How to Find the Coordinates of an Icosahedron's Faces?

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The discussion focuses on obtaining the coordinates (X, Y, Z) of the faces of an icosahedron programmatically, avoiding manual data entry. The data structure in use is GeodesicSphere::Faces::Vertex::Coordinates, which organizes vertex coordinates for each face. The challenge lies in efficiently assigning values to multiple indices since adjacent faces share vertices. A proposed solution involves designating one vertex as a 'north pole' and systematically traversing the icosahedron to sort the faces, using two vertices from the previous face to calculate the new one. Additionally, a reference is made to the latitude of non-pole nodes being arctan(1/2) and a suggestion to consult Wikipedia for Cartesian coordinates of the icosahedron for further insights.
BJRowlett
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Ive posted this on gamedev.net, and figured Id post here aswell.

What I am trying to figure out, is how to get the coordinates (X, Y, Z) of the faces of an Icosahedron, without having to manually enter all the data.

This wouldn't be as hard, but the data structure for the information is as follows:
GeodesicSphere::Faces::Vertex::Coordinates

An example of this is:
(GeoSphere.FaceIndex[0].A.X, GeoSphere.FaceIndex[0].A.Y, GeoSphere.FaceIndex[0].A.Z)
(GeoSphere.FaceIndex[0].B.X, GeoSphere.FaceIndex[0].B.Y, GeoSphere.FaceIndex[0].B.Z)
(GeoSphere.FaceIndex[0].C.X, GeoSphere.FaceIndex[0].C.Y, GeoSphere.FaceIndex[0].C.Z)
Which holds the X,Y,Z values of the all the vertices on the first Face.

With this, I will have to assign the values to multiple indices coordinates, because adjacent faces share vertices. And I haven't found an easy way to do so.

Also, finding a efficient way of sorting the faces, my idea was to select one node to be a 'north pole' and select any face, then work around the icosahedron, moving down a level, and repeating. This way, I can always use two of the vertices from the previous face, and just calculate the new one.

If you have any ideas that would work better, it would be greatly appreciated.


~BJRowlett
 
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I do recall the latitude of the non-pole nodes are arctan( 1/2 )
 
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