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I have a set of rotations:

$R(t) \in R^{3x3}, t = 1, 2, ... T$.

I can extract the orientation of a body $\theta (t)$ from the rotation matrix $R(t)$. I am interested to estimate the angular velocity $\omega (t)$ and angular acceleration $\alpha (t)$. I have performed spline quaternion interpolation. I imagine that I can use the following formula to estimate $\omega (t)$:

$\omega = 2 \frac{dq}{dt} * \hat{q} $,

where $\hat{q}$ is the inverse of $q$. What is the formula for computing $\alpha (t)$?

crow
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1 Answers1

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So you know the quaternion's first derivative $$ \dot q = \frac{1}{2} \omega q $$ which is how you got to your equation $$ \omega = 2 \dot q \hat q $$ The quaternion's second derivative is $$ \ddot q = \frac{1}{2}(\dot \omega q + \omega \dot q) $$ and substituting the first derivative above gives $$ \ddot q = \frac{1}{2}(\dot \omega q + \omega \frac{1}{2} \omega q) $$ $$ \ddot q = \frac{1}{2} \dot \omega q + \frac{1}{4} \omega \omega q $$ Which you can use to get the acceleration $\dot \omega$ $$ \dot \omega = 2 (\ddot q \hat q - (\dot q \hat q )^2) $$

Carser
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  • Thanks for the derivation. Regarding the quantity $(\dot{q}\hat{q})^2$ do I have to use the exponential property of quaternions? – crow May 24 '16 at 07:10
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    This requires you to know $\dot{q}$ in order to find $\omega$ though. How do you get the quantity $\dot{q}$ from the sequence of quaternions? – adamconkey Sep 27 '17 at 17:23
  • Also, $\omega$ is 3-dimensional. You can interpret it as a quaternion with 0 scalar part, but how do you enforce that it gets 0 scalar part when you compute it in this way as the scaled product of two quaternions? – adamconkey Sep 27 '17 at 20:23
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    I estimated $\dot{q}, \ddot{q}$ by fitting quaternion splines and evaluating the derivatives. – crow Sep 29 '17 at 13:54