Path of a simple turning car? I am planning to build a small car that needs to travel through three specific points. Obviously, the car will need to travel in a partial circle to do this. In order to do this, I need to calculate the angle for the front wheels. I do not plan on using Ackermann steering, instead, the whole axle will turn slightly with the wheels attached. The angles will all be under ~15 degrees, and as such, both wheels will remain a relatively similar distance from the back axle < ~1-inch difference on a ~10-inch car.
Knowing the length of the car (I'd assume wheel center to wheel center would be needed), as well as the axles length (the front and back axle may or may not end up being different lengths, ie. tricycle vs car), and the angle of rotation of the wheels how can I determine an equation for the path the car will travel through? Additionally, would the fact that the entire axle is tilting slightly, meaning the wheels are not equidistance from the back be significant? If so, is it possible to determine the above calculations knowing that value as well?
Thank you! 
 A: If you prolong the axes of the rear and of the front wheels, the point where they meet is the center of rotation.   
Since the rear axle is fixed and steering is obtained by rotating the whole front axle, we get a scheme as the one shown.

It is clear from the picture that, when the steering angle $\beta$ has a large value, the rear and front turning radii
are sensibly different: an effect that is well learnt by fresh drivers when parking backwards.
The (ideal) kynematics of the car obeys to the following set of equations
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  {\bf u}_{\,T}  = \left( {\cos \alpha ,\sin \alpha } \right)\quad {\bf u}_{\,H}
  = \left( {\cos \left( {\alpha  + \beta } \right),\sin \left( {\alpha  + \beta } \right)} \right) \hfill \cr 
  {\bf n}_{\,T}  = \left( { - \sin \alpha ,\cos \alpha } \right)\quad {\bf n}_{\,H}
  = \left( { - \sin \left( {\alpha  + \beta } \right),\cos \left( {\alpha  + \beta } \right)} \right) \hfill \cr 
  {\bf H} = {\bf T} + L\,{\bf u}_{\,T}  \hfill \cr 
  0 = {\bf n}_{\,T}  \cdot \,{d \over {dt}}{\bf T} = {\bf n}_{\,H}  \cdot {d \over {dt}}{\bf H}
 = {\bf n}_{\,H}  \cdot \left( {{d \over {dt}}{\bf T} + L\,{d \over {dt}}{\bf u}_{\,T} } \right) \hfill \cr}  \right.
}$$
where $\alpha$ and $\beta$ and the other parameters are function of time.
Now, if we assume that the car has a rear traction which provides, through a differential gear, a constant
average speed $v$ at the center of the axle $T$, then
$$
{d \over {dt}}{\bf T} = v\,{\bf u}_{\,T} 
$$
and we get an equation linking $\alpha$ and $\beta$ as
$$
\eqalign{
  & 0 = {\bf n}_{\,H}  \cdot \left( {{d \over {dt}}{\bf T} + L\,{d \over {dt}}{\bf u}_{\,T} } \right) =   \cr 
  &  = \left( { - \sin \left( {\alpha  + \beta } \right),\cos \left( {\alpha  + \beta } \right)} \right) \cdot \left( {v\,\left( {\cos \alpha ,\sin \alpha } \right)
 + L\left( { - \sin \alpha ,\cos \alpha } \right){{d\alpha } \over {dt}}} \right) =   \cr 
  &  =  - v\sin \left( {\alpha  + \beta } \right)\cos \alpha  + L{{d\alpha } \over {dt}}\sin \left( {\alpha  + \beta } \right)\sin \alpha  +   \cr 
  &  + v\cos \left( {\alpha  + \beta } \right)\sin \alpha  + L{{d\alpha } \over {dt}}\cos \left( {\alpha  + \beta } \right)\cos \alpha  =   \cr 
  &  =  - v\sin \beta  + L{{d\alpha } \over {dt}}\cos \beta  \cr} 
$$
i.e.:
$$ \bbox[lightyellow] {  
{{d\alpha } \over {dt}} = {v \over L}\tan \beta 
}$$
For a constant $\beta$ and $\alpha(0)=0$ we get
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  \alpha (t) = \left( {{v \over L}\tan \beta } \right)t = \omega \,t \hfill \cr 
  {d \over {dt}}{\bf T} = v\,{\bf u}_{\,T}  = v\,\left( {\cos \left( {\omega \,t} \right),\sin \left( {\omega \,t} \right)} \right) \hfill \cr 
  {\bf T}(t) = {\bf T}_{\,0}  + {v \over \omega }\left( {\sin \left( {\omega \,t} \right),1 - \cos \left( {\omega t} \right)} \right) \hfill \cr 
  {\bf H}(t) = {\bf T}(t) + L\,\left( {\cos \left( {\omega \,t} \right),\sin \left( {\omega \,t} \right)} \right) \hfill \cr}  \right.
}$$
The trajectories of $T$ and $H$ are concentric circles

