I have added you to our white list, welcome to the community. Moved this into the OpenGL Forum.
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You could transform feedback. Run a geometry shader after the tesselation shader, record the generated vertices in a separate vbo and read the data back to CPU memory.
Thank you for your suggestion. This would return the list of tessellated triangles.
However I am using a halfedge datastructure to do distance computation on the tessellated result. That datastructure cannot be generated out of a list of loose triangles. Therefore I am computing all tessellated points and incrementally fill up the halfedge list.
Because all GPU vendors have to produce the same tessellation output I thought the equation for the first tessellation point as marked on my attached image is not a secret. Can you share it?
I did some research on OpenGL forums and found a useful hint in https://www.opengl.org/registry/specs/ARB/tessellation_shader.txt
[..] Otherwise, for each corner of the outer
triangle, an inner triangle corner is produced at the intersection of two
lines extended perpendicular to the corner's two adjacent edges running
through the vertex of the subdivided outer edge nearest that corner. [..]
How about non-equlateral triangles? Well, I wasn't the first to ask: https://www.opengl.org/discussion_boards/showthread.php/171470-Triangle-tessellation-in-GL_ARB_tessellation_shade?highlight=tessellation
[..] It really only needs to be done for one triangle, the one with vertices (1, 0, 0), (0, 1, 0), (0, 0, 1). Which is an equilateral triangle where the algorithm works correctly. The resulting vertices can be used as barycentric coordinates for any triangle. So basically, for each tessellation-level parameters, the tessellation only needs to be done once, and the resulting data can be used for any other triangle. [..]
This is not yet the final mathematical equation but a bit of linear algebra will help: build two perpendicular vectors and let them intersect, store uvw coordinates for reuse with any triangle later.
By simplifying the triangle in question it is possible to discard the 3rd dimension and to reduce the system of linear equations to one linear equation:
Point IntersectionFinder::intersectionPoint( const int tesslevel ) const
// given equilateral triangle with:
// A in origin
// length AB = 1
// z coordinate == 0
// AB == 1 --> B = (1;0)
// equilateral: C.x = 0.5 * AB and C.y = AB * sin(60deg) --> ( 0.5; sin(60) )
// observation: iff triangle is equilateral
// all intersections for tessellation meet on ray
// from A through midpoint BC
// because perpendicular rays which meet in intersection points
// always have the same length
// mid( BC ).x = 0.5 * AB + 0.5 BC
// mid( BC ).y = 0.5 * sin(60) --> ( 0.75; 0.5 * sin60 )
// slope( A, mid( BC ) ): due to A being in origin slope m = mid(BC).y / mid(BC).x
// --> 0.5 * sin60 / 0.75 == 1/2 * sin( 60deg ) / 3/4
// --> m = 1/2 * 4/3 * sin( 60deg );
const Scalar m = ( 1.0 / 2.0 ) * ( 4.0 / 3.0 ) * sin( 60.0 * DEG2RAD );
// calculate tessellation intersection point i depending on tessLevel t
// i.x = AB / t with AB = 1 --> i.x(t) = 1 / t
// i.y = i.x * slope --> i.y = m * 1 / t
// i.y(t) = m/t;
double x = 1.0 / static_cast<double>( tesslevel );
double y = m / static_cast<double>( tesslevel );
double z = 0.0;
return Point( x, y, z );
The calculated intersection point is only valid in the given equilateral triangle. Now this point needs to be converted from cartesian coordinates to triangle independent barycentric coordinates. Equations for that are here: http://gamedev.stackexchange.com/questions/23743/whats-the-most-efficient-way-to-find-barycentric-coordinates.