Rotate Vector by Quaternion

Introduction

Previously we were able to convert Euler angles to a Quaternion, and a Quaternion to a Matrix. In game programming we want fast 3D rotations, thus this lesson outlines how to rotate a vector by a Quaternion.

References

Matrix Method

Key Concepts

The key concepts of this part of the lesson are:

Review of previous concepts:
- Create a Quaternion from Euler Angles
- Create a Matrix from a Quaternion
- Multiply a Vector by a Matrix

Review Concepts

From Lesson 2.5 we learned that a Quaternion can be created from Euler angles as follows:

$Q=\left[\begin{array}{c}cos(\frac{Y}{2})\\\left(\begin{array}{c}0\\sin(\frac{Y}{2})\\0\end{array}\right)\end{array}\right]\left(\left[\begin{array}{c}cos(\frac{P}{2})\\\left(\begin{array}{c}sin(\frac{P}{2})\\0\\0\end{array}\right)\end{array}\right]\left[\begin{array}{c}cos(\frac{R}{2})\\\left(\begin{array}{c}0\\0\\sin(\frac{R}{2})\end{array}\right)\end{array}\right]\right)$

Also, from Lesson 2.5 we learned that the conversion of a Quaternion to a matrix was done by:

From Lesson 2.1 we learned that a vector multiplied by a matrix was done by:

$\left[\begin{array}{ccc}M_{11}&M_{12}&M_{13}\\M_{21}&M_{22}&M_{23}\\M_{31}&M_{32}&M_{33}\end{array}\right]\times{\left[\begin{array}{c}V_x\\V_y\\V_z\end{array}\right]}=\left[\begin{array}{c}V_xM_{11}+V_yM_{12}+V_zM_{13}\\V_xM_{21}+V_yM_{22}+V_zM_{23}\\V_xM_{31}+V_yM_{32}+V_zM_{33}\end{array}\right]={\left[\begin{array}{c}V_x\\V_y\\V_z\end{array}\right]}'$

Substitution gives us:

$R_{Q}=\left[\begin{array}{ccc}1-2(Q_{y}^2+Q_{z}^2)&2(Q_{x}Q_{y}-Q_{w}Q_{z})&2(Q_{x}Q_{z}+Q_{w}Q_{y})\\2(Q_{x}Q_{y}+Q_{w}Q_{z})&1-2(Q_{x}^2+Q_{z}^2)&2(Q_{y}Q_{z}-Q_{w}Q_{x})\\2(Q_{x}Q_{z}-Q_{w}Q_{y})&2(Q_{y}Q_{z}+Q_{w}Q_{x})&1-2(Q_{x}^2+Q_{y}^2)\end{array}\right]\times{\left[\begin{array}{c}V_x\\V_y\\V_z\end{array}\right]}={\left[\begin{array}{c}V_x\\V_y\\V_z\end{array}\right]}'$

Example Given the Euler angles of Roll = 5^o, Pitch = -10^o, and Yaw = 15^o, the resulting Quaternion would be:

$Q=\left[\begin{array}{c}0.98623585\\-0.08065606\\0.133679\\0.05444693\end{array}\right]$

The matrix would be (rounded to 4 decimal places):

$R_Q=\left[\begin{array}{ccc}0.9583&-0.1290&0.2549\\0.0858&0.9811&0.1736\\-0.2725&-0.1445&0.9513\end{array}\right]$

Now the final multiplication, given $V=\left[\begin{array}{c}2\\-3\\4\end{array}\right]$ is:

$\left[\begin{array}{ccc}0.9583&-0.1290&0.2549\\0.0858&0.9811&0.1736\\-0.2725&-0.1445&0.9513\end{array}\right]\times{\left[\begin{array}{c}2\\-3\\4\end{array}\right]}=\left[\begin{array}{c}3.3231\\-2.0769\\3.6937\end{array}\right]$

Direct (Optimized) Method

The reference for this lesson outlines the following equation for multiplying a vector by a Quaternion:

$V'=Q\times{V}\times{\bar{Q}}$

The product of two Quaternions is given by (given the two quaternions A and B, where a Quaternion is defined as $Q=(Q_r,Q_{xyz})$ :

$AB=(A_rB_r-A_{xyz}\cdot{B_{xyz}},A_rB_{xyz}+B_rA_{xyz}+A_{xyz}\times{B_{xyz}})$

Also V must be treated like a Quaternion (i.e., $V_q=(0,V)$ ). The second reference goes through all the detail to get a faster method using the following steps: