Calculate the shape of a Tetrahedron given the dihedral angles

Question

Let's suppose we have 5 of the 6 dihedral angles of a tetrahedron, and the vertices A,B,C,D. Where $A = (0,0,0)$ and $B = (1,0,0)$. We also limit $C$ to be of the form: $(x_1,x_2,0)$. and lastly $D = (x_3,x_4,x_5)$. Given the 5 known values of 5 of the dihedral angles, we should be able to calculate the 6th dihedral angle and the values $x_i$. But the math gets fairly complicated and I have not been able to find a correct solution.

So the question is: what are these values?

Any help would be greatly appreciated.

Answer 1

As you did, take $A = (0, 0, 0), B = (1, 0, 0), C = (x_1, x_2 , 0) $, and finally $D = (x_3, x_4, x_5) $.

There are four faces: $ABC$, $ABD$, $BCD$, $CAD$, with unit outward normal vectors $n_1, n_2, n_3, n_4 $. Let assume that $x_5 \gt 0 $, so we can take

$ n_1 = (0, 0, -1) $

The minus sign comes from the fact that we're taking the unit normal to point outward of the tetrahedron.

Now suppose $\theta_1$ is the first given dihedral angle between face $ABC$ and $ ABD $. Let $n_2$ be the outward unit normal vector of $ABD$, then $n_2$ is a rotation of $n_1$ about the $x$ axis by an angle of $\pi - \theta_1$, therefore,

$n_2 = (0, \sin( \pi - \theta_1) , - \cos( \pi - \theta_1) ) = (0, - \sin(\theta_1), \cos( \theta_1) ) $

Let $n_3$ be the outward unit normal vector of $BCD$, then $n_3$, and let $\theta_2$ be the dihedral angle between $ABC$ and $BCD$, then $n_2$ can be viewed as a rotation about the $x$ axis by $\pi - \theta_2$ followed by a rotation by $\phi_1 \in (\dfrac{\pi}{2}, \pi ) $ about the $z$ axis applied to the vector $n_1$. This leads to the expression for $n_2$:

$n_3 = R_z( \phi_1 ) R_x (\pi - \theta_2 ) [0, 0, -1]^T $

Hence,

$ n_3 = Rz ( \phi_1) [ 0, - \sin(\theta_2) , \cos ( \theta_2 ) ]^T \\ = [ \sin(\phi_1) \sin(\theta_2) , - \cos(\phi_1) \sin(\theta_2) , \cos(\theta_2) ]^T $

In exactly the same way, we deduce that $n_4$ the outward normal to face $CAD$ is

$n_4 = [ \sin(\phi_2) \sin(\theta_3) , - \cos(\phi_2) \sin(\theta_3) , \cos(\theta_3) ]^T $

where $ \phi_2 \in ( - \pi , - \dfrac{\pi}{2} ) $

These are the expressions for all the normals, and they are all unit vectors pointing out of the tetrahedron surface.

Next, we want to find $\phi_1 $ and $\phi_2$ using the given angles $\theta_4$ (the dihedral angle between $ABD$ and $BCD$ ) and $\theta_5$ (the dihedral angle between $ABD$ and $CAD$ ).

For that we have the following:

$ \cos(\pi - \theta_4) = - \cos(\theta_4) = n_2 \cdot n_3 $

And similarly,

$ - \cos(\theta_5) = n_2 \cdot n_4 $

It is straightforward to solve these two equations for $\phi_1$ and $\phi_2$ given that $\phi_1 \in (\dfrac{\pi}{2} , \pi )$ and $ \phi_2 \in (- \pi , - \dfrac{ \pi}{2} ) $

Now the angle $\theta_6$ between $BCD$ and $CAD$ follows, it is equal to

$ \theta_6 = \pi - \cos^{-1}( n_3 \cdot n_4 ) $

In addition,

$ \angle CBA = \pi - \phi_1 $

$ \angle CAB = \pi + \phi_2 $

$ \angle ACB = - \pi + \phi_1 - \phi_2 $

It follows that

$ AC = AB \left(\dfrac{ \sin \angle CBA }{\sin \angle ACB } \right) $

Therefore,

$ x_1 = AC \cos \angle CAB $

$ x_2 = AC \sin \angle CAB $

To find $(x_3, x_4, x_5)$ we just intersect the three planes $ABD$, $BCD$ and $CAD$.