3
$\begingroup$

I'm kind of confused about the normalization of a dual quaternion.

From the texts (e.g. Kavan 2008), for a dual quaternion $\hat{q} = a + \varepsilon b $, the norm is defined as $$ ||\hat{q}|| = ||a|| + \varepsilon \frac{a \cdot b}{||a||} $$

I'm interested in getting a unit dual quaternion, which I understand is defined to be when $||a|| = 1$ and the dual part is 0 (i.e. $a \cdot b = 0$).

I notice that most people perform the normalization by simply dividing both $a$ and $b$ by $||a||$ (e.g. https://stackoverflow.com/questions/23174899/properly-normalizing-a-dual-quaternion). However, from what it looks like to me, this only ensures that $||a|| = 1$, and does not do anything to ensure that $a \cdot b = 0$ (i.e. to get the dual part to 0).

Is this normalization of a dual quaternion not the same as trying to normalize a dual quaternion into a unit dual quaternion?

I tried looking into what normalisation meant: $$ \frac{\hat{q}}{||\hat{q}||} = \frac{a + \varepsilon b}{||a|| + \varepsilon \frac{a \cdot b}{||a||}} * \frac{||a|| - \varepsilon \frac{a \cdot b}{||a||}}{||a|| - \varepsilon \frac{a \cdot b}{||a||}} \\ = \frac{a}{||a||} + \varepsilon \left[ \frac{b}{||a||} - \frac{a}{||a||} \left(\frac{a}{||a||} \cdot \frac{b}{||a||} \right)\right] $$

So it feels like the normalisation of only dividing both parts by $||a||$ applies only if $a\cdot b=0$ already?

What then does using the full normalised dual part $\left[ \frac{b}{||a||} - \frac{a}{||a||} \left(\frac{a}{||a||} \cdot \frac{b}{||a||} \right)\right]$ mean? I tried looking into whether I could prove whether that full normalised dual part was equal to 0 (for the definition of a unit dual quaternion) but couldn't do so.

$\endgroup$
2

1 Answer 1

2
$\begingroup$

I agree with you that I also do not believe the answer about dividing real and dual part by $\|a\|$ is correct. To normalize a random quaternion $\hat{q} = a + b \varepsilon$, you have to compute the exact way you wrote in your question: $$ \begin{align*} u & = u_a + u_b \varepsilon \\ & = \frac{\hat{q}}{||\hat{q}||} = \frac{a + \varepsilon b}{||a|| + \varepsilon \frac{a^\top b}{||a||}} \frac{||a|| - \varepsilon \frac{a^\top b}{||a||}}{||a|| - \varepsilon \frac{a^\top b}{||a||}} \\ & = \frac{a}{||a||} + \varepsilon \left[ \frac{b}{||a||} - \frac{a}{||a||} \left(\frac{a^\top b }{||a||^2} \right)\right] \end{align*} $$

This new dual quaternion is a unit dual quaternion because it satisfies two criteria:

  1. $\|u_a\| = \sqrt{u_a^\top u_a} = 1$. $$ \sqrt{u_a^\top u_a} = \sqrt{\frac{a^\top a}{\|a\|^2}} = 1 $$

  2. $u_a^\top u_b = 0$. $$ u_a^\top u_b = \frac{a^\top}{\|a\|}\left[ \frac{b}{||a||} - \frac{a}{||a||} \left(\frac{a^\top b }{||a||^2} \right)\right] = \frac{a^\top b}{\| a\|^2} - \frac{a^\top a(a^\top b)}{\| a\|^2 \| a\|^2} = \frac{a^\top b}{\| a\|^2} - 1\frac{a^\top b}{ \| a\|^2} = 0 $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .