# Derivation for angular acceleration from quaternion profile

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

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

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

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.

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

## 1 Answer

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:

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