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Advanced Engineering Dynamics

R. Valéry Roy

Section 1.6 Orientation of a Rigid Body: Quaternions

Quaternions are to spatial rotations what complex numbers are to planar rotations. The set of quaternions  1 , denoted as \(\mathbb{H}\text{,}\) is the set of elements \(Q\) defined as
\begin{equation*} Q = q_0 + \bq = q_0 + q_1 \be_1 + q_2 \be_2 + q_3 \be_3 \end{equation*}
given four real numbers \(q_0, q_1, q_2 , q_3\) and a right-handed basis \((\be_1,\be_2,\be_3)\) of orthonormal vectors of a referential \(\cE\text{.}\) Hence a quaternion \(Q\in \mathbb{H}\) is defined by its scalar part \(q_0\) and its vector part \(\text{Vect}(Q)= \bq = q_1 \be_1 + q_2 \be_2 + q_3 \be_3\text{.}\) A pure quaternion \(Q = 0 +\bq\) is a quaternion whose scalar part is zero. The set of pure quaternions forms a subset of \(\mathbb{H}\) denoted \(\mathbb{H}_0\text{.}\)
Quaternion algebra is governed by the following fundamental rules:
\begin{equation} \begin{array}{c} \be_1^2 = \be_2^2 = \be_3^2 = \be_1\be_2\be_3 = -1 \\ \be_3 = \be_1\be_2 = - \be_2\be_1 \\ \be_2= \be_3\be_1 = - \be_1\be_3 \\ \be_1 = \be_2\be_3 = -\be_3\be_2 \\ \end{array}\tag{1.6.1} \end{equation}
known as Hamilton rules. Quaternions form a 4-dimensional vector space over \(\mathbb{R}\text{:}\) given two elements \(Q, Q'\) of \(\mathbb{H}\) and a real scalar \(\lambda\text{,}\) addition and scalar multiplication can be defined as
\begin{equation*} Q + Q' = (q_0 + q'_0 ) + (q_1 + q'_1 )\be_1 + (q_2 + q'_2 )\be_2 + (q_3 + q'_3)\be_3 \end{equation*}
\begin{equation*} \lambda Q = \la q_0 + \la q_1 \be_1 + \la q_2 \be_2 + \la q_3 \be_3. \end{equation*}
Quaternions can be multiplied according to the rules (1.6.1). It can be shown that the product \(QQ' = (q_0 + \bq)(q'_0 + \bq')\) of two quaternions is given by
\begin{equation} QQ' = q_0 q'_0 - \bq \cdot \bq' + q_0 \bq' + q'_0 \bq + \bq \times \bq'\tag{1.6.2} \end{equation}

Remark 1.6.1.

An equivalent way to define quaternions is as elements \((q_0,q_1,q_2,q_3)\) of \(\mathbb{R}^4\) with the following addition and multiplication rules
\begin{equation*} (q_0,q_1,q_2,q_3)+ (q^{\prime}_0,q^{\prime}_1,q^{\prime}_2,q^{\prime}_3)= (q_0+q^{\prime}_0,q_1+q^{\prime}_1,q_2+q^{\prime}_2,q_3+q^{\prime}_3) \end{equation*}
\begin{align*} (q_0,q_1,q_2,q_3)\times (q^{\prime}_0,q^{\prime}_1,q^{\prime}_2,q^{\prime}_3)\amp = (q_0q^{\prime}_0 -q_1q^{\prime}_1-q_2q^{\prime}_2-q_3 q^{\prime}_3,\\ \amp\qquad q_0q^{\prime}_1+ q_1 q^{\prime}_0+q_2q^{\prime}_3 - q_3 q^{\prime}_2,\\ \amp\qquad q_0 q^{\prime}_2 + q_2 q^{\prime}_0 − q_1q^{\prime}_3 + q_3 q^{\prime}_1,\\ \amp\qquad q_3 q^{\prime}_0 + q_0 q^{\prime}_3 + q_1q^{\prime}_2 − q_2 q^{\prime}_1) \end{align*}
Note that the multiplication of quaternions is not commutative: formula (1.6.2) shows that in general \(QQ' \neq Q'Q\text{.}\) Also note that the product of two pure quaternions \(Q = 0+\bq\) and \(Q'=0+\bq'\) of \(\mathbb{H}_0\) is not a pure quaternion:
\begin{equation*} QQ' = (0+\bq)(0+\bq') = -\bq \cdot \bq' + \bq\times\bq' \end{equation*}
The inverse of a quaternion can be defined by first defining the conjugate \(\conjQ\) of a quaternion \(Q\text{:}\)
\begin{equation*} \conjQ = \overline{q_0 + \bq} = q_0 - \bq = q_0 - q_1 \be_1 - q_2 \be_2 - q_3 \be_3 \end{equation*}
Note that for \(Q\in\mathbb{H}_0\text{,}\) \(\conjQ = - Q\) and that \(\overline{Q Q'} = \conjQ' \conjQ\) for any two arbitrary quaternions \(Q\) and \(Q'\text{.}\) The norm \(|Q|\) of a quaternion \(Q\) is then defined by \((Q\conjQ)^{1/2}=(\conjQ Q)^{1/2}\text{.}\) Formula (1.6.2) shows that this expression has a meaning since
\begin{equation*} |Q|^2 = Q\conjQ = \conjQ Q= q_0^2 + q_1^2 + q_2^2 + q_3^2 \end{equation*}
Then the inverse of a non-zero quaternion \(Q\) is the quaternion \(Q^{-1}\) defined as \(Q Q^{-1}= Q^{-1} Q =1\text{:}\) it can be shown that \(Q^{-1}\) is given by
\begin{equation*} Q^{-1} = \frac{\conjQ}{|Q|^2} \end{equation*}
A unit quaternion is a quaternion \(Q\) of norm \(|Q| =1\text{.}\) It can be readily shown that all unit quaternions can be written in the form
\begin{equation*} Q = \cos\theta + \sin\theta \, \bu \end{equation*}
where angle \(\theta\) satisfies \(0 \leq \theta \lt 2\pi\) and \(\bu\) is a unit vector. If \(Q\) is a unit quaternion, its inverse is given by \(Q^{-1}= \conjQ\text{.}\)

Remark 1.6.2.

Unit quaternions \(Q = \cos\theta + \sin\theta \, \bu\) can be written in the form \(Q= e^{\theta \bu}\) where the exponential \(e^Q\) of quaternion \(Q\) is formally defined as the series \(1+ Q+ Q^2/2 + \cdots Q^n/n! + \cdots\text{.}\) See Exercise 1.8.13.

Example 1.6.3.

a. Given two quaternions \(Q_1\) and \(Q_2\text{,}\) show that \(Q_1 Q_2 -Q_2 Q_1 = 2 \bq_1 \times \bq_2\text{.}\)
b. Given two pure quaternions \(Q_1 = 0+\bq_1\) and \(q_2 = 0+ \bq_2\text{,}\) show that
\begin{equation*} Q_1 Q_2 +Q_2 Q_1 = -2 \bq_1 \cdot \bq_2 \end{equation*}
Solution.
a.According to (1.6.2) we can write
\begin{equation*} Q_1 Q_2 = q_{10} q_{20} - \bq_1 \cdot \bq_2 + q_{10} \bq_2 + q_{20} \bq_1 + \bq_1 \times \bq_2 \end{equation*}
\begin{equation*} Q_2 Q_1 = q_{10} q_{20} - \bq_1 \cdot \bq_2 + q_{10} \bq_2 + q_{20} \bq_1 + \bq_2 \times \bq_1 \end{equation*}
which leads to
\begin{equation*} Q_1 Q_2 - Q_2 Q_1 = 2 \bq_1 \times \bq_2 \end{equation*}
b. For two pure quaternions we have
\begin{equation*} Q_1 Q_2 =- \bq_1 \cdot \bq_2 + \bq_1 \times \bq_2 , \qquad Q_2 Q_1 =- \bq_1 \cdot \bq_2 + \bq_2 \times \bq_1 \end{equation*}
leading to
\begin{equation*} Q_1 Q_2 + Q_2 Q_1 = -2 \bq_1 \cdot \bq_2 \end{equation*}
We see in particular that the square of a pure quaternion \(Q=0+\bvv\) is the scalar \(Q^2 = - \bvv\cdot\bvv = -\bvv^2\text{.}\)
The usefulness of quaternions stems from their connection to rotations as shown by the following theorem.

Proof.

