Go-GL"项目"给出意外结果的方法

时间:2016-07-20 03:14:14

标签: opengl go go-gl

Go-GL的Project方法给了我意想不到的大屏幕坐标。

总结:

// Screen is 800x600.
projection := mgl32.Perspective(
    mgl32.DegToRad(45), // Field of view (45 degrees).
    800.0 / 600.0,      // Aspect ratio.
    0.1,                // Near Z at 0.1.
    10)                 // Far Z at 10.
camera := mgl32.LookAtV(
    mgl32.Vec3{0, 0.1, 10}, // Camera out on Z and slightly above.
    mgl32.Vec3{0, 0, 0},    // Looking at the origin.
    mgl32.Vec3{0, 1, 0}     // Up is positive Y.
model := mgl32.Ident4()     // Simple model matrix, to avoid confusion.
modelView := camera.Mul4(model) // The model-view matrix (== camera, here).

// Okay, so what does the origin translate to? Expect center-of-screen, with arbitrary-seeming depth.
origin := mgl32.Vec3{0, 0, 0}
screenOrigin := mgl32.Project(origin, modelView, projection, 0, 0, 800, 600)
fmt.Printf("Origin: (%v, %v, %v)", screenOrigin[0], screenOrigin[1], screenOrigin[2])

// What about the point 5 to the right of the origin?
// Expect increased X, but still less than screenWidth.
// Y and Z should be the same as for the origin.
// In my actual program, I drew (-1,-1,-1)-(1,1,1) cube at (5,0,0) and
// it is completely visible in the window.
test := mgl32.Vec3{5, 0, 0}
screenTest := mgl32.Project(test, modelView, projection, 0, 0, 800, 600)
fmt.Printf("Test:   (%v, %v, %v)", screenTest[0], screenTest[1], screenTest[2])

结果?

Origin: (400, 300, 5.500255)
Test:   (4021.32, 300, 5.500255)

4021.32?这就是离开屏幕的方式!

我几乎消除了我能想到的所有变量,除了立方体渲染我用作提示basicall。我的代码基于此,但我移动了相机和多维数据集:https://github.com/go-gl/examples/tree/master/gl41core-cube

如果我使用(0,0,0)作为输入向量并将Ident4()转换为+ X中的5个单位(这是有意义的),我会得到相同的结果。

那么我做错了什么?根据立方体的位置,我会想到X = 625。

1 个答案:

答案 0 :(得分:3)

出于好奇,我查看了mgl32的源代码。似乎project方法完全错误:

obj4 := obj.Vec4(1)

vpp := projection.Mul4(modelview).Mul4x1(obj4)
win[0] = float32(initialX) + (float32(width)*(vpp[0]+1))/2
win[1] = float32(initialY) + (float32(height)*(vpp[1]+1))/2
win[2] = (vpp[2] + 1) / 2

正如人们可能注意到的那样,在矩阵乘法之后缺少透视分割,因此只要w != 1.0f,该函数就不起作用。

在第一个例子中,由于x和y在矩阵乘法后为零,因此无法注意到误差,因此除以w不会产生差异(z除外)。在第二个示例中,x坐标不再为零,并且错误可见。