Looking for triangles attached to a vertice

Questions & Answers
1

Hey guys,

taking a mesh into consideration.
If I want to see all triangles a vertice is attached to, I did the following.

Loop start until triangle.length
    if myVertice == mesh.vertices[triangle*] // found the same vertice in anouther or the same triangle*

This works, but I don’t like the idea of itirating to all Vertices for triangle length every time.
Is there another way to solve this Problem?
Thanks, Sinith

2

In general you want to cache the various arrays that the mesh object provides in a local variable when you need to access them more than once.

// C#
var verts = mesh.vertices;
var triangles = mesh.triangles;
for(int I = 0; I < triangles.length; I++)
{
    if (vertex == verts[triangles*])*

// …

}

3

If this actually causes you performance problems, you can always use the solution of " use more memory".

-At startup: build an array the same size as your mesh vertex array.
-Each element of this array stores a set/list of triangles the vertex belongs to.
-Once built, you can simply index to this array (using the vertex index) to get the list of triangles the vertex belongs to, without any need to loop.

4

Hey guys, thx for the quick answers.

I am still in the design phase and going through some ideas and tests before blueprinting everything.

Well the idea is to do a 3d pathfinding on a big sphere and I need this - find the vertices near me - feature. I want to have several objects doing this pathfinding simultaneously and they have to check on every vertices the next near vertices… So I guess this can cause some performance issues. So I could just set it up and then go for “just more memory” solution…

What I have been doing the last hour was to create a sphere from scratch. Here I have the solution to assign each vertices a latitude and a longitud. Just like earth I can calculate very easy the vertices next to the one the characters are currently on.

I just though maybe there is a neat little function that finds the vertices and triangles of near the current one.

Or maybe I just asked the question wrong.

Reading out Glurth answer now… this is a good approach as well. would definitely save up some time. I’ll look into that.

Cheers and again thx for the answers!
Singt