3D Rotation Representations

Euler-angles

3D rotations can be represented by a sequence of 2D rotations about each axis respectively. Typically, the order is $Z$ , $Y$ , and then $X$ ; the angles are denoted by $α, β$ and $γ$ .

Gimbal Lock

When two axes become aligned, applying a rotation around these axes is equivalent to each other, resulting in degree of freedom loss.

Suppose we have $β = \pm \frac{π}{2}$ . If we fix $α + γ$ , the result is unchanged.

Rotation Matrices

A matrix $R \in R^{3 \times 3}$ can be used to represent a 3D rotation. However, $R$ should be in the rotation group $SO (3)$ , and thus satisfy

\begin{aligned} R^{T} R & = R R^{T} = I, \\ det (R) & = 1. \end{aligned}

With the above constraints, a rotation matrix has actually 3 degrees of freedom.

We can also combine the rotation matrix $R$ and a translation vector $t$ into a transformation matrix

T = [\begin{matrix} R & t \\ 0 & 1 \end{matrix}] \in R^{4 \times 4} .

When applying this to a vector, note that this vector should be in homogeneous coordinates.

Axis-angles

According to Euler's rotation theorem, any 3D rotation can be specified using two parameters: a unit vector $ω$ that defines the axis of the rotation, and the angle $θ$ that describes the magnitude of the rotation about that axis. The rotated vector can be represented using Rodrigues' formula, which has 3 representations.

Vector Notation

a^{'} = \cos θ a + \sin θ (ω \times a) + (1 - \cos θ) (ω \cdot a) ω .

proof

Decompose $a$ into projection on and rejection of $ω$
$a = a_{n} + a_{t}$
where
$a_{n} ∥ ω, a_{t} ⊥ ω$
We have
$\begin{aligned} a_{n} & = (ω \cdot a) ω \\ a_{t} & = a - a_{n} \end{aligned}$

Building orthogonal basis using $ω$ , $a_{t}$ , and $ω \times a_{t}$ , we can calculate the rotated vector
$\begin{aligned} a^{'} & = \sin θ (ω \times a_{t}) + \cos θ a_{t} + a_{n} \\ = \sin θ (ω \times a - {ω \times a_{n}}^{0}) + \cos θ (a - a_{n}) + a_{n} \\ = \cos θ a + \sin θ (ω \times a) + (1 - \cos θ) (ω \cdot a) ω \end{aligned}$

Matrix Notation

R (ω, θ) = I + \sin θ [ω]_{\times} + (1 - \cos θ) [ω]_{\times}^{2},

where $[ω]_{\times}$ is the skew symmetric matrix of the vector $ω$ .

proof

$\begin{aligned} a^{'} & = a - (1 - \cos θ) a + (1 - \cos θ) (ω \cdot a) ω + \sin θ (ω \times a) \\ = a - (1 - \cos θ) (a - (ω \cdot a) ω) + \sin θ (ω \times a) \\ = a + (1 - \cos θ) ((ω \cdot a) ω - (ω \cdot ω) a) + \sin θ (ω \times a) \\ = a + (1 - \cos θ) (ω \times (ω \times a)) + \sin θ (ω \times a) \\ = a + (1 - \cos θ) [ω]_{\times}^{2} a + \sin θ [ω]_{\times} a \\ = \underset{R}{\underset{⏟}{(I + (1 - \cos θ) [ω]_{\times}^{2} + \sin θ [ω]_{\times})}} a \end{aligned}$

Exponential Notation

R (ω, θ) = e^{[ω]_{\times} θ} .

