reconstructing space curves from curvature and torsion

Question

Given $\kappa (s)$ and $\tau (s)$ and a frenet apparatus $\lbrace T_0,N_0,B_0 \rbrace$, how can you reconstruct a space curve?

I know I need to use the frenet-serret equations, but I can't put a finger on it. Thanks.

Answer 1

You need a starting point $\gamma_0$ of your curve $\gamma$. Then you can solve the initial value problem: $$\begin{align}\gamma'(s) &= T(s), \\ T'(s) &= \kappa(s) N(s), \\ N'(s) &= -\kappa(s)T(s) + \tau(s)B(s), \\ B'(s) &= -\tau(s)N(s), \\ \gamma(0) = 0,\, T(0) &= T_0,\, N(0) = N_0,\, B(0) = B_0\end{align}$$

Answer 2

What you want to do is reconstruct the curve from it's so-called "natural equation". This page shows you how, or you can find this in almost any decent book on differential geometry.

The location of the start point and the initial Frenet frame will be enough. You reconstruct the curve by several integration steps, and the initial conditions provide the constants of integration. Without them, you don't know the position and orientation of the curve.

You won't be able to get explicit equations for the curve (unless you have explicit equations for the curvature and torsion); your answer will be in terms of the integrals I mentioned.

Answer 3

As already noted by previous answerers, in general one is unable or unwilling to obtain closed-form solutions for the differential equations that result from Frenet-Serret (except in the simplest cases, like helices). In practice, what one often does is to resort to numerical methods (e.g. Runge-Kutta) if all one wants is to see the curve with a given curvature and torsion function. The initial conditions for these differential equations are obtained by placing your Frenet trihedral in some convenient position. One common practice is to place the inital trihedral such that the curve passes through the origin, and the three vectors are coincident with the axes.