在 DirectX 中取消投影屏幕坐标时出现意外结果
Unexpected result when unprojecting screen coordinates in DirectX
为了能够确定用户是否点击了我的任何 3D 对象，我尝试将点击的屏幕坐标转换为矢量，然后我用它来检查是否有任何三角形被击中.为此，我使用了 DirectX 提供的 XMVector3Unproject 方法，并在 C++/CX 中实现了所有内容。
我面临的问题是，取消投影屏幕坐标所产生的矢量与我预期的完全不同。下图说明了这一点：
单击发生时的光标位置（以黄色突出显示）在左侧的等距视图中可见。我一单击，取消投影产生的矢量就会出现在图像中指示的模型后面，作为穿透模型的白线。因此，它不是从光标位置开始并进入等距视图中的屏幕，而是出现在完全不同的位置。
当我在等距视图中水平移动鼠标并单击时，然后垂直移动鼠标并单击会出现以下图案。两张图片中的所有线条都代表点击产生的矢量。为了更好的可见性，模型已被删除。
从上图可以看出，所有矢量似乎都来自同一位置。如果我更改视图并重复该过程，则会出现相同的图案，但矢量的原点不同。
这是我用来提出这个问题的代码片段。首先，我使用以下代码接收光标位置，并将其与绘图区域的宽度和高度一起传递给我的“SelectObject”方法：
void Demo::OnPointerPressed(Object^ sender, PointerEventArgs^ e)
{
Point currentPosition = e->CurrentPoint->Position;
if(m_model->SelectObject(currentPosition.X, currentPosition.Y, m_renderTargetWidth, m_renderTargetHeight))
{
m_RefreshImage = true;
}
}
“SelectObject”方法如下所示：
bool Model::SelectObject(float screenX, float screenY, float screenWidth, float screenHeight)
{
XMMATRIX projectionMatrix = XMLoadFloat4x4(&m_modelViewProjectionConstantBufferData->projection);
XMMATRIX viewMatrix = XMLoadFloat4x4(&m_modelViewProjectionConstantBufferData->view);
XMMATRIX modelMatrix = XMLoadFloat4x4(&m_modelViewProjectionConstantBufferData->model);
XMVECTOR v = XMVector3Unproject(XMVectorSet(screenX, screenY, 5.0f, 0.0f),
0.0f,
0.0f,
screenWidth,
screenHeight,
0.0f,
1.0f,
projectionMatrix,
viewMatrix,
modelMatrix);
XMVECTOR rayOrigin = XMVector3Unproject(XMVectorSet(screenX, screenY, 0.0f, 0.0f),
0.0f,
0.0f,
screenWidth,
screenHeight,
0.0f,
1.0f,
projectionMatrix,
viewMatrix,
modelMatrix);
// Code to retrieve v0, v1 and v2 is omitted
if(Intersects(rayOrigin, XMVector3Normalize(v - rayOrigin), v0, v1, v2, depth))
{
return true;
}
}
最终，DirectX::TriangleTests 命名空间的 Intersects 方法使用计算出的向量来检测三角形是否被击中。我省略了上面截取的代码，因为它与这个问题无关。
为了渲染这些图像，我使用了一个正交投影矩阵和一个可以围绕其生成视图矩阵的局部 x 轴和 y 轴旋转的相机。世界矩阵始终保持不变，即它只是一个单位矩阵。
视图矩阵计算如下（基于Frank Luna的书3D游戏编程中的例子）：
void Camera::SetViewMatrix()
{
XMFLOAT3 cameraPosition;
XMFLOAT3 cameraXAxis;
XMFLOAT3 cameraYAxis;
XMFLOAT3 cameraZAxis;
XMFLOAT4X4 viewMatrix;
// Keep camera's axes orthogonal to each other and of unit length.
m_cameraZAxis = XMVector3Normalize(m_cameraZAxis);
m_cameraYAxis = XMVector3Normalize(XMVector3Cross(m_cameraZAxis, m_cameraXAxis));
// m_cameraYAxis and m_cameraZAxis are already normalized, so there is no need
// to normalize the below cross product of the two.
m_cameraXAxis = XMVector3Cross(m_cameraYAxis, m_cameraZAxis);
// Fill in the view matrix entries.
