HCR - Rotation Matrix Descriptor 'Quaternion'

NOTE:IMPORTANTE ==There is no possibility that 2 different quaternions give the same final rotation.==

Let’s take an example: in the angle and axis descriptor formulas, we find that:

$$\theta ={\cos }^{-1}\left(\frac{r_{1,1}+r_{2,2}+r_{3,3}-1}{2}\right)$$

Now if we consider $r_{1,1}+r_{2,2}+r_{3,3}=1$ then $\theta$ can be both equal to $0$ or $\pi$, so we have and indetermination on the angle.

Let’s now define $\eta$ and $ϵ$ with respect to $\theta$ and $\vec{v}$ as:\begin{align} &\eta = \cos\left(\frac{\theta}{2}\right) \\[7px] &\varepsilon = \sin\left(\frac{\theta}{2}\right)\cdot \vec v \end{align} Where:

• $\vec{v}$ is a vector ($3\times 1$) and so is $\epsilon$.
• While $\eta$ is a scalar ($1\times 1$).

Notice that:${\eta }^2+{\epsilon }_x^2+{\epsilon }_y^2+{\epsilon }_z^2=1$And:$\begin{array}{c}\eta =\frac{1}{2}\sqrt{r_{1,1}+r_{2,2}+r_{3,3}}\\ \epsilon =\left[\begin{array}{l}\mathrm{sign}(r_{3,2}-r_{2,3})\sqrt{1+r_{1,1}-r_{2,2}-r_{3,3}}\\ \mathrm{sign}(r_{1,3}-r_{3,1})\sqrt{1+r_{2,2}-r_{1,1}-r_{3,3}}\\ \mathrm{sign}(r_{2,1}-r_{1,2})\sqrt{1+r_{3,3}-r_{1,1}-r_{2,2}}\end{array}\right]\end{array}$So by using $\eta$ and $\epsilon$ as constraints we can then find the rotation matrix $R$ :

$$R=\left[\begin{array}{lcc}{r}_{1,1}& r_{1,2}& r_{1,3}\\ r_{2,1}& r_{2,2}& r_{2,3}\\ r_{3,1}& r_{3,2}& r_{3,3}\end{array}\right]$$

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Quaternions re-explained

(As everyone knows them)

• Youtube - ‘Quaternions and 3d rotation, explained interactively’ by ‘3Blue1Brown’

Given a point $\overline{p}$ an axis of rotation $\overline{v}$ and the 2 angles of rotation $\theta$ and $\phi$ we can define:

$$q=\left(\cos \left(\frac{\theta }{2}\right)+\sin \left(\frac{\phi }{2}\right)\right)\cdot (\overline{v}\cdot [i,j,k])$$

Where:

• $i^2=j^2=k^2=ijk=-1$

So once we have defined $q$ the new point ${\overline{p}}^{\prime }$ location after the rotation is given by:

$${\overline{p}}^{\prime }=q\cdot p\cdot q^{-1}$$

Where:

$$\begin{array}{lcc}\text{if:}& q& =w_0+x_{0}i+y_{0}j+z_{0}k\\ \text{then:}& q^{-1}& =w_0-x_{0}i-y_{0}j-z_{0}k\end{array}$$