Ackermann Steering Angle

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 $$vec{v} = frac{dvec{q}}{dt} = frac{dvec{p}}{dt} + l, frac{dvec{E}_1}{dt}$$ 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, $frac{dvec{p}}{dt} = s , vec{E}_1$ and begin{align} frac{dvec{E}_1}{dt} &= frac{d}{dt}Big(cos(theta), vec{e}_1 + sin(theta), vec{e}_2Big) = -,sin(theta),frac{dtheta}{dt}, vec{e}_1 + cos(theta) ,frac{dtheta}{dt},vec{e}_2 &= frac{dtheta}{dt},Big(-,sin(theta), vec{e}_1 + cos(theta),vec{e}_2Big) &= frac{dtheta}{dt}, vec{E}_2end{align} which yields $$ v , cos(phi), vec{E}_1 + v , sin(phi), vec{E}_2 = ,,vec{v},, = frac{dvec{p}}{dt} + l, frac{dvec{E}_1}{dt} = s , vec{E}_1 + l,frac{dtheta}{dt}, vec{E}_2$$ i.e. $$ v , cos(phi), vec{E}_1 + v , sin(phi), vec{E}_2 = s , vec{E}_1 + l,frac{dtheta}{dt}, vec{E}_2$$ or component-wise begin{align} &v , cos(phi) = s &v , sin(phi) = l,frac{dtheta}{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{dtheta}{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 $$frac{dtheta}{dt} = frac{v}{l} , sin(phi) = frac{s}{l,cos(phi)} , sin(phi) = frac{s}{l} , tan(phi) $$ 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{dtheta}{dt} =frac{s}{l} , tan(phi) end{align}