Obvious constraints for the bond angles of a 3D polygon

Question

I have a 3D geometry question following my previous question.

Suppose we have a 3D polygon, determined by the cartesian coordinates of its points $(x_1, .... x_n)$, and edges $((x_1,x_2), \dots, (x_{n-1},x_n), (x_n, x_1))$. We define the bond lengths as the lengths of the edges , and the bond angles as the angles $\widehat{x_{i-1}x_ix_{i+1}}$ and $\widehat{x_{n}x_1x_{2}}$

Is there an obvious constraint that the bond angles should satisfy in 3D ? In 2D, we know that their sum should be $\pi(n-2$). Wondering if there is such an obvious constraint for 3D case? I cannot think of anything but explicitly writing the equations and this is ugly. Thank you !!

Answer 1

Let $S$ be your sum and $\theta_i = \pi - \widehat{x_{i-1}x_ix_{i+1}}$. The quantity $$K \stackrel{def}{=} \sum \limits_{i=1}^n \theta_i = \pi n - S$$ is essentially the polygonal version of total absolute curvature of a closed space curve.

Approximating your 3D polygon by closed smooth space curves,

• Fenchel's theorem tell us $K \ge 2\pi$. Furthermore, $K = 2\pi$ when and only when the polygon is planar and convex.
• If the polygon forms a knot, then Fáry–Milnor theorem tell us $K \ge 4\pi$.

This means in general, $S$ is bounded from above by $\pi(n-2)$ or $\pi(n-4)$ depends on whether your polygon is unknotted or not. In particular, if $S > \pi(n-4)$, then your polygon is unknotted.