In Jack B. Kuipers' Quaternions and Rotation Sequences page 194-195, section 8.7.1 explains how to decompose a tracking sequence quaternion $q = q_0 + \vec{i}q_1+\vec{j}q_2+\vec{k}q_3$ into two other quaternions $a^3 = a_0 + \vec{k}a_3$ and $b^2 = b_0 + \vec{j}b_2$ (the 3 and 2 denote axes of rotation, not quaternion exponentiation). Equation 8.10 states that

$$q = q_0 + \vec{i}q_1+\vec{j}q_2+\vec{k}q_3 = a^3b^2$$ $$ = a_0b_0 - \vec{i}a_3b_2 + \vec{j}a_0b_2+\vec{k}a_3b_0$$

which makes sense in the context of quaternion multiplication. However, he then states

In Equation 8.10, it is easily verified, that $$q_0q_1 + q_2q_3 = 0 \tag{8.11}$$

Where did this constraint come from?

Edit: This text continues by tabulating this factorization for all combinations of $a^ib^j$ for $i,j \in [1,2,3]$. Associated with each factorization is an associated constraint of the form $q_lq_m \pm q_oq_p=0$. Where do these come from?


It turns out that the constraint is nothing more complicated than seeing that $$q_0q_1 = -a_0b_0a_3b_2$$ while $$q_2q_3 = a_0b_0a_3b_2$$

and, therefore, for all a and b subject to the forms above, this constraint equation holds.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.