C++ - 使用 DirectX11 进行光线转换

Question

我的光线转换计算遇到了很多麻烦，无法完全弄清楚问题出在哪里。我正在使用 DirectX 的数学库来创建 vector 等。

到目前为止，这是我的代码:

auto normalizedDeviceCoords = glm::vec2((float(a_frontend->GetMousePos().x) / float(a_frontend->GetWidth())) * 2.f - 1.f, 
                                        -((float(a_frontend->GetMousePos().y) / float(a_frontend->GetHeight())) * 2.f - 1.f));
auto mouseOrigin = DirectX::XMVectorSet(normalizedDeviceCoords.x, normalizedDeviceCoords.y, 0.0f, 1.0f);
auto mouseEnd = DirectX::XMVectorSet(normalizedDeviceCoords.x, normalizedDeviceCoords.y, 1.0f, 1.0f);

auto viewproj = DirectX::XMMatrixMultiply(a_frontend->GetViewMatrix(), a_frontend->GetProjMatrix());
auto determinant = DirectX::XMMatrixDeterminant(a_camera.ViewProj());
auto inverseviewproj = DirectX::XMMatrixInverse(&determinant, a_camera.ViewProj());
auto rayOrigin = DirectX::XMVector4Transform(mouseOrigin, inverseviewproj);
auto rayEnd = DirectX::XMVector4Transform(mouseEnd, inverseviewproj);
auto raySubtraction = DirectX::XMVectorSubtract(rayEnd, rayOrigin);
auto rayDirection = DirectX::XMVector3Normalize(raySubtraction);

auto planeNormal = DirectX::XMVectorSet(0.f, 1.f, 0.f, 0.f);
auto pointOnPlane = DirectX::XMVectorSet(0.f, -0.1f, 0.f, 0.f);

DirectX::XMFLOAT3 denominator;
DirectX::XMStoreFloat3(&denominator, DirectX::XMVector3Dot(planeNormal, rayDirection));
if (fabs(denominator.x) <= 0.0001f)
{
    return;
}

auto pointMinusRay = DirectX::XMVectorSubtract(pointOnPlane, rayOrigin);

DirectX::XMFLOAT3 t;
auto almostT = DirectX::XMVector3Dot(pointMinusRay, planeNormal);
DirectX::XMStoreFloat3(&t, DirectX::XMVectorScale(almostT, 1.f / denominator.x));

if (t.x < 0)
{
    return;
}

auto rayDirectionLength = DirectX::XMVector3Length(rayDirection);
auto rayPoint = DirectX::XMVectorAdd(rayOrigin, DirectX::XMVectorScale(rayDirection, t.x));

我已经检查过归一化的设备坐标是正确的，光线方向的长度也是 1，并且光线方向看起来是正确的(仅靠数字很难判断)。

任何信息都会有用。我已经在此处和其他论坛上查看了一些有关堆栈溢出的信息，但一无所获。

Answer 1

我的队友能够解决这个问题。他刚刚开始使用 DirectX Math 函数 Vector3Unproject。这将采用鼠标坐标并将它们转换为世界坐标，同时使用原点和终点(因此采用 z 为 0 然后为 1 的鼠标坐标)。然后将它们彼此相减，得到一个指向世界某处的 vector 。然后，我们能够采用该方向并根据相机的当前 y 位置除以方向 vector y 的负值来转换它。这是我们现在拥有的所有代码:

DirectX::XMVECTOR mouseNear = DirectX::XMVectorSet((float)a_mousePos.x, (float)a_mousePos.y, 0.0f, 0.0f);
DirectX::XMVECTOR mouseFar = DirectX::XMVectorSet((float)a_mousePos.x, (float)a_mousePos.y, 1.0f, 0.0f);
DirectX::XMVECTOR unprojectedNear = DirectX::XMVector3Unproject(mouseNear, 0, 0, a_width, a_height, a_nearZ, a_farZ,
                                                                a_projection, a_view, DirectX::XMMatrixIdentity());
DirectX::XMVECTOR unprojectedFar = DirectX::XMVector3Unproject(mouseFar, 0, 0, a_width, a_height, a_nearZ, a_farZ,
                                                                a_projection, a_view, DirectX::XMMatrixIdentity());
DirectX::XMVECTOR result = DirectX::XMVector3Normalize(DirectX::XMVectorSubtract(unprojectedFar, unprojectedNear));
DirectX::XMFLOAT3 direction;
DirectX::XMStoreFloat3(&direction, result);
return direction;

//Get the distance to the ground.
DirectX::XMFLOAT3 cameraPosition = a_camera.GetPosition();

//Get the point on the ground.
cameraPosition.x += direction.x * (cameraPosition.y / -direction.y);
cameraPosition.z += direction.z * (cameraPosition.y / -direction.y);

希望这对将来的某个人有所帮助。