I have a set of equations that describe the geometry of a vehicle in low speed turning (a single track vehicle with the assumption that the tyres go in the direction they are pointing). A constraint is applied to an outer point on the body which must follow a circular path of a given radius. A further constraint defines the ratio of steering by the rear wheel to that of the front wheel.
The question that's prompted the equations is what is the front steer angle that forces the body point to follow the given radius? The arrangement is as follows:
Known parameters are the body point radius, $R$, the ratio of rear to front steer, $k = \frac \beta \alpha$, the wheelbase, $L=a+b$, and the distances between the body point and the rear axle, $x$ and $y$.
Starting with: $$R^2=(T+y)^2+(b+x)^2$$
I then define T and b: $$T=\frac{\sin(90-\alpha)\!\cdot\!FR}{\sin(90)}$$ $$b=L-\frac{\sin(\alpha)\!\cdot\!FR}{\sin(90)}$$
FR can be defined as: $$FR=\frac{\sin(90-\beta)\!\cdot\!L}{\sin(\alpha+\beta)}$$
Which is: $$FR=\frac{\sin(90-k\alpha)\!\cdot\!L}{\sin(\alpha+k\alpha)}$$
So T and b are: $$T=\frac{\sin(90-\alpha)\!\cdot\!\sin(90-k\alpha)\!\cdot\!L}{\sin(90)\!\cdot\!\sin(\alpha+k\alpha)}$$ $$b=L-\frac{\sin(\alpha)\!\cdot\!\sin(90-k\alpha)\!\cdot\!L}{\sin(90)\!\cdot\!\sin(\alpha+k\alpha)}$$
From here I can redefine $R^2$ using only the known values plus $\alpha$ and I can show the equation works by manually plugging in a value for $\alpha$. However, I would like to be able to rearrange the equation to solve for $\alpha$ but it seems beyond my abilities. I'd be grateful for any direction on how I might do that.
Thanks.