Axis-angle, quaternion, and homogeneous transforms

Axis-angle parameterisation

To solve the gimbal lock problem, we can adopt a redundant parameterisation using an axis $𝒌=\left[{𝑘}_{𝑥},{𝑘}_{𝑦},{𝑘}_{𝑧}\right]$ with $\Vert 𝒌\Vert =1$ and an angle $𝜃$, resulting in four parameters overall.

Frame 0 in gray and frame 1 in blue

The rotation ${𝑹}_{𝒌}\left(𝜃\right)$ can be found by transforming the coordinates of $𝒌$:

1. Given frame $1$ with axes ${𝑥}_1,{𝑦}_1,{𝑧}_1$ (with ${𝑧}_1$ coincident with $𝒌$), we can rotate frame $0$ by $𝛼$ about the $𝑧$-axis and by $𝛽$ around the $𝑦$-axis.

   $${𝑹}_1^0={𝑹}_{𝑧}\left(𝛼\right){𝑹}_{𝑦}\left(𝛽\right)$$

2. If a transformation $𝑨$ is expressed in frame $1$ as ${𝑞}_1=𝑨{𝑝}_1$, then:

   $${𝑞}_0={\left({𝑹}_0^1\right)}^{-1}𝑨{𝑹}_0^1{𝑝}^0={𝑹}_{𝑧}\left(𝛼\right){𝑹}_{𝑦}\left(𝛽\right)𝑨{\left({𝑹}_{𝑧}\left(𝛼\right){𝑹}_{𝑦}\left(𝛽\right)\right)}^{-1}{𝑝}^0$$

   which leads to:

   $${𝑹}_{𝒌}\left(𝜃\right)={𝑹}_{𝑧}\left(𝛼\right){𝑹}_{𝑦}\left(𝛽\right){𝑹}_{𝑧}\left(𝜃\right){𝑹}_{𝑦}\left(-𝛽\right){𝑹}_{𝑧}\left(-𝛼\right)$$

3. Using trigonometry we can find $𝛼$ and $𝛽$:

   $$\begin{array}{c}sin\left(𝛼\right)=\frac{{𝑘}_{𝑦}}{\sqrt{{𝑘}_{𝑥}^2+{𝑘}_{𝑦}^2}}\\ cos\left(𝛼\right)=\frac{{𝑘}_{𝑥}}{\sqrt{{𝑘}_{𝑥}^2+{𝑘}_{𝑦}^2}}\\ sin\left(𝛽\right)=\sqrt{{𝑘}_{𝑥}^2+{𝑘}_{𝑦}^2}\\ cos\left(𝛽\right)={𝑘}_{𝑧}\end{array}$$

The rotation matrix is:

$${𝑹}_{𝒌}\left(𝜃\right)=\left[\begin{array}{ccc}{𝑘}_{𝑥}^2\left(1-{𝑐}_{𝜃}\right)+{𝑐}_{𝜃}& {𝑘}_{𝑥}{𝑘}_{𝑦}\left(1-{𝑐}_{𝜃}\right)-{𝑘}_{𝑧}{𝑠}_{𝜃}& {𝑘}_{𝑥}{𝑘}_{𝑧}\left(1-{𝑐}_{𝜃}\right)+{𝑘}_{𝑦}{𝑠}_{𝜃}\\ {𝑘}_{𝑥}{𝑘}_{𝑦}\left(1-{𝑐}_{𝜃}\right)+{𝑘}_{𝑧}{𝑠}_{𝜃}& {𝑘}_{𝑦}^2\left(1-{𝑐}_{𝜃}\right)+{𝑐}_{𝜃}& {𝑘}_{𝑦}{𝑘}_{𝑧}\left(1-{𝑐}_{𝜃}\right)-{𝑘}_{𝑥}{𝑠}_{𝜃}\\ {𝑘}_{𝑥}{𝑘}_{𝑧}\left(1-{𝑐}_{𝜃}\right)-{𝑘}_{𝑦}{𝑠}_{𝜃}& {𝑘}_{𝑦}{𝑘}_{𝑧}\left(1-{𝑐}_{𝜃}\right)+{𝑘}_{𝑥}{𝑠}_{𝜃}& {𝑘}_{𝑧}^2\left(1-{𝑐}_{𝜃}\right)+{𝑐}_{𝜃}\end{array}\right]$$

We specified four parameters ($𝒌$ and $𝜃$), but there’s the constraint ${𝑘}_{𝑥}^2+{𝑘}_{𝑦}^2+{𝑘}_{𝑧}^2=1$, so the system actually has three degrees of freedom.

We can see that ${𝑹}_{-𝒌}\left(-𝜃\right)={𝑹}_{𝒌}\left(𝜃\right)$.

The axis-angle parameterisation isn’t globally invertible. However, it is locally invertible, which means that unlike the Euler one, there are no singularities such as gimbal lock.

