# Slurping up SLERP

For work recently, we had to deal with functions on the surface of the sphere. While conceptually simple, this actually introduces more complexity to computations. Even something as simple as distance calculations are now different due to the curvature.

Most of these concepts are really well defined by Wikipedia or other resources. The only qualm I have with the source is the fact that there is no standardization of notation: the usage of $\theta$ and $\varphi$ are almost different and this, frankly, costed me a lot of grief.

However, the concept of shortest point interpolation (so called slerp) between two points on the sphere was really not well explained I think. There is a Scipy implementation, and a short Wikipedia article but both never really discussed the proof of why the formula

\begin{align}
S(p_0, p_1, t) = \frac{\sin((1 – t)\theta)}{\sin \theta} p_0 + \frac{\sin(t\theta)}{\sin(\theta)}p_1
\end{align}

where $\theta = \arccos(p_0 \cdot p_1)$ is actually on the sphere. I was using this for a plotting purpose, so I didn’t care about the “interpolation” aspect, but see this post for a derivation from that point of view.

Let’s quickly give detailed proofs on why this formula works, and to prove that for any $0\le t \le 1$ that we get a point on the sphere that lies on the great circle between the two points $p_0, p_1$. On a sphere of radius 1, $\theta$ is in fact the great circle distance between $p_0$ and $p_1$.

To this end, we have to show that the points defined by slerp is of norm 1, and also lies on the plane defined by the origin, and $p_0, p_1$ (definition of the great circle). This second point is easy to do, we simply have to show that $S(p_0, p_1, t) \cdot (p_0 \times p_1) = 0$ (e.g. the slerp points are orthogonal to the vector which defines the plane). This equality is true due to the simple equaliy $a \cdot (a \times b) = 0$.

Thus the slerp points lies on the plane defined by the great circle, it remains to show that the norm is actually 1. This can be done with just the sine subtraction formula and the usual Pythagorean identity. First note
\begin{align*}
\left\Vert S(p_0, p_1, t) \right\Vert^2 &= \frac{\sin^2((1-t)\theta) + \sin^2(t \theta) + 2 \sin((1-t)\theta) \sin(t\theta) \cos(\theta) } {\sin^2(\theta)}
\end{align*}
by the fact $p_0, p_1$ are norm 1.
Expanding the first term, we have
\begin{align*}
\sin^2((1-t)\theta) = \sin^2(\theta) \cos^2(t\theta) + \cos^2(\theta) \sin^2(t\theta) – 2 \sin(\theta) \cos(\theta) \sin(t\theta) \cos(t\theta).
\end{align*}
and note that the cross term can be expanded using the same sine difference formula
\begin{align*}
2 \sin((1-t)\theta) \sin(t\theta) \cos(\theta) &= 2 \sin(\theta)\cos(\theta)\sin(t\theta)\cos(t\theta) – 2 \cos^2(\theta) \sin^2(t\theta).
\end{align*}
Thus, cancelling the common terms, we have that the numerator is equal to
\begin{align*}
\sin^2(\theta) \cos^2(t\theta) + \cos^2(\theta) \sin^2(t\theta) + \sin^2(t \theta)- 2 \cos^2(\theta) \sin^2(\theta) = \\
\sin^2(\theta) \cos^2(t\theta) + (\cos^2(\theta) +1- 2 \cos^2(\theta)) \sin^2(t\theta) = \\
\sin^2(\theta) \cos^2(t\theta) + (\sin^2(\theta)) \sin^2(t\theta) = \sin^2(\theta)
\end{align*}
which equals the denominator.

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