float x = -XMVectorGetX(XMVector3Dot(m_cameraPosition, m_cameraXAxis));
float y = -XMVectorGetX(XMVector3Dot(m_cameraPosition, m_cameraYAxis));
float z = -XMVectorGetX(XMVector3Dot(m_cameraPosition, m_cameraZAxis));
XMStoreFloat3(&cameraPosition, m_cameraPosition);
XMStoreFloat3(&cameraXAxis , m_cameraXAxis);
XMStoreFloat3(&cameraYAxis , m_cameraYAxis);
XMStoreFloat3(&cameraZAxis , m_cameraZAxis);
viewMatrix(0, 0) = cameraXAxis.x;
viewMatrix(1, 0) = cameraXAxis.y;
viewMatrix(2, 0) = cameraXAxis.z;
viewMatrix(3, 0) = x;
viewMatrix(0, 1) = cameraYAxis.x;
viewMatrix(1, 1) = cameraYAxis.y;
viewMatrix(2, 1) = cameraYAxis.z;
viewMatrix(3, 1) = y;
viewMatrix(0, 2) = cameraZAxis.x;
viewMatrix(1, 2) = cameraZAxis.y;
viewMatrix(2, 2) = cameraZAxis.z;
viewMatrix(3, 2) = z;
viewMatrix(0, 3) = 0.0f;
viewMatrix(1, 3) = 0.0f;
viewMatrix(2, 3) = 0.0f;
viewMatrix(3, 3) = 1.0f;
m_modelViewProjectionConstantBufferData->view = viewMatrix;
}
它受两种围绕相机的 x 轴和 y 轴旋转相机的方法的影响：
void Camera::ChangeCameraPitch(float angle)
{
XMMATRIX rotationMatrix = XMMatrixRotationAxis(m_cameraXAxis, angle);
m_cameraYAxis = XMVector3TransformNormal(m_cameraYAxis, rotationMatrix);
m_cameraZAxis = XMVector3TransformNormal(m_cameraZAxis, rotationMatrix);
}
void Camera::ChangeCameraYaw(float angle)
{
XMMATRIX rotationMatrix = XMMatrixRotationAxis(m_cameraYAxis, angle);
m_cameraXAxis = XMVector3TransformNormal(m_cameraXAxis, rotationMatrix);
m_cameraZAxis = XMVector3TransformNormal(m_cameraZAxis, rotationMatrix);
}
世界/模型矩阵和投影矩阵计算如下：
void Model::SetProjectionMatrix(float width, float height, float nearZ, float farZ)
{
XMMATRIX orthographicProjectionMatrix = XMMatrixOrthographicRH(width, height, nearZ, farZ);
XMFLOAT4X4 orientation = XMFLOAT4X4
(
1.0f, 0.0f, 0.0f, 0.0f,
0.0f, 1.0f, 0.0f, 0.0f,
0.0f, 0.0f, 1.0f, 0.0f,
0.0f, 0.0f, 0.0f, 1.0f
);
XMMATRIX orientationMatrix = XMLoadFloat4x4(&orientation);
XMStoreFloat4x4(&m_modelViewProjectionConstantBufferData->projection, XMMatrixTranspose(orthographicProjectionMatrix * orientationMatrix));
}
void Model::SetModelMatrix()
{
XMFLOAT4X4 orientation = XMFLOAT4X4
(
1.0f, 0.0f, 0.0f, 0.0f,
0.0f, 1.0f, 0.0f, 0.0f,
0.0f, 0.0f, 1.0f, 0.0f,
0.0f, 0.0f, 0.0f, 1.0f
);
XMMATRIX orientationMatrix = XMLoadFloat4x4(&orientation);
XMStoreFloat4x4(&m_modelViewProjectionConstantBufferData->model, XMMatrixTranspose(orientationMatrix));
}
坦率地说，我还不明白我面临的问题。如果任何有更深入见解的人能给我一些提示，告诉我需要在哪里应用更改，以便从非投影计算的矢量从光标位置开始并移动到屏幕中，我将不胜感激。
编辑 1:
我认为这与我的相机位于世界坐标 (0, 0, 0) 的事实有关。相机围绕其局部 x 轴和 y 轴旋转。据我了解，相机创建的视图矩阵构建了图像投影到的平面。如果是这样的话，它就可以解释为什么光线位于某种 "unexpected" 位置。
我的假设是我需要将相机移出中心，使其位于对象的外部。但是，如果简单地修改相机的成员变量m_cameraPosition，我的模型就会完全变形。
有没有人愿意提供帮助？
我在 Model::SetViewMatrix() 函数中使用了 XMMatrixLookAtRH API 来计算视图矩阵并得到了 v 和 rayOrigin 向量的合适值。
例如：
XMStoreFloat4x4(
&m_modelViewProjectionConstantBufferData->view,
XMMatrixLookAtRH(m_cameraPosition, XMVectorSet(0.0f, 1.0f, 0.0f, 0.0f),
XMVectorSet(1.0f, 0.0f, 0.0f, 0.0f))
);
虽然我无法在屏幕上可视化输出，但我通过在控制台应用程序中计算简单值来检查结果，矢量值似乎是正确的。请检查您的申请并确认。
注意：你必须给出焦点和向上方向矢量参数才能使用XMMatrixLookAtRH API 而不是您当前的方法。
感谢您的提示，卡皮尔。我尝试了 XMMatrixLookAtRH 方法，但无法使用该方法更改相机的俯仰/偏航，因此我放弃了该方法并想出自己生成矩阵。
解决我问题的方法是在将模型、视图和投影矩阵传递给 XMVector3Unproject 之前使用 XMMatrixTranspose 转置它们。因此，与其使用如下代码
XMMATRIX projectionMatrix = XMLoadFloat4x4(&m_modelViewProjectionConstantBufferData->projection);
XMMATRIX viewMatrix = XMLoadFloat4x4(&m_modelViewProjectionConstantBufferData->view);
XMMATRIX modelMatrix = XMLoadFloat4x4(&m_modelViewProjectionConstantBufferData->model);
XMVECTOR rayBegin = XMVector3Unproject(XMVectorSet(screenX, screenY, -m_boundingSphereRadius, 0.0f),
0.0f,
0.0f,
screenWidth,
screenHeight,
0.0f,
1.0f,
projectionMatrix,
viewMatrix,
modelMatrix);
需要
XMMATRIX projectionMatrix = XMMatrixTranspose(XMLoadFloat4x4(&m_modelViewProjectionConstantBufferData->projection));
XMMATRIX viewMatrix = XMMatrixTranspose(XMLoadFloat4x4(&m_modelViewProjectionConstantBufferData->view));
XMMATRIX modelMatrix = XMMatrixTranspose(XMLoadFloat4x4(&m_modelViewProjectionConstantBufferData->model));
XMVECTOR rayBegin = XMVector3Unproject(XMVectorSet(screenX, screenY, -m_boundingSphereRadius, 0.0f),
0.0f,
0.0f,
screenWidth,
screenHeight,
0.0f,
1.0f,
projectionMatrix,
viewMatrix,
modelMatrix);
我不完全清楚为什么我需要在将矩阵传递给 unproject 方法之前转置它们。但是，我怀疑这与我移动相机时遇到的问题有关。 this posting.
