What is the minimal G-force curve in 2-dimensional space? Given two parallel roads, which need to be connected, what shape of curve would produce the minimum overall horizontal G-force(s) on travelers?

Is it a $sin$ or $cos$ wave?
Is it a basic cubic function?
Is it something else?
I'm working on an engineering project, not actually involving roads, but the road analogy is easier to understand than my actual project (it involves more complex topics, like aerodynamics, which would just confuse the problem needlessly).
 A: The answer is two arcs. One arc with a constant gee loading in one direction and then flipping to the opposite direction. This is called the bang-bang method, and it is no very smooth, but the gee forces never exceed the specified maximum.
Given a path $y(x)$ the instantaneous radius of curvature at each x is
$$ \rho = \frac{ \left(1+ \left(\frac{{\rm d}y}{{\rm d}x}\right)^2 \right)^\frac{3}{2} }{ \frac{{\rm d}^2 y}{{\rm d}x^2} } $$
The lateral acceleration is $a_L = \frac{v^2}{\rho}$ so we are comparing paths using the parameter $\gamma = \frac{L}{\rho}$
Here are some possible curves (use $L$ for the transition length, and $W$ for the step width)
$$ \begin{align} y(x) &= \tfrac{W}{2} \sin \left(\frac{\pi x}{L} \right)  & \text{harmonic}\\
y(x) &= \begin{cases} -\frac{L^2-W^2}{4 W} + \sqrt{ \left( \frac{(L^2+W^2)^2}{16 W^2}-\left( \frac{L}{2}+x \right)^2 \right)} & x<0 \\ \frac{L^2-W^2}{4 W} - \sqrt{ \left( \frac{(L^2+W^2)^2}{16 W^2}-\left( \frac{L}{2}-x \right)^2 \right)} & x>0 \end{cases} & \text{arcs} \\
y(x) &=- \tfrac{W}{2} \frac{{\rm erf}\left(\frac{2 \pi x}{L} \right)}{{\rm erf}(\pi)} & \text{smooth} \end{align} $$
Above ${\rm erf}(x)$ is the error-function 
$$\begin{align} \frac{L}{\rho(x)} & = \frac{4\pi^2 L^2 W \sin\left( \frac{\pi x}{L} \right)}{ \left( \pi^2 W^2 \cos^2 \left(\frac{\pi x}{L}\right)+4 L^2\right)^\frac{3}{2}} & \text{harmonic} \\
\frac{L}{\rho} & = \pm \frac{4 L W}{L^2+W^2} & \text{arcs} \\
\frac{L}{\rho(x)} & = \frac{ 16 L W x \pi^\frac{5}{2} {\rm e}^\frac{8 \pi^2 x^2}{L^2} {\rm erf}(\pi)^2}{\left( L^2 {\rm e}^\frac{8 \pi^2 x^2}{L^2} {\rm erf}(\pi)^2+4 \pi W^2\right)^\frac{3}{2}} & \text{smooth} \end{align} $$
The peak for the harmonic is $\frac{L}{\rho} = \frac{\pi^2 W}{2 L}$ at $x=\frac{L}{2}$ which is always a higher value than the arcs solution.
The peak for the smooth is not easy to find analytically, but for some test cases I looked at it was much higher than the arcs solution.
