Android OpenGL ES 2.0屏幕坐标为世界坐标

Sta*_*kJP 20 android coordinate-transformation opengl-es-2.0

我正在构建一个使用OpenGL ES 2.0的Android应用程序,而且我遇到了问题.我正在尝试将屏幕坐标(用户触摸的位置)转换为世界坐标.我尝试过阅读和玩弄GLU.gluUnProject,但我做错了或者只是不明白.

这是我的尝试......

public void getWorldFromScreen(float x, float y) {
    int viewport[] = { 0, 0, width , height};

    float startY = ((float) (height) - y);
    float[] near = { 0.0f, 0.0f, 0.0f, 0.0f };
    float[] far = { 0.0f, 0.0f, 0.0f, 0.0f };

    float[] mv = new float[16];
    Matrix.multiplyMM(mv, 0, mViewMatrix, 0, mModelMatrix, 0);

    GLU.gluUnProject(x, startY, 0, mv, 0, mProjectionMatrix, 0, viewport, 0, near, 0);
    GLU.gluUnProject(x, startY, 1, mv, 0, mProjectionMatrix, 0, viewport, 0, far, 0);

    float nearX = near[0] / near[3];
    float nearY = near[1] / near[3];
    float nearZ = near[2] / near[3];

    float farX = far[0] / far[3];
    float farY = far[1] / far[3];
    float farZ = far[2] / far[3];
}
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我得到的数字似乎不对,这是使用这种方法的正确方法吗?它适用于OpenGL ES 2.0吗?我应该在这些计算之前使模型矩阵成为单位矩阵(Matrix.setIdentityM(mModelMatix,0))吗?

作为后续,如果这是正确的,我如何选择输出Z?基本上,我总是知道我希望世界坐标处于什么距离,但GLU.gluUnProject中的Z参数似乎是近平面和远平面之间的某种插值.它只是一个线性插值?

提前致谢

Ero*_*rol 22

/**
    * Calculates the transform from screen coordinate
    * system to world coordinate system coordinates
    * for a specific point, given a camera position.
    *
    * @param touch Vec2 point of screen touch, the
      actual position on physical screen (ej: 160, 240)
    * @param cam camera object with x,y,z of the
      camera and screenWidth and screenHeight of
      the device.
    * @return position in WCS.
    */
   public Vec2 GetWorldCoords( Vec2 touch, Camera cam)
   {  
       // Initialize auxiliary variables.
       Vec2 worldPos = new Vec2();

       // SCREEN height & width (ej: 320 x 480)
       float screenW = cam.GetScreenWidth();
       float screenH = cam.GetScreenHeight();

       // Auxiliary matrix and vectors
       // to deal with ogl.
       float[] invertedMatrix, transformMatrix,
           normalizedInPoint, outPoint;
       invertedMatrix = new float[16];
       transformMatrix = new float[16];
       normalizedInPoint = new float[4];
       outPoint = new float[4];

       // Invert y coordinate, as android uses
       // top-left, and ogl bottom-left.
       int oglTouchY = (int) (screenH - touch.Y());

       /* Transform the screen point to clip
       space in ogl (-1,1) */       
       normalizedInPoint[0] =
        (float) ((touch.X()) * 2.0f / screenW - 1.0);
       normalizedInPoint[1] =
        (float) ((oglTouchY) * 2.0f / screenH - 1.0);
       normalizedInPoint[2] = - 1.0f;
       normalizedInPoint[3] = 1.0f;

       /* Obtain the transform matrix and
       then the inverse. */
       Print("Proj", getCurrentProjection(gl));
       Print("Model", getCurrentModelView(gl));
       Matrix.multiplyMM(
           transformMatrix, 0,
           getCurrentProjection(gl), 0,
           getCurrentModelView(gl), 0);
       Matrix.invertM(invertedMatrix, 0,
           transformMatrix, 0);       

       /* Apply the inverse to the point
       in clip space */
       Matrix.multiplyMV(
           outPoint, 0,
           invertedMatrix, 0,
           normalizedInPoint, 0);

       if (outPoint[3] == 0.0)
       {
           // Avoid /0 error.
           Log.e("World coords", "ERROR!");
           return worldPos;
       }

       // Divide by the 3rd component to find
       // out the real position.
       worldPos.Set(
           outPoint[0] / outPoint[3],
           outPoint[1] / outPoint[3]);

       return worldPos;       
   }
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这里进一步解释算法.

  • 我的解决方案都是用C++编写的.只是扫描这个答案,它看起来对我来说.所以,我不会发布答案.:] (2认同)