To show this, we first find that for any \(Q\in\mathbb{H}\) and any \(\bV\in\mathbb{H}_0 \) we have
\begin{equation*} Q\bV \conjQ = (q_0^2 - \bq^2) \bV + 2 (\bq\cdot\bV)\bq + 2 q_0 \bq\times\bV \end{equation*}
by two consecutive uses of (1.6.2). Hence \(Q\bV \conjQ\) is an element of \(\mathbb{H}_0\text{.}\) Then with \(Q = \cos\frac{\al}{2} + \sin\frac{\al}{2} \, \bu\) we find
\begin{equation*} \cR_Q (\bV) = \cos\al \bV + (1-\cos\al) (\bu\cdot\bV)\bu + \sin\al \bu\times \bV \end{equation*}
Upon using the identity \(\bu\times(\bu\times\bV) = (\bu\cdot\bV)\bu -\bV\) we find
\begin{equation*} \cR_Q (\bV) = \bV + \sin\al \bu\times\bV + (1-\cos\al) \bu\times(\bu\times\bV) \end{equation*}
This is exactly the expression of the mapping of vector \(\bV\) by rotation \(\cR_{\al ,\bu}\) as given by Rodrigues formula (1.5.2).

Remark 1.6.5.

The correspondence between unit quaternions and rotations is not one-to-one since both \(Q\) and \(-Q\) correspond to the same rotation.
Hence rotations of vectors can be performed by simple quaternion multiplications as illustrated by the following example.

Example 1.6.6.

Find the mapping of \(\bV = \be_3\) by the rotation of angle \(\al =\pi/3\) about \(\bu = (\be_1+\be_2+\be_3)/\sqrt{3}\text{.}\)
Solution.
First we define the quaternion \(Q= \cos\frac{\al}{2} + \sin\frac{\al}{2} \, \bu\)
\begin{equation*} Q = \frac{\sqrt{3}}{2} + \frac{\be_1+\be_2+\be_3}{2\sqrt{3}} \end{equation*}
Then \(\cR_Q (\bV)\) is given as the product \(Q \bV \conjQ\text{:}\)
\begin{align*} \cR_Q (\bV) \amp = (\frac{\sqrt{3}}{2} + \frac{\be_1+\be_2+\be_3}{2\sqrt{3}} ) \be_3 (\frac{\sqrt{3}}{2} - \frac{\be_1+\be_2+\be_3}{2\sqrt{3}} )\\ \amp = \frac{\sqrt{3}}{6}(-1+\be_1-\be_2+3 \be_3 )(\frac{\sqrt{3}}{2} - \frac{\be_1+\be_2+\be_3}{2\sqrt{3}} ) \\ \amp =\frac{2}{3}\be_1 -\frac{1}{3}\be_2 +\frac{2}{3}\be_3 \end{align*}
This property of quaternions is illustrated in the following examples.

Example 1.6.8.

Consider the sequence of two rotations \(\cR_{\al_1 , \be_3}\) and \(\cR_{\al_2 , \be_3}\text{:}\)
\begin{equation*} \bV \xrightarrow{\cR_{\al_1 , \be_3}} \bV_1 \xrightarrow{\cR_{\al_2 , \be_3}} \bV_2 \end{equation*}
By using quaternions, find the equivalent rotation which maps \(\bV\) to \(\bV_2\text{.}\)
Solution.
We first introduce the unit quaternions \(Q_1 = \cos\frac{\al_1}{2} + \sin\frac{\al_1}{2} \be_3\) and \(Q_2 = \cos\frac{\al_2}{2} + \sin\frac{\al_2}{2} \be_3\text{.}\) Then \(\bV_2 = \cR_Q (\bV) \) with \(Q\) given by
\begin{align*} Q \amp = Q_2 Q_1 = (\cos\frac{\al_1}{2} + \sin\frac{\al_1}{2} \be_3)(\cos\frac{\al_2}{2} + \sin\frac{\al_2}{2} \be_3)\\ \amp = \cos\frac{\al_1}{2}\cos\frac{\al_2}{2} - \sin\frac{\al_1}{2} \sin\frac{\al_2}{2} + (\cos\frac{\al_1}{2}\sin\frac{\al_2}{2}+\cos\frac{\al_2}{2} \sin\frac{\al_1}{2})\be_3\\ \amp = \cos\frac{\al_1+\al_2}{2} + \sin\frac{\al_1+\al_2}{2}\be_3 \end{align*}
As expected this corresponds to the rotation \(\cR_{\al_1+\al_2 , \be_3}\text{.}\)

Example 1.6.9.

Consider the Euler sequence of three rotations \(\cR_{\psi , \be_3}\) , \(\cR_{\te , \bu_1}\) and \(\cR_{\phi , \bhb_3}\) which maps a basis \((\be_1,\be_2,\be_3)\) of \(\cE\) to \((\bhb_1,\bhb_1,\bhb_3)\) of a rigid body \(\cB\text{:}\)
\begin{equation*} (\be_1,\be_2,\be_3) \xrightarrow{\cR_{\psi , \be_3}} (\bu_1 ,\bu_2,\be_3) \xrightarrow{\cR_{\te , \bu_1}} (\bu_1,\bv_2,\bz_3) \xrightarrow{\cR_{\phi , \bhb_3}} (\bhb_1,\bhb_2,\bhb_3) \end{equation*}
Find the equivalent rotation using quaternions.
Solution.
The three unit quaternions associated with \(\cR_{\psi , \be_3}\text{,}\) \(\cR_{\te , \bu_1}\) and \(\cR_{\phi , \bhb_3}\) are
\begin{equation*} Q_\psi = \cos\frac{\psi}{2} + \sin\frac{\psi}{2} \be_3, \quad Q_\te = \cos\frac{\te}{2} + \sin\frac{\te}{2} \bu_1,\quad Q_\phi = \cos\frac{\phi}{2} + \sin\frac{\phi}{2} \bhb_3 \end{equation*}
In order to find the product \(Q_\phi Q_\te Q_\psi\text{,}\) the three quaternions must be expressed on the same basis. It is most convenient to use basis \((\bu_1 ,\bu_2,\be_3)\text{,}\) then we must express \(Q_\phi\) as
\begin{equation*} Q_\phi = \cos\frac{\phi}{2} + \sin\frac{\phi}{2} (\cos\te \be_3 -\sin\te \bu_2) \end{equation*}
First, we find the product \(Q_\te Q_\psi\) using (1.6.2)
\begin{align*} Q_\te Q_\psi \amp = (\cos\frac{\te}{2} + \sin\frac{\te}{2} \bu_1)(\cos\frac{\psi}{2} + \sin\frac{\psi}{2} \be_3)\\ \amp = \cos\frac{\psi}{2}\cos\frac{\te}{2} + \cos\frac{\psi}{2}\sin\frac{\te}{2} \bu_1 - \sin\frac{\psi}{2}\sin\frac{\te}{2} \bu_2 + \sin\frac{\psi}{2}\cos\frac{\te}{2} \be_3 \end{align*}
Then we find the product \(Q_\phi Q_\te Q_\psi\)
\begin{align*} Q_\phi Q_\te Q_\psi \amp = c_{\phi /2} c_{\psi/2}c_{\te/2} - s_{\psi/2} s_{\phi/2} s_\te s_{\te/2} - s_{\psi/2} s_{\psi/2} c_\te c_{\te/2} \\ \amp \qquad +(c_{\psi/2}s_{\te/2} c_{\psi/2} - s_{\psi/2}s_{\phi/2} s_\te c_{\te/2} + s_{\psi/2} s_{\psi/2} c_\te s_{\te/2})\bu_1 \\ \amp \qquad -(s_{\psi/2}s_{\te/2} c_{\phi/2} + c_{\psi/2}s_{\phi/2} s_\te c_{\te/2} - c_{\psi/2} s_{\phi/2} c_\te s_{\te/2})\bu_2 \\ \amp \qquad +(s_{\psi/2}c_{\te/2} c_{\phi/2} + c_{\psi/2}s_{\phi/2} c_\te c_{\te/2} + c_{\psi/2} s_{\phi/2} s_\te s_{\te/2})\be_3 \end{align*}
using the notation \(c_\te = \cos\te\text{,}\) \(s_\te = \sin\te\text{,}\) etc. After simplifications, this gives
\begin{equation*} Q_\phi Q_\te Q_\psi = c_{\te/2} c_{(\psi+\phi)/2} + s_{\te/2} c_{(\psi+\phi)/2} \bu_1 - s_{\te/2} s_{(\psi+\phi)/2} \bu_2 + c_{\te/2} s_{(\psi+\phi)/2} \be_3 \end{equation*}
We can then express \(Q_\phi Q_\te Q_\psi\) on basis \((\be_1,\be_2,\be_3)\) to find
\begin{align*} Q_\phi Q_\te Q_\psi \amp = \cos\frac{\te}{2} \cos \left(\frac{\psi+\phi}{2}\right) +\sin\frac{\te}{2} \cos \left(\frac{\psi-\phi}{2}\right) \be_1 + \sin\frac{\te}{2} \sin \left(\frac{\psi-\phi}{2}\right) \be_2 \\ \amp \qquad +\cos\frac{\te}{2} \sin \left(\frac{\psi+\phi}{2}\right) \be_3 \end{align*}
We can equate \(Q_\phi Q_\te Q_\psi\) to unit quaternion \(\cos\frac{\alpha}{2} + \sin\frac{\alpha}{2} \widehat{\bU}\) where angle \(\alpha\) and unit vector \(\widehat{\bU}\) are given by
\begin{equation*} \cos\frac{\alpha}{2} =\cos\frac{\te}{2} \cos \left(\frac{\psi+\phi}{2}\right) \end{equation*}
\begin{equation*} \sin\frac{\alpha}{2} \widehat{\bU} =\sin\frac{\te}{2} \cos \left(\frac{\psi-\phi}{2}\right) \be_1 + \sin\frac{\te}{2} \sin \left(\frac{\psi-\phi}{2}\right) \be_2 +\cos\frac{\te}{2} \sin \left(\frac{\psi+\phi}{2}\right) \be_3 \end{equation*}
The sequence of three rotations \(\cR_{\psi , \be_3}\text{,}\) \(\cR_{\te , \bu_1}\) and \(\cR_{\phi , \bhb_3}\) is then equivalent to the rotation \(\cR_{\alpha ,\widehat{\bU}}\text{.}\)
Quaternions are also helpful to derive interesting properties of rotations. The following result relates to the so-called rotation vector of a rotation \(\cR_{\al,\bu}\) defined as the vector \(\rot = \tan\frac{\al}{2} \, \bu \text{.}\)