在 Whosebug 上已经描述了这个问题
我还没有设法解决那个问题。简单地转置视图矩阵并不能解决它。不过，我的主要问题已经解决，我的模型终于可以点击了。
如果有人要补充说明为什么需要转置矩阵或为什么移动相机会扭曲模型，请继续 post 评论或回答。
我能够使用 XMMatrixLookAtRH 方法获得 v 和 rayOrigin 向量的相等值以及使用此代码的自定义视图矩阵，而无需矩阵转置操作：
#include <directxmath.h>
using namespace DirectX;
XMVECTOR m_cameraXAxis;
XMVECTOR m_cameraYAxis;
XMVECTOR m_cameraZAxis;
XMVECTOR m_cameraPosition;
XMMATRIX gView;
XMMATRIX gView2;
XMMATRIX gProj;
XMMATRIX gModel;
void SetViewMatrix()
{
XMVECTOR lTarget = XMVectorSet(2.0f, 2.0f, 2.0f, 1.0f);
m_cameraPosition = XMVectorSet(1.0f, 1.0f, 1.0f, 1.0f);
m_cameraZAxis = XMVector3Normalize(XMVectorSubtract(m_cameraPosition, lTarget));
m_cameraXAxis = XMVector3Normalize(XMVector3Cross(XMVectorSet(1.0f, -1.0f, -1.0f, 0.0f), m_cameraZAxis));
XMFLOAT3 cameraPosition;
XMFLOAT3 cameraXAxis;
XMFLOAT3 cameraYAxis;
XMFLOAT3 cameraZAxis;
XMFLOAT4X4 viewMatrix;
// Keep camera's axes orthogonal to each other and of unit length.
m_cameraZAxis = XMVector3Normalize(m_cameraZAxis);
m_cameraYAxis = XMVector3Normalize(XMVector3Cross(m_cameraZAxis, m_cameraXAxis));
// m_cameraYAxis and m_cameraZAxis are already normalized, so there is no need
// to normalize the below cross product of the two.
m_cameraXAxis = XMVector3Cross(m_cameraYAxis, m_cameraZAxis);
// Fill in the view matrix entries.
float x = -XMVectorGetX(XMVector3Dot(m_cameraPosition, m_cameraXAxis));
float y = -XMVectorGetX(XMVector3Dot(m_cameraPosition, m_cameraYAxis));
float z = -XMVectorGetX(XMVector3Dot(m_cameraPosition, m_cameraZAxis));
XMStoreFloat3(&cameraPosition, m_cameraPosition);
XMStoreFloat3(&cameraXAxis, m_cameraXAxis);
XMStoreFloat3(&cameraYAxis, m_cameraYAxis);
XMStoreFloat3(&cameraZAxis, m_cameraZAxis);
viewMatrix(0, 0) = cameraXAxis.x;
viewMatrix(1, 0) = cameraXAxis.y;
viewMatrix(2, 0) = cameraXAxis.z;
viewMatrix(3, 0) = x;
viewMatrix(0, 1) = cameraYAxis.x;
viewMatrix(1, 1) = cameraYAxis.y;
viewMatrix(2, 1) = cameraYAxis.z;
viewMatrix(3, 1) = y;
viewMatrix(0, 2) = cameraZAxis.x;
viewMatrix(1, 2) = cameraZAxis.y;
viewMatrix(2, 2) = cameraZAxis.z;
viewMatrix(3, 2) = z;
viewMatrix(0, 3) = 0.0f;
viewMatrix(1, 3) = 0.0f;
viewMatrix(2, 3) = 0.0f;
viewMatrix(3, 3) = 1.0f;
gView = XMLoadFloat4x4(&viewMatrix);
gView2 = XMMatrixLookAtRH(m_cameraPosition, XMVectorSet(2.0f, 2.0f, 2.0f, 1.0f),
XMVectorSet(1.0f, -1.0f, -1.0f, 0.0f));
//m_modelViewProjectionConstantBufferData->view = viewMatrix;
printf("yo");
}
void SetProjectionMatrix(float width, float height, float nearZ, float farZ)
{
XMMATRIX orthographicProjectionMatrix = XMMatrixOrthographicRH(width, height, nearZ, farZ);
XMFLOAT4X4 orientation = XMFLOAT4X4
(
1.0f, 0.0f, 0.0f, 0.0f,
0.0f, 1.0f, 0.0f, 0.0f,
0.0f, 0.0f, 1.0f, 0.0f,
0.0f, 0.0f, 0.0f, 1.0f
);
XMMATRIX orientationMatrix = XMLoadFloat4x4(&orientation);
gProj = XMMatrixTranspose( XMMatrixMultiply(orthographicProjectionMatrix, orientationMatrix));
}
void SetModelMatrix()
{
XMFLOAT4X4 orientation = XMFLOAT4X4
(
1.0f, 0.0f, 0.0f, 0.0f,
0.0f, 1.0f, 0.0f, 0.0f,
0.0f, 0.0f, 1.0f, 0.0f,
0.0f, 0.0f, 0.0f, 1.0f
);
XMMATRIX orientationMatrix = XMMatrixTranspose( XMLoadFloat4x4(&orientation));
gModel = orientationMatrix;
}
bool SelectObject(float screenX, float screenY, float screenWidth, float screenHeight)
{
XMMATRIX projectionMatrix = gProj;
XMMATRIX viewMatrix = gView;
XMMATRIX modelMatrix = gModel;
XMMATRIX viewMatrix2 = gView2;
XMVECTOR v = XMVector3Unproject(XMVectorSet(screenX, screenY, 5.0f, 0.0f),
0.0f,
0.0f,
screenWidth,
screenHeight,
0.0f,
1.0f,
projectionMatrix,
viewMatrix,
modelMatrix);
XMVECTOR rayOrigin = XMVector3Unproject(XMVectorSet(screenX, screenY, 0.0f, 0.0f),
0.0f,
0.0f,
screenWidth,
screenHeight,
0.0f,
1.0f,
projectionMatrix,
viewMatrix,
modelMatrix);
// Code to retrieve v0, v1 and v2 is omitted
auto diff = v - rayOrigin;
auto diffNorm = XMVector3Normalize(diff);
XMVECTOR v2 = XMVector3Unproject(XMVectorSet(screenX, screenY, 5.0f, 0.0f),
0.0f,
0.0f,
screenWidth,
screenHeight,
0.0f,
1.0f,
projectionMatrix,
viewMatrix2,
modelMatrix);
XMVECTOR rayOrigin2 = XMVector3Unproject(XMVectorSet(screenX, screenY, 0.0f, 0.0f),
0.0f,
0.0f,
screenWidth,
screenHeight,
0.0f,
1.0f,
projectionMatrix,
viewMatrix2,
modelMatrix);
auto diff2 = v2 - rayOrigin2;
auto diffNorm2 = XMVector3Normalize(diff2);
printf("hi");
return true;
}
int main()
{
SetViewMatrix();
SetProjectionMatrix(1000, 1000, 0.0f, 1.0f);
SetModelMatrix();
SelectObject(500, 500, 1000, 1000);
return 0;
}
请使用此代码检查您的申请并确认。您会看到代码与您之前的代码相同。唯一增加的是相机参数的初始值，使用 XMMatrixLookAtRH 方法计算 SetViewMatrix() 中的第二个视图矩阵，并使用 SelectObject().
