Introduction to rotation for computer graphics

2d rotation

A 2d rotation has only one parameter $θ$ , when the basis vectors $\hat{i} = [1, 0]$ and $\hat{j} = [0, 1]$ are rotated by an angle $θ$

p = \cos θ \hat{i} + \sin θ \hat{j} q = - \sin θ \hat{i} + \cos θ \hat{j}

Which builds the rotation matrix

R (θ) = [\begin{matrix} p \\ q \end{matrix}] = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]

When a vector $v$ is transformed by this matrix we know that the vector will be a linear combination of the basis which are $p$ and $q$

\begin{aligned} v^{'} = vR (θ) & = v_{x} p + v_{y} q \\ = v_{x} [\begin{array}{c} \cos θ & \sin θ \end{array}] + v_{y} [\begin{array}{c} - \sin θ & \cos θ \end{array}] \\ = {[\begin{array}{c} v_{x} \cos θ - v_{y} \sin θ \\ v_{x} \sin θ + v_{y} \cos θ \end{array}]}^{T} \end{aligned}

Using a matrix to encode this operation

\begin{aligned} v^{'} = vR (θ) & = [\begin{array}{c} v_{x} & v_{y} \end{array}] [\begin{array}{c} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array}] \\ = {[\begin{array}{c} v_{x} \cos θ - v_{y} \sin θ \\ v_{x} \sin θ + v_{y} \cos θ \end{array}]}^{T} \end{aligned}

3d rotation

About cardinal axes

R_{x} (α) = [\begin{matrix} 1 & 0 & 0 \\ 0 & \cos α & - \sin α \\ 0 & \sin α & \cos α \end{matrix}]

R_{y} (β) = [\begin{matrix} \cos β & 0 & \sin β \\ 0 & 1 & 0 \\ - \sin β & 0 & \cos β \end{matrix}]

R_{z} (γ) = [\begin{matrix} \cos γ & - \sin γ & 0 \\ \sin γ & \cos γ & 0 \\ 0 & 0 & 1 \end{matrix}]

About an arbitrary axis

Given an axis $\hat{n}$ and an amount of rotation around it $θ$ our goal is to find a rotation matrix that rotates about $\hat{n}$ by th angle $θ$

v^{'} = R (\hat{n}, θ) v

The basic idea is to solve this problem in a plane perpendicular to $\hat{n}$ which becomes a 2d problem

Separate $v$ in two vectors, a vector parallel to $\hat{v}$ called $v_{∥}$ and a vector perpendicular to $\hat{v}$ called $v_{⊥}$ such that $v_{∥} + v_{⊥} = v$

\begin{aligned} v_{∥} & = (v \cdot \hat{n}) \hat{n} \\ v_{⊥} & = v - v_{∥} \end{aligned}

After the rotation it’s obvious that the $v_{∥}$ component will be the same and only the vector $v_{⊥}$ will be rotated

A plane can be defined with two vectors that lie on it, since we have $v_{⊥}$ and we also know the normal of the plane (which is $\hat{n}$ ) any vector perpendicular to both vectors will also lie in the plane, we can use the cross product to find this vector

w = \hat{n} \times v_{⊥}

The length of $w$ is

\begin{aligned} ‖ w ‖ & = ‖ \hat{n} ‖ ‖ v_{⊥} ‖ \sin 90^{°} \\ = ‖ v_{⊥} ‖ \end{aligned}

Which means that $w$ has the same length as $v_{⊥}$ , note that even though they have the same length they don’t necessarily have unit length

$w$ and $v_{⊥}$ form now a 2d coordinate space where the $x$ -axis is $v_{⊥}$ and the $y$ -axis is $v_{⊥}$

Let $v_{⊥}^{'}$ be a vector that is the result of rotating $v_{⊥}$ by an angle $θ$ , we can find the projection of it onto the $x$ -axis and the $y$ -axis as follows

\begin{aligned} v_{⊥, x}^{'} & = (‖ v_{⊥} ‖ \cos θ) \hat{v_{⊥}} = \cos θ v_{⊥} \\ v_{⊥, y}^{'} & = (‖ v_{⊥} ‖ \sin θ) \hat{w} = \sin θ w \end{aligned}

