Sphere arch parameterization I have an engineering problem that I can't get right.
I have a unit vector in point A and a unit vector in point B. I know that the two points are connected by a spherical arch. That is, they both lie on the same sphere, and the two unit vectors both point along this sphere (tangential to the sphere). I also know the arch length between the two points, but I am not given the actual positions of the points (need to set a reference).
I need a parameterization in one variable f(t) of the curve between the two points. Figure: Parametrize curve along sphere from Q to P
From what I figure, it should be defined by the central angle between the points (https://en.wikipedia.org/wiki/Great-circle_distance) and the radius R of the sphere, which I have both found. But I can't figure out how to make a parameterization f(t) = [x(t), y(t), z(t)].
EDIT:
Note that I am not given the positions A and B, only the unit vector in these points which is assumed to point tangential to the sphere and the arch length between them. This is a real life problem, and the apparatus gives direction of motion at t_0 and direction of motion at t_1 and the distance travelled. It is assumed to travel in spherical motions with a large R (I can adjust for the R later). So I can naturally set the position of A to (0,0,0), but I am not given the position of B.
All help is appreciated:) Kind regards,
Fredrik
 A: Original answer, with some extra explanations As was pointed out in the comments, you should specify whether the arcs are parts of great circles. The link in your questions discusses the great-circle distance, hence I assume this is indeed what you mean.
A general parametrization of a circle.
A general parametrization of a (full) circle of radius $R$ in $\mathbb{R}^3$, with the origin as a center, is
$$
 f : [0, 2\pi] \to \mathbb{R}^3 : t \mapsto R\cos t\, u + R\sin t\, v, \tag{1}
$$
where $u$ and $v$ are two orthonormal vectors.
Note that $f(0)=u$ and $f(\pi/2)$. You can also verify that $f$ is indeed
a circle:
$$
\begin{align}
 \|f(t)\|^2 &= R^2 \cos^2 t (u\cdot u)+ R^2 \sin^2 t (v\cdot v)- 2 R^2 \cos t \sin t  (u \cdot v) \\
& = R^2.
\end{align}
$$
Where does the expression come from?
This expression is obtained performing a suitable rotation of the parametrization of the "standard circle":
$$
(R\cos t, R\sin t, 0) = R\cos t\, \mathbf{e}_1 + R\sin t\, \mathbf{e}_2. \tag{2}
$$
We have just replace the $\mathbf{e}_i$ by two other vectors.
What do we need to do?
We need to look for two orthonormal vectors $u$ and $v$ and plug these into expression $(1)$. (Using expression $(2)$ and calculating a suitable rotation in $\mathbb{R}^3$ involves more work.)
If we are given two unit vectors $A$ and $B$ that determine the two positions on the sphere: we are going to look for a vector $C$ such that

*

*$C$ is a unit vector

*$C$ lies in the plane spanned by $A$ and $B$ (we assume they span a plane).

*$C$ is orthogonal to $A$.

Let $\theta = d/R$ be the angle from $A$ to $B$.
Then $C$ becomes
$$
     C = \frac{B - (B\cdot A) A}{\|B - (B\cdot A)A\|} = \frac{B - \cos \theta A}{\sin\theta}.
$$
The parametrization then becomes
$$
 f : [0,\theta] \to \mathbb{R}^3 : t \mapsto R\cos t\, A + R\sin t\, C,
$$
As an exercise, you can verify that $f(\theta) = B$.
Addendum: answer if $A$ and $B$ describe tangent vectors
After my first answer, the OP clarified that $A$ and $B$ are not positions, but they are the (unit) directions of the spherical arc.
We can still use expression $(1)$. When we derive $(1)$, we get that
the speed unit vector at $t$ is $-\sin t u + \cos t v$. Evaluating  $t_0$ and $t_1$ gives
$$
\begin{align*}
  - \sin t_0 u + \cos t_0 v &= A,\\
  - \sin t_1 u + \cos t_1 v &= B. 
\end{align*}
$$
Solving this system (together with $\|u\|=1$, $\|v\|=1$ and $u\cdot v = 0$),  give $u$ and $v$.
Alternatively, since the movement is circular, $A$ and $B$ lie in the plane of movement. Hence the vector $A\times B$ is  normal to the plane that contains the arc.
So taking the cross product of $A$ and $B$ with $A\times B/\|A\times B\|$, should give you (up to a minus sign) the position vectors $u$ and $v$.