中的两个视图矩阵计算向量
无需转置
我不需要转置任何矩阵。 Projection 和 Model 矩阵不需要转置，因为它们都是对角矩阵，转置它们会得到相同的矩阵。我认为也不需要 View 矩阵的转置。 XMMatrixLookAtRH 解释的公式 here provides the view matrix exactly like yours. Also, the sample project given here 在检查交集时不转置其矩阵。您可以下载并检查示例项目。
可能的问题来源
1) 初始化：我唯一没能看到的代码是你对 m_cameraZAxis、m_cameraXAxis、nearZ、farZ 参数等的初始化。另外，我没有使用你的相机旋转功能。如您所见，我已经通过使用位置、目标和方向向量进行计算来初始化相机。请检查您 m_cameraZAxis 的初始计算是否符合我的示例代码。
2) LH/RH look：确保没有意外混淆 left-hand 和 right-hand 查找代码中的任何位置。
3) 检查您的旋转代码（ChangeCameraPitch 或 ChangeCameraYaw）是否意外创建了 不正交 的相机轴。您正在使用相机的 Y 轴 作为 ChangeCameraYaw 中的输入和 ChangeCameraPitch 中的输出。但是 Y 轴 被叉积或 X 轴和 Z 轴重置为 SetViewMatrix。所以 Y 轴的早期值可能会丢失。
祝你申请成功！请告诉您是否找到了问题的正确解决方案和根本原因。
As mentioned the issue was not fully resolved even though clicking now works. The issue with the distortion of the model when moving the camera, which I suspected is related, was still present. What I meant with "the model gets distorted" is visible in the following illustration:
The left image shows how the model looks when the camera is located in the center of the world, i.e. (0, 0, 0) while the right image shows what happens when I move the camera in negative y-axis方向。 As can be seen the model widens at the bottom and gets smaller at the top which is the same behavior described in the link I already provided above.
What I eventually did to resolve both issues is:
- Transpose the matrices before passing them to XMVector3Unproject (already described above)
- Transposed my view matrix by changing the code of the
SetViewMatrix method (code see below)
The SetViewMatrix method now looks as follows:
void Camera::SetViewMatrix()
{
XMFLOAT3 cameraPosition;
XMFLOAT3 cameraXAxis;
XMFLOAT3 cameraYAxis;
XMFLOAT3 cameraZAxis;
XMFLOAT4X4 viewMatrix;
// Keep camera's axes orthogonal to each other and of unit length.
m_cameraZAxis = XMVector3Normalize(m_cameraZAxis);
m_cameraYAxis = XMVector3Normalize(XMVector3Cross(m_cameraZAxis, m_cameraXAxis));
// m_cameraYAxis and m_cameraZAxis are already normalized, so there is no need
// to normalize the below cross product of the two.
m_cameraXAxis = XMVector3Cross(m_cameraYAxis, m_cameraZAxis);
// Fill in the view matrix entries.
float x = -XMVectorGetX(XMVector3Dot(m_cameraPosition, m_cameraXAxis));
float y = -XMVectorGetX(XMVector3Dot(m_cameraPosition, m_cameraYAxis));
float z = -XMVectorGetX(XMVector3Dot(m_cameraPosition, m_cameraZAxis));
//XMStoreFloat3(&cameraPosition, m_cameraPosition);
XMStoreFloat3(&cameraXAxis, m_cameraXAxis);
XMStoreFloat3(&cameraYAxis, m_cameraYAxis);
XMStoreFloat3(&cameraZAxis, m_cameraZAxis);
viewMatrix(0, 0) = cameraXAxis.x;
viewMatrix(0, 1) = cameraXAxis.y;
viewMatrix(0, 2) = cameraXAxis.z;
viewMatrix(0, 3) = x;
viewMatrix(1, 0) = cameraYAxis.x;
viewMatrix(1, 1) = cameraYAxis.y;
viewMatrix(1, 2) = cameraYAxis.z;
viewMatrix(1, 3) = y;
viewMatrix(2, 0) = cameraZAxis.x;
viewMatrix(2, 1) = cameraZAxis.y;
viewMatrix(2, 2) = cameraZAxis.z;
viewMatrix(2, 3) = z;
viewMatrix(3, 0) = 0.0f;
viewMatrix(3, 1) = 0.0f;
viewMatrix(3, 2) = 0.0f;
viewMatrix(3, 3) = 1.0f;
m_modelViewProjectionConstantBufferData->view = viewMatrix;
}
So I just exchanged row and column coordinates. Note that I had to make sure that my ChangeCameraYaw method gets called before my ChangeCameraPitch method. This is necessary because the orientation of the model otherwise is not as I want it.
There is also another approach that could be used. Instead of transposing the view matrix by exchanging the row and column coordinates and transposing it before passing it to XMVector3Unproject I could use the row_major keyword in the vertex shader together with the view matrix:
cbuffer ModelViewProjectionConstantBuffer : register(b0)
{
matrix model;
row_major matrix view;
matrix projection;
};
I came across this idea in this blog post. The keyword row_major influences on how the shader compiler interprets the matrix in memory. The same could also be achieved by changing the order of the vector * matrix multiplication in the vertex shader, i.e. using pos = mul(view, pos); instead of pos = mul(pos, view);
That's pretty much it. The two issues are indeed interconnected but using what I posted in this question I was able to resolve both issues so I'm accepting my own reply as answer to this question. Hope it helps someone in the future.
