How to find parametric equation for a curve with varying radius of curvature

Question

I would like to find the parametric equation for a curve starting at radius of curvature 10 at angle 0 degrees and ending at radius of curvature 100 at 90 degrees. The equation for change in radius of curvature along the path will be specified.

One can imagine this as a 90 degree arc of a circle, except the radius of the circle is changing along the path. This is similar to the concept of Euler curves, except the curvature only changes linearly for Euler curves.

Any suggestions on how to approach this problem?

Edit: Thanks for the many replies! Actually my question is that the rate of change of radius of curvature can be any function. In my case, I'd like it to be a tanh function going from radius r1 to radius r2. The radii I mentioned above are also arbitrary values. I am looking for an analytical or numerical way to approach the generalized problem.

Answer 1

There is a simple curve, the ellipse

$$\begin{cases}x&=&a \cos(t)\\ y&=&b\sin(t)\end{cases} \ \ \ \iff \ \ \ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 \ \ \ \ $$

providing an answer to your issue. Indeed, its curvature is

$$K=\dfrac{ab}{(a^2 \sin^2 t + b^2 \cos^2 t)^{3/2}}\tag{1}$$

(see example 1 in this reference), therefore with max. and min. radii of curvature :

$$a^2/b \ \ \text{ in} \ \ A(a,0) \ \ \ \ \text{ and } \ \ \ \ b^2/a \ \ \text{ in} \ \ B(0,b).$$

It remains to solve : $$a^2/b=10 \ \ \ \text{ and } \ \ \ \ b^2/a=100...$$

giving

$$a=10^{4/3} \approx 21.54 \ \ \ \text{ and } \ \ \ \ b=10^{5/3}\approx 46.42.$$

Edit : the locus of centers of curvature, called the evolute, of an ellipse is an elongated astroid : see the very nice figures there

Answer 2

Natural/intrinsic differential equations need to be used. The Whewell's equation binding tangent rotation $\phi$ and arc length $s$ is more suitable compared to Cesàro equation $(\kappa, s)$ and it can be used for a given variation of curvature thus :

$$\frac{1}{\kappa} = R= \frac{ds}{d \phi} = a \tanh (\phi) \tag1 $$

Integrate using arbitrary boundary conditions ( does not matter because the entire track can be subjected to any Euclidean motion, i.e., rotation and shift in plane) arc length $ s=0 ,\, \phi= -\frac{\pi}{4}$

$$s= a\,log\,\dfrac{\cosh \phi}{\cosh \pi/4} ;\; \cosh \phi= \cosh \pi/4 \cdot e^{s/a}\tag2 $$

and parametric coordinates

$$ x= \int \cos\phi \;ds,\,y= \int \sin\phi \;ds,\,\tag3 $$

Designated constant $a$ is obtained (by numerical solution) of two simultaneous equations for$(\phi_i, a) $ $$ R_1= a \tanh \phi_i,\; R_2= a\tanh (\phi_i+ \pi/2)\tag4 $$

Numerical integration gives the required bent curve.

EDIT1:

The unmodified $\tanh(x)$ has a maximum of $45^\circ$ against the required $90^\circ$ tangent rotation. It does not result in the required rotation, so is an unsuitable choice for specifying the curve.

An Ellipse rotates $90^\circ$ for curvature ratio $ (a/b)^3 $ in a quadrant. Also the director circle is the solution of perpendicular tangents in any ellipse.

Other examples can be suggested: the Cornu spiral and the Involute can accommodate any large rotation between perpendicular tangents.