DirectX11：沿z轴平移会导致顶点变形

Question

我正在渲染一个矩形棱镜并翻译它。然而，当我将它从相机中移开时，有时模型会做一些意想不到的事情;它会延伸或根本不翻译。它似乎都取决于顶点的z坐标。如果模型的正面从2.0f开始，则模型转换得很好。但是，如果前部处于最小Z距离（1.0f），则模型将伸展，并且不会翻译相同的面部。如果前面在1.0f后面，模型将根本不显示在屏幕上。

这是我的模型数据：

前三个浮点数是位置，接下来三个是正常值，最后两个是紫外线对

VertexData cubeData[] = 
{

    //back
    {-0.5f,-0.5f,1.0f,0.0f,0.0f,1.0f,0.0f,0.0f},
    {-0.5f, 0.5f,1.0f,0.0f,0.0f,1.0f,0.0f,0.0f},
    { 0.5f, 0.5f,1.0f,0.0f,0.0f,1.0f,0.0f,0.0f},
    {-0.5f,-0.5f,1.0f,0.0f,0.0f,1.0f,0.0f,0.0f},
    { 0.5f,-0.5f,1.0f,0.0f,0.0f,1.0f,0.0f,0.0f},
    { 0.5f, 0.5f,1.0f,0.0f,0.0f,1.0f,0.0f,0.0f},

    //front
    {-0.5f,-0.5f,-1.0f,0.0f,0.0f,-1.0f,0.0f,0.0f},
    {-0.5f, 0.5f,-1.0f,0.0f,0.0f,-1.0f,0.0f,0.0f},
    { 0.5f, 0.5f,-1.0f,0.0f,0.0f,-1.0f,0.0f,0.0f},
    {-0.5f,-0.5f,-1.0f,0.0f,0.0f,-1.0f,0.0f,0.0f},
    { 0.5f,-0.5f,-1.0f,0.0f,0.0f,-1.0f,0.0f,0.0f},
    { 0.5f, 0.5f,-1.0f,0.0f,0.0f,-1.0f,0.0f,0.0f},

    //left
    {-0.5f,-0.5f, 1.0f,-1.0f,0.0f,0.0f,0.0f,0.0f},
    {-0.5f, 0.5f, 1.0f,-1.0f,0.0f,0.0f,0.0f,0.0f},
    {-0.5f,-0.5f,-1.0f,-1.0f,0.0f,0.0f,0.0f,0.0f},
    {-0.5f, 0.5f,-1.0f,-1.0f,0.0f,0.0f,0.0f,0.0f},
    {-0.5f, 0.5f, 1.0f,-1.0f,0.0f,0.0f,0.0f,0.0f},
    {-0.5f,-0.5f,-1.0f,-1.0f,0.0f,0.0f,0.0f,0.0f},

    //right
    {0.5f,-0.5f, 1.0f,1.0f,0.0f,0.0f,0.0f,0.0f},
    {0.5f, 0.5f, 1.0f,1.0f,0.0f,0.0f,0.0f,0.0f},
    {0.5f,-0.5f,-1.0f,1.0f,0.0f,0.0f,0.0f,0.0f},
    {0.5f, 0.5f,-1.0f,1.0f,0.0f,0.0f,0.0f,0.0f},
    {0.5f, 0.5f, 1.0f,1.0f,0.0f,0.0f,0.0f,0.0f},
    {0.5f,-0.5f,-1.0f,1.0f,0.0f,0.0f,0.0f,0.0f}
};

我使用for循环来改变每个顶点的z坐标

std::vector<VertexData> vertexDataCube;
float distance = 0.0f;
for(int i = 0;i < 24;i++)
{
    cubeData[i].z += distance;
    vertexDataCube.push_back(cubeData[i]);
}

这是程序的输出

距离 = 0.0f

距离 = 2.0f

距离 = 1.0f

距离 = -1.0f

在这些图片的每一个中，顶点已向前翻译了5个单位，但就像我上面说的那样，如果距离为负，这似乎不会改变输出，并且如果距离为0.0f。

以下是我用来设置模型，视图和投影矩阵的代码。

DirectX::XMFLOAT3 look,pos,up;
look = DirectX::XMFLOAT3(0.0f,0.0f,100.0f);
pos = DirectX::XMFLOAT3(0.0f,0.0f,1.0f);
up = DirectX::XMFLOAT3(0.0f,1.0f,0.0f);

XMStoreFloat4x4(&constBufferData.mModel,DirectX::XMMatrixIdentity());
XMStoreFloat4x4(&constBufferData.mView,DirectX::XMMatrixLookToLH(DirectX::XMLoadFloat3(&pos),DirectX::XMLoadFloat3(&look),DirectX::XMLoadFloat3(&up)));
XMStoreFloat4x4(&constBufferData.mPerspective,DirectX::XMMatrixPerspectiveFovLH(3.14159f/4.0f,WINDOW_WIDTH/WINDOW_HEIGHT,1.0f,100.0f));

然后在我的主循环中，我根据用户输入翻译模型矩阵，但是现在我只需将其翻译为5.0f。

XMStoreFloat4x4(&constBufferData.mModel,DirectX::XMMatrixTranslation(0.0f,0.0f,5.0f));

Answer 1

您的屏幕截图绝对有意义：

1. 距离= 0：您在立方体内部并看到内部的部分内容。由于近剪裁平面，很可能没有绘制两个平面。

2. 立方体相距2个单位，大小为1个单位。因此，你看到它的一面。如果你稍微旋转一下，你也会看到另外两面（试试吧！）。

3. 距离是1个单位。立方体尺寸也是1个单位：立方体正好在（数学上接触）相机前面。因此它覆盖整个屏幕。

4. 立方体在相机后面。没什么好看的。这就是呈现的内容：背景。

Imho所有的屏幕截图都显示了我希望他们展示的内容。