BorlandTalk.com Borland discussion newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  find adjacent triangle coords to a regular icosahedron trian          BorlandTalk.com Forum Index -> Delphi Graphics View previous topic :: View next topic   Author Message Paul Nicholls Guest Posted: Thu Apr 05, 2007 7:40 am    Post subject: find adjacent triangle coords to a regular icosahedron trian Hi all, I already have code that can create an array of triangles for a N subdivided regular icosahedron; N=0 -> 20 faces, N=1 -> 80 faces, etc. I am now wondering if given this information: icosahedron face = (p0, p1, p2) where p0-p2 are 3d counterclockwise coordinates of the triangle face looking from the outside of the icosahedron icosahedron origin = ox,oy,oz one face edge = p0->p1, p1->p2, or p2->p0 how can I find the missing third point for the adjacent equalateral triangle using the chosen face edge? I already have two of the points using the two edge points that I have chosen... I think it has something to do with the dihedral angle but I am not sure how to compute the missing face coord using this. thanks in advance, Cheers, Paul. "The plastic veneer of civilization is easily melted in the heat of the moment" - Paul Nicholls. paul_nicholls (AT) hotmail (DOT) NOSPAM.com Remove ".NOSPAM" to reply. Back to top Nils Haeck Guest Posted: Sun Apr 08, 2007 8:39 pm    Post subject: Re: find adjacent triangle coords to a regular icosahedron t Hi Paul, You obviously have a list of face triangles, and their vertices P0 P1 and P2. So in order to find a neighbour triangle, you just have to search the list of triangles for another triangle having Pi and Pj. If you store the neighbour pointers in the triangle, you only have to do this once. Having a current, always up-to-date list of these neighbour pointers also helps out later, when you want to do e.g. searches over the surface. It only costs you 3x pointer size per triangle to store these neighbour pointers. This shouldn't be a big thing unless you work with millions of triangles. If I misunderstood, please let me know. Nils "Paul Nicholls" <paul_nicholls (AT) hotmail (DOT) NOSPAM.com> schreef in bericht news:46146197$1 (AT) newsgroups (DOT) borland.com... Quote: Hi all, I already have code that can create an array of triangles for a N subdivided regular icosahedron; N=0 -> 20 faces, N=1 -> 80 faces, etc. I am now wondering if given this information: icosahedron face = (p0, p1, p2) where p0-p2 are 3d counterclockwise coordinates of the triangle face looking from the outside of the icosahedron icosahedron origin = ox,oy,oz one face edge = p0->p1, p1->p2, or p2->p0 how can I find the missing third point for the adjacent equalateral triangle using the chosen face edge? I already have two of the points using the two edge points that I have chosen... I think it has something to do with the dihedral angle but I am not sure how to compute the missing face coord using this. thanks in advance, Cheers, Paul. "The plastic veneer of civilization is easily melted in the heat of the moment" - Paul Nicholls. paul_nicholls (AT) hotmail (DOT) NOSPAM.com Remove ".NOSPAM" to reply. Back to top Paul Nicholls Guest Posted: Wed Apr 11, 2007 8:11 am    Post subject: Re: find adjacent triangle coords to a regular icosahedron t Nils Haeck wrote: Quote: Hi Paul, You obviously have a list of face triangles, and their vertices P0 P1 and P2. So in order to find a neighbour triangle, you just have to search the list of triangles for another triangle having Pi and Pj. If you store the neighbour pointers in the triangle, you only have to do this once. Having a current, always up-to-date list of these neighbour pointers also helps out later, when you want to do e.g. searches over the surface. It only costs you 3x pointer size per triangle to store these neighbour pointers. This shouldn't be a big thing unless you work with millions of triangles. If I misunderstood, please let me know. Nils "Paul Nicholls" <paul_nicholls (AT) hotmail (DOT) NOSPAM.com> schreef in bericht news:46146197$1 (AT) newsgroups (DOT) borland.com... Hi all, I already have code that can create an array of triangles for a N subdivided regular icosahedron; N=0 -> 20 faces, N=1 -> 80 faces, etc. I am now wondering if given this information: icosahedron face = (p0, p1, p2) where p0-p2 are 3d counterclockwise coordinates of the triangle face looking from the outside of the icosahedron icosahedron origin = ox,oy,oz one face edge = p0->p1, p1->p2, or p2->p0 how can I find the missing third point for the adjacent equalateral triangle using the chosen face edge? I already have two of the points using the two edge points that I have chosen... I think it has something to do with the dihedral angle but I am not sure how to compute the missing face coord using this. thanks in advance, Cheers, Paul. "The plastic veneer of civilization is easily melted in the heat of the moment" - Paul Nicholls. paul_nicholls (AT) hotmail (DOT) NOSPAM.com Remove ".NOSPAM" to reply. Hi Nils That is fine if I want to keep the list of triangles and search in them for the neighbouring triangles joined to the specified triangle. I instead want to see if I can calculate any neighbouring triangle to a single specified triangle on the surface of a regular icosahedron just as if I had looked in the equivalent list of linked neighbouring triangles. This will reduce memory requirements if I am looking at a very high subdivision level of the icosahedron - I am going to try some fractal planet coding and I plan on using the nearest triangle to the view point as the 'seed' and create a list of triangles around this one that fit in the view frustum. I am trying to avoid holding a large list of triangles, many that I probably won't need to render at all depending on the zoom. So given: 1. a 'seed' equalatral triangle on the surface of the icosahedron 2. the radius of the icosahedron 3. one of the 3 edges of the seed triangle I want to calculate the neighbour triangle's missing vertex any ideas? My first idea was this: A / \ / \ / \ / \ B/----m----\C | | | | | | | | O A,B,C are the counter clockwise triangle vertices on the surface of the icosahedron m is the midpoint on the chosen edge I want to find the neighbouring triangle to, in this case edge B-C o is the origin of the icosahedron I was thinking that perhaps I could just mirror point A around the axis O-m 180 degrees... do you think this sounds feasible? cheers, Paul. Back to top Nils Haeck Guest Posted: Mon Apr 16, 2007 12:42 am    Post subject: Re: find adjacent triangle coords to a regular icosahedron t Since B and C are both on the sphere, you can mirror A in the plane through points 0-B-C. What you do is find the projection of A on OBC, as well as the height above, then construct a point A' from the projection and the height, but now in direction below. I'm not sure why you could just use a limited number of triangles on a sphere to start with, then subdivide the triangles that you're interested in to have in more detail. Subdivision is easy; find the sphere through the points ABC, and use it's midpoint to construct 3 new triangles. Perhaps you don't want to do this because the resulting triangles are too sharp? Nils Back to top Paul Nicholls Guest Posted: Mon Apr 16, 2007 4:30 pm    Post subject: Re: find adjacent triangle coords to a regular icosahedron t Nils Haeck wrote: Quote: Since B and C are both on the sphere, you can mirror A in the plane through points 0-B-C. What you do is find the projection of A on OBC, as well as the height above, then construct a point A' from the projection and the height, but now in direction below. I'm not sure why you could just use a limited number of triangles on a sphere to start with, then subdivide the triangles that you're interested in to have in more detail. Subdivision is easy; find the sphere through the points ABC, and use it's midpoint to construct 3 new triangles. Perhaps you don't want to do this because the resulting triangles are too sharp? Nils Hi Nils, I have tried to calculate the new triangle adjacent to the given triangle, but this only seems to work with no subdivisions of the original icosahedron, not when 1 or more subdivisions are in effect... I am going to have to try what you suggested and start with a smaller subset of triangles that I will then subdivide further. cheers, Paul. 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