A problem popped recently to the mental horizon: how close a predicted trajectory of a ship comes to a real trajectory? For a moment, let us consider only the unit sphere and unit vectors and we happily identify points $p$ of a unit sphere centered to origo $O$ with vectors $p-O$... Also, we identify a point $p= (\theta,\phi)$ expressed by its latitude $\phi$ and longitude $\theta$ with its 3D map $p\rightarrow q$ image $q= (\cos\theta\cos\phi,\, \cos\theta\sin\phi,\, \sin\theta)$. So the reader, be aware. (And if the notation gets you baffled, check the notation page...)
Let us assume that an arc $[p_0,b]$ is a part of the predicted trajectory and $p_1$ is the ground truth (the pun intended). $p_0$ can be the last reference point and $b= \hat{p}_1$ is the prediction for $p_1$. The problem can be presented as finding the arc distance of a point $p_1$ from a great circle $C$ defined by the arc $[p_0,b]\subset C$. This great circle $C$ can be identified with a unit vector $c$, which is a normal of the plane of $C$.
OK, $c\perp b$ and $c\perp p_0$, that is: $c= \pm(p_0\times b)^0$ and we can find a vector $a\in [p_0,b]\subseteq C$ which is a projection of $p_1$ on the plane $C$ with a suitable scaling so that $\|a\|=1$: \[a= (p_1 - (p_1\cdot c) c)^0.\;\; (1)\] Now, arc lengths shown in Fig. 1 are: \[\alpha_\perp= |[a,p_1]|= \pi/2-\cos^{-1}(c\cdot p_1)\;\; (2) \\ \alpha_\parallel= |[a,p_0]|= \cos^{-1}(a\cdot p_0).\;\; (3)\].
Eq. (2) is useful in estimating the relative randezvous error $e_p=|[a,p_1]|/|[p_0,p_1]|$ and Eq. (3) leads to the relative time error $e_t= |[p_0,a]|/|[p_0,b]|-1$. The estimate of $e_t$ assumes time information was not a part of the teaching of the predictor, which is not always the case.
Now, the question arises... Should we use spherical geometry here? Would computations of $e_p$ and $e_t$ become faster, or would the numerical accuracy be better, or both? The latter is important, since it is easy to have numerical instability at distances below 20 km. As it will be revealed soon (after the Midsummer Feast, I hope), the answer is a surprisingly close call. In general, a lot of trigonometric trickery can be substituted by vector formulas (using either geometric algebra or vector algebra) without much loss in computational efficiency, when operations happen in the modern processing environments and intermediate results have enough memory to dwell in...
But while you are waiting, here is the way where all points get projected to a plane defined by points $p_0$,$p_1$ and $b$: $l_\perp= \sqrt{1-\cos(\beta)^2}\|p_1-p_0\|$ and $l_\parallel= \cos(\beta)\,\|p_1-p_0\|$, where $\cos(\beta)= (p_1-p_0)^0\cdot (b-p_0)^0$.
It yields quite good results as long as the distances are under 200 km ($|[u,v]|/\|u-v\| - 1 < 32\times10^{-3}$).
