I'm using quaternions to solve Euler's equations of motion. I have the substitution that $\dot{q} = \frac{1}{2}q\bigodot\omega$, where $\omega$ is the angular velocity of the rotating frame. Hence $\omega = 2q^*\bigodot \dot{q}$ where $q^*$ is the conjugate of $q$. In Euler's quation, the inertia tensor normally multiplies the $\omega$ vector like so: $\omega \times I\omega$. I understand that in the setting of quaternions the vector $\omega$ becomes $\omega_{[quat]} = [0, \omega_x\bf{i}, \omega_y\bf{j}, \omega_z\bf{k}]$ and is said to have no real part. But how do we express the inertia tensor for this new convention to work? I was considering treating it as a rotation matrix and converting to a quaternion itself but this doesn't work when det$(I)\neq1 $ where $I$ is the inertia tensor.

Any help greatly appreciated.


Render the following as quaternion imaginary units:




And the real unit matrix:


with $i$ as the imaginary unit in ordinary complex numbers. Then for all real numbers $a,b,c,d$ the combination


will reproduce all the algebraic properties of quaternions with "real part" $a$ and "imaginary parts" $b,c,d$. For instance, $(\sigma_x)(\sigma_y)=-(\sigma_y)(\sigma_x)=\sigma_z$ and $(\sigma_x)^2=(\sigma_y)^2=(\sigma_z)^2=-\sigma_0$.

  • $\begingroup$ Thanks for this response, although I'm not entirely sure how I would use this to transform $I$ into a quaternion. Are you suggesting I essentially express the quaternions as matrices instead? Thanks $\endgroup$
    – liamw
    Nov 2 '18 at 1:40
  • $\begingroup$ That was how I interpreted it based on the title. Maybe something was unclear? $\endgroup$ Nov 2 '18 at 1:41
  • $\begingroup$ Sorry, to rephrase it: I have an inertia tensor that is 3x3 and normally multiplies a 3x1 vector. However I've transformed my 3x1 vector into a quaternion by adding 0 real part. What must I do to the inertia tensor for this to still be a calculable multiplication. $\endgroup$
    – liamw
    Nov 2 '18 at 2:26

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