Expressing $v_{⊥}^{'}$ as a linear combination of the basis

v_{⊥}^{'} = \cos θ v_{⊥} + \sin θ w

Reconstructing the solution from the observations above

\begin{aligned} v_{∥} & = (v \cdot \hat{n}) \hat{n} \\ v_{⊥} & = v - v_{∥} \\ = v - (v \cdot \hat{n}) \hat{n} \\ w & = \hat{n} \times v_{⊥} \\ = \hat{n} \times (v - v_{∥}) \\ = \hat{n} \times v - \hat{n} \times v_{∥} \\ = \hat{n} \times v \end{aligned}

Finally

\begin{aligned} v^{'} & = v_{⊥}^{'} + v_{∥}^{'} \\ = \cos θ v_{⊥} + \sin θ w + (v \cdot \hat{n}) \hat{n} \\ = \cos θ (v - (v \cdot \hat{n}) \hat{n}) + \sin θ (\hat{n} \times v) + (v \cdot \hat{n}) \hat{n} \\ = \cos θ v - \cos θ (v \cdot \hat{n}) \hat{n} + \sin θ (\hat{n} \times v) + (v \cdot \hat{n}) \hat{n} \\ (1) & = \cos θ v + \sin θ (\hat{n} \times v) + (1 - \cos θ) (v \cdot \hat{n}) \hat{n} \end{aligned}

Now we can compute what the basis vectors are after the transformation above (by using each of the basis vectors as $v$ on $(1)$ ) to construct a rotation matrix

\begin{aligned} p & = [\begin{array}{c} 1 \\ 0 \\ 0 \end{array}] p^{'} = [\begin{array}{c} n_{x}^{2} (1 - \cos θ) + \cos θ \\ n_{x} n_{y} (1 - \cos θ) + n_{z} \sin θ \\ n_{x} n_{z} (1 - \cos θ) - n_{z} \sin θ \end{array}] \\ q & = [\begin{array}{c} 0 \\ 1 \\ 0 \end{array}] q^{'} = [\begin{array}{c} n_{y} n_{x} (1 - \cos θ) - n_{z} \sin θ \\ n_{y}^{2} (1 - \cos θ) + \cos θ \\ n_{y} n_{z} (1 - \cos θ) + n_{x} \sin θ \end{array}] \\ r & = [\begin{array}{c} 0 \\ 0 \\ 1 \end{array}] r^{'} = [\begin{array}{c} n_{z} n_{x} (1 - \cos θ) + n_{y} \sin θ \\ n_{z} n_{y} (1 - \cos θ) - n_{x} \sin θ \\ n_{z}^{2} (1 - \cos θ) + \cos θ \end{array}] \end{aligned}

Constructing the matrix from these vectors

R (\hat{n}, θ) = [\begin{matrix} p^{'} & q^{'} & r^{'} \end{matrix}]

3d rotations using quaternions

A complex rotor is a unit norm complex number which rotates another complex number by the angle $θ$ and has the form

e^{i θ} = \cos θ + i \sin θ

Hamilton had hoped that a unit-norm quaternion $q$ could be used to rotate a vector which is stored as a pure quaternion $p$ , the unit norm quaternion is given by

\begin{aligned} (2) & q & = [s, λ \hat{n}] s, λ \in R, \hat{n} \in R^{3} \\ | \hat{n} | & = 1 \\ s^{2} + λ^{2} & = 1 \end{aligned}

p = [0, v] v \in R^{3}

Let’s compute the product $p^{'} = q p$

\begin{aligned} p^{'} & = q p \\ = [s, λ \hat{n}] [0, v] \\ (3) & = [- λ \hat{n} \cdot v, s v + λ \hat{n} \times v] \end{aligned}

Special case

What if $\hat{n}$ is perpendicular to $v$ ? Then the scalar quantity of $(3)$ is zero and we are left with the pure quaternion

\begin{matrix} (4) & p^{'} = [0, s v + λ \hat{n} \times v] given that \hat{n} is perpendicular to n \end{matrix}

