我有：

在屏幕坐标x，y（0,0是屏幕中间，1,1是左上角） 屏幕尺寸 在摄像头位置矢量 在摄像头的外观载体 投影矩阵 模型视图矩阵 Y = 0平面正常的（0,1,0,0） 的Y = 0平面位置（0,0,0,0）

和我希望得到在y = 0平面的，其中我点击我的窗口中的位置（x，0，Z）（应该是一条线 - 面相交，但考虑到相机性能）。

烦人，我没有访问GLU要求unprojecting等。只是基本向量和矩阵计算。我并不真的需要确切的code因为如此，只是技术 - 作为一个线平面交叉点是很容易做到的。找到，它从眼通过屏幕上的点线是困难的部分。

我还以为是只使用由摄像机位置投射了一缕相机的外观载体，但是这并没有考虑到鼠标的坐标。所以，我需要考虑到相机的视野呢？

解决方案

  //获得的视线，通过鼠标光标线

闪烁视[4];
GLdouble mvmatrix [16]，projmatrix [16];
闪烁真的; / * OpenGL的y坐标位置* /
GLdouble WX，WY，WZ; / *返回世界X，Y，Z COORDS * /
GLdouble WX2，WY2，WZ2; / *返回世界X，Y，Z COORDS * /
glGetIntegerv（GL_VIEWPORT，视口）;
glMatrixMode（GL_MODELVIEW）;
glLoadIdentity（）;
glGetDoublev（GL_MODELVIEW_MATRIX，mvmatrix）;
glGetDoublev（GL_PROJECTION_MATRIX，projmatrix）;
/ *注视[3]是在像素窗口的高度* /
真的=视[3]  - （闪烁）point.y  -  1;
gluUnProject（（GLdouble）point.x，（GLdouble）真的，0.0，
    mvmatrix，projmatrix，视口，和放大器; WX，和放大器; WY，和放大器; WZ）;
//的printf（设为z世界COORDS = 0.0是（％F，％F，％F）\ N，
// WX，WY，WZ）;
gluUnProject（（GLdouble）point.x，（GLdouble）真的，1.0，
    mvmatrix，projmatrix，视口，和放大器; WX2，和放大器; WY2，和放大器; WZ2）;
//的printf（设为z世界COORDS = 1.0是（％F，％F，％F）\ N，
// WX，WY，WZ）;

//视线交汇与Y = 0平面线

双F = WY /（WY2  -  WY）;
双X2D = WX  -  F *（WX2  -  WX）;
双z2d = WZ  -  F *（WZ2  -  WZ）;
 

point.x，point.y是鼠标屏幕共同ords，0.0是左上角。

在code假设Y = 0平面填充视口。 （你是看不起世界，从一个狭窄的飞机端口，看不到任何的天空。）如果Y = 0平面不填写的视口，你要测试的x，y位置是天上。（WY2 - WY＆LT;小的值），​​并做适当的事情为你的应用

I have:

screen coordinates x,y (0,0 being the middle of the screen, 1,1 being top left) screen dimensions camera location vector camera look vector projection matrix ModelView matrix y=0 Plane normal of (0,1,0,0) y=0 plane location of (0,0,0,0)

And am looking to get the location (x,0,z) on the y=0 plane of where i click on my window (should be a line - plane intersection, but has to take into account the camera properties).

Annoyingly, I don't have the access to the GLU calls for unprojecting and the like. Just basic vector and matrix calculations. I don't really need the exact code as such, but just the technique - as a line plane intersection is easy enough to do. Finding the line that goes from the eye through the point on the screen is the hard part.

I thought it was just using the camera look vector for a ray projected from the camera location, but this doesn't take into account the mouse co-ordinates. So do i need to take into account the camera FOV too?

解决方案

// get line of sight through mouse cursor

GLint viewport[4];
GLdouble mvmatrix[16], projmatrix[16];
GLint realy;  /*  OpenGL y coordinate position  */
GLdouble wx, wy, wz;  /*  returned world x, y, z coords  */
GLdouble wx2, wy2, wz2;  /*  returned world x, y, z coords  */
glGetIntegerv (GL_VIEWPORT, viewport);
glMatrixMode(GL_MODELVIEW);
glLoadIdentity();
glGetDoublev (GL_MODELVIEW_MATRIX, mvmatrix);
glGetDoublev (GL_PROJECTION_MATRIX, projmatrix);
/*  note viewport[3] is height of window in pixels  */
realy = viewport[3] - (GLint) point.y - 1;
gluUnProject ((GLdouble) point.x, (GLdouble) realy, 0.0,
    mvmatrix, projmatrix, viewport, &wx, &wy, &wz);
//printf ("World coords at z=0.0 are (%f, %f, %f)\n",
//  wx, wy, wz);
gluUnProject ((GLdouble) point.x, (GLdouble) realy, 1.0,
    mvmatrix, projmatrix, viewport, &wx2, &wy2, &wz2);
//printf ("World coords at z=1.0 are (%f, %f, %f)\n",     
//  wx, wy, wz);

// line of sight intersection with y = 0 plane

double f = wy / ( wy2 - wy );
double x2d = wx - f * (wx2 - wx );
double z2d = wz - f * (wz2 - wz );

point.x, point.y are the mouse screen co-ords, 0.0 being top left.

The code assumes that the y = 0 plane fills the viewport. ( You are looking down on the world from a narrow aircraft port and cannot see any of the sky. ) If the y=0 plane does NOT fill the view port, you have to test for the x,y location being 'in the sky' ( wy2 - wy < small value ) and do something appropriate for your application.