About a week ago I had to get uv coordinates accurately transferred on to a new set of points.
I knew these points lay on the existing surface, and that a nearest neighbour solution would cause problems, because a point can be closer to a different vert than the primitive it sits on. I also knew I had to be able to correctly interpolate quads and triangles.

After consulting the web I found out that you could use the triangle’s Barycentric coordinates and the quadratic’s filtered uv position to correctly interpolate the vertices that define a primative (or face). Both methods return a set of uv coordinates that help define the point’s relative position on that sampled face.

Remember that with this particular uv set I mean the uv set relative to a certain primitive or face. Not the uv set that defines a vert’s position in 2d space (relative to other vertices).

In the case of a triangle (defined by: A,B,C), the uv set specifies the weight of every vert, say: (U: 0.2, V: 0.5, W: 0.3). Note that for simplicity the last value is often left out because if the point lies within the triangle, the seperate weights always add up to one. Coming back on the previous weights, we can write the third weight (W) as: 1-(0.2+0.5).

You could also say that the uv coordinates describe the distance you have to walks over the edge of the triangle starting from vert A to vert B and C. If the value of u or v (where u = location on edge: A-B, and v = location on edge: C-A) ends up to be higher or lower than one, the point lies outside of the triangle.

The point’s position (P) can now easily be calculated: P = (B*U) + (C*V) + (A*W)

For a quadratic primitive the uv coordinates don’t describe the vert weights directly. The set only describes the relative position of the point based on the primitive’s local uv space. But by using a simple bilinear interpolation we can use these coordinates to calculate that point’s position (or interpolated uv values). We start by defining two new points in the u axis, and use these points to find the final weight in the v axis.

Say we have a square defined by: A, B, C and D and a uv set that defines a point in this square with the coordinates: (U)0,2 and (V)0,5. It’s a right sided primitive. We can figure out the two new points (P1 and P2) in the u direction by using the primtive’s edges: P1 = (B-A)*U, P2= (D-C)*U. But this only get’s us half way. We have the interpolated u value as a set of 2 points, but we can use these points as an edge to get the final position (PF) (or whatever data you see fit): PF = (P2-P1)*V. And that’s it…

But this doesn’t explain how to get those uv coordinates. Houdini has some build in functions for this that can be accessed with VEX, C++ and Python. For this example I will stick with Python. But you could also write your own. A good example on finding the Barycentric coordinates of a triangle can be found here. Finding the coordinates for a quadratic primitive is more diffifcult and depends on the way the primitive is described. Doing a simple google search on: ‘parametric surface’ will get you half way. This wiki article is also a good starting point.

Now for some code. I wrote a UV Interpolator in Python that corrrectly interpolates the uv attribute from a set of vertices. I already know the primitve to sample, so I don’t perform any unnecessary ray casts. The operator works with both quads and triangles. The function that is used to get the uv coordinates local to the primitive is: hou.Prim.nearestToPosition(Point). In vex these coordinates are generated when doing an intersect operation.