Smooth path between two points with given turn rate and initial and final direction

Question

basic smooth path

I'm trying to model the fastest route for a particle to go from one point, heading in a given direction to another point, heading in another direction. This route can have a maximum turn rate that is a given value, theta. In the picture, point/direction A is in blue, point/direction B is in green and the smooth route between them is in red.

I have been attempting to model it with circles and allowing for a reasonable amount of distance to get on the appropriate direction before getting to point B but it is all very ad hoc.

Is there a way to do this more rigorously?

Answer 1

Given some fixed initial and final conditions ($r$ vector, $x,y,\theta,\rho$ scalars): $$ r_0=(x_0,y_0), \dot r_0=(\dot x_0,\dot y_0)\\ r_f=(x_f,y_f), \dot r_f=(\dot x_f,\dot y_f) $$

and the orbit constraint: $$ |\frac{dr}{dt}|=\dot \theta = \omega_0 $$ The problem is fairly complicated. Suppose you are requested to be at the same vector speed, but just to move to the right, or worst, to keep your position but moving into the opposite direction. You will have to introduce one turn (or a complicated eye shaped turn) just to displace the particle into the requested state.

Please confirm.

I think the best solution for the problem could be forcing the solution into a predefined shape including two circles and a straight line, which is the solution for the simplest cases not involving the longest eye-shaped curve for repositioning the particle. Under that case, the problem "could" be easier to solve. But still complicated.

Check, which i am sure you don t want to program:

• Optimal Control,
• Hamilton Jacobi Bellman Equations

Answer 2

If you have a limited radial acceleration, the fastest way to turn through a given angle is a tiny radius. The radial acceleration is $\omega ^2 r$ where $\omega$ is the angular velocity and $r$ is the radius of the turn. Keeping $r$ small lets $\omega$ get large. The time to make a $180^\circ=\pi$ radian turn is $\frac \pi \omega$ so you want small $r$ and high $\omega$. At the limit, in this model you can make the turn instantly at $r=0$.