Quaterions are a generalization of complex numbers that have three distinct imgaginary components instead of one (in addition to the one real component). Like the space of complex numbers can be represented with the 2-D plane, quaterions can be represented as points in 4-D space. The analogies do not end there: the unit circle in the complex plane is equivalent to the set of orientation angles $\phi$ in 2-D (there is just one orientation angle in 2-D). The 3-D unit sphere in the 4-D space of quaternions is equivalent to the orientation of a rigid body in 3-D space. Alternative ways to describe orientation in 3-D are the three Euler angles $\psi$, $\theta$, $\phi$. The three Euler angles are related to the yaw, pitch, and roll angles or ships and aircrafts.

What’s the motivation to use a unit complex number instead of an angle or to use an unit quaternion instead of a set of three Euler angles? Other reasons besides a fancy in theory, are that complex numbers and quaternions can be used out of sheer lazyness. For instance, when using an angle to describe orientation in 2-D one has to worry about periodic boundary conditions, that is the value of the angle where it jumps from 360 degrees back to 0 degrees (or equivalently from +180 degrees to -180 degrees). If you ever have tried to write down a function for the difference between two angles that can handle the periodicity of its two arguments, and immediately fell into an extended episode of procrastination instead of fixing that ugly bug, you know why we need a solution for us lazy people. Using complex numbers, the solution is simply

\[\operatorname{angle-diff}(a,b) = \arg(\exp(ia)/\exp(ib))\]

where $\exp(ia)$ can be implemented with the indentity $\exp(ia) = \cos(a) + i \sin(a)$ and $\arg$ can be implemented with the $\mathrm{atan2}$ function on a computer. It’s easy to imagine that the situation is exacerbated in 3-D. (Which i’m too lazy to explain here in detail).

Another application is to compute an interpolated angle $\phi^\prime$ that is located between two reference angles $\phi_1$ and $\phi_2$. The easy way out is to take the average of the two corresponding complex numbers. Since the average will be in inside the unit circle and not on the unit cicle, on has to place it back on the unit circle by normalizing the average.

\[\tilde{z}(t^\prime) = (t_1 - t^\prime) z(t_1) + (t_2 - t^\prime) z(t_2)\] \[z(t^\prime) = \frac{\tilde{z}(t^\prime)}{\left| \tilde{z}(t^\prime) \right|}\]

The same recipe applies when encoding 3-D rotatations as unit quaternions.

With all these obscure operations on the unit cicle or the unit sphere of quaternions, it’s easy to loose sight of the meaning of distance in these spaces. What does it mean for two numbers on the complex unit circle to have a separation of, say 2? For the complex unit circle the interpretation is more or less clear. To second order, the chord length is appromately the arc length (which is then directly related to the angle). See https://en.wikipedia.org/wiki/Circular_segment.

\[\frac{1}{2} \sin(\Delta \phi) = \frac{1}{2} \left| a - b \right|\]

For small distances, one can lazily approximate $\sin x\approx x$ to find

\[\Delta \phi \approx \left| a - b \right|\]

What is the corresponding result for quaternions? Does the Euclidean distance between two unit quaterions also correspond to some angle? But which angle, since there are three Euler angles?

Any rotation \(P\) can be transformed into \(Q\) by applying the rotation $Q(P^{-1})$. If you consider this in axis-angle form, you can call the magnitude of the angle as the size of the difference between the two angles.

In quaternion language,

\[p=a+bi+cj+dk\]

and

\[q=w+xi+yj+zk.\] \[p^{-1} = a-bi-cj-dk,\]

this gives the distance as \(2 \arccos(wa+xb+yc+zd).\) You don’t have to worry about the sign of the angle because you’ll get one for $Q(P^{-1})$ and the other for $P(Q^{-1})$. If you define the range of your arccos as -180 to 180, two angles only have the same cosine if they are opposites.

In axis-angle, this translates to 2 times the arccos of

\[\cos(\theta_p/2)\cos(\theta_q/2) + \mathbf{e}_p\cdot \mathbf{e}_q \sin(\theta_p/2)\sin(\theta_q/2),\]

where \(\theta_p\) / \(\theta_q\) are the angles of the two rotations and \(e_p.e_q\) is the dot product of the two axes. This is only simple when \(e_p=e_q\) so the dot product is $1$, and you get

\[\cos(\theta_p/2)\cos(\theta_q/2)+\sin(\theta_p/2)\sin(\theta_q/2),\]

which is

\[\cos(\theta_p/2-\theta_q/2),\]

so 2 arccos that is \(+/-(\theta_p-\theta_q)\) as expected. When the axes are perpendicular, you get a distance of \(2\arccos(\cos(\theta_p/2)\cos(\theta_q/2)),\) which I don’t think has a simpler form. When the axes are different, you can use some trig identities to get 2 times the arccos of

\[\frac{1}{2}(1+\mathbf{e}_p\cdot\mathbf{e}_q)\cos(\theta_p/2- \theta_q/2)+(1-\mathbf{e}_p\cdot\mathbf{e}_q)\cos(\theta_p/2+\theta_q/2)),\]

but I don’t think further simplification is possible. This is easier to deal with using quaternions, but I don’t know if it’s useful for your purposes. This also answers my question on what two rotations have maximum possible distance as two 180 degree separate rotations around the same axis.

The distance between two rotation quaternions is always \(\cos(\theta/2).\) So the difference between the rotation axes is never taken into account in the distance. It’s only and exactly the angle.

Here is the argument:

For unit quaternions $a$, $x$, $y$, we have

\[\left|ax-ay\right|^2 = \left|x-y\right|^2\]

Proof:

\[\left|ax-ay\right|^2 = (\overline{ax}-\overline{ay})(ax-ay)\] \[=\left|ax\right|^2 + \left|ay\right|^2 - \overline{x}\,\overline{a}ay - \overline{y}\,\overline{a}ax\] \[= 2 - \overline{x} \left|a\right|^2 y - \overline{y} \left|a\right|^2 x = 2 - \overline{x}y - \overline{y}x\] \[= \left|x-y\right|^2\]

Lets chose $a=x^{-1}$ and set $z=x^{-1}y$

thus

\[\left|x-y\right|^2 = \left|1-z\right|^2 = (1-\overline{z})(1-z)\] \[= 1 + \left|z\right|^2 - \overline{z} - z = 2 - 2 \mathrm{Re}(z) = 2 - 2 \cos(\theta/2)\]

where I have used $\left

q \right

= 1$ for all rotations.

For small angles theta we even have:

\[\cos(\theta/2) = 1 - \theta^2/8\]

Therefore

\[\left|x-y\right|^2 \approx 2 - 2 (1 - \theta^2/8) = \theta^2/4\]

and

\[|x-y| \approx \theta/2\]