Let’s analyze the vector part of $(4)$ (which is now a 3d entity because it’s a pure quaternion), since $\hat{n}$ is perpendicular to $v$ then the vector $\hat{n} \times v$ will have a norm equal to $‖ \hat{n} \times v ‖ = ‖ \hat{n} ‖ ‖ v ‖ \sin 90^{°}$ and also since $\hat{n}$ is a unit vector then $‖ \hat{n} \times v ‖ = ‖ v ‖$ which means that we have two orthogonal vectors with the same length

To rotate the vector $v$ about $\hat{n}$ let’s transform $v$ to the 2d space whose basis vectors are $v$ and $\hat{n} \times v$ and perform the rotation there which is trivially $[\cos θ, \sin θ]$ , therefore all we have to do in $(4)$ is make the scalar quantities multiplying each vector equal the projection of the rotated vector over the basis

p^{'} = [0, \cos θ v + \sin θ \hat{n} \times v]

Which makes the quaternion $q$ have the form

\begin{aligned} (5) & q & = [\cos θ, \sin θ \hat{n}] \end{aligned}

And it acts as a rotor only when $\hat{n}$ is perpendicular to $v$

Important notes/facts about orthogonal quaternions

If $q$ is a rotor about the unit vector $\hat{n}$ by an angle $θ$ whose vector term is perpendicular to the pure quaternion $p$
- $q p$ and $p q^{- 1}$ rotate $p$ by an angle $θ$ about $\hat{n}$
- $p q$ and $q^{- 1} p$ rotate $p$ by an angle $- θ$ about $\hat{n}$
- Each of these products leave $p^{'}$ unscaled (because $q$ is a unit norm quaternion)

General case

Let’s use $(2)$ as the starting point, note that this time its vector part it’s not necessarily perpendicular to the pure quaternion $p$ , the product $q p$ yields

\begin{aligned} q p & = [s, λ \hat{n}] [0, v] \\ = [- λ \hat{n} \cdot v, s v + λ \hat{n} \times v] \end{aligned}

Note that the term $- λ \hat{n} \cdot v$ does not vanish since for the general case $\hat{n}$ and $v$ are no longer perpendicular, what’s more important is that the product $q p$ is no longer a pure quaternion, multiplying a vector by a non-orthogonal quaternion has converted some of the vector information into the quaternion’s scalar component

What happens if we post-multiply $q p$ by $q^{- 1}$ , could it reverse the operation? (Note that since $q$ is a norm quaternion $q^{- 1} = q^{*}$ )

q p q^{- 1} = [- λ \hat{n} \cdot v, s v + λ \hat{n} \times v] [s, - λ \hat{n}]

Let’s first check if doing this multiplication makes the scalar component vanish

\begin{aligned} q p q^{- 1} & = [- λ s \hat{n} \cdot v - (s v + λ \hat{n} \times v) \cdot (- λ \hat{n}), \dots] \\ = [- λ s \hat{n} \cdot v + (s v) \cdot (λ \hat{n}) + (λ \hat{n} \times v) \cdot (λ \hat{n}), \dots] \\ = [- λ s \hat{n} \cdot v + (s v) \cdot (λ \hat{n}) + 0, \dots] since \hat{n} is perpendicular to \hat{n} \times v \\ = [- λ s \hat{n} \cdot v + λ s v \cdot \hat{n}), \dots] \\ = [0, \dots] \end{aligned}

Indeed it magically made the scalar component vanish! Now let’s look at the vector component of $q p q^{- 1}$

\begin{aligned} q p q^{- 1} & = [0, s (s v + λ \hat{n} \times v) + (- λ \hat{n} \cdot v) (- λ \hat{n}) + (s v + λ \hat{n} \times v) \times (- λ \hat{n})] \\ = [0, s^{2} v + s λ (\hat{n} \times v) + λ^{2} (\hat{n} \cdot v) \hat{n} - s λ (v \times \hat{n}) - λ^{2} (\hat{n} \times v \times \hat{n})] \end{aligned}

Let’s expand the cross product

(\hat{n} \times v) \times \hat{n} = (\hat{n} \cdot \hat{n}) v - (v \cdot \hat{n}) \hat{n} = v - (v \cdot \hat{n}) \hat{n}

