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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!

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    $\begingroup$ I think $sr$ is the unit steradian of solid angles (i.e., $sr$ is not a product of $s$ and $r$). $\endgroup$ – user614671 Dec 13 '18 at 23:39
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For the mere requested values you might want to have a look here.

A more descriptive way on the derivation of these values might be found already in the old German book of Max Brückner, "Vielecke und Vielflache, Theorie und Geschichte", Leipzig, Teubner Verlag (1900), at page 139.

--- rk

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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 $$\phi_2 = \pi - \arcsin \frac {2\sqrt {1 - v \,}} 3.$$

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