How to get the necessary local yaw angles to produce a circle?

Question

This is a geometry question

Say I want to generate the points (x,y) for a circle as you see in the figure

However the only way I can describe these points is through local yaw angles.(the green arrows)

The equations to generate (x,y) from yaw angles are:

$\Delta_t$ is a constant, and the only thing I can give is φ The rest are calculated.(of course the first point is known - the origin)

So my question is, what are the values of the local yaw φ angles in order to construct a circle?

EDIT： Made a mistake using phi small and capital in the equation. I will rewrite it with the first one being theta

The equation is more understandable as

$\theta_{n+1}=\sum\phi_t\Delta t_n$

and the only thing I can specify is $\phi_t$

Answer 1

Let $R$ be the required radius, and let's assume that the motion is around the origin (0,0) and that it starts from point $(0,R)$.

At time $n$ we are at point $x_n, y_n$, and (if we our algorithm is correct) we have $x_n^2 + y_n^2 = R^2$.

Our goal is to find $\phi_n$ such that $x_{n+1}, y_{n+1}$ stays on the circle, i.e., $x_{n+1}^2 + y_{n+1}^2 = R^2$. By plugging this into the formula from the original post we get $$ \left(v_n^2 \Delta t_n^2 \cos^2\phi_n + 2v_n \Delta t_n \cos \phi_n x_n + x_n^2 \right) + $$ $$ +\left(v_n^2 \Delta t_n^2 \sin^2\phi_n + 2v_n \Delta t_n \sin \phi_n y_n + y_n^2 \right) = R^2 \;\;\;\Longrightarrow$$ $$ v_n^2 \Delta t_n^2 + 2v_n\Delta t_n \left(x_n\cos\phi_n + y_n\sin\phi_n \right) = 0 \;\;\;\Longrightarrow$$ $$ x_n\cos\phi_n + y_n\sin\phi_n = -\frac{v_n\Delta t_n}{2} $$ Since $x_n = R\cos\theta_n$ and $y_n = R\sin\theta_n$ we get $$ x_n\cos\phi_n + y_n\sin\phi_n = R \cos\left(\theta_n-\phi_n \right) \;\;\;\Longrightarrow $$ $$ \cos\left(\theta_n - \phi_n \right) = -\frac{v_n\Delta t_n}{2R} \;\;\;\Longrightarrow $$$$ \phi_n = \theta_n - \arccos\frac{v_n\Delta t_n}{2R} $$