What is the solid angle at a vertex in a snub cube?

Question

I was able to find or derive an expression for most other Platonic or Archimedean solids, but for the snub cube I was not able to find a value nor find an expression anywhere.

Answer 1

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 $$\boxed{\Omega=8\sin^{-1}\left(\frac{(2x-1)\sqrt{3x^2-1}}{\sqrt{3}(4x^2-1)}\right)+2\sin^{-1}\left(\frac{(2x-1)\sqrt{2x^2-1}}{4x^2-1}\right) \space sr}$$ where, $x$ is a parametric variable of an 8th degree polynomial obtained from HCR's Theory of Polygon, given as follows $$128x^8-352x^6+256x^4-72x^2+7=0\ \ \quad \forall \ x>1$$ After $5$ successive iterations using Newton-Raphson's method, the value of $x$ can be fairly approximated as follows
$x\approx 1.343713374$

Now, substituting the above value of $x$, one should get the approximate value of solid angle subtended by the snub cude at any of its 24 vertices ( all lying on the spherical surface) as follows

$$\Omega=\left.8\sin^{-1}\left(\frac{(2x-1)\sqrt{3x^2-1}}{\sqrt{3}(4x^2-1)}\right)+2\sin^{-1}\left(\frac{(2x-1)\sqrt{2x^2-1}}{4x^2-1}\right)\right|_{x\approx 1.343713374} \\ \approx 3.589629551 \space sr$$

For solid angles subtended by all the Archimedean solids, kindly go through Table of solid angles subtended by Archimedean solids at their vertices by HCR