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+--- Thread: [OT]: looking for 3D geometry help! (/thread-14975.html)
[OT]: looking for 3D geometry help! - cyrille de brébisson - 05-12-2020 06:34 AM
Hello,
Sorry for the off topic item here...
I am looking for help on a 3D geometry question...
The 3D figure that I am looking at is formed by 3 uniforlmy spread concentric 3D rings: something similar to the pictures here:
https://www.google.com/search?sxsrf=ALeKk02OWwACh3HPHXf-ErVRfHWaB0gtfQ:1589264548300&source=univ&tbm=isch&q=sudiball&client=firefox-b-d&sa=X&ved=2ahUKEwjS5Ovm163pAhWnzYUKHR_eC_AQsAR6BAgJEAE
The question is: what is the angle between the vertical (vertical here being the main axis of the "ball") plan passing by the horizontal diameter of one of the ring and the plan of said ring (I know, the question is hard to parse, explaining 3D positions in text is hard)!
I know, form measurement that the angle is around 55°, BUT I am looking for the analytic answer... which I have not yet figured out!
Thanks a lot,
Cyrille
RE: [OT]: looking for 3D geometry help! - ijabbott - 05-12-2020 09:16 AM
So it's basically part of a regular, spherical octahedron?
I think the angle you are looking for is the dihedral angle between the base of an equilateral square pyramid (the base bisects a regular octahedron) and one of the sides, which is \( \tan^{-1}(\sqrt 2) \).
Alternatively, the dihedral angle between faces of a regular octahedron is \( \cos^{-1}(-\frac{1}{3}) \), and the angle you want is half of that: \( \frac{\cos^{-1}(-\frac{1}{3})}{2} \).
At least I assume that's the angle you want because it is approx 54.74° whereas its complement is approx 35.26°.
RE: [OT]: looking for 3D geometry help! - cyrille de brébisson - 05-13-2020 06:14 AM
Hello,
Thanks so much, that is exactly what I was looking for.
Cyrille