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Dear Mathematicians I need your help for a new sculpture!

I will attach images but first imagine 2 hexagons - where one is rotated 30 deg. They are separated by 12 equilateral triangles. I need to confirm the angle between the 2 faces as show in the image. My computer render calculation makes it 145.222 deg but I need to confirm this is absolutely correct before proceeding to fabrication!

Your help would be greatly appreciated, thank you! Pete

This shows the angle I need to confirm:

enter image description here

This is the computer design angle that I need to confirm:

enter image description here

This is something like what the final sculpture will look like!

enter image description here

This shows how I made it:

enter image description here

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    $\begingroup$ Looks correct. I find the angle you want equals to $\cos^{-1}\left(\frac{1-2\sqrt{3}}{3}\right) \sim 145.22189133^\circ$, matching your own estimate. $\endgroup$ Aug 20 at 10:20
  • $\begingroup$ Thank you for for your generosity in helping me with the problem. It is greatly appreciated. You can see my art at petemoorhouse.co.uk if of interest. Have a great day! $\endgroup$ Aug 20 at 11:24
  • $\begingroup$ I don't see what this question has to do with platonic solids. $\endgroup$ 21 hours ago
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You are requesting a dihedral angle of a (regular) hexagonal antiprism. This is calculated as follows. Consider a hexagon with vertices $v_0, \ldots, v_5$, where $$v_k = \left(\cos \frac{\pi}{3} k, \sin \frac{\pi}{3} k, d \right),$$ and another hexagon with vertices $w_0, \ldots, w_5$ where $$w_k = \left(\cos \frac{\pi}{6} (2k+1), \sin \frac{\pi}{6} (2k+1), -d\right).$$

The distance these hexagons are separated by is $2d$. In order for the lateral triangles to be equilateral, the distance between, say, $v_0$ and $w_0$, must be $1$. This implies $$1 = \left(1 - \cos \frac{\pi}{6}\right)^2 + \left(0 - \sin \frac{\pi}{6}\right)^2 + (2d)^2,$$ hence $$d = \frac{\sqrt{\sqrt{3} - 1}}{2}.$$ The dihedral angle $\theta$ between triangular faces is therefore given by the equation $$\cos \theta = \frac{\vec a \cdot \vec b}{|\vec a| |\vec b|},$$ where $$\vec a = w_5 - \frac{v_0 + w_0}{2} = \left(\frac{\sqrt{3} - 2}{4}, -\frac{3}{4}, -d\right), \\ \vec b = v_1 - \frac{v_0 + w_0}{2} = \left(-\frac{\sqrt{3}}{4}, \frac{2\sqrt{3} - 1}{4}, d \right).$$ This gives $$\vec a \cdot \vec b = - \frac{\sqrt{3}}{4} - d^2, \\ |\vec a| |\vec b| = 1 - \frac{\sqrt{3}}{4} + d^2,$$ hence $$\theta = \arccos \frac{1 - 2 \sqrt{3}}{3} \approx 145.2218913319^\circ. \approx 145^\circ 13' 18.80879''.$$ The equivalent radian measure is approximately $2.534600149715126$ radians.

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  • $\begingroup$ Thank you for for your generosity in helping me with the problem. It is greatly appreciated. You can see my art at petemoorhouse.co.uk if of interest. Have a great day! $\endgroup$ Aug 20 at 11:25
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What you are looking for is the dihedral angle of a hexagonal antiprism. I found this site that contains the complicated formula: $$\cos \alpha = 1-\frac23\left(\frac{3-\tan^2{\frac{90}n}}{4} +\frac{\sin^2 \frac{270}{n}}{ \sin^2 \frac{180}{n}}\right)$$

The results table on that page only goes up to $n=5$, but putting in $n=6$ I get that the angle $\alpha$ is $145.22189133...$, so your software gives the correct angle rounded to three decimal places.

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  • $\begingroup$ Thank you for for your generosity in helping me with the problem. It is greatly appreciated. You can see my art at petemoorhouse.co.uk if of interest. Have a great day! $\endgroup$ Aug 20 at 11:26

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