为了能够确定用户是否点击了我的任何 3D 对象，我尝试将点击的屏幕坐标转换为矢量，然后我用它来检查是否有任何三角形被击中.为此，我使用了 DirectX 提供的 XMVector3Unproject 方法，并在 C++/CX 中实现了所有内容。
我面临的问题是，取消投影屏幕坐标所产生的矢量与我预期的完全不同。下图说明了这一点：
当我在等距视图中水平移动鼠标并单击时，然后垂直移动鼠标并单击会出现以下图案。两张图片中的所有线条都代表点击产生的矢量。为了更好的可见性，模型已被删除。
这是我用来提出这个问题的代码片段。首先，我使用以下代码接收光标位置，并将其与绘图区域的宽度和高度一起传递给我的“SelectObject”方法：
void Demo::OnPointerPressed(Object^ sender, PointerEventArgs^ e)
{
Point currentPosition = e->CurrentPoint->Position;
if(m_model->SelectObject(currentPosition.X, currentPosition.Y, m_renderTargetWidth, m_renderTargetHeight))
{
m_RefreshImage = true;
}
}
“SelectObject”方法如下所示：
bool Model::SelectObject(float screenX, float screenY, float screenWidth, float screenHeight)
{
XMMATRIX projectionMatrix = XMLoadFloat4x4(&m_modelViewProjectionConstantBufferData->projection);
XMMATRIX viewMatrix = XMLoadFloat4x4(&m_modelViewProjectionConstantBufferData->view);
XMMATRIX modelMatrix = XMLoadFloat4x4(&m_modelViewProjectionConstantBufferData->model);
XMVECTOR v = XMVector3Unproject(XMVectorSet(screenX, screenY, 5.0f, 0.0f),
0.0f,
0.0f,
screenWidth,
screenHeight,
0.0f,
1.0f,
projectionMatrix,
viewMatrix,
modelMatrix);
XMVECTOR rayOrigin = XMVector3Unproject(XMVectorSet(screenX, screenY, 0.0f, 0.0f),
0.0f,
0.0f,
screenWidth,
screenHeight,
0.0f,
1.0f,
projectionMatrix,
viewMatrix,
modelMatrix);
// Code to retrieve v0, v1 and v2 is omitted
if(Intersects(rayOrigin, XMVector3Normalize(v - rayOrigin), v0, v1, v2, depth))
{
return true;
}
}
最终，DirectX::TriangleTests 命名空间的 Intersects 方法使用计算出的向量来检测三角形是否被击中。我省略了上面截取的代码，因为它与这个问题无关。
为了渲染这些图像，我使用了一个正交投影矩阵和一个可以围绕其生成视图矩阵的局部 x 轴和 y 轴旋转的相机。世界矩阵始终保持不变，即它只是一个单位矩阵。
视图矩阵计算如下（基于Frank Luna的书3D游戏编程中的例子）：
void Camera::SetViewMatrix()
{
XMFLOAT3 cameraPosition;
XMFLOAT3 cameraXAxis;
XMFLOAT3 cameraYAxis;
XMFLOAT3 cameraZAxis;
XMFLOAT4X4 viewMatrix;
// Keep camera's axes orthogonal to each other and of unit length.
m_cameraZAxis = XMVector3Normalize(m_cameraZAxis);
m_cameraYAxis = XMVector3Normalize(XMVector3Cross(m_cameraZAxis, m_cameraXAxis));
// m_cameraYAxis and m_cameraZAxis are already normalized, so there is no need
// to normalize the below cross product of the two.
m_cameraXAxis = XMVector3Cross(m_cameraYAxis, m_cameraZAxis);
// Fill in the view matrix entries.
float x = -XMVectorGetX(XMVector3Dot(m_cameraPosition, m_cameraXAxis));
float y = -XMVectorGetX(XMVector3Dot(m_cameraPosition, m_cameraYAxis));
float z = -XMVectorGetX(XMVector3Dot(m_cameraPosition, m_cameraZAxis));
XMStoreFloat3(&cameraPosition, m_cameraPosition);
XMStoreFloat3(&cameraXAxis , m_cameraXAxis);
XMStoreFloat3(&cameraYAxis , m_cameraYAxis);
XMStoreFloat3(&cameraZAxis , m_cameraZAxis);
viewMatrix(0, 0) = cameraXAxis.x;
viewMatrix(1, 0) = cameraXAxis.y;
viewMatrix(2, 0) = cameraXAxis.z;
viewMatrix(3, 0) = x;
viewMatrix(0, 1) = cameraYAxis.x;
viewMatrix(1, 1) = cameraYAxis.y;
viewMatrix(2, 1) = cameraYAxis.z;
viewMatrix(3, 1) = y;
viewMatrix(0, 2) = cameraZAxis.x;
viewMatrix(1, 2) = cameraZAxis.y;
viewMatrix(2, 2) = cameraZAxis.z;
viewMatrix(3, 2) = z;
viewMatrix(0, 3) = 0.0f;
viewMatrix(1, 3) = 0.0f;
viewMatrix(2, 3) = 0.0f;
viewMatrix(3, 3) = 1.0f;
m_modelViewProjectionConstantBufferData->view = viewMatrix;
}
它受两种围绕相机的 x 轴和 y 轴旋转相机的方法的影响：
void Camera::ChangeCameraPitch(float angle)
{
