Euler Parameters

1) Preface

In geometry, Euler's rotation theorem states that, in three-dimensional space, any displacement of a rigid body such that a point on the rigid body remains fixed, is equivalent to a single rotation about some axis that runs through the fixed point. It also means that the composition of two rotations is also a rotation. The axis of rotation is known as an Euler axis (source Wikipedia)

So, according to the above mentioned Euler's rotation theorem, any 3D rotation (or sequence of rotations) can be specified using two parameters: a unit vector that defines an axis of rotation; and an angle μ describing the magnitude of the rotation about that axis, rotation to be considered positive according to the right hand rule.

Let’s assume now to have a fixed reference system XYZ, and consider a versor V (our rotation axis) forming with the three axis, the three angles α β and γ.

Suppose to have an additional reference system (X’Y’Z’) in origin coincident with XYZ.

We want to apply a rotation of an angle μ to X’Y’Z’ around the vector V, and to calculate the relative transformation matrix. The scope is very simple: have an instrument to calculate the attitude (the angular position) of an object (an aircraft) once the rotation is applied. In fact we will learn that knowing this transformation matrix, allow us to know the asset of the aircraft.

The problem could seem very complex, but we can reduce the difficulties with some smart ideas.

If we assume to be able to rotate the X’Y’Z’ reference system (by means of the transformation matrix A) in such e way that the X’ axis is coincident with the vector V, we would be in a good position. From now on it would be sufficient to make a rotation around the X’ axis (by means of the transformation matrix R) to have the rotation around V: note that this transformation would be very simple. In order to obtain the our final result would be now sufficient to apply the inverse transformation by means of the matrix A^-1

2) The transformation matrix

Now let’s discover the matrix A but firstly let’s choose the way to superimpose X’ to V:

1. first rotate X’Y’Z’ reference system around the Z≡Z’ axis until the plane formed by V and X’ becomes perpendicular the XY≡X’Y’ plane
2. second rotate the X’Y’Z’ reference system around the Y’ axis until X’ is coincident with V.

It’s to be noted that in this way the Y’ axis is restricted on the XY plane, and being the X’ axis superimposed to V, the Z’ axis is perpendicular to V.

The generic orthogonal transformation matrix have the form:

$A=\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right)$

but from matrix theory we know that:

• the rows represent the unit vector of each axis of the new reference system respect to the old one
• each row represents a unit vector perpendicular to the unit vectors represented by the other rows and their components are the director cosines of the unit vector
• each column represents a unit vector perpendicular to the unit vector represented by the other columns

As consequence of the above properties we have:

$a_{11}=\cos (\alpha )$    $a_{12}=\cos (\beta )$    $a_{13}=\cos (\gamma )$

The fact that Y’ lies on the XY plane implies that $a_{23}=0$ (the third component of its unit vector is null)

Considering the last column we can write:   $a_{13}^2+a_{23}^2+a_{33}^2=1$ and

$a_{33}^2+{\cos }^2(\gamma )=1$  which leads to   $a_{33}=\pm \sin (\gamma )$

The resulting matrix becomes

$A=\left(\begin{array}{ccc}\cos \alpha & \cos \beta & \cos \gamma \\ a_{21}& a_{22}& 0\\ a_{31}& a_{32}& \pm \sin \gamma \end{array}\right)$

Let’s now multiply first the column by the last column

$\cos (\alpha )\ast \cos (\gamma )\pm a_{31}\ast \sin (\gamma )=0$ which leads to $a_{31}=\mp \cos \alpha \ast \cot \gamma$

redoing the same operation between the second and last columns

$\cos \beta \ast \cos \gamma \pm a_{32}\ast \sin \gamma =0$   which leads to   $a_{32}=\mp \cos \beta \ast \mathit{cotg}\gamma$

The matrix A now becomes

$A=\left(\begin{array}{ccc}\cos \alpha & \cos \beta & \cos \gamma \\ a_{21}& a_{22}& 0\\ \mp \cos \alpha \ast \mathit{cotg}\gamma & \mp \cos \beta \ast \mathit{cotg}\gamma & \pm \sin \gamma \end{array}\right)$

We need now to compute the last two members, $a_{21}$ and $a_{22}$ .

One simple relation comes considering that the second row is a vector unit, that is:

$a_{21}^2+a_{22}^2=1$

considering the first and second rows they represent two perpendicular unity vector and thus

$a_{21}\ast \cos (\alpha )+a_{22}\ast \cos (\beta )=0$  extracting a21 from this relation and substituting it in the above

$a_{22}^2+\frac{{\cos }^2(\beta )}{{\cos }^2(\alpha )}\ast a_{22}^2=1$    $a_{22}^2(1+\frac{{\cos }^2(\beta )}{{\cos }^2(\alpha )})=1$

$a_{22}^2\frac{({\cos }^2\beta +{\cos }^2\alpha )}{{\cos }^2\alpha }=1$   consider that from row 1 of matrix A

${\cos }^2\beta +{\cos }^2\alpha =1-{\cos }^2\gamma ={\sin }^2\gamma$ and thus  $a_{22}^2\ast \frac{{\sin }^2\gamma }{{\cos }^2\alpha }=1$

