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A non-axis-constant check of the reconstruction formula: for $q = (\cos \frac{t}{2} + φ (e_{1}) \sin \frac{t}{2}) (\cos \frac{t}{2} + φ (e_{2}) \sin \frac{t}{2})$ one gets $ω (t) = (\cos t, 1, \sin t)$

Example

On $I = R$ define

$u (t) : = \cos (t / 2) + φ (e_{1}) \sin (t / 2), v (t) : = \cos (t / 2) + φ (e_{2}) \sin (t / 2), q (t) : = u (t) v (t),$

where $e_{1}, e_{2}, e_{3}$ is the standard basis of $R^{3}$ , $φ$ is the identification of The quaternion rotation map $ρ$ : conjugation on the imaginary quaternions, the unit sphere $S^{3}$ , and quaternion lifts, and the product is the quaternion product. Then:

$q : R \to S^{3}$ is of class $C^{1}$ .
$\overset{ˉ}{u} u^{'} = \frac{1}{2} φ (e_{1})$ and $\overset{ˉ}{v} v^{'} = \frac{1}{2} φ (e_{2})$ , and

$q (t)^{- 1} q^{'} (t) = \overset{ˉ}{v} (t) (\overset{ˉ}{u} (t) u^{'} (t)) v (t) + \overset{ˉ}{v} (t) v^{'} (t) = \frac{1}{2} φ (\cos t e_{1} + e_{2} + \sin t e_{3}) .$

Consequently $R = ρ \circ q$ is a $C^{1}$ rotation path whose body angular velocity is $ω (t) = \cos t e_{1} + e_{2} + \sin t e_{3}$ , that is, $ω (t) = (\cos t, 1, \sin t)$ in coordinates; this $ω$ is neither constant nor along a fixed axis.

Facts & Assumptions

Given: The basis vectors $e_{1}, e_{2}, e_{3}$ , the curves $u, v, q$ above, 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.

[L1]

$(\sin x)^{'} = \cos x$ , $(\cos x)^{'} = - \sin x$ , $\sin^{2} x + \cos^{2} x = 1$ , the chain rule, and $C^{1}$ closure under composition (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 $p = s + φ (w)$ , $q^{'} = t + φ (z)$ one has $p q^{'} = (s t - ⟨ w, z ⟩_{2}) + φ (s z + t w + w \times z)$ ; for $C^{1}$ quaternion curves the product rule $(p q)^{'} = p^{'} q + p q^{'}$ holds; $N (p q) = N (p) N (q)$ , $N (s + φ (w)) = s^{2} + ∥ w ∥_{2}^{2}$ , and a unit quaternion satisfies $q^{- 1} = \overset{ˉ}{q}$ because $\hat{N (q)} = q \overset{ˉ}{q} = 1_{H}$ (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 $j i = - k$ , The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The cross product in $R^{3}$ ).

[L3]

For $N (q) = 1$ and $q = \cos (θ / 2) + φ (\hat{n}) \sin (θ / 2)$ , $∥ \hat{n} ∥_{2} = 1$ , one has $ρ (q) (φ (z)) = φ (\cos θ z + (1 - \cos θ) ⟨ \hat{n}, z ⟩_{2} \hat{n} + \sin θ (\hat{n} \times z))$ ; and $ρ (p) (φ (z)) = p φ (z) \overset{ˉ}{p}$ for $p \in S^{3}$ (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).

[L4]

For a $C^{1}$ path $q : R \to S^{3}$ the body angular velocity of $ρ \circ q$ is $ω = 2 φ^{- 1} (q^{- 1} q^{'})$ , equivalently the unique vector with $q^{- 1} q^{'} = \frac{1}{2} φ (ω)$ (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 $ω = 2 φ^{- 1} (q^{- 1} q^{'})$ clauses 1-3).

Verification

technique · direct; compute the two factors of $q^{-1}q'$ and apply one axis-angle rotation

1.1L1L2algebra

Both $u$ and $v$ are $C^{1}$ and take values in $S^{3}$ , since each has the form $\cos (t / 2) + φ (\hat{n}) \sin (t / 2)$ with $\hat{n}$ unit and $N = s^{2} + ∥ w ∥_{2}^{2} = 1$ by [L2]; by the multiplicativity of [L2], $N (q) = N (u) N (v) = 1$ , so $q$ takes values in $S^{3}$ , and $q$ is $C^{1}$ as a product of $C^{1}$ curves. This proves clause 1.

2.1step 1.1L1L2algebra

Directly from the definitions and [L1], $u^{'} = - \frac{1}{2} \sin (t / 2) + \frac{1}{2} φ (e_{1}) \cos (t / 2)$ and $v^{'} = - \frac{1}{2} \sin (t / 2) + \frac{1}{2} φ (e_{2}) \cos (t / 2)$ . Multiplying with the formula of [L2] and using $\overset{ˉ}{u} = \cos (t / 2) - φ (e_{1}) \sin (t / 2)$ , $\overset{ˉ}{v} = \cos (t / 2) - φ (e_{2}) \sin (t / 2)$ gives $\overset{ˉ}{u} u^{'} = \frac{1}{2} (- \cos (t / 2) \sin (t / 2) + \sin (t / 2) \cos (t / 2)) + \frac{1}{2} φ (e_{1}) (\cos^{2} (t / 2) + \sin^{2} (t / 2)) = \frac{1}{2} φ (e_{1})$ , and the identical computation with $e_{2}$ gives $\overset{ˉ}{v} v^{'} = \frac{1}{2} φ (e_{2})$ ; the two scalar parts vanish by the cancellation of the cross terms, and $φ (e_{j})^{2} = - 1$ was used.

2.2step 1.1L2L3algebra

The unit quaternion $v (t)$ has the axis-angle form $v (t) = \cos (t / 2) + φ (e_{2}) \sin (t / 2)$ , while $\overset{ˉ}{v} (t) = \cos (- t / 2) + φ (e_{2}) \sin (- t / 2)$ ; applying [L3] with $θ = - t$ and $\hat{n} = e_{2}$ gives $ρ (\overset{ˉ}{v}) (φ (e_{1})) = φ (\cos t e_{1} + (1 - \cos t) ⟨ e_{2}, e_{1} ⟩_{2} e_{2} + \sin (- t) (e_{2} \times e_{1})) = φ (\cos t e_{1} + \sin t e_{3})$ , because $⟨ e_{2}, e_{1} ⟩_{2} = 0$ and $e_{2} \times e_{1} = - e_{3}$ by [L2].

3.1step 2.1step 2.2L2L3algebra

Since $q = u v$ , the product rule and $\overline{u v} = \overset{ˉ}{v} \overset{ˉ}{u}$ of [L2] give $q^{- 1} q^{'} = \overset{ˉ}{q} q^{'} = \overset{ˉ}{v} \overset{ˉ}{u} (u^{'} v + u v^{'}) = \overset{ˉ}{v} (\overset{ˉ}{u} u^{'}) v + \overset{ˉ}{v} v^{'}$ . By [step 2.1], $\overset{ˉ}{v} (\overset{ˉ}{u} u^{'}) v = \frac{1}{2} \overset{ˉ}{v} φ (e_{1}) v$ , and since $v \in S^{3}$ the identity $ρ (\overset{ˉ}{v}) (φ (e_{1})) = \overset{ˉ}{v} φ (e_{1}) v$ of [L3] gives $\overset{ˉ}{v} φ (e_{1}) v = φ (ρ (\overset{ˉ}{v}) (e_{1}))$ ; substituting [step 2.2] and $\overset{ˉ}{v} v^{'} = \frac{1}{2} φ (e_{2})$ yields $q^{- 1} q^{'} = \frac{1}{2} φ (\cos t e_{1} + e_{2} + \sin t e_{3})$ , which proves clause 2.

4.1step 3.1L4∎

By [L4] the body angular velocity of $R = ρ \circ q$ is the vector $ω$ with $q^{- 1} q^{'} = \frac{1}{2} φ (ω)$ , which by [step 3.1] is $ω (t) = \cos t e_{1} + e_{2} + \sin t e_{3}$ ; its component along $e_{2}$ is constant while the other two oscillate, so its direction is not constant and no fixed axis contains all values. This proves clause 3.

Remarks

The example checks the reconstruction formula on a product of two one-parameter rotations about different axes; the resulting $ω$ is nonconstant, so the example is not a disguised fixed-axis case. The computation is finite and uses only the product form, the multiplicativity, and one evaluation of the axis-angle formula.
No dynamical input is used: $q$ is an explicitly given curve and no equation of motion is solved. The example asserts nothing about which curves arise from any mechanical system.

A non-axis-constant check of the reconstruction formula: for $q = (\cos \frac{t}{2} + φ (e_{1}) \sin \frac{t}{2}) (\cos \frac{t}{2} + φ (e_{2}) \sin \frac{t}{2})$ one gets $ω (t) = (\cos t, 1, \sin t)$

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

A non-axis-constant check of the reconstruction formula: for q=(cos⁡t2+φ(e1)sin⁡t2)(cos⁡t2+φ(e2)sin⁡t2)q=(cos2t​+φ(e1​)sin2t​)(cos2t​+φ(e2​)sin2t​) one gets ω(t)=(cos⁡t,1,sin⁡t)ω(t)=(cost,1,sint)

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

A non-axis-constant check of the reconstruction formula: for $q = (\cos \frac{t}{2} + φ (e_{1}) \sin \frac{t}{2}) (\cos \frac{t}{2} + φ (e_{2}) \sin \frac{t}{2})$ one gets $ω (t) = (\cos t, 1, \sin t)$