Given a profile of unit quaternions $q(t)$ that represents the orientation of a body over time, I like to get the angular acceleration $\dot \omega (t)$. I tried to find a formula myself, but I get two different results for two different derivations.

Derivation 1

Starting from the kinematic equations


$$\dot{\omega}=2(\ddot{q}\hat{q}+\dot q \hat{\dot{q}})$$

And using the property of conjugate quaternions:

$$\dot\omega=2\ddot q \hat q$$

Derivation 2

Again, taking the kinematic equation

$$\dot q = \frac 1 2 \omega q$$

But now taking the derivative of $q$ w.r.t. $t$ gives

$$ \ddot q=\frac 1 2 (\dot \omega q+\omega\dot q) $$

After substituting the formula for $\omega$

$$ \ddot q=\frac{1}{2} (\dot \omega q+2 \dot q\hat q \dot q)$$

Finally, the result after multiplying both sided by $\hat q$ and rearranging is

$$\dot \omega = 2\ddot q\hat q -2(\dot q \hat q)^2$$

Which of these derivations is correct, and why? I have already consulted several sources, where I seem to find both formulas. For example here for derivation 1 and here for derivation 2. I'm new to using quaternions, so maybe I'm missing some mathematical concept.

  • 1
    $\begingroup$ I think this question would be much more appropriate for Physics Stack Exchange. $\endgroup$ – Wojowu Feb 12 at 10:55
  • $\begingroup$ Thanks, I will do. However, I found more useful related posts on Mathematics stack exchange. $\endgroup$ – jochim Feb 12 at 11:01

I am not a physicist but, to my understanding, both derivations are equal.

In your first derivation you get:

$\dot\omega = 2\ddot q \hat q + 2 \dot q \hat{\dot q} $

In your second derivation you get the same. Departing from the unit quaternion condition:

$q \hat q = 1$

$0 = \frac{d}{dt}(q \hat q) = \dot q \hat q + q \hat{\dot q}$

We get the identity:

$\dot q \hat q = - q \hat{\dot q}$

So the square $(\dot q\hat q)^2 = - \dot q \hat{\dot q}$ because:

$(\dot q\hat q)^2 = \dot q\hat q (\dot q \hat q) = -\dot q \hat q q \hat{\dot q} = - \dot q \hat{\dot q}$

And you get again:

$\dot\omega = 2\ddot q \hat q + 2 \dot q \hat{\dot q} $


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