Yes, can be regarded as the portion of area belongs to the corresponding point divided by the total area of triangle.
0x484884
When I first saw this, the solution I came up with was to first subtract xj from x (to move xj to the origin) and then do a change of basis to represent x in terms of xi - xj xk - xj. Then we could use the coefficient for xi - xj to weight xi, xk - xj to weight xk, and the remainder (since they should sum to 1 as any point in the triangle is a convex combination of the vertices) to weight xj. Would this be equivalent to these strategies? It seems like it should be reasonably efficient...
keenan
@0x484884 Not sure I totally follow your idea. Maybe you can write out a derivation of barycentric coordinates using your approach, to show that it's equivalent to the other two?
Yes, can be regarded as the portion of area belongs to the corresponding point divided by the total area of triangle.
When I first saw this, the solution I came up with was to first subtract xj from x (to move xj to the origin) and then do a change of basis to represent x in terms of xi - xj xk - xj. Then we could use the coefficient for xi - xj to weight xi, xk - xj to weight xk, and the remainder (since they should sum to 1 as any point in the triangle is a convex combination of the vertices) to weight xj. Would this be equivalent to these strategies? It seems like it should be reasonably efficient...
@0x484884 Not sure I totally follow your idea. Maybe you can write out a derivation of barycentric coordinates using your approach, to show that it's equivalent to the other two?