Proof.

Rotation \(\cR_{\gamma , \bw}\) corresponds to quaternion \(Q_w\) given by
\begin{align*} Q_w \amp = Q_v Q_u = (c_{{\beta}/{2}} + s_{{\beta}/{2}} \bv)(c_{{\al}/{2}} + s_{{\al}/{2}} \bu)\\ \amp = c_{{\beta}/{2}}c_{{\al}/{2}} -s_{{\beta}/{2}}s_{{\al}/{2}} (\bu\cdot\bv) + c_{{\beta}/{2}}s_{{\al}/{2}} \bu + c_{{\al}/{2}}s_{{\beta}/{2}} \bv + s_{{\beta}/{2}}s_{{\al}/{2}} \bv \times \bu \end{align*}
We then equate the real parts and the vector parts to obtain two equations:
\begin{equation*} \cos{\frac{\ga}{2}} = \cos{\frac{\beta}{2}}\cos{\frac{\al}{2}} -\sin{\frac{\beta}{2}}\sin{\frac{\al}{2}} (\bu\cdot\bv) \end{equation*}
\begin{equation*} \sin{\frac{\ga}{2}} \bw = \cos{\frac{\beta}{2}}\sin{\frac{\al}{2}} \bu + \cos{\frac{\al}{2}}\sin{\frac{\beta}{2}} \bv + \sin{\frac{\beta}{2}}\sin{\frac{\al}{2}} \bv \times \bu \end{equation*}
We factor \(\cos{\frac{\beta}{2}}\cos{\frac{\al}{2}}\) in the second equation:
\begin{equation*} \sin{\frac{\ga}{2}} \bw =\cos{\frac{\beta}{2}}\cos{\frac{\al}{2}} (\rot_u + \rot_v + \rot_v \times \rot_u) \end{equation*}
then divide by \(\cos{\frac{\ga}{2}}\) and use the first equation to obtain the final result
\begin{equation*} \rot_w = \frac{\cos{\frac{\beta}{2}}\cos{\frac{\al}{2}}}{\cos{\frac{\ga}{2}}}(\rot_u + \rot_v + \rot_v \times \rot_u) = \frac{1}{1- \rot_u \cdot \rot_v} (\rot_u + \rot_v + \rot_v \times \rot_u) \end{equation*}
Given a unit quaternion \(Q=q_0 + \bq = q_0 + q_1 \be_1 + q_2 \be_2 +q_3 \be_3\text{,}\) we can find the matrix \([\cR_{\al, \bu}]_{_E}\) of the rotation associated with quaternion \(Q\) on basis \((\be_1,\be_2,\be_3)\) in terms of \((q_0,q_1,q_2,q_3)\text{.}\) According to Rodrigues formula, operator \(\cR_{\al, \bu}\) takes the form
\begin{equation*} \cR_{\al, \bu} = \cI + (\sin\al) \cU + (1-\cos\al) \cU^2 = \cI + 2q_0 \cV + 2 \cV^2 \end{equation*}
where \(\cU\) is the operator \(\bV \mapsto \bu \times \bV\) and \(\cV\) is the operator given by
\begin{equation*} \cV: \quad \bV \mapsto \sin\frac{\al}{2} \bu\times \bV = \bq \times \bV \end{equation*}
We can use our previous derivations to find the matrix of \(\cR_{\al, \bu}\) on basis \((\be_1,\be_2,\be_3)\text{:}\)
