Post Reply 
[OT]: looking for 3D geometry help!
05-12-2020, 06:34 AM
Post: #1
[OT]: looking for 3D geometry help!
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=ALeK...AR6BAgJEAE

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

Although I work for the HP calculator group, the views and opinions I post here are my own. I do not speak for HP.
Find all posts by this user
Quote this message in a reply
05-12-2020, 09:16 AM (This post was last modified: 05-12-2020 02:07 PM by ijabbott.)
Post: #2
RE: [OT]: looking for 3D geometry help!
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°.

— Ian Abbott
Find all posts by this user
Quote this message in a reply
05-13-2020, 06:14 AM
Post: #3
RE: [OT]: looking for 3D geometry help!
Hello,

Thanks so much, that is exactly what I was looking for.

Cyrille

Although I work for the HP calculator group, the views and opinions I post here are my own. I do not speak for HP.
Find all posts by this user
Quote this message in a reply
Post Reply 




User(s) browsing this thread: