Given the position of the vehicle ($x,y$) at different time points, the speed of the vehicle (m/s), the direction the vehicle is facing (heading — in degrees), the track width of the vehicle, and the wheelbase of the vehicle, how can I calculate the steering angle of the vehicle?
I tried following the approximation detailed in Ackerman Steering relating to vehicle heading but got very weird inner and outer steering angles.
If anyone can give me a few hints, I would greatly appreciate it!
I use a single-Track model
(„METHODE ZUR ERSTELLUNG UND ABSICHERUNG EINER MODELLBASIERTEN SOLLVORGABE FÜR FAHRDYNAMIKREGELSYSTEME Michael Graf“)
the velocity components given in local system $~x~,y~$are:
$$\begin{bmatrix} v_x \\ v_y \\ \end{bmatrix}=v\begin{bmatrix} \cos(\beta) \\ \sin(\beta) \\ \end{bmatrix}$$
where $~\beta~$ is the side slip angle and v the vehicle velocity.
the sideslip angle $\alpha_v$ at the front wheel is:
$$\alpha_v=\delta-\beta-\frac{l_v\,\dot{\psi}}{v_x}=\delta-\beta-\frac{l_v\,\dot{\psi}}{v\,\cos(\beta)}\tag 1$$
where $\delta$ is the steering angle.
because the side slip angle $~\beta\mapsto 0~$ and the sideslip angle $~\alpha_v\mapsto 0~$ , you obtain from equation (1)
$$0=\delta-\frac{l_v\,\dot{\psi}}{v}\tag 2$$
solving equation (2) for $~\dot{\psi}~$ :
$$\dot{\psi}=\frac{v}{l_v}\,\delta$$
Edit:
$$\alpha_v=\delta-\kappa=\delta-\left(\beta+\arctan{\frac{\dot{\psi}\,l_v}{v_x}}\right) $$
Here is a derivation. Let us denote by $l$ the segment of length $l$ representing the wheel base, and by $w$ the segment of length $w$ representing the track of the rear wheels, perpendicular to $l$. Their intersection is the point $P$.
Choose a world Cartesian coordinate system $O\, \vec{e}_1 \, \vec{e}_2$ and another coordinate system $P \, \vec{E}_1 \, \vec{E}_2$ attached to the vehicle, cantered at the point $P$ and with two vectors $\vec{E}_1$ and $\vec{E}_2$ of length one, such that vector $\vec{E}_1$ is aligned with the wheel base $l$ and vector $\vec{E}_2$ is aligned with the track $w$. Observe that $\vec{E}_1$ is perpendicular to $\vec{E}_2$.
Consider the vector $\vec{p} = \vec{OP}$, which is the position vector of point $P$ with respect to the world system $O\, \vec{e}_1 \, \vec{e}_2$. Decompose $$\vec{OP} = \vec{p} = x\, \vec{e}_1 + y\, \vec{e}_2$$
Let $\theta$ be the angle between the vectors $\vec{e}_1$ and $\vec{E}_1$, i.e. $\theta$ is the angle between the horizontal axis $O\, \vec{e}$ and the line $P\, \vec{E}_1$. Then, since $\vec{E}_1$ is of length one, we can decompose it in the world system as $$\vec{E}_1 = \cos(\theta)\, \vec{e}_1 + \sin(\theta)\, \vec{e}_2$$ Since $\vec{E}_2$ is perpendicular to $\vec{E}_1$ $$\vec{E}_2 = - \,\sin(\theta)\, \vec{e}_1 + \cos(\theta)\, \vec{e}_2$$ The position and orientation of the vehicle, which change with time $t$, are uniquely determined by the functions \begin{align} &x = x(t)\\ &y = y(t)\\ &\theta = \theta(t) \end{align}
