Assembly: Microsoft.Xna.Framework (in microsoft.xna.framework.dll)
public static float Barycentric ( float value1, float value2, float value3, float amount1, float amount2 )
- The coordinate on one axis of vertex 1 of the defining triangle.
- The coordinate on the same axis of vertex 2 of the defining triangle.
- The coordinate on the same axis of vertex 3 of the defining triangle.
- The normalized barycentric (areal) coordinate b2, equal to the weighting factor for vertex 2, the coordinate of which is specified in value2.
- The normalized barycentric (areal) coordinate b3, equal to the weighting factor for vertex 3, the coordinate of which is specified in value3.
About Barycentric Coordinates
Given a triangle with vertices V1, V2, and V3, any point P on the plane of that triangle can be specified by three weighting factors b1, b2, and b3, each of which indicates how much relative influence the corresponding triangle vertex contributes to the location of the point, as specified in the following formulas.
Px = (b1 * V1x) + (b2 * V2x) + (b3 * V3x); Py = (b1 * V1y) + (b2 * V2y) + (b3 * V3y); Pz = (b1 * V1z) + (b2 * V2z) + (b3 * V3z);
Such triple-weighting factors b1, b2, and b3 are called barycentric coordinates.
Barycentric coordinates express relative weights, meaning that (k * b1), (k * b2), and (k * b3) are also coordinates of the same point as b1, b2, and b3 for any positive value of k.
If a set of barycentric coordinates is normalized so that: b1 + b2 + b3 = 1, the resulting coordinates are unique for the point in question, and are known as areal coordinates. When normalized in this way, only two coordinates are needed, say b2 and b3, since b1 is known to equal (1 − b2 − b3).
What MathHelper Barycentric Does
On an axis a (a being the x-, y-, or z-axis), the MathHelper.Barycentric method takes three triangle vertex coordinates on that axis (V1a, V2a, and V3a), and two areal coordinates b2 and b3 of some point P (b2 is the amount1 argument, and b3 is the amount2 argument). The b2 coordinate relates to vertex V2, and the b3 coordinate relates to V3.
Barycentric then calculates the Cartesian coordinate of Pa as follows:
Pa = ( (1 - b2 - b3) * V1a ) + (b2 * V2a) + (b3 * V3a);
Thus, to calculate the x-axis Cartesian coordinate of P, you would pass the x-coordinates of the triangle vertices to Barycentric together with the approproiate areal coordinates of P.