XMMATRIX rotationMatrix = XMMatrixRotationAxis(m_cameraXAxis, angle);
m_cameraYAxis = XMVector3TransformNormal(m_cameraYAxis, rotationMatrix);
m_cameraZAxis = XMVector3TransformNormal(m_cameraZAxis, rotationMatrix);
}
void Camera::ChangeCameraYaw(float angle)
{
XMMATRIX rotationMatrix = XMMatrixRotationAxis(m_cameraYAxis, angle);
m_cameraXAxis = XMVector3TransformNormal(m_cameraXAxis, rotationMatrix);
m_cameraZAxis = XMVector3TransformNormal(m_cameraZAxis, rotationMatrix);
}
世界/模型矩阵和投影矩阵计算如下：
void Model::SetProjectionMatrix(float width, float height, float nearZ, float farZ)
{
XMMATRIX orthographicProjectionMatrix = XMMatrixOrthographicRH(width, height, nearZ, farZ);
XMFLOAT4X4 orientation = XMFLOAT4X4
(
1.0f, 0.0f, 0.0f, 0.0f,
0.0f, 1.0f, 0.0f, 0.0f,
0.0f, 0.0f, 1.0f, 0.0f,
0.0f, 0.0f, 0.0f, 1.0f
);
XMMATRIX orientationMatrix = XMLoadFloat4x4(&orientation);
XMStoreFloat4x4(&m_modelViewProjectionConstantBufferData->projection, XMMatrixTranspose(orthographicProjectionMatrix * orientationMatrix));
}
void Model::SetModelMatrix()
{
XMFLOAT4X4 orientation = XMFLOAT4X4
(
1.0f, 0.0f, 0.0f, 0.0f,
0.0f, 1.0f, 0.0f, 0.0f,
0.0f, 0.0f, 1.0f, 0.0f,
0.0f, 0.0f, 0.0f, 1.0f
);
XMMATRIX orientationMatrix = XMLoadFloat4x4(&orientation);
XMStoreFloat4x4(&m_modelViewProjectionConstantBufferData->model, XMMatrixTranspose(orientationMatrix));
}
坦率地说，我还不明白我面临的问题。如果任何有更深入见解的人能给我一些提示，告诉我需要在哪里应用更改，以便从非投影计算的矢量从光标位置开始并移动到屏幕中，我将不胜感激。
编辑 1:
我认为这与我的相机位于世界坐标 (0, 0, 0) 的事实有关。相机围绕其局部 x 轴和 y 轴旋转。据我了解，相机创建的视图矩阵构建了图像投影到的平面。如果是这样的话，它就可以解释为什么光线位于某种 "unexpected" 位置。
我的假设是我需要将相机移出中心，使其位于对象的外部。但是，如果简单地修改相机的成员变量m_cameraPosition，我的模型就会完全变形。
有没有人愿意提供帮助？
我在 Model::SetViewMatrix() 函数中使用了 XMMatrixLookAtRH API 来计算视图矩阵并得到了 v 和 rayOrigin 向量的合适值。
例如：
XMStoreFloat4x4(
&m_modelViewProjectionConstantBufferData->view,
XMMatrixLookAtRH(m_cameraPosition, XMVectorSet(0.0f, 1.0f, 0.0f, 0.0f),
XMVectorSet(1.0f, 0.0f, 0.0f, 0.0f))
);
虽然我无法在屏幕上可视化输出，但我通过在控制台应用程序中计算简单值来检查结果，矢量值似乎是正确的。请检查您的申请并确认。
注意：你必须给出焦点和向上方向矢量参数才能使用XMMatrixLookAtRH API 而不是您当前的方法。
感谢您的提示，卡皮尔。我尝试了 XMMatrixLookAtRH 方法，但无法使用该方法更改相机的俯仰/偏航，因此我放弃了该方法并想出自己生成矩阵。
解决我问题的方法是在将模型、视图和投影矩阵传递给 XMVector3Unproject 之前使用 XMMatrixTranspose 转置它们。因此，与其使用如下代码
XMMATRIX projectionMatrix = XMLoadFloat4x4(&m_modelViewProjectionConstantBufferData->projection);
XMMATRIX viewMatrix = XMLoadFloat4x4(&m_modelViewProjectionConstantBufferData->view);
XMMATRIX modelMatrix = XMLoadFloat4x4(&m_modelViewProjectionConstantBufferData->model);
XMVECTOR rayBegin = XMVector3Unproject(XMVectorSet(screenX, screenY, -m_boundingSphereRadius, 0.0f),
0.0f,
0.0f,
screenWidth,
screenHeight,
0.0f,
1.0f,
projectionMatrix,
viewMatrix,
modelMatrix);
需要
XMMATRIX projectionMatrix = XMMatrixTranspose(XMLoadFloat4x4(&m_modelViewProjectionConstantBufferData->projection));
XMMATRIX viewMatrix = XMMatrixTranspose(XMLoadFloat4x4(&m_modelViewProjectionConstantBufferData->view));
XMMATRIX modelMatrix = XMMatrixTranspose(XMLoadFloat4x4(&m_modelViewProjectionConstantBufferData->model));
XMVECTOR rayBegin = XMVector3Unproject(XMVectorSet(screenX, screenY, -m_boundingSphereRadius, 0.0f),
0.0f,
0.0f,
screenWidth,
screenHeight,
0.0f,
1.0f,
projectionMatrix,
viewMatrix,
modelMatrix);
我不完全清楚为什么我需要在将矩阵传递给 unproject 方法之前转置它们。但是，我怀疑这与我移动相机时遇到的问题有关。 this posting.