The velocity of the point $P$ relative to $O\, \vec{e}_1\,\vec{e}_2$ is $$\frac{d \vec{p}}{dt} = \frac{dx}{dt}\, \vec{e}_1 + \frac{dy}{dt}\, \vec{e}_2$$ If we denote the magnitude of this velocity (the magnitude is called speed) by $$s = s(t) = \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$$, the restriction that the rear wheels do not slip implies that the orthogonal projection of the velocity $\frac{d \vec{p}}{dt}$ along the the segment $w$ (which coincides with the line $P \, \vec{E}_2$) is zero. Therefore $\frac{d \vec{p}}{dt}$ is always aligned with the vector $\vec{E}_1$ and therefore $$\frac{d \vec{p}}{dt} = s\, \vec{E}_1$$ or in more detail $$\frac{d \vec{p}}{dt} = \frac{dx}{dt}\, \vec{e}_1 + \frac{dy}{dt}\, \vec{e}_2 = s\, \cos(\theta)\, \vec{e}_1 + s\, \sin(\theta)\, \vec{e}_2$$ which component-wise yields \begin{align} &\frac{dx}{dt} = s\, \cos(\theta)\\ &\frac{dy}{dt} = s\, \sin(\theta) \end{align} Our next step is to look at the steering. Denote the other end of the segment $l$, representing the wheel base, by $Q$ (that is the end of segment $l$, opposite to point $P$). As with $P$, let $\vec{q} = \vec{OQ}$ be the position vector of point $Q$ in the world coordinates. By vector addition $$\vec{OQ} = \vec{OP} + \vec{PQ}$$ i.e. $$\vec{q} = \vec{p} + l\, \vec{E}_1$$ The velocity of $\vec{q}$ is $$\vec{v} = \frac{d\vec{q}}{dt} = \frac{d\vec{p}}{dt} + l\, \frac{d\vec{E}_1}{dt}$$ If $v = |\vec{v}|$ is the magnitude (i.e. speed) of $Q$ in the world system, on one hand we can decompose $$\vec{v} = v \, \cos(\phi)\, \vec{E}_1 + v \, \sin(\phi)\, \vec{E}_2$$ On the other hand, $\frac{d\vec{p}}{dt} = s \, \vec{E}_1$ and \begin{align} \frac{d\vec{E}_1}{dt} &= \frac{d}{dt}\Big(\cos(\theta)\, \vec{e}_1 + \sin(\theta)\, \vec{e}_2\Big) = -\,\sin(\theta)\,\frac{d\theta}{dt}\, \vec{e}_1 + \cos(\theta) \,\frac{d\theta}{dt}\,\vec{e}_2\\ &= \frac{d\theta}{dt}\,\Big(-\,\sin(\theta)\, \vec{e}_1 + \cos(\theta)\,\vec{e}_2\Big)\\ &= \frac{d\theta}{dt}\, \vec{E}_2\end{align} which yields $$ v \, \cos(\phi)\, \vec{E}_1 + v \, \sin(\phi)\, \vec{E}_2 = \,\,\vec{v}\,\, = \frac{d\vec{p}}{dt} + l\, \frac{d\vec{E}_1}{dt} = s \, \vec{E}_1 + l\,\frac{d\theta}{dt}\, \vec{E}_2$$ i.e. $$ v \, \cos(\phi)\, \vec{E}_1 + v \, \sin(\phi)\, \vec{E}_2 = s \, \vec{E}_1 + l\,\frac{d\theta}{dt}\, \vec{E}_2$$ or component-wise \begin{align} &v \, \cos(\phi) = s\\ &v \, \sin(\phi) = l\,\frac{d\theta}{dt} \end{align} So putting together the component-wise equations of the velocities at $P$ and $Q$ we get the differential equations \begin{align} &\frac{dx}{dt} = s\, \cos(\theta)\\ &\frac{dy}{dt} = s\, \sin(\theta)\\ &\frac{d\theta}{dt} = \frac{v}{l} \, \sin(\phi) \\ &v \, \cos(\phi) = s \end{align} By solving the fourth equation for $v = \frac{s}{\cos(\phi)}$ and plugging the result in the third equations $$\frac{d\theta}{dt} = \frac{v}{l} \, \sin(\phi) = \frac{s}{l\,\cos(\phi)} \, \sin(\phi) = \frac{s}{l} \, \tan(\phi) $$ we obtain the system of differential equations get the differential equations \begin{align} &\frac{dx}{dt} = s\, \cos(\theta)\\ &\frac{dy}{dt} = s\, \sin(\theta)\\ &\frac{d\theta}{dt} =\frac{s}{l} \, \tan(\phi) \end{align}
