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.

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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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?

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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.

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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.

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Any Optometrics grating is available with either standard or CW-type construction for higher  damage thresh­old performance. They can also be produced on substrates with a lower coefficient of thermal expansion for better stability. No damage threshold minimums apply to gratings with an overcoat.

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The numerical aperture of an optical system is a dimensionless number that characterizes the range of angles over which the system can accept or emit light.

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.

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Echelle replica gratings are special low period gratings designed for use in high orders only. The maximum possible resolution is obtained but, in general, it is necessary to use a second grating or prism to separate out the overlapping diffracted orders. Echelle replica gratings are an ideal solution for high-resolution spectroscopy due to their high dispersion in high orders.

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.

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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 $C:[x_3,y_3,z_3]$.

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!

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 $K\approx 0.928191378"$.

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