# Snub cube's angles I am trying to build a snub cube. I have made $$6$$ squares and $$32$$ equilateral triangles (out of perler beads if you're curious). I am trying to figure out the angles at which I adjoin the squares to the triangles, and the triangles to other triangles.

I have found a few formulas, but I think I am a bit overwhelmed by the vocabulary used and do not understand what the listed variables are.

H. Rajpoot says "There is a general expression of the solid angle subtended by the snub cube at any of its $$24$$ vertices is given by the general expression \begin{align}\Omega&=2\sin^{-1}\left(\frac{(1-\sqrt{1-K^2})-\sqrt{2K^2-1}}{K^2\sqrt{2}}\right)+8\sin^{-1}\left(\frac{(1-\sqrt{1-K^2})-\sqrt{4K^2-1}}{2K^2\sqrt{3}}\right)\\&\approx 3.589629551 \space sr,\end{align} where $$K\approx 0.928191378"$$.

and Felix Marin says that the formula to find the angles is $$\cos\left(\vphantom{\Large A}\angle{\rm ABC}\right) = {\left(\vec{A} - \vec{B}\right)\cdot\left(\vec{C} - \vec{B}\right) \over \left\vert\vec{A} - \vec{B}\right\vert\;\left\vert\vec{C} - \vec{B}\right\vert}$$ where $$A$$, $$B$$, and $$C$$ are are vectors $$A:[x_1,y_1,z_1]$$, $$B:[x_2,y_2,z_2]$$, and $$C:[x_3,y_3,z_3]$$.

I suppose, I am completely overwhelmed. I have a sight feeling that finding the 'subtended angle' is not the same as the angle I am trying to find. Is that true? What is $$s$$? $$r$$? Why are $$A$$, $$B$$, & $$C$$ vectors and how do I know which vectors to use?

I saw online, here that the the coordinates for the vertices of a snub cube are all the even permutations of $$(±1, ±1/t, ±t)$$ with an even number of plus signs, along with all the odd permutations with an odd number of plus signs, where $$t ≈ 1.83929$$ is the tribonacci constant.

Are these the values I am supposed to use to find the vectors to use the second equation? Is there an easier way to do this? I fell like I have way over-complicated this.

edit: okay, I found this website that says the square-triangle angle is $$142$$ degrees, $$59$$ minutes and the triangle-triangle angle is $$153$$ degrees, $$14$$ minutes. Would still be stoked to know how on earth to figure this out on my own. thanks!

• I think $sr$ is the unit steradian of solid angles (i.e., $sr$ is not a product of $s$ and $r$). – user614671 Dec 13 '18 at 23:39

A way to obtain the coordinates of the vertices is given here. To find the coordinates of $$B$$, rotate $$A = (1, v, w)$$ by $$\pi/2$$ around the $$z$$-axis to map the blue face to the yellow face, then by $$\pi/2$$ around the $$x$$-axis, then by $$\pi/2$$ around the $$y$$-axis: $$(1, v, w) \to (-v, 1, w) \to (-v, -w, 1) \to (1, -w, v) = B.$$ To find the coordinates of $$C$$, repeat the first two steps above and rotate by $$\pi$$ around the $$z$$-axis to obtain $$(-v, -w, 1) \to (v, w, 1) = C.$$ Then an outward normal to the triangular face is $$\mathbf n = ((1, -w, v) - (v, w, 1)) \times ((1, v, w) - (v, w, 1))$$ and the dihedral angle between a square and an adjacent triangular face is $$\phi_1 = \arccos \frac {\mathbf n \cdot (-1, 0, 0)} {|\mathbf n|} = \pi - \arcsin \sqrt {\frac {n_y^2 + n_z^2} {n_x^2 + n_y^2 + n_z^2}}.$$ The rational function under the square root simplifies to at most a quadratic polynomial in $$v$$ since $$v$$ is a root of a cubic polynomial, giving $$\phi_1 = \pi - \arcsin \sqrt {\frac {2 v} 3}.$$ Similarly, the angle between two adjacent triangular faces is $$\phi_2 = \pi - \arcsin \frac {2\sqrt {1 - v \,}} 3.$$