Quaternion and Euler angles small angle proof

Question

Let's start with a quaternion $q = \begin{bmatrix} q1 & q2 & q3 & q4 \end{bmatrix}^T$. Where $q_4$ is the scalar part, which is equal to:

\begin{equation} q_4 = cos(\frac{\alpha}{2}) \end{equation}

where $\alpha$ is the rotation angle around Euler's eigenaxis.

Now if we have a 3-2-1 Euler rotation sequence (with angles $\psi$, $\theta$ and $\phi$), the transformation from Euler angles to quaternions is as follows:

\begin{equation} \begin{bmatrix} q_1 \\[1em] q_2 \\[1em] q_3 \\[1em] q_4 \end{bmatrix} = \begin{bmatrix} \text{sin}\frac{\phi}{2} \text{cos}\frac{\theta}{2} \text{cos}\frac{\psi}{2} - \text{cos}\frac{\phi}{2} \text{sin}\frac{\theta}{2} \text{sin}\frac{\psi}{2}\\[1em] \text{cos}\frac{\phi}{2} \text{sin}\frac{\theta}{2} \text{cos}\frac{\psi}{2} + \text{sin}\frac{\phi}{2} \text{cos}\frac{\theta}{2} \text{sin}\frac{\psi}{2}\\[1em] \text{cos}\frac{\phi}{2} \text{cos}\frac{\theta}{2} \text{sin}\frac{\psi}{2} - \text{sin}\frac{\phi}{2} \text{sin}\frac{\theta}{2} \text{cos}\frac{\psi}{2}\\[1em] \text{cos}\frac{\phi}{2} \text{cos}\frac{\theta}{2} \text{cos}\frac{\psi}{2} + \text{sin}\frac{\phi}{2} \text{sin}\frac{\theta}{2} \text{sin}\frac{\psi}{2} \end{bmatrix} \end{equation}

I've tested numerically that for small Euler angles ($\psi$, $\theta$, $\phi$), that $\alpha = \sqrt{\psi^2+\theta^2+\phi^2}$. I've also tried to do it algebraically but I always seem to get stuck.

Can someone help me with an algebraic proof?

Answer 1

What are you trying to prove? Every representation of the attitude is equivalent to a parameterization of $SO_3$ whose Lie algebra is the "cross product matrix" $$\begin{bmatrix} 0 & -x_3 & x_2\\ x_3 & 0 &-x_1 \\ -x_2 & x_1 & 0\\ \end{bmatrix} $$ so you better get that (plus ones on the diagonal) when you substitute the linearized attitude representation into the DCM. For the quaternions you can just use the Euler symmetric parameters, like you were, to find $q \mapsto [1,\mathbf{e}]$ so you just use the DCM as a function of the quaternions and eliminate all the terms that look like $e_ie_j$. For the Euler angles you can just linearize all of the sines and cosines in the 321 sequence and get the same cross product matrix with the right angles in the right places (for a 321 sequence), then read off what Euler angles have to correspond to what linearized quaternion components.

Alternatively, if you never want to deal with DCMs (I don't blame you) parameterize the quaternions with the Euler angles using $$q = e^{i\phi}e^{j\theta}e^{k\psi}$$ and linearize the exponentials individually. The results should be the same.

Answer 2

I've derived the proof, it was not that hard:

We know that for small angles $\phi$, $\theta$, and $\psi$ that the quaternion looks like:

\begin{equation} [q_1~~q_2~~q_3~~q_4]^T = [\phi/2~~\theta/2~~\psi/2~~1]^T \end{equation}

Also for small angles $\alpha$:

\begin{equation} \begin{bmatrix} \mathbf{q} \\ q_4\end{bmatrix} = \begin{bmatrix} \hat{\mathbf{e}} \sin \frac{\alpha}{2}\\ \cos \frac{\alpha}{2} \end{bmatrix} \approx \begin{bmatrix} \hat{\mathbf{e}} \frac{\alpha}{2} \\ 1\end{bmatrix} \end{equation}

The norm of the vector part of the quaternion shows us:

\begin{equation} ||2\mathbf{q}|| = ||\alpha \hat{\mathbf{e}}|| = \alpha \end{equation}

(because the unit vector has norm 1)

Combining this with the previous result, gives:

\begin{equation} ||2 \mathbf{q} || = \sqrt{\phi^2 +\theta^2+\psi^2} \end{equation}

Thus $\alpha = \sqrt{\phi^2 +\theta^2+\psi^2}$