5 replies to this topic

[313230]

Which is more?

 

Is that number of vertices always triples the number of triangles?

Thanks

[rdragon1]

No, vertices can be shared between adjacent triangles, as long as all attributes match (position, normal, uv, whatever). For example, vertices along a 'hard edge' (not shared normals) are not shared between adjacent triangles. At maximum, numVertices = numTriangles*3, but there is as much sharing as possible.

[RulerOfNothing]

You always have more vertices than triangles (assuming all your geometry is made up of triangles), but it is possible to reuse vertices between triangles (by having all triangles with a common vertex or by having consecutive triangles share 2 of their vertices) so there can be less than three vertices per triangle.

[Álvaro]

You always have more vertices than triangles (assuming all your geometry is made up of triangles)

Even that's not true. An icosahedron has 12 vertices but 20 triangles.

[MJP]

You always have more vertices than triangles (assuming all your geometry is made up of triangles)

 

Not only is that false, but it's actually opposite of the common case. Typically you have more triangles than vertices.

[313230]

Many thanks