $θ$ gives $e (0) = 1$ and $q (0) = q_{0}$ ; the angular velocity remains $θ^{'} (t) \hat{n}$ . The initial value $q_{0}$ may be any unit quaternion, so left multiplication by $q_{0}$ produces either initial lift sign.

A time-varying rate about a fixed axis: $q (t) = q_{0} (\cos ((θ (t) - θ (0)) / 2) + φ (\hat{n}) \sin ((θ (t) - θ (0)) / 2))$ solves the lift equation with $ω (t) = θ^{'} (t) \hat{n}$

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

A time-varying rate about a fixed axis: q(t)=q0(cos⁡((θ(t)−θ(0))/2)+φ(n^)sin⁡((θ(t)−θ(0))/2))q(t)=q0​(cos((θ(t)−θ(0))/2)+φ(n^)sin((θ(t)−θ(0))/2)) solves the lift equation with ω(t)=θ′(t)n^ω(t)=θ′(t)n^

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

A time-varying rate about a fixed axis: $q (t) = q_{0} (\cos ((θ (t) - θ (0)) / 2) + φ (\hat{n}) \sin ((θ (t) - θ (0)) / 2))$ solves the lift equation with $ω (t) = θ^{'} (t) \hat{n}$