2010-08-21 87 views
-1

我有这个算法,我发现这里只有一两件事让我为难一下:帮助理解这一点?

Clear the stencil buffer to 1. 
    Pick an arbitrary vertex v0, probably somewhere near the polygon to reduce floating-point errors. 
    For each vertex v[i] of the polygon in clockwise order: 
     let s be the segment v[i]->v[i+1] (where i+1 will wrap to 0 when the last vertex is reached) 
     if v0 is to the "right" of s: 
      draw a triangle defined by s, v[i], v[i+1] that adds 1 to the stencil buffer 
     else 
      draw a triangle defined by s, v[i], v[i+1] that subtracts 1 from the stencil buffer 
    end for 
    fill the screen with the desired color/texture, testing for stencil buffer values >= 2. 

By "right of s" I mean from the perspective of someone standing on v[i] and facing v[i+1]. This can be tested by using a cross product: 

cross(v0 - v[i], v[i+1] - v[i]) > 0 

是困惑我的是,我需要绘制由S,V [I],V [定义的三角形的部分我+ 1]。如果S是段v [i] - > v [i + 1]那么这是如何实现的?

感谢

+1

是的,它看起来像一个错误,因为* s *是由v [i]和[i + 1]定义的段。我的镜头是应该有* v0 *而不是* s *。如果你已经提供了关于这个算法应该做什么的更多信息,它也会有很大的帮助。它有影子卷吗? – Rekin 2010-08-21 15:16:35

+0

不,我正在尝试使用模具而不是GLUTESS绘制凹面多边形 – jmasterx 2010-08-21 15:18:20

回答

1

如果我没有记错,你必须绘制三角形分别是V0 - v [I] - v [I + 1]

+0

完美谢谢! – jmasterx 2010-08-21 15:22:00

0

如果我正确地读你的榜样,s是间段2个顶点是三角形的边缘。所以它以顺时针的方式在三角形周围“散步”,填充顶点。

“卷绕” - 顺时针或逆时针 - 确定三角形的法线。

0

烨,它看起来像一个错误,因为s的段由v [I]v [I + 1]定义。这在绘制凹多边形的情况下是有意义的。用v0,v [i],v [i + 1]