lemma

If $ω = [a, b, c]^{T}$ , letting $K = [ω]_{\times}$ , then
$K^{3} = - ‖ ω ‖_{2}^{2} K$

proof

\begin{aligned} K & = [\begin{array}{c} 0 & - c & b \\ c & 0 & - a \\ - b & a & 0 \end{array}] \\ K^{2} & = [\begin{array}{c} - b^{2} - c^{2} & a b & c a \\ a b & - c^{2} - a^{2} & b c \\ c a & b c & - a^{2} - b^{2} \end{array}] \\ K^{3} & = [\begin{array}{c} 0 & (a^{2} + b^{2} + c^{2}) c & - (a^{2} + b^{2} + c^{2}) b \\ - (a^{2} + b^{2} + c^{2}) c & 0 & (a^{2} + b^{2} + c^{2}) a \\ (a^{2} + b^{2} + c^{2}) b & - (a^{2} + b^{2} + c^{2}) a & 0 \end{array}] \\ = - ‖ ω ‖_{2}^{2} K \end{aligned}

Therefore, if $ω$ is a unit vector, $K^{3} = - K$

proof

Using Taylor's series for $K = [ω]_{\times}$ , note that $K^{3} = - K$ from the lemma proved above
$\begin{aligned} e^{K θ} & = I + K θ + \frac{(K θ)^{2}}{2!} + \frac{(K θ)^{3}}{3!} + \frac{(K θ)^{4}}{4!} + \dots \\ = I + K θ + K^{2} \frac{θ^{2}}{2!} - K \frac{θ^{3}}{3!} - K^{2} \frac{θ^{4}}{4!} + \dots \\ = I + K (θ - \frac{θ^{3}}{3!} + \dots) + K^{2} (\frac{θ^{2}}{2!} - \frac{θ^{4}}{4!} + \dots) \\ = I + \sin θ K + (1 - \cos θ) K^{2}, \end{aligned}$
which is the same as the formula in matrix notation.

Quaternions

Quaternions are generalized complex numbers of the form:

q = w + x i + y j + z k,

where

i^{2} = j^{2} = k^{2} = i j k = - 1.

Quaternion products are not reflexible, i.e.
$\begin{aligned} i j & = k, \\ j i & = - k . \end{aligned}$

A quaternion can also be represented as a scalar with a 3D vector:

q = {[\begin{matrix} w & v \end{matrix}]}^{T},

where $v = {[\begin{matrix} x & y & z \end{matrix}]}^{T}$ .

Quaternion Product

The product of two quaternions is

q_{1} \circ q_{2} = [\begin{matrix} w_{1} w_{2} - v_{1} \cdot v_{2} \\ w_{1} v_{2} + w_{2} v_{1} + v_{1} \times v_{2} \end{matrix}] = [\begin{matrix} w_{1} w_{2} - x_{1} x_{2} - y_{1} y_{2} - z_{1} z_{2} \\ w_{1} x_{2} + x_{1} w_{2} + y_{1} z_{2} - z_{1} y_{2} \\ w_{1} y_{2} - x_{1} z_{2} + y_{1} w_{2} + z_{1} x_{2} \\ w_{1} z_{2} + x_{1} y_{2} - y_{1} x_{2} + z_{1} w_{2} \end{matrix}] .

Rotation Quaternions

A quaternion can also be used to represent a rotation:

q = [\begin{matrix} \cos \frac{θ}{2} \\ \sin \frac{θ}{2} ω \end{matrix}] .

Rotation of vector $a$ can be defined as

{\tilde{a}}^{'} = q \tilde{a} q^{- 1},

where $\tilde{a}$ is a pure quaternion:

\tilde{a} = [\begin{matrix} 0 \\ a \end{matrix}] .

