三角形扇的纹理坐标

Question

我试图“填充”我绘制的几何几何体的表面。我正在使用GL_TRIANGLE_FAN原语。 （例如：1个枢纽（中心）点和12个其他点）。我已经计算了0-1区间内每个顶点的纹理坐标。但结果我得到了这个，它有点困惑。我希望获得像image那样的结果。请帮忙，这里有什么问题？ 如何在此类三角测量中计算正确的纹理坐标（GL_TRIANGLE_FAN） 在image红点是我的要点

*代码 - 代码段：

 assert(("CROSS type intersection needs only 5 vertices : center point and "
         "rest points in anticlockwise order", (lp->size() > 5) && (lp->size() < 5)));
 osg::Vec3 vAlong_1,vAlong_2;
 vAlong_1 = (*lp)[1] - (*lp)[4];
 vAlong_2 = (*lp)[1] - (*lp)[2];
 eps = ((*lp)[2] - (*lp)[4]).length() * 0.2 / 2;
 vAlong_1.normalize();
 vAlong_2.normalize();

 _edgeCoords->push_back((*lp)[0]);
 _edgeCoords->push_back((*lp)[1]);
 if (CMF::euclidDistance((*lp)[0],(*lp)[1]) <= CMF::euclidDistance((*lp)[0],(*lp)[2])) {
    float cosAlpha = -(vAlong_1 * vAlong_2);
    float extraLength = ((*lp)[2] - (*lp)[1]).length() * cosAlpha;
     _edgeCoords->push_back((*lp)[1] + vAlong_1 * (eps + extraLength));
     _edgeCoords->push_back((*lp)[2] + vAlong_1 * eps);
 } else {
     float cosAlpha = (vAlong_1 * vAlong_2);
     float extraLength = ((*lp)[2] - (*lp)[1]).length() * cosAlpha;
     _edgeCoords->push_back((*lp)[1] + vAlong_1 * eps);
     _edgeCoords->push_back((*lp)[2] + vAlong_1 * (eps + extraLength));
 }
_edgeCoords->push_back((*lp)[2]);
if (CMF::euclidDistance((*lp)[0],(*lp)[2]) <= CMF::euclidDistance((*lp)[0],(*lp)[3])) {
   float cosAlpha = -(vAlong_1 * vAlong_2);
   float extraLength = ((*lp)[3] - (*lp)[2]).length() * cosAlpha;
    _edgeCoords->push_back((*lp)[2] - vAlong_2 * (eps + extraLength));
    _edgeCoords->push_back((*lp)[3] - vAlong_2 * eps);
} else {
    float cosAlpha = (vAlong_1 * vAlong_2);
    float extraLength = ((*lp)[3] - (*lp)[2]).length() * cosAlpha;
    _edgeCoords->push_back((*lp)[2] - vAlong_2 * eps);
    _edgeCoords->push_back((*lp)[3] - vAlong_2 * (eps + extraLength));
}
_edgeCoords->push_back((*lp)[3]);
if (CMF::euclidDistance((*lp)[0],(*lp)[3]) <= CMF::euclidDistance((*lp)[0],(*lp)[4])) {
   float cosAlpha = -(vAlong_1 * vAlong_2);
   float extraLength = ((*lp)[4] - (*lp)[3]).length() * cosAlpha;
    _edgeCoords->push_back((*lp)[3] - vAlong_1 * (eps + extraLength));
    _edgeCoords->push_back((*lp)[4] - vAlong_1 * eps);
} else {
    float cosAlpha = (vAlong_1 * vAlong_2);
    float extraLength = ((*lp)[4] - (*lp)[3]).length() * cosAlpha;
    _edgeCoords->push_back((*lp)[3] - vAlong_1 * eps);
    _edgeCoords->push_back((*lp)[4] - vAlong_1 * (eps + extraLength));
}
_edgeCoords->push_back((*lp)[4]);
if (CMF::euclidDistance((*lp)[0],(*lp)[1]) <= CMF::euclidDistance((*lp)[0],(*lp)[4])) {
   float cosAlpha = -(vAlong_1 * vAlong_2);
   float extraLength = ((*lp)[4] - (*lp)[1]).length() * cosAlpha;
    _edgeCoords->push_back((*lp)[4] + vAlong_2 * eps);
    _edgeCoords->push_back((*lp)[1] + vAlong_2 * (eps + extraLength));
} else {
    float cosAlpha = (vAlong_1 * vAlong_2);
    float extraLength = ((*lp)[4] - (*lp)[1]).length() * cosAlpha;
    _edgeCoords->push_back((*lp)[4] + vAlong_2 * (eps + extraLength));
    _edgeCoords->push_back((*lp)[1] + vAlong_2 * eps);
}
_edgeCoords->push_back((*lp)[1]);

_tCoords->push_back(osg::Vec2(0.5,0.5));
_tCoords->push_back(osg::Vec2(0.666,0.666));
_tCoords->push_back(osg::Vec2(0.666,1.0));
_tCoords->push_back(osg::Vec2(0.333,1.0));
_tCoords->push_back(osg::Vec2(0.333,0.666));
_tCoords->push_back(osg::Vec2(0.0,0.666));
_tCoords->push_back(osg::Vec2(0.0,0.333));
_tCoords->push_back(osg::Vec2(0.333,0.333));
_tCoords->push_back(osg::Vec2(0.333,0.0));
_tCoords->push_back(osg::Vec2(0.666,0.0));
_tCoords->push_back(osg::Vec2(0.666,0.333));
_tCoords->push_back(osg::Vec2(1.0,0.333));
_tCoords->push_back(osg::Vec2(1.0,0.666));
_tCoords->push_back(osg::Vec2(0.666,0.666));

Answer 1

尝试保持2d位置始终等于每个顶点的纹理坐标。这将确保您的几何图形显示为纹理的无失真切口。然后，您可以根据需要重新缩放和居中网格，而不会通过将变换应用于顶点位置来扭曲纹理。

这样做的一种方法是创建一个推动单个顶点的函数，接受顶点的2d坐标和要应用的任何变换。然后，该函数将2d坐标作为texcoords推，然后转换它们并将结果推送到位置。