I expected this issue to be much easier to solve than it turned out to be.
A vehicle starts at point A and moves through point B in geometric space. It travels at a constant speed v, and starts out with angle $\theta$ (where 0 degrees is northward). The vehicle needs to have a certain angle change per second to reach point B given these parameters.
I've been having trouble determining the length of the traveled arc. Is there an exact method to determine the required angular velocity?
On
Let $\theta$ be the launch angle (from vertical). You need angular velocity I.e. $\omega=\dfrac{v}{r}$.
You already have $v$. Now, we only need to determine $r$, the radius of the circle, whose arc is $AB$.
Let the center of the circle be $O$. We have $OA=OB=r$ (let $AB=d$ )and $\angle BAO=\angle ABO=\theta$.
Using all this given data, and applying cosine formula, you get $$\cos\theta = \frac{d^2+r^2-r^2}{2dr}\implies d=2r\cos\theta$$ or :$$r=\frac{d}{2}\sec \theta$$
Hence $$\omega =\frac vr= \frac{2v}{AB}\cos \theta$$
The vehicle’s velocity is always tangent to its path, which in this case is a circular arc. A tangent to a circle is perpendicular to the circle’s radius at the point of tangency, which means that the angle of arc subtended by the vehicle’s path from $A$ to $B$ is equal to its net change in heading. By symmetry, the angle between the ray from $A$ through $B$ and the vehicle’s final heading is equal and opposite to the angle made by its initial heading. Call the latter angle $\phi$, so that the total change in heading from $A$ to $B$ is $2\phi$. If you can compute the bearing from $A$ to $B$ directly, then you just need to subtract $\theta$ from this bearing to get $\phi$, otherwise you can compute it from the dot product or cross product of the initial velocity vector with $\overrightarrow{AB}$.
Now, for a circle of radius $R$, the arc length corresponding to a central angle of $2\phi$ is $2R\phi$, while the corresponding chord length is $2R\sin\phi$, which makes the ratio of chord length to arc length ${\sin\phi\over\phi}$. Assuming that $A$ and $B$ are distinct so that $\sin\phi\ne0$, the distance traveled by the vehicle is ${\phi\over\sin\phi}AB$, from which we can compute $\omega = {2v\over AB}\sin\phi$. Note that this formula preserves the sign of $\phi$: In other words, it can tell the difference between a left and right turn to reach $B$.