woodworker wanting to know formulas for odd shaped boxes

Solution 1:

My linear algebra is a bit rusty, but I think it provides a means to solve this if you have Cartesian coordinates for all the verticies.

For a face $A$, with points in Cartesian coordinates $P_{A,1}$, $P_{A,2}$, $P_{A,3}$ going clockwise around it, $\phi_{A,2}$, the internal angle of the face at $P_{A,2}$ is the arccosine of the cross product of the norms of $P_{A,1}$ and $P_{A,3}$ seen as vectors from $P_{A,2}$

$$ \phi_{A,2}=\cos^{-1}\left(\left\|P_{A,1}-P_{A,2}\right\|\times\left\|P_{A,3}-P_{A,2}\right\|\right) $$

For a face $A$, with points in Cartesian coordinates $P_{A,1}$, $P_{A,2}$, $P_{A,3}$ going clockwise around it (any three points on the face if it has more than 3, as long as they are going clockwise), the normal vector $\vec{n}_A$ (the direction the face is facing) is the norm of the dot product of $P_{A,1}$ and $P_{A,3}$ seen as vectors from $P_{A,2}$.

$$ \vec{n}_A=\left\|\left(P_{A,1}-P_{A,2}\right)\cdot\left(P_{A,3}-P_{A,2}\right)\right\| $$

The angle $\theta_{AB}$ between faces $A$ and $B$ with normals $\vec{n}_A$ and $\vec{n}_B$ is the supplement of the angle between the normals, which is the the arccosine of their cross product.

$$ \theta_{AB}=\pi-\cos^{-1}\left(\vec{n}_A\times\vec{n}_B\right) $$

The mitre angle for their shared edge (assuming equal thicknesses of material) is half that $$\frac{\theta_{AB}}{2}$$