Therefore

\begin{aligned} q p q^{- 1} & = [0, s^{2} v + s λ (\hat{n} \times v) + λ^{2} (\hat{n} \cdot v) \hat{n} - s λ (v \times \hat{n}) - λ^{2} (v - (v \cdot \hat{n}) \hat{n})] \\ = [0, s^{2} v + s λ (\hat{n} \times v) + λ^{2} (\hat{n} \cdot v) \hat{n} - s λ (v \times \hat{n}) - λ^{2} v + λ^{2} (v \cdot \hat{n}) \hat{n})] \\ = [0, s^{2} v + 2 s λ (\hat{n} \times v) + λ^{2} (\hat{n} \cdot v) \hat{n} - λ^{2} v + λ^{2} (v \cdot \hat{n}) \hat{n})] \\ = [0, (s^{2} - λ^{2}) v + 2 s λ (\hat{n} \times v) + 2 λ^{2} (v \cdot \hat{n}) \hat{n}] \end{aligned}

Let’s make $s = \cos θ$ and $λ = \sin θ$ just like in $(5)$ (it worked as a rotor when it was orthogonal to $p$ , it might work with the general case too)

q p q^{- 1} = [0, (\cos^{2} θ - \sin^{2} θ) v + 2 \cos θ \sin θ (\hat{n} \times v) + 2 \sin^{2} θ (v \cdot \hat{n}) \hat{n}]

Which involves double-angle terms, replacing these terms with double angle-identities

q p q^{- 1} = [0, \cos 2 θ v + \sin 2 θ (\hat{n} \times v) + (1 - \cos 2 θ) (v \cdot \hat{n}) \hat{n}]

The product created a pure quaternion equal to $v$ rotated by an angle $2 θ$ , if we want to rotate $v$ by an angle $θ$ we must build a half angle $θ$ quaternion $q$ (note above that $q$ was equal to $(5)$ )

\begin{matrix} (6) & q = [\cos \frac{1}{2} θ, \sin \frac{1}{2} θ \hat{n}] \end{matrix}

Using $(6)$ the product is

q p q^{- 1} = [0, \cos θ v + \sin θ (\hat{n} \times v) + (1 - \cos θ) (v \cdot \hat{n}) \hat{n}]

Note that the vector part of $q p q^{- 1}$ is identical to $(1)$

Quaternion difference and dot product

Let $a$ and $b^{'}$ be two unit norm quaternions (rotors that have the same form as $(6)$ ), the quaternion to rotate from $a$ to $b$ is given by $d a = b$ and is known as quaternion difference, finding the value of $d$ given that we know $a$ and $b$

\begin{aligned} d a & = b \\ d (a a^{*}) & = b a^{*} since a is a unit norm quaternion its inverse is equal to its conjugate \\ d & = b a^{*} \end{aligned}

Expanding the product

\begin{aligned} d & = [s_{b}, b] [s_{a}, - a] \\ = [s_{b} s_{a} + b \cdot a, - s_{b} a + s_{a} b - b \times a] \end{aligned}

Note that the scalar part of this quaternion is equal to the inner product (a generalization of the dot product to abstract vector spaces) between two quaternions

d = [⟨ a, b ⟩, - s_{b} a + s_{a} b - b \times a]

Remembering that a rotor is given by $(6)$ we can relate the inner product between rotor quaternions with the scalar quantity of $(6)$ and interpret it geometrically just like the dot product between two vectors in 3d/2d space but this time noticing that the dot product gives the cosine of half the angle between the quaternions

a \cdot b = \cos \frac{θ}{2}

This means that the angle between $a$ and $b$ is equal to

θ = 2 \arccos (a \cdot b)

Or using the half angle formulas

\begin{aligned} \cos^{2} \frac{θ}{2} & = \frac{1}{2} (1 + \cos θ) \\ (a \cdot b)^{2} & = \frac{1}{2} (1 + \cos θ) \\ \cos θ & = 2 (a \cdot b)^{2} - 1 \\ θ & = \arccos (2 (a \cdot b)^{2} - 1) \end{aligned}

The second formula works for all the cases as noted here (the first one doesn’t work when $a \cdot b < 0$ )