Finally   $a_{22}=\pm \cos \alpha \ast \mathit{csc}\gamma$

similar considerations lead to

$a_{21}=\mp \cos \beta \ast \mathit{csc}\gamma$

The sign ambiguities can be solved adding the requirements that for α = 0 the matrix reduces to an identity matrix

The matrix A takes now the final form

$A=\left(\begin{array}{ccc}\cos \alpha & \cos \beta & \cos \gamma \\ -\cos \beta \ast \mathit{csc}\gamma & \cos \alpha \ast \mathit{csc}\gamma & 0\\ -\cos \alpha \ast \mathit{cotg}\gamma & -\cos \beta \ast \mathit{cotg}\gamma & \sin \gamma \end{array}\right)$

Let’s now consider, as stated at the beginning, the rotation of an angle μ around the X’ axis:

the corresponding matrix R takes the very easy form:

R = $\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \mu & \sin \mu \\ 0& -\sin \mu & \cos \mu \end{array}\right)$

The matrix A^-1 to remove the effect of matrix A, is very simple too.

Considering Matrix theory, being matrix A unitary and orthogonal, the inverse of A^-1 is simply the transposed of A, that means that the rows of A^-1 are the columns of A, i.e.:

A^-1 = $\left(\begin{array}{ccc}\cos \alpha & -\cos \beta \ast \mathit{csc}\gamma & -\cos \alpha \ast \mathit{cotg}\gamma \\ \cos \beta & \cos \alpha \ast \mathit{csc}\gamma & -\cos \beta \ast \mathit{cotg}\gamma \\ \cos \gamma & 0& \sin \gamma \end{array}\right)$

Now we have all the components to find the matrix T = A^-1 R A describing the rotation of an angle μ around the vector V of the reference system X’Y’Z’ (we omit the tedious math transformations):

T = $\left(\begin{array}{ccc}1-2{\sin }^2\frac{\mu }{2}{\sin }^2\alpha & 2({\sin }^2\frac{\mu }{2}\cos \alpha \cos \beta +\sin \frac{\mu }{2}\cos \frac{\mu }{2}\cos \gamma )& 2({\sin }^2\frac{\mu }{2}\cos \alpha \cos \gamma -\sin \frac{\mu }{2}\cos \frac{\mu }{2}\cos \beta )\\ 2({\sin }^2\frac{\mu }{2}\cos \alpha \cos \beta -\sin \frac{\mu }{2}\cos \frac{\mu }{2}\cos \gamma )& 1-2{\sin }^2\frac{\mu }{2}{\sin }^2\beta & 2({\sin }^2\frac{\mu }{2}\cos \beta \cos \gamma +\sin \frac{\mu }{2}\cos \frac{\mu }{2}\cos \alpha )\\ 2({\sin }^2\frac{\mu }{2}\cos \alpha \cos \gamma +\sin \frac{\mu }{2}\cos \frac{\mu }{2}\cos \beta )& 2({\sin }^2\frac{\mu }{2}\cos \beta \cos \gamma -\sin \frac{\mu }{2}\cos \frac{\mu }{2}\cos \alpha )& 1-2{\sin }^2\frac{\mu }{2}{\sin }^2\gamma \end{array}\right)$ (1)

The matrix (1) is extremely important and even if could seem strange or too complicated, has its logic. Let’s try to discover the logic by making the following substitution:

$a_1=\cos \frac{\mu }{2}$

$a_2=\cos \alpha \sin \frac{\mu }{2}$

$a_3=\cos \beta \sin \frac{\mu }{2}$

$a_4=\cos \gamma \sin \frac{\mu }{2}$

These four parameters are called the Euler parameters. It may be seen from their definition that they obey the relationship:

$a_1^2+a_2^2+a_3^2+a_4^2=1$

We can say that Euler in the 18^th century anticipated what Hamilton formulated precisely in the 19^th century: the quaternions. We will introduce quaternions in another post, showing later how to find their components, and from the components how to derive the attitude angles.

Matrix T takes now the form:

T = $\left(\begin{array}{ccc}{a}_1^2+a_2^2-a_3^2-a_4^2& 2(a_1{a}_4+a_2{a}_3)& 2(a_2{a}_4-a_1{a}_3)\\ 2(a_2{a}_3-a_1{a}_4)& a_1^2-a_2^2+a_3^2-a_4^2& 2(a_3{a}_4+a_1{a}_2)\\ 2(a_2{a}_4+a_1{a}_3)& 2(a_3{a}_4-a_1{a}_2)& a_1^2-a_2^2-a_3^2+a_4^2\end{array}\right)$ (2)

This means that if I have a vector $\overline{A}$ whose coordinates on the reference system X, Y, Z are $\left(x_a,y_a,z_a\right)$ , the same vector in the reference system X’, Y’, Z’ will have the coordinates

$\left(\begin{array}{c}x{\text{'}}_a\\ y{\text{'}}_a\\ z{\text{'}}_a\end{array}\right)=T\left(\begin{array}{c}{x}_a\\ y_a\\ z_a\end{array}\right)$    3)

This result is very important and will be used in the following findings