Misunderstanding about Reynolds number I am confused about Reynolds number. I am trying to use the formula to see if NASCAR cars have laminar or turbulent flow. But I am not sure how exactly to use the equation in order to do such calculation. 
 A: Use the Reynolds number equation:
$Re={vL \over \nu}$
where $\nu\approx 1.5\times10^{-5}m^2/s$ is the kinematic viscosity for air. If you enter this into the equation, you end up with 
$Re\approx 67000{v\over{m/s}}{L\over m}$
i.e. for a race car traveling at 40m/s and with a length of 4m it comes out to be around 10 million, which is certainly $Re>>1$, i.e. in a regime far beyond the requirement for the Stokes formula to be a good approximation. 
A: The Reynolds number gives a ratio between forces of inertial origin and those of viscous origin. For a given geometry of the problem, increasing the Reynolds number will lead to turbulent flow from a certain threshold. However, this threshold is strongly dependent on the geometry: this is actually common knowledge, a better design (more "aerodynamic" we say in common language) will lead to lower drag and delayed transition to turbulence.
So calculating the Reynolds number just gives a rough indication whether the flow will be turbulent or not, unless you find $Re<1$ which would mean that inertial forces are negligible and turbulence can never occur (uncommon for human-scale and common fluids).
However, if you do a wind-tunnel experiment with a reduced-size model having the exact geometry of your car (and size $L_R$ say), and determine the Reynolds number $Re^*= V_R^* L_R/\nu$ from which developed turbulence appears, then you can determine the velocity at which the full size car $L_F$ will create turbulence: it will simply be the velocity $V_F=V_F^*$ giving the same Reynolds, $Re = V_F L_F/\nu = Re^*$. 
That said, full size cars are never so well designed that you have laminar flow around them, so yes, expect turbulence around a racing car.
