Question

我的部分代码通过存储xyz位置，xyz刻度和四元数来存储等效的4x3矩阵。以下代码段：

class tTransform
{

    // data
    tVector4f    m_Position;
    tQuaternion  m_Rotation;
    tVector4f    m_Scale;

};

我想将这些对象中的两个相乘，（好像它是一个矩阵乘法），并且我想知道是否有更快/更好的方法来做到比将每个对象转换为矩阵，做到这样倍增，然后再次提取结果位置，旋转和缩放？

Answer 1

健康警告，因为这是来自内存并且完全未经测试。您需要定义或替换tQuaternion和tVector4 s。

的运算符

class tTransform
{

    // data
    tVector4f    m_Position;
    tQuaternion  m_Rotation;
    tVector4f    m_Scale;

public:
    // World = Parent * Local (*this == parent)
    tTransform operator * (const tTransform& localSpace)
    {
        tTransform worldSpace;
        worldSpace.m_Position = m_Position + 
                                m_Rotation * (localSