# Quaternion and Euler angles small angle proof

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:

$$q_4 = cos(\frac{\alpha}{2})$$

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:

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

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?

• "Rotation calculation via quaternions has come to replace the use of direction cosines in aerospace applications through their reduction of the required calculations, and their ability to minimize round-off errors. Also, in computer graphics the ability to perform spherical interpolation between quaternions with relative ease is of value" (From Euler's rotation theorem, in Wikipedia). I fear your question would be not very comfortable to answer. – Piquito Jul 8 '15 at 16:55
• This answer is not very helpful – MichaelDeSanta Jul 8 '15 at 18:46
• Agreed, Michael. Piquito is obviously not a working aerospace engineer! (Euler angles show up ALL THE TIME!) – Mortified Through Math Jun 4 '16 at 0:33

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.

• I just wanted to prove that for small angles, the rotation angle around the Euler axis is a good representation for the combined Euler angles. For example when presenting small errors in control systems, I want to capture the error in 1 parameter, not in 3. – MichaelDeSanta Jun 5 '16 at 8:37

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:

$$[q_1~~q_2~~q_3~~q_4]^T = [\phi/2~~\theta/2~~\psi/2~~1]^T$$

Also for small angles $\alpha$:

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

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

$$||2\mathbf{q}|| = ||\alpha \hat{\mathbf{e}}|| = \alpha$$

(because the unit vector has norm 1)

Combining this with the previous result, gives:

$$||2 \mathbf{q} || = \sqrt{\phi^2 +\theta^2+\psi^2}$$

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

• All of this is certainly true and follows trivially from the understanding Lie theoretic ideas behind parameterizing attitude ($SO_3$). Essentially your (correct) argument is since the Lie algebra of a Lie group is not only unique but also uniquely parameterized, each parameterization of the attitude must be simply related in the linear limit. You have then inferred this relation for Euler angles and the axis-angle representation. – Mortified Through Math Jun 5 '16 at 19:26