Quaternion products are equivalent to Rodrigue's formula.

proof

From the definition of quaternion product,
$q \tilde{a} = [\begin{matrix} - \sin \frac{θ}{2} (ω \cdot a) \\ \cos \frac{θ}{2} a + \sin \frac{θ}{2} (ω \times a) \end{matrix}] .$
We also know that
$q^{- 1} = [\begin{matrix} \cos \frac{θ}{2} \\ - \sin \frac{θ}{2} ω \end{matrix}] .$
Therefore
$\begin{aligned} q \tilde{a} q^{- 1} & = [\begin{array}{c} - {\sin \frac{θ}{2} \cos \frac{θ}{2} (ω \cdot a) + \sin \frac{θ}{2} \cos \frac{θ}{2} (ω \cdot a)}^{0} + \sin^{2} \frac{θ}{2} ({(ω \times a) \cdot ω}^{0}) \\ \sin^{2} \frac{θ}{2} (ω \cdot a) ω + \cos^{2} \frac{θ}{2} a + \sin \frac{θ}{2} \cos \frac{θ}{2} (ω \times a) - \sin \frac{θ}{2} \cos \frac{θ}{2} (a \times ω) - \sin^{2} \frac{θ}{2} (\underset{a - (a \cdot ω) ω}{\underset{⏟}{(ω \times a) \times ω}}) \end{array}] \\ = [\begin{array}{c} 0 \\ 2 \sin^{2} \frac{θ}{2} (ω \cdot a) ω + (\cos^{2} \frac{θ}{2} - \sin^{2} \frac{θ}{2}) a + 2 \sin \frac{θ}{2} \cos \frac{θ}{2} (ω \times a) \end{array}] \\ = [\begin{array}{c} 0 \\ \cos θ a + \sin θ (ω \times a) + (1 - \cos θ) (ω \cdot a) ω \end{array}] . \end{aligned}$

Double Cover Issue

Quaternions have double cover issue that $q$ and $- q$ represents the same rotation. If we want to sum to quaternion, we need to calculate if the quaternions are in the same hemisphere.

Dual Quaternions

A dual quaternion has the form

\hat{q} = q_{r} + ϵ q_{d},

where $q_{r}$ and $q_{d}$ are ordinary quaternions. $q_{r}$ is the real part, and $q_{d}$ is the dual part. $ϵ$ is the dual unit satisfying $ϵ^{2} = 0$ .

The addition of two dual quaternions is component-wise.
The multiplication of two dual quaternions satisfies

{\hat{q}}_{1} {\hat{q}}_{2} = q_{r 1} q_{r 2} + ϵ (q_{r 1} q_{d 2} + q_{r 2} q_{d 1}) .

6D Rotation Representation

Also see On the Continuity of Rotation Representations in Neural Networks.

Represent a 3D rotation with two 3D vectors $a_{1}$ and $a_{2}$ , which makes sure continuity in the representation space.

Conversion Between Representations

From Euler-angles to Rotation Matrices

Given Euler-angles $α, β, γ$ , we can define the rotation matrix about each axis:

\begin{aligned} R_{x} & = [\begin{array}{c} 1 & 0 & 0 \\ 0 & \cos α & - \sin α \\ 0 & \sin α & \cos α \end{array}]; \\ R_{y} & = [\begin{array}{c} \cos β & 0 & \sin β \\ 0 & 1 & 0 \\ - \sin β & 0 & \cos β \end{array}]; \\ R_{z} & = [\begin{array}{c} \cos γ & - \sin γ & 0 \\ \sin γ & \cos γ & 0 \\ 0 & 0 & 1 \end{array}] . \end{aligned}

The resulting rotation matrix is

R (α, β, γ) = R_{x} (α) R_{y} (β) R_{z} (γ) .

From Rotation Matrices to Quaternions

Given a rotation matrix $R \in SO (3)$ , where

R = [\begin{matrix} R_{11} & R_{12} & R_{13} \\ R_{21} & R_{22} & R_{23} \\ R_{31} & R_{32} & R_{33} \end{matrix}],

we want to compute the quaternion

q = [w, x, y, z] .

First compute $tr (R) = R_{11} + R_{22} + R_{33}$ .
If $tr (R) > 0$ , then

\begin{aligned} w & = \frac{1}{2} \sqrt{tr (R) + 1} \\ x & = \frac{1}{4 w} (R_{32} - R_{23}) \\ y & = \frac{1}{4 w} (R_{13} - R_{31}) \\ z & = \frac{1}{4 w} (R_{21} - R_{12}) \end{aligned}

If $tr (R) \leq 0$ , then we find the largest diagonal element.
If $R_{11}$ is the largest,

\begin{aligned} x & = \frac{1}{2} \sqrt{1 + R_{11} - R_{22} - R_{33}} \\ w & = \frac{1}{4 x} (R_{32} - R_{23}) \\ y & = \frac{1}{4 x} (R_{12} + R_{21}) \\ z & = \frac{1}{4 x} (R_{13} + R_{31}) \end{aligned}

If $R_{22}$ is the largest,

\begin{aligned} y & = \frac{1}{2} \sqrt{1 + R_{22} - R_{11} - R_{33}} \\ w & = \frac{1}{4 y} (R_{13} - R_{31}) \\ x & = \frac{1}{4 y} (R_{12} + R_{21}) \\ z & = \frac{1}{4 y} (R_{23} + R_{32}) \end{aligned}

If $R_{33}$ is the largest,

\begin{aligned} z & = \frac{1}{2} \sqrt{1 + R_{33} - R_{11} - R_{22}} \\ w & = \frac{1}{4 z} (R_{21} - R_{12}) \\ x & = \frac{1}{4 z} (R_{13} + R_{31}) \\ y & = \frac{1}{4 z} (R_{23} + R_{32}) \end{aligned}

From Quaternions and Translations to Dual Quaternions

Given a quaternion $q = [w, x, y, z]$ and a translation vector $[t_{x}, t_{y}, t_{z}]$ , the real part of the dual quaternion is simply

q_{r} = q .

The dual part is calculated via quaternion multiplication

q_{d} = \frac{1}{2} [0, t_{x}, t_{y}, t_{z}] * q .

The resulting dual quaternion is

\hat{q} = q_{r} + ϵ q_{d} .

From 6D Representations to Rotation Matrices

Given 3D vectors $a_{1} \neq λ a_{2}$ , the resulting rotation matrix is

R = [\begin{matrix} ∣ & ∣ & ∣ \\ b_{1} & b_{2} & b_{3} \\ ∣ & ∣ & ∣ \end{matrix}],

where

\begin{aligned} b_{1} & = N (a_{1}), \\ b_{2} & = N (a_{2} - (a_{2} \cdot b_{1}) b_{1}), \\ b_{3} & = b_{1} \times b_{2}, \\ N (x) & := \frac{x}{‖ x ‖} . \end{aligned}

Summary

Methods	Descriptions	Pros	Cons
Euler-angles	Use rotation angles about three principle axes to represent a rotation	- Intuitive to represent - Use only three parameters	- Gimbal lock: one degree of freedom is lost if two of the three rotational axes align - Hard and unintuitive to interpolate - Do not commute under composition - Non-uniqueness: infinite number of angle choices for a rotation
Rotation Matrices	Use a $3 \times 3$ or $4 \times 4$ (with translation) matrix to represent a rotation	- Trivial to compute and apply - Easy to combine two rotations - Can represent rotation and translation in one matrix	- Redundancy in parameters - Normalization required - Hard to interpolate - Hard to visualize
Axis-angles	Use a unit axis $ω$ and a rotation angle $θ$ to represent a rotation	- Straightforward	- Non-uniqueness: infinite number of angle choices for a rotation - Hard to interpolate
Quaternions	Encode the axis-angle into a quaternion $[\cos \frac{θ}{2}, \sin \frac{θ}{2} ω]$	- Provide direct ways for smooth interpolation - Use only four parameters	- Unintuitive to understand - Combining rotations requires non-linear operations
Dual Quaternions	Use two quaternions to represent both rotation and translation	- Unify rotation and translation into a single framework - Lower computational cost than using matrices - Require fewer parameters than matrices - Easy to concatenate transforms	- Unintuitive to understand - Not widely adopted