Looking for rounded corner plane curve with certain properties (SIDESTEPPED) For a project involving simulating traffic lights, I am currently looking for a formula to get a rounded 90-degree corner (to describe the path of a turning car) with certain properties:


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*Defined in terms of elementary functions


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*Cartesian, polar, parametric OK

*Implicit OK if solvable in terms of elementary functions


*Relevant portion fits exactly into square with unit side length (for simplicity, let's assume (0,0) to (1,1), with the endpoints of the relevant portion at (0,1) and (1,0))

*Symmetric about diagonal through "center" of turn (the line $y=x$ for the above simplifying assumptions)

*$\frac{dy}{dx}=0$ at one of the endpoints, and $\infty$ at the other (doesn't matter which one's which, as one can flip the result over the line $y=1-x$ to get the other case)

*Curvature increases until intersection with symmetry axis, and then decreases again

*Zero curvature at endpoints

*Curvature function of $x$, $\theta$, $t$, etc. is continuous (but it can have a cusp in the middle)

*Cusps of physical curve only at endpoints (if they exist)

*Constants should be expressible in terms of combinations of elementary functions of rational numbers and/or predefined values (so $2$, $\sqrt2$, and $\pi$, or even $\sqrt\pi$, are OK, but not something like $2.015287329...$, because I want this to be exact)

*Has arclength formula defined in terms of elementary functions

*Arclength formula is invertible (i.e. what value of $x$, $\theta$, $t$, etc. do I need in order to get an arc of length $\ell$?)


I've tried at least ten or twenty different curves from several different approaches (including polynomials, superellipses, a variant of Cassini ovals with four foci, fitting polar curves to $sec(x)$ and $csc(x)$, quartic Bézier curves, and finding other parametric equations that have zero second derivative at $t=0$, then flipping them across $t=\frac12$), and there's only one that I've tried ($x=\cos^5(t)$, $y=\sin^5(t)$) that has an actual arclength formula ($\frac5{8\sqrt3}(-\cos(2t)\sqrt3\sqrt{1+3\cos^2(2t)}+\ln(\sqrt{1+3\cos^2(2t)}-\cos(2t)\sqrt3))$), but it's not invertible, as there are $t$s both inside and outside of the $\ln$. (However, I may have missed a viable example.) I've tried searching online, without finding what I'm looking for (although it could be that I'm searching for the wrong thing).
 A: The Euler spiral (also known as the Cornu spiral or clothoid) is the natural choice for a smooth transition curve, where by smooth we mean one that will not cause an abrupt change in acceleration (i.e., a  jerk).
The Euler spiral is usually expressed in terms of the Fresnel integral, which can be expressed in closed form as follows,
$$z(s)=\int_0^s e^{i \pi s^2/2} ds=\frac{1+i}{2} erf\left(\frac{1-i}{2} \sqrt{\pi} \cdot s\right)$$
where $s$ is the arc length. Now it's also known that a spiral can be expressed in terms of it curvature, $\kappa$, namely,
$$z=\int e^{i\int \kappa(s)ds}ds$$
It follows that the natural equation of this spiral  is $\kappa \propto s$, or in term of the radius of curvature, $\rho$
$$\rho s=\text{constant}$$
It is this property that we are interested in. When traversing a bend we seek to assure that the centrifugal force changes continuously in order to avoid a jerk. If this easement is not applied there would be an abrupt change in the acceleration at the point of transition from a straight path to a curved one.
