I have a 3D geometry related question. Suppose you have a polygon in 3D, determined by the cartesian coordinates of its points $(x_1, .... x_n)$, and edges $((x_1,x_2), \dots, (x_{n-1},x_n), (x_n, x_1))$. Suppose that you fix:
- $\forall i \in [1,n-1]$ , the distances $|x_{i+1} - x_i|$, and ${|x_1 - x_n|}$
- $\forall i \in [1,n-1]$ , the bond angles $\widehat{x_{i-1}x_ix_{i+1}}$ and $\widehat{x_{n}x_1x_{2}}$
Does this uniquely determine the shape of the possible 3D polygon ?
Some thoughts
Since we need $n$ points to define our polygon, we need to determine $3n$ variables (the $x$, $y$ and $z$ variables).
The constraints on distances give us $n$ equations.
The constraints on the bond angles give us $n$ equations.
And because our 3D polygon can rotate and translate in space, we center it around $(0,0,0)$, and fix a vector of spherical coordinates $(\phi, \psi)$ for it. So that is $5$ more equations. So in total we have $2n + 5$ equations.
My questions is : did I miss some constraints ? Why am I obtaining $2n+5$ and not $3n$ ? It feels intuitive that fixing the bond length and bond angles should give a unique 3D shape, could you give me your thoughts on this ?
For those familiar with torsion angles
Instead of characterizing the polygon with cartesian coordinates, we can use spherical coordinates as above, with distances $|x_{i+1} - x_i|$, bond angles, and torsion angles. Another way of putting the question is, does determining the distances and bond angles determine the torsion angles.
Thank you for your answers !
The answer is no.
Let us consider the case of an initialy planar polygon with $10$ sides (see figure below) arranged so as to form a star with 5 branches with angles $\widehat{x_{i-1}x_ix_{i+1}}$ alternately equal to $2 \pi /3$ and $-\pi/3$.
One can turn it into a skew polygon for example by rotating vertex $x_2$ in a space rotation by any angle around line $x_1x_3$ without modifying angle $\widehat{x_{1}x_2x_{3}}$, and without modifying any length...