\begin{align} [\cR_{\al , \bu}]_{_E} \amp = \begin{pmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{pmatrix} + 2 q_0 \begin{pmatrix} 0 \amp -q_3 \amp q_2 \\ q_3 \amp 0 \amp -q_1 \\ -q_2\amp q_1 \amp 0 \end{pmatrix} +2 \begin{pmatrix} q_0^2 + q_1^2-1 \amp q_1 q_2 \amp q_1 q_3 \\ q_1 q_2 \amp q_0^2 +q_2^2-1 \amp q_2 q_3 \\ q_1 q_3 \amp q_2 q_3 \amp q_0^2 + q_3^2-1 \end{pmatrix}\tag{1.6.4}\\ \amp = \begin{pmatrix} 2q_0^2+2q_1^2 -1\amp 2 (q_1 q_2 -q_0q_3) \amp 2(q_1 q_3 +q_0 q_2) \\ 2(q_1 q_2 +q_0 q_3) \amp 2q_0^2+ 2q_2^2-1 \amp 2(q_2 q_3 -q_0 q_1) \\ 2(q_1 q_3- q_0 q_2) \amp 2(q_2 q_3+q_0 q_1) \amp 2q_0^2+ 2q_3^2 -1 \end{pmatrix}\tag{1.6.5} \end{align}
Conversely, given a rotation matrix \([r_{ij}]_{_E}\text{,}\) we can find the corresponding quaternion \(Q= q_0 + q_1 \be_1 + q_2 \be_2 +q_3 \be_3\text{.}\) The previous derivations show that the symmetric and skew-symmetric parts of matrix \([r_{ij}]_{_E}\) are given by
\begin{equation*} [\frac{r_{ij}+ r_{ji}}{2}]_{_E} = [\cI + 2 \cV^2]_{_E}\quad , \qquad [\frac{r_{ij}- r_{ji}}{2}]_{_E} = [2q_0 \cV ]_{_E} \end{equation*}
Equality of the skew-symmetric parts gives three equations
\begin{equation} q_0 q_1 = \frac{1}{4} (r_{32}- r_{23}), \quad q_0 q_2 = \frac{1}{4} (r_{13}- r_{31}), \quad q_0 q_3 = \frac{1}{4} (r_{21}- r_{12})\tag{1.6.6} \end{equation}
Equality of the symmetric parts yields four equations
\begin{equation} \begin{array}{c} q_0^2=\frac{1}{4}(1+r_{11}+r_{22}+r_{33})\\ q_1^2=\frac{1}{4}(1+r_{11}-r_{22}-r_{33})\\ q_2^2=\frac{1}{4}(1-r_{11}+r_{22}-r_{33})\\ q_3^2=\frac{1}{4}(1-r_{11}-r_{22}+r_{33}) \end{array}\tag{1.6.7} \end{equation}
after taking into account \(q_0^2+q_1^2+q_2^2+q_3^2=1\text{.}\) Since both \(\pm Q\) give the same rotation, one of two possible values for \(q_0\) can be found from the first equation of (1.6.7). Then the values of \((q_1,q_2,q_3)\) are found uniquely from (1.6.6).
Despite having been discovered long ago, quaternions have witnessed renewed interest in many fields. The representation of rotations \cR_{\te,\bu} by unit quaternions e^{\te \bu} offers a more compact and economical way (with just 4 real scalars) to compute their mappings than with their matrix representations. Furthermore, quaternions, unlike Euler angles, do not suffer from singular behavior at particular configurations (the so-called “gimbal locks”). These features alone make them superior to other methods.
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