这是我在 xtk 中取消投影的函数。如果您发现错误,请告诉我。现在有了结果点和相机位置,我应该能够找到我的交点。为了使下一步的计算更快,我将在拾取事件中调用它,因此我只需尝试与给定对象的交集。如果我有时间,我也会尝试测试边界框。
注意:最后几行不是必需的,我可以处理射线而不是点。
X.camera3D.prototype.unproject = function (x,y) {
// get the 4x4 model-view matrix
var mvMatrix = this._view;
// create the 4x4 projection matrix from the flatten gl version
var pMatrix = new X.matrix(4,4);
for (var i=0 ; i<16 ; i++) {
pMatrix.setValueAt(i - 4*Math.floor(i/4), Math.floor(i/4), this._perspective[i]);
}
// compute the product and inverse it
var mvpMatrxix = pMatrix.multiply(mwMatrix); /** Edit : wrong product corrected **/
var inverse_mvpMatrix = mvpMatrxix.getInverse();
if (!goog.isDefAndNotNull(inverse_mvpMatrix)) throw new Error("Could not inverse the transformation matrix.");
// check if x & y are map in [-1,1] interval (required for the computations)
if (x<-1 || x>1 || y<-1 || y>1) throw new Error("Invalid x or y coordinate, it must be between -1 and 1");
// fill the 4x1 normalized (in [-1,1]⁴) vector of the point of the screen in word camera world's basis
var point4f = new X.matrix(4,1);
point4f.setValueAt(0, 0, x);
point4f.setValueAt(1, 0, y);
point4f.setValueAt(2, 0, -1.0); // 2*?-1, with ?=0 for near plan and ?=1 for far plan
point4f.setValueAt(3, 0, 1.0); // homogeneous coordinate arbitrary set at 1
// compute the picked ray in the world's basis in homogeneous coordinates
var ray4f = inverse_mvpMatrix.multiply(point4f);
if (ray4f.getValueAt(3,0)==0) throw new Error("Ray is not valid.");
// return in not-homogeneous coordinates to compute the 3D direction vector
var point3f = new X.matrix(3,1);
point3f.setValueAt(0, 0, ray4f.getValueAt(0, 0) / ray4f.getValueAt(3, 0) );
point3f.setValueAt(1, 0, ray4f.getValueAt(1, 0) / ray4f.getValueAt(3, 0) );
point3f.setValueAt(2, 0, ray4f.getValueAt(2, 0) / ray4f.getValueAt(3, 0) );
return point3f;
};
Edit
Here,在我的存储库中,您可以在camera3D.js和renderer3D.js中找到用于在xtk中进行高效3D拾取的函数。