Reconstructing curve from curvature as a function of arclength

Question

Let's say we have the curve on the left, for which we can can calculate the curvature as a function of the arc-length using:

$$\kappa = \frac{\frac{d^2y}{dx^2}}{(1 +(\frac{dy}{dx})^2)^{3/2}}$$

But now assume we lose the left figure, and we only have the following information:

• $\kappa(l)$ + sign indicating curvature direction
• $x(l=0)$
• $y(l=0)$

Could we reconstruct the curve using this information?

Answer 1

I seem to have found an answer, adapted from this question:

$$ \alpha (l)=\left( \int \cos\theta (l)dl +a,\int \sin\theta (l)dl +b\right) $$ where

$$ \theta (l) = \int \kappa(l)dl+\phi $$

Here $a=x(0)$, $b=y(0)$ and $\phi$ is the initial heading.

Answer 2

I'm going to add an answer that I like a little better based on the book Differential Geometry by E. Kreyszig, Dover (1991), because I think it's useful.

Let $\mathbf{x}(s)$ be a vector curve in $\mathbb R^2$ parameterized by the scalar arc length $s$ $$\mathbf{x}(s) = \left(x(s),y(s)\right).$$ Then the first derivative of $\mathbf{x}(s)$ with respect to $s$ is $$\mathbf{\dot x}(s) = \frac{d \mathbf{x}(s)}{ds} = \mathbf{t}(s),$$ where $\mathbf{t}(s)$ is the unit tangent vector to the curve $\mathbf{x}(s)$, pointing towards the direction of increasing $s$. The second derivative of $\mathbf{x}(s)$ is given by $$\mathbf{\ddot x}(s) = \frac{d^2 \mathbf{x}(s)}{ds^2} = \mathbf{\dot t}(s) = \kappa(s) \mathbf{n}(s),$$ where $\kappa(s)$ is the signed curvature of the curve and $\mathbf{n}(s)$ is the unit normal vector pointing towards the interior of the circle (of radius of curvature $r(s) = 1/\kappa(s)$) circumscribed by the curve. The normal vector to the curve is also given by rotating the tangent vector by $\pi/2$ in the anticlockwise direction $$\mathbf{n}(s) = \left(-\dot y(s),\dot x(s)\right).$$

Thus, we have a set of coupled non-linear but otherwise ordinary differential equations to solve (numerically) for the original curve, given the boundary conditions $$\mathbf{x}(0) = \mathbf{x}_0,\quad \mathbf{\dot x}(0) = \mathbf{t}_0.$$

The relationship to what was given in the previous answer is the turning angle. The turning angle is the angle at which the tangent vector points with respect to the $+x$ axis $$\mathbf{t}(s) = \left(\cos{\psi(s)},\sin{\psi(s)}\right).$$ Then combining this definiton for $\mathbf{t}(s)$ and the definition of signed curvature we obtain $$\kappa(s) = \frac{d \psi(s)}{ds}.$$

Answer 3

Yes, by direct integration, if $\phi$ the slope/rotation is given/available. Used $s$ in place of $l$ to denote arc length.

$$\kappa = \dfrac{\dfrac{d^2y}{dx^2}}{[1 +(\dfrac{dy}{dx})^2]^{3/2}} $$

$$\text{Integrate, using}\; x= {\cos \phi}{\;ds\;} ,\; y= {\sin \phi}{\;ds\;\;}$$

we can also express curvature directly using intrinsic / natural $ s$ as parameter:

$$\kappa=\dfrac{d\phi}{ds} = K $$ A sketch, approximate/schematic slightly shifted, starting at the origin.

The curve at right shows curvature of the given curve you gave, at right.

The red curved line modification is drawn to more exactly indicate an inflection at R.There is another inflection at I.