# Why is the view header showing triangle count when my mesh is all quads?

I know for a fact I don't have any tris so this thing is wrong, however, I notice that it's the same number as edges. I also find it hard to believe I only have 2 more vertices than I do faces. It takes 4 vertices to make one face and after that 2 for every adjoining face.

• Sry for the noob question but I can't find the answer in the manual May 10, 2014 at 3:30

Without getting too technical, in 3d, there is no such thing as a true quad. Quads and ngons are mere abstractions to make objects easier to manipulate (imagine modeling a car or face using triangles only). At the very core however, every object is comprised of triangles. Blender and other modeling packages abstract these away but they are there.

You might have seen proof of this if you've ever badly transformed a face for example, if you were to paint all 4 vertices of a plane and move a vertex out of place, you will see evidence that there is supporting geometry somewhere holding that "quad" together.

To test, just add a plane and check the header again, you will see that it shows 2 tris, this is accurate because a flat plane in its simplest form is two triangles. If you were to cut it diagonally with say the knife tool, it would still show 2 because now it is still 2 triangles.

Now if you were to create an actual triangle, it would show 1.

In a perfect sphere without any actual triangles, say a cube morphed into a sphere (by subdividing and using To Sphere AltShiftS), the amount of triangles will always be twice that of faces, because remember, by default a quad is 2 triangles.

This might not be the case for edges however, that count depends on how uniform your object is, toplogy etc.

• Ok gotcha. Thanks for the answer. I had read that I should avoid tris and ngons so I was worried I was doing something horrible by having over 3000 tris lol May 10, 2014 at 4:59
• Also suddenly something makes sense. For example in the ear of my image there are pats where the quad looks like its folded in half, because the vertices don't allow for a solid face. Now I realize that triangles make it possible May 10, 2014 at 5:02

About the relation between vertex, edge and face counts: For a simple closed surface, essentially a "deformed" sphere as in your example, but without handles, such as (the surface of) a donut, we have the Euler formula:

vertices + faces = edges + 2

Additionally, in your example we have essentially as many vertices as faces. Just associate each face with its top left vertex, this will pair off most faces and most vertices successfully. Only in a few weird places you will have some vertices left out or being the top left of two faces (or it is unclear where top left is meant to be). But if vertices are approximately the same as faces, then by Euler's equation, edges are approximately twice as numerous as either of them. Now if we subdivide all your faces into triangles, we have triangle = 2 faces because all your faces are in fact quadrangles. Consequently triangles are approximately the same as edges .

The reason you have fewer vertices than expected is because vertices that are in the exact same place as other vertices can often be treated as just one vertex.

This is where index tracking comes in. It's the sort of thing that GPU's rely on to reduce calculations since many of the calculations done on one vertex are also valid for the vertices which are also in the same spot.

Now the information that has already been calculated for one vertex can be fetched from cache memory and reused for all vertices that have matching index numbers.