In today’s post, I explain the use of quaternions for rotations.

Background

What’s a quaternion?

A quaternion is a 4-element vector, typically represented as $\begin{bmatrix} w, i, j, k \end{bmatrix}$. ($w$ can go on either the start or the end of the vector.)

In our very limited context, a quaternion can be used to represent a vector, and a rotation about that vector.

With this system, we can represent any rotation.

Why not Euler angles?

Euler angles (roll, pitch, and yaw) have three issues I care about:

• Gimbal lock - when applying rotations, if two axes become aligned, they’re the same axis now and a degree of rotation freedom is lost permanently :(
• Interpolation troubles - because each axis interpolates by itself, the most direct path isn’t always taken. This matters a LOT for control systems.
• Poorly documented conventions - Since it matters in what order you apply each Euler angle (roll, pitch, yaw), it’s necessary to define and stick to a convention. There is no universal convention, which opens up the possibility of misinterpretation.

Quaternions solve all of these troubles, while being mathematically beautiful.

How do I make one?

A detailed analysis of quaternion math is both beyond the scope of this article and my knowledge. Therefore, I will stick to practicals.

To represent a rotation of $\theta$ about a unit vector $\begin{bmatrix} u_x, u_y, u_z \end{bmatrix}$, use the half-angle theorem:

$$q=\begin{bmatrix} \cos\frac{\theta}{2} & u_x\sin\frac{\theta}{2} & u_y\sin\frac{\theta}{2} & u_z\sin\frac{\theta}{2} \end{bmatrix}$$

which can also be represented as:

$$q = \cos\frac{\theta}{2} + u_x\sin\frac{\theta}{2}\hat{i} + u_y\sin\frac{\theta}{2}\hat{j} + u_z\sin\frac{\theta}{2}\hat{k}$$

which will be important very shortly.

How do I rotate a vector using a quaternion?

You need a base vector, $v$, and a rotation to apply, $q$. The rotated quaternion, $v'$, is defined as follows:

$$v' = qvq^*$$

($q^*$ is the conjugate of $q$, which just means that the imaginary components $i, j, k$ of $q$ have their signs flipped. This conjugate multiplication also ensures we end up with a rotated Cartesian vector, and not another quaternion!)

This begs a VERY important question:

How do I multiply two quaternions together?!

It’s easy to see how to add quaternions together (you do it piecewise, like ordinary vectors), but surely piecewise multiplication will leave a horrifying mess of $ij$s, $wk$s, et cetera. This is where Sir William Rowan Hamilton innovated when he invented quaternions, and quaternion multiplication is aptly done via the “Hamilton product”:

Multiply the quaternions piecewise, and then simplify using the following brilliant identities:

$$i^2=j^2=k^2=1$$ $$ij=k, ji=-k$$ $$jk=i, kj=-i$$ $$ki=j, ik=-j$$

The result is your new rotated quaternion!

Rotation Example

Say that we wanted to rotate a vector $v=\begin{bmatrix} 1 & 0 & 0 \end{bmatrix}$ by 90 degrees around the $y$ axis $\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}$ (as in this Stack Exchange post):

What we’re solving for is $v' = qvq^*$, so we need to find $q$ and $q^*$ and then conduct a couple Hamilton products.

Our desired rotation quaternion is defined with half angles:

$$q = \cos\frac{90^\circ}{2} + 0\hat{i} + \sin\frac{90^\circ}{2}\hat{j} + 0\hat{k}$$ $$q = 0.707+0.707\hat{j}$$

We’ll need the conjugate later, which is just $q$ with its imaginary parts negated:

$$q^* = 0.707-0.707\hat{j}$$

The first Hamilton product we’ll carry out is $qv$, as follows:

$$(0.707+0.707\hat{j})(1\hat{i})$$ $$0.707\hat{i}+0.707\hat{j}\hat{i}$$

which, using the identities above, simplifies to:

$$0.707\hat{i}-0.707\hat{k}$$

Now we can solve for $v'$ by multiplying by $q^*$:

$$v'=(0.707\hat{i}-0.707\hat{k})(0.707-0.707\hat{j})$$ $$v'= (0.707\hat{i})(0.707) - (0.707\hat{i})(0.707\hat{j}) - (0.707\hat{k})(0.707) + (0.707\hat{k})(0.707\hat{j})$$ $$v'= 0.5\hat{i} - 0.5\hat{i}\hat{j} - 0.5\hat{k} + 0.5\hat{k}\hat{j}$$

Simplifying once again using the quaternion identities:

$$v'= 0.5\hat{i} - 0.5\hat{k} - 0.5\hat{k} - 0.5\hat{i}$$ $$v'= -\hat{k} = \begin{bmatrix}0 & 0 & -1\end{bmatrix}$$

This matches up perfectly with what we expect: the $x$ axis rotated by 90 degrees around the $y$ axis is equal to the negative $z$ axis.

Thanks for reading!