在 Whosebug 上已经描述了这个问题我还没有设法解决那个问题。简单地转置视图矩阵并不能解决它。不过，我的主要问题已经解决，我的模型终于可以点击了。
如果有人要补充说明为什么需要转置矩阵或为什么移动相机会扭曲模型，请继续 post 评论或回答。
我能够使用 XMMatrixLookAtRH 方法获得 v 和 rayOrigin 向量的相等值以及使用此代码的自定义视图矩阵，而无需矩阵转置操作：
#include <directxmath.h>
using namespace DirectX;
XMVECTOR m_cameraXAxis;
XMVECTOR m_cameraYAxis;
XMVECTOR m_cameraZAxis;
XMVECTOR m_cameraPosition;
XMMATRIX gView;
XMMATRIX gView2;
XMMATRIX gProj;
XMMATRIX gModel;
void SetViewMatrix()
{
XMVECTOR lTarget = XMVectorSet(2.0f, 2.0f, 2.0f, 1.0f);
m_cameraPosition = XMVectorSet(1.0f, 1.0f, 1.0f, 1.0f);
m_cameraZAxis = XMVector3Normalize(XMVectorSubtract(m_cameraPosition, lTarget));
m_cameraXAxis = XMVector3Normalize(XMVector3Cross(XMVectorSet(1.0f, -1.0f, -1.0f, 0.0f), m_cameraZAxis));
XMFLOAT3 cameraPosition;
XMFLOAT3 cameraXAxis;
XMFLOAT3 cameraYAxis;
XMFLOAT3 cameraZAxis;
XMFLOAT4X4 viewMatrix;
// Keep camera's axes orthogonal to each other and of unit length.
m_cameraZAxis = XMVector3Normalize(m_cameraZAxis);
m_cameraYAxis = XMVector3Normalize(XMVector3Cross(m_cameraZAxis, m_cameraXAxis));
// m_cameraYAxis and m_cameraZAxis are already normalized, so there is no need
// to normalize the below cross product of the two.
m_cameraXAxis = XMVector3Cross(m_cameraYAxis, m_cameraZAxis);
// Fill in the view matrix entries.
float x = -XMVectorGetX(XMVector3Dot(m_cameraPosition, m_cameraXAxis));
float y = -XMVectorGetX(XMVector3Dot(m_cameraPosition, m_cameraYAxis));
float z = -XMVectorGetX(XMVector3Dot(m_cameraPosition, m_cameraZAxis));
XMStoreFloat3(&cameraPosition, m_cameraPosition);
XMStoreFloat3(&cameraXAxis, m_cameraXAxis);
XMStoreFloat3(&cameraYAxis, m_cameraYAxis);
XMStoreFloat3(&cameraZAxis, m_cameraZAxis);
viewMatrix(0, 0) = cameraXAxis.x;
viewMatrix(1, 0) = cameraXAxis.y;
viewMatrix(2, 0) = cameraXAxis.z;
viewMatrix(3, 0) = x;
viewMatrix(0, 1) = cameraYAxis.x;
viewMatrix(1, 1) = cameraYAxis.y;
viewMatrix(2, 1) = cameraYAxis.z;
viewMatrix(3, 1) = y;
viewMatrix(0, 2) = cameraZAxis.x;
viewMatrix(1, 2) = cameraZAxis.y;
viewMatrix(2, 2) = cameraZAxis.z;
viewMatrix(3, 2) = z;
viewMatrix(0, 3) = 0.0f;
viewMatrix(1, 3) = 0.0f;
viewMatrix(2, 3) = 0.0f;
viewMatrix(3, 3) = 1.0f;
gView = XMLoadFloat4x4(&viewMatrix);
gView2 = XMMatrixLookAtRH(m_cameraPosition, XMVectorSet(2.0f, 2.0f, 2.0f, 1.0f),
XMVectorSet(1.0f, -1.0f, -1.0f, 0.0f));
//m_modelViewProjectionConstantBufferData->view = viewMatrix;
printf("yo");
}
void SetProjectionMatrix(float width, float height, float nearZ, float farZ)
{
XMMATRIX orthographicProjectionMatrix = XMMatrixOrthographicRH(width, height, nearZ, farZ);
XMFLOAT4X4 orientation = XMFLOAT4X4
(
1.0f, 0.0f, 0.0f, 0.0f,
0.0f, 1.0f, 0.0f, 0.0f,
0.0f, 0.0f, 1.0f, 0.0f,
0.0f, 0.0f, 0.0f, 1.0f
);
XMMATRIX orientationMatrix = XMLoadFloat4x4(&orientation);
gProj = XMMatrixTranspose( XMMatrixMultiply(orthographicProjectionMatrix, orientationMatrix));
}
void SetModelMatrix()
{
XMFLOAT4X4 orientation = XMFLOAT4X4
(
1.0f, 0.0f, 0.0f, 0.0f,
0.0f, 1.0f, 0.0f, 0.0f,
0.0f, 0.0f, 1.0f, 0.0f,
0.0f, 0.0f, 0.0f, 1.0f
);
XMMATRIX orientationMatrix = XMMatrixTranspose( XMLoadFloat4x4(&orientation));
gModel = orientationMatrix;
}
bool SelectObject(float screenX, float screenY, float screenWidth, float screenHeight)
{
XMMATRIX projectionMatrix = gProj;
XMMATRIX viewMatrix = gView;
XMMATRIX modelMatrix = gModel;
XMMATRIX viewMatrix2 = gView2;
XMVECTOR v = XMVector3Unproject(XMVectorSet(screenX, screenY, 5.0f, 0.0f),
0.0f,
0.0f,
screenWidth,
screenHeight,
0.0f,
1.0f,
projectionMatrix,
viewMatrix,
modelMatrix);
XMVECTOR rayOrigin = XMVector3Unproject(XMVectorSet(screenX, screenY, 0.0f, 0.0f),
0.0f,
0.0f,
screenWidth,
screenHeight,
0.0f,
1.0f,
projectionMatrix,
viewMatrix,
modelMatrix);
// Code to retrieve v0, v1 and v2 is omitted
auto diff = v - rayOrigin;
auto diffNorm = XMVector3Normalize(diff);
XMVECTOR v2 = XMVector3Unproject(XMVectorSet(screenX, screenY, 5.0f, 0.0f),
0.0f,
0.0f,
screenWidth,
screenHeight,
0.0f,
1.0f,
projectionMatrix,
viewMatrix2,
modelMatrix);
XMVECTOR rayOrigin2 = XMVector3Unproject(XMVectorSet(screenX, screenY, 0.0f, 0.0f),
0.0f,
0.0f,
screenWidth,
screenHeight,
0.0f,
1.0f,
projectionMatrix,
viewMatrix2,
modelMatrix);
auto diff2 = v2 - rayOrigin2;
auto diffNorm2 = XMVector3Normalize(diff2);
printf("hi");
return true;
}
int main()
{
SetViewMatrix();
SetProjectionMatrix(1000, 1000, 0.0f, 1.0f);
SetModelMatrix();
SelectObject(500, 500, 1000, 1000);
return 0;
}
请使用此代码检查您的申请并确认。您会看到代码与您之前的代码相同。唯一增加的是相机参数的初始值，使用 XMMatrixLookAtRH 方法计算 SetViewMatrix() 中的第二个视图矩阵，并使用 SelectObject().
