关于用面积求线段交点的问题

WHGStudio

请教一个问题，实时碰撞检测算法技术一书中，用面积求两线段交点，求告之参数 t 的推导过程
// Test if segments ab and cd overlap. If they do, compute and return
// intersection t value along ab and intersection position p
int Test2DSegmentSegment(Point a, Point b, Point c, Point d, float &t, Point &p)
{
　　// Sign of areas correspond to which side of ab points c and d are
　　float a1 = Signed2DTriArea(a, b, d); // Compute winding of abd (+ or -)
　　float a2 = Signed2DTriArea(a, b, c); // To intersect, must have sign opposite of a1
　　// If c and d are on different sides of ab, areas have different signs
　　if (a1 * a2 < 0.0f)
　　{
　　　　// Compute signs for a and b with respect to segment cd
　　　　float a3 = Signed2DTriArea(c, d, a); // Compute winding of cda (+ or -)
　　　　// Since area is constant a1 - a2 = a3 - a4, or a4 = a3 + a2 - a1
　　　　// float a4 = Signed2DTriArea(c, d, b); // Must have opposite sign of a3
　　　　float a4 = a3 + a2 - a1;
　　　　// Points a and b on different sides of cd if areas have different signs
　　　　if (a3 * a4 < 0.0f)
　　　　{
　　　　　　// Segments intersect. Find intersection point along L(t) = a + t * (b - a).
　　　　　　// Given height h1 of a over cd and height h2 of b over cd,
　　　　　　// t = h1 / (h1 - h2) = (b*h1/2) / (b*h1/2 - b*h2/2) = a3 / (a3 - a4),
　　　　　　// where b (the base of the triangles cda and cdb, i.e., the length
　　　　　　// of cd) cancels out.
　　　　　　t = a3 / (a3 - a4);// 此公式是怎么推导出来的？
　　　　　　p = a + t * (b - a);
　　　　　　return 1;
　　　　}
　　}
　　// Segments not intersecting (or collinear)
　　return 0;
}

stuartc

t = h1 / (h1 - h2) = (b*h1/2) / (b*h1/2 - b*h2/2) = a3 / (a3 - a4)
那个b指的是点c和点d的距离。h1是三角形cda的高，h2是三角形cdb的高。因为这两个三角形有相同的底，所以可以用面积来算，免得把h1和h2算出来。

WHGStudio

stuartc 发表于 2013-5-26 18:02
t = h1 / (h1 - h2) = (b*h1/2) / (b*h1/2 - b*h2/2) = a3 / (a3 - a4)
那个b指的是点c和点d的距离。h1是三 ...

关键是怎么推导出 t = h1 / ( h1- h2)的？为什么参数t可以用h1和h2来表示？

stuartc

那是因为相似三角形的原因。把那两个三角形的高画出来，然后那两个里面的直角三角形相似，对应边的长成比例。设l1、l2为斜边的长就能推导出t = l1 / (l1 + l2)。因为 l1/l2 = h1/h2，也可以用h1和h2来表示。