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I have a 3D model which already has every buffer I can dream : positions, normals, uvs, etc. I would like to modify its geometry (moving an existing vertex or creating a new one) to, for instance, chamfer my edges. I can easily compute the new position and the new normal (3D vectors), but I wonder how to compute the new UVs (2D vector), or rather, deduce them from the adjacent vertices UVs.

It could be cool if someone knows how to do it in a generic way. But maybe my specific use case provides more useful data:

enter image description here

In my specific use case, the new vertex (D) will be on the bisector of the angle α (BAC), which I know. Moreover, as I know the 3D position of every vertex, I have all distances AC, AB and AD.

Seems like I should revise my geometry lessons.. Is trigonometry enough to solve this ?

Thanks in advance !

MrBooks
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2 Answers2

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In this specific case you can use ABC triangle to find D. Compute intersection point of Bisector and BC (let's call it E), find its UVs and AE distance. Then interpolate between A and E based on those distances. If the D point is outside of the triangle it may still work in some situations.

In the more general approach I'd include some point (say F) that is above your image, but not shown. Basically you'd want to find the smallest polygon that encloses your new point and use it to compute new UVs.

kolenda
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  • E(u,v) = (B(u,v) + C(u,v)) / 2;; D(u,v) = A(u,v) + (E(u,v) - A(u,v)) * (||AD|| / ||AE||);; Hey you're right, this should work too ! – MrBooks Jan 02 '18 at 15:15
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Finally found this solution that perfectly answers the question !

Using barycentric interpolation it's easy to retrieve the new UVs from the new point position. Here is the generic code in javascript using BabylonJS for vector classes and operations:

var f1 = p1.subtract(f);
var f2 = p2.subtract(f);
var f3 = p3.subtract(f);

var va = BABYLON.Vector3.Cross(p1.subtract(p2), p1.subtract(p3));
var va1 = BABYLON.Vector3.Cross(f2, f3);
var va2 = BABYLON.Vector3.Cross(f3, f1);
var va3 = BABYLON.Vector3.Cross(f1, f2);

var a = va.length();
var a1 = va1.length() / a * Math.sign(BABYLON.Vector3.Dot(va, va1));
var a2 = va2.length() / a * Math.sign(BABYLON.Vector3.Dot(va, va2));
var a3 = va3.length() / a * Math.sign(BABYLON.Vector3.Dot(va, va3));

return uv1.scale(a1).addInPlace(uv2.scale(a2)).addInPlace(uv3.scale(a3));

With:

  • p1, p2, p3 three Vector3 of the triangle point positions
  • f a Vector3 of the new point position
  • uv1, uv2, uv3 three Vector2 of the triangle point uvs

This works for any point on the plane of the triangle p1p2p3, points outside the triangle included.

Obviously, this doesn't work if the three points are aligned, they must form a triangle with an area greater than zero.

MrBooks
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