无需转置
我不需要转置任何矩阵。 Projection 和 Model 矩阵不需要转置，因为它们都是对角矩阵，转置它们会得到相同的矩阵。我认为也不需要 View 矩阵的转置。 XMMatrixLookAtRH 解释的公式 here provides the view matrix exactly like yours. Also, the sample project given here 在检查交集时不转置其矩阵。您可以下载并检查示例项目。
可能的问题来源
1) 初始化：我唯一没能看到的代码是你对 m_cameraZAxis、m_cameraXAxis、nearZ、farZ 参数等的初始化。另外，我没有使用你的相机旋转功能。如您所见，我已经通过使用位置、目标和方向向量进行计算来初始化相机。请检查您 m_cameraZAxis 的初始计算是否符合我的示例代码。
2) LH/RH look：确保没有意外混淆 left-hand 和 right-hand 查找代码中的任何位置。
3) 检查您的旋转代码（ChangeCameraPitch 或 ChangeCameraYaw）是否意外创建了 不正交 的相机轴。您正在使用相机的 Y 轴 作为 ChangeCameraYaw 中的输入和 ChangeCameraPitch 中的输出。但是 Y 轴 被叉积或 X 轴和 Z 轴重置为 SetViewMatrix。所以 Y 轴的早期值可能会丢失。
祝你申请成功！请告诉您是否找到了问题的正确解决方案和根本原因。
As mentioned the issue was not fully resolved even though clicking now works. The issue with the distortion of the model when moving the camera, which I suspected is related, was still present. What I meant with "the model gets distorted" is visible in the following illustration:
The left image shows how the model looks when the camera is located in the center of the world, i.e. (0, 0, 0) while the right image shows what happens when I move the camera in negative y-axis方向。 As can be seen the model widens at the bottom and gets smaller at the top which is the same behavior described in the link I already provided above.
What I eventually did to resolve both issues is:
- Transpose the matrices before passing them to XMVector3Unproject (already described above)
- Transposed my view matrix by changing the code of the
SetViewMatrixmethod (code see below)
The SetViewMatrix method now looks as follows:
void Camera::SetViewMatrix()
{
XMFLOAT3 cameraPosition;
XMFLOAT3 cameraXAxis;
XMFLOAT3 cameraYAxis;
XMFLOAT3 cameraZAxis;
XMFLOAT4X4 viewMatrix;
// Keep camera's axes orthogonal to each other and of unit length.
m_cameraZAxis = XMVector3Normalize(m_cameraZAxis);
m_cameraYAxis = XMVector3Normalize(XMVector3Cross(m_cameraZAxis, m_cameraXAxis));
// m_cameraYAxis and m_cameraZAxis are already normalized, so there is no need
// to normalize the below cross product of the two.
m_cameraXAxis = XMVector3Cross(m_cameraYAxis, m_cameraZAxis);
// Fill in the view matrix entries.
float x = -XMVectorGetX(XMVector3Dot(m_cameraPosition, m_cameraXAxis));
float y = -XMVectorGetX(XMVector3Dot(m_cameraPosition, m_cameraYAxis));
float z = -XMVectorGetX(XMVector3Dot(m_cameraPosition, m_cameraZAxis));
//XMStoreFloat3(&cameraPosition, m_cameraPosition);
XMStoreFloat3(&cameraXAxis, m_cameraXAxis);
XMStoreFloat3(&cameraYAxis, m_cameraYAxis);
XMStoreFloat3(&cameraZAxis, m_cameraZAxis);
viewMatrix(0, 0) = cameraXAxis.x;
viewMatrix(0, 1) = cameraXAxis.y;
viewMatrix(0, 2) = cameraXAxis.z;
viewMatrix(0, 3) = x;
viewMatrix(1, 0) = cameraYAxis.x;
viewMatrix(1, 1) = cameraYAxis.y;
viewMatrix(1, 2) = cameraYAxis.z;
viewMatrix(1, 3) = y;
viewMatrix(2, 0) = cameraZAxis.x;
viewMatrix(2, 1) = cameraZAxis.y;
viewMatrix(2, 2) = cameraZAxis.z;
viewMatrix(2, 3) = z;
viewMatrix(3, 0) = 0.0f;
viewMatrix(3, 1) = 0.0f;
viewMatrix(3, 2) = 0.0f;
viewMatrix(3, 3) = 1.0f;
m_modelViewProjectionConstantBufferData->view = viewMatrix;
}
So I just exchanged row and column coordinates. Note that I had to make sure that my ChangeCameraYaw method gets called before my ChangeCameraPitch method. This is necessary because the orientation of the model otherwise is not as I want it.
There is also another approach that could be used. Instead of transposing the view matrix by exchanging the row and column coordinates and transposing it before passing it to XMVector3Unproject I could use the row_major keyword in the vertex shader together with the view matrix:
cbuffer ModelViewProjectionConstantBuffer : register(b0)
{
matrix model;
row_major matrix view;
matrix projection;
};
I came across this idea in this blog post. The keyword row_major influences on how the shader compiler interprets the matrix in memory. The same could also be achieved by changing the order of the vector * matrix multiplication in the vertex shader, i.e. using pos = mul(view, pos); instead of pos = mul(pos, view);
That's pretty much it. The two issues are indeed interconnected but using what I posted in this question I was able to resolve both issues so I'm accepting my own reply as answer to this question. Hope it helps someone in the future.