Quaternions

The inverse of a real number $𝑥$ is ${𝑥}^{-1}=\frac{1}{𝑥}$, which, when multiplied by $𝑥$, gives 1.

For complex numbers we can find a similar concept: a number $𝑧$ multiplied by its inverse $\frac{\overline{𝑧}}{{\Vert 𝑧\Vert }^2}$ still gives 1.

There is no equivalent concept in ${ℝ}^3$, but it exists in ${ℝ}^4$.

A quaternion is an element of a vector space over the reals with the form $𝑞={𝑞}_0+{𝑞}_1𝒊+{𝑞}_2𝒋+{𝑞}_3𝒌$ where $\left\{1,𝒊,𝒋,𝒌\right\}$ is the basis set.

Tip

A shorthand notation distinguishes the scalar part from the vector part:

$$𝑞=\left[{𝑞}_0,𝒒\right]={𝑞}_0+𝒒$$

Quaternion product

Let’s define an axiom for quaternion bases:

$${𝑖}^2={𝑗}^2={𝑘}^2=𝑖𝑗𝑘=-1$$

from which we can derive the multiplication rules between any two basis elements:

$$\begin{array}{c}𝑖𝑗=𝑘𝑗𝑖=-𝑘\\ 𝑗𝑘=𝑖𝑘𝑗=-𝑖\\ 𝑘𝑖=𝑗𝑖𝑘=-𝑗\end{array}$$

To compute the product of quaternions we can use the shorthand notation:

$$𝑝𝑞=\left({𝑝}_0+𝒑\right)\left({𝑞}_0+𝒒\right)={𝑝}_0{𝑞}_0+{𝑝}_0𝒒+{𝑞}_0𝒑+𝒑𝒒$$

Using the rules from the axiom above, the product of the vector parts can be simplified as $𝒑𝒒=-𝒑\cdot 𝒒+𝒑\times 𝒒$. Thus the full product, with scalar and vector parts grouped, is:

$$𝑝𝑞=\left({𝑝}_0{𝑞}_0-𝒑\cdot 𝒒\right)+\left({𝑝}_0𝒒+{𝑞}_0𝒑+𝒑\times 𝒒\right)$$

Quaternion conjugate

For a quaternion $𝑞={𝑞}_0+{𝑞}_1𝒊+{𝑞}_2𝒋+{𝑞}_3𝒌$, the conjugate is:

$$\overline{𝑞}={𝑞}_0-{𝑞}_1𝒊-{𝑞}_2𝒋-{𝑞}_3𝒌$$

The real number $\Vert 𝑞\Vert =\sqrt{𝑞\overline{𝑞}}=\sqrt{{𝑞}_0^2+{𝑞}_1^2+{𝑞}_2^2+{𝑞}_3^2}$ is the norm of the quaternion.

Inverse of a quaternion

A quaternion (except 0) has a unique inverse ${𝑞}^{-1}=\frac{\overline{𝑞}}{{\Vert 𝑞\Vert }^2}$.

Quaternion rotations

Let’s go back to complex numbers. Recall that they can also be expressed in polar coordinates as $𝑧=𝜌\left(cos\left(𝜃\right)+𝑗sin\left(𝜃\right)\right)$.

Multiplying two complex numbers rotates the resulting vector, especially when both have unit norm:

$${𝑧}_1{𝑧}_2={𝜌}_1\left(cos\left({𝜃}_1\right)+𝑗sin\left({𝜃}_1\right)\right){𝜌}_2\left(cos\left({𝜃}_2\right)+𝑗sin\left({𝜃}_2\right)\right)={𝜌}_1{𝜌}_2\left(cos\left({𝜃}_1+{𝜃}_2\right)+𝑗sin\left({𝜃}_1+{𝜃}_2\right)\right)$$

Let’s consider a unit-length quaternion $𝑞=\left[𝜂,𝜺\right]$. We can express it as $𝑞=\left[cos\left(\frac{𝜃}{2}\right),sin\left(\frac{𝜃}{2}\right)𝒌\right]$ where $𝒌$ is a unit vector defined as $𝒌=\frac{𝜺}{\Vert 𝜺\Vert }$ and $𝜃=2{atan}_2\left(\Vert 𝜺\Vert ,𝜂\right)$.

The quaternion $𝑞$ represents a rotation of angle $𝜃$ about the axis defined by the vector $𝒌$.

Let’s take a pure quaternion $𝑝=\left[0,𝒑\right]$. The vector ${𝑝}^{\prime }$ rotated by $𝑞$ can be computed using ${𝑝}^{\prime }=𝑞𝑝\overline{𝑞}$ which yields a quaternion of the form $\left[0,{𝒑}^{\prime }\right]$.

$${𝒑}^{\prime }=\left({𝜂}^2-{\Vert 𝜺\Vert }^2\right)𝒑+2𝜂\left(𝜺\times 𝒑\right)+2\left(𝜺\cdot 𝒑\right)𝜺$$

Substituting $𝜃$ and $𝒌$, we obtain Rodrigues’ rotation formula that describes the rotation of the vector $𝒑$:

$${𝒑}^{\prime }=cos\left(𝜃\right)𝒑+sin\left(𝜃\right)\left(𝒌\times 𝒑\right)+\left(1-cos\left(𝜃\right)\right)\left(𝒌\cdot 𝒑\right)𝒌$$

Note

The formula is equivalent to applying the rotation from the matrix given by the axis-angle parameterisation.

The parameters of a unit-length quaternion correspond to three degrees of freedom, which can be expressed as:

• $𝜂$, ${𝜀}_{𝑥}$, ${𝜀}_{𝑦}$, ${𝜀}_{𝑧}$, with the constraint ${𝜂}^2+{𝜀}_{𝑥}^2+{𝜀}_{𝑦}^2+{𝜀}_{𝑧}^2=1$;
• $𝜃$, ${𝑘}_{𝑥}$, ${𝑘}_{𝑦}$, ${𝑘}_{𝑧}$, with the constraint ${𝑘}_{𝑥}^2+{𝑘}_{𝑦}^2+{𝑘}_{𝑧}^2=1$;

Finding parameters from the rotation

Given a rotation matrix $𝑹=\left[\begin{array}{ccc}{𝑟}_{1,1}& {𝑟}_{1,2}& {𝑟}_{1,3}\\ {𝑟}_{2,1}& {𝑟}_{2,2}& {𝑟}_{2,3}\\ {𝑟}_{3,1}& {𝑟}_{3,2}& {𝑟}_{3,3}\end{array}\right]$, we can find the parameters:

$$\begin{array}{c}𝜂=\frac{1}{2}\sqrt{{𝑟}_{1,1}+{𝑟}_{2,2}+{𝑟}_{3,3}+1}\\ 𝜺=\frac{1}{2}\left(\begin{array}{c}sgn\left({𝑟}_{3,2}-{𝑟}_{2,3}\right)\sqrt{{𝑟}_{1,1}-{𝑟}_{2,2}-{𝑟}_{3,3}+1}\\ sgn\left({𝑟}_{1,3}-{𝑟}_{3,1}\right)\sqrt{-{𝑟}_{1,1}+{𝑟}_{2,2}-{𝑟}_{3,3}+1}\\ sgn\left({𝑟}_{2,1}-{𝑟}_{1,2}\right)\sqrt{-{𝑟}_{1,1}-{𝑟}_{2,2}+{𝑟}_{3,3}+1}\end{array}\right)\end{array}$$

Composition of rotations

Let’s consider two rotations associated with the unit-length quaternions ${𝑞}_0$ and ${𝑞}_1$. The first rotation transforms a vector $𝒑$ into ${𝒑}^{\prime }$ and the second transforms ${𝒑}^{\prime }$ into ${𝒑}^{″}$.

$${𝒑}^{″}={𝑞}_1{𝑝}^{\prime }\overline{{𝑞}_1}={𝑞}_1\left({𝑞}_0𝑝\overline{{𝑞}_0}\right)\overline{{𝑞}_1}=\left({𝑞}_1{𝑞}_0\right)𝑝\overline{\left({𝑞}_1{𝑞}_0\right)}$$

Note

The conjugate of a product is the product of the conjugates in reverse order.

The composition of rotations is the product of their corresponding quaternions.

Translations

A point ${𝒑}^1$ in frame $1$ can be expressed in frame $0$ as ${𝒑}^0={𝒐}_1^0+{𝑹}_1^0{𝒑}^1$.

The inverse relation is found by isolating ${𝒑}^1$:

$${𝒑}^1={𝑹}_0^1{𝒑}^0-{𝑹}_0^1{𝒐}_1^0$$

Homogeneous coordinates

We need a compact way to represent both rotation and translation.

To do that, we create the homogeneous representation $\widetilde{𝒑}$ of $𝒑$ by appending a coordinate with value 1.

The homogeneous transformation matrix is defined as:

$${𝑨}_1^0=\left[\begin{array}{cc}{𝑹}_1^0& {𝒐}_1^0\\ {\mathrm{𝟎}}^{𝑇}& 1\end{array}\right]$$

which can then be used as usual to apply transformations by multiplying with vectors:

$${\widetilde{𝒑}}^0={𝑨}_1^0{\widetilde{𝒑}}^1$$

The inverse transformation can be found by inverting the matrix:

$${𝑨}_0^1={\left({𝑨}_1^0\right)}^{-1}=\left[\begin{array}{cc}{\left({𝑹}_1^0\right)}^{𝑇}& -{\left({𝑹}_1^0\right)}^{𝑇}{𝒐}_1^0\\ {\mathrm{𝟎}}^{𝑇}& 1\end{array}\right]$$

Unlike rotation matrices, homogeneous transformations are not orthogonal, so ${𝑨}^{-1}\ne {𝑨}^{𝑇}$. But they can still be composed using matrix multiplication.