Consider the rotation which maps a basis \((\bha_1 ,\bha_2 ,\bha_3)\) of referential \(\cA\) to a basis \((\bhb_1 ,\bhb_2 ,\bhb_3)\) of a referential \(\cB\) in motion relative to \(\cA\text{.}\) To this rotation corresponds to a time-varying unit quaternion which we denote as \(Q_{B/A}(t)\text{.}\) A vector \(\bvv_0\) fixed in \(\cA\) is mapped into a vector \(\bvv\) fixed in \(\cB\text{.}\) If we define the pure quaternions \(V_0 = 0+\bvv_0\) and \(V(t) =0+\bvv\text{,}\) the mapping of \(\bvv_0\) onto \(\bvv\) corresponds to the quaternion relationship
\begin{equation*}
V (t) = Q_{B/A}(t) V_0 \conjQ_{B/A}(t)
\end{equation*}
where
\(\conjQ_{B/A}\) is the conjugate of
\(Q_{B/A}\text{.}\) We can find the derivative of
\(\bvv\) relative to
\(\cA\) according to
(3.1.2):
\begin{equation*}
\left( \frac{d \bvv}{ dt } \right)_{\cA} = \bom _{\cB/ \cA} \times \bvv
\end{equation*}
Let
\(\Omega_{B/A}\) be the pure quaternion defined as
\(0+\bom _{\cB/ \cA}\text{.}\) Then, the cross-product
\(\bom _{\cB/ \cA} \times \bvv \) can be written as the quaternion
\(\frac{1}{2}(\Omega_{B/A} V - V \Omega_{B/A})\) (see
Example 1.6.3).
This gives the quaternion relationship between \(V_0\) and \(V\text{:}\)
\begin{equation*}
\frac{d}{dt} ( Q_{B/A} V_0 \conjQ_{B/A}) = \frac{1}{2}(\Omega_{B/A} V - V
\Omega_{B/A})
\end{equation*}
A unit quaternion \(Q\) satisfies \(Q\conjQ =\conjQ Q= 1\) and time-differentiation (relative to \(\cA\)) leads to \(d \conjQ /dt = - \conjQ (dQ /dt) \conjQ \text{.}\) So the last equation becomes
\begin{equation*}
\frac{d Q_{B/A} }{dt} V_0 \conjQ_{B/A} - Q_{B/A}V_0 \conjQ_{B/A}
\frac{d Q_{B/A} }{dt} \conjQ_{\cB/\cA} = \frac{1}{2}(\Omega_{B/A} V - V \Omega_{B/A})
\end{equation*}
Then using \(V_0 = \conjQ_{B/A} V Q_{B/A}\) we obtain
\begin{equation*}
\frac{d Q_{B/A} }{dt} \conjQ_{B/A} V - V \frac{dQ_{B/A} }{dt}
\conjQ_{B/A} = \frac{1}{2}(\Omega_{B/A} V -V \Omega_{B/A})
\end{equation*}
which can be written in the form
\begin{equation*}
\left(
\Omega_{B/A} -2 \, \frac{dQ_{B/A} }{dt} \conjQ_{B/A}
\right) V
-
V
\left(
\Omega_{B/A} -2 \frac{dQ_{B/A} }{dt} \conjQ_{B/A}
\right) = 0
\end{equation*}
or, in vector form, as
\begin{equation*}
\left(
\bom _{\cB/ \cA} -2 \, \frac{dQ_{B/A} }{dt} \conjQ_{B/A}
\right) \times \bvv = 0
\end{equation*}
Since this last equation holds true for all \(\bvv\text{,}\) we obtain:
Theorem 3.4.1. Quaternion formulation of the Angular Velocity.
If quaternion \(Q_{B/A}(t)\) corresponds to the rotation operator which maps a basis of referential \(\cA\) to a basis of referential \(\cB\text{,}\) then the angular velocity \(\bom _{\cB/ \cA}\) of referential \(\cB\) relative to referential \(\cA\) is given by the pure quaternion
\begin{equation}
\Omega_{B/A}= 0+\bom _{\cB/ \cA} = 2 \,\frac{dQ_{B/A} }{dt} \conjQ_{B/A}\tag{3.4.1}
\end{equation}
where the time-differentiation is performed relative to \(\cA\text{.}\)
It is also possible to express angular velocity
\(\bom _{\cB/ \cA} \) in terms of the derivative of Euler parameters: substitution of
\(Q_{B/A} = \cos\frac{\al}{2} + \sin\frac{\al}{2} \bu\) in equation
(3.4.1) gives
\begin{equation}
\bom _{\cB/ \cA} = \dal \bu + \sin\al \frac{d\bu }{dt} + (1-\cos\al)\bu \times \frac{d\bu }{dt}\tag{3.4.2}
\end{equation}
where \((\al,\bu)\) are the equivalent angle/direction of the rotation \(\cR_{\al,\bu}\) which maps a basis of \(\cA\) to a basis of \(\cB\text{.}\)
This result can also be recast in terms of rotation vector \(\rot = \tan\frac{\al}{2} \bu\) and its time-derivative
\begin{equation}
\bom _{\cB/ \cA} = \frac{2}{1+ \rot^2}
\Big( \frac{d\rot }{dt} + \rot \times \frac{d\rot}{dt}
\Big) \tag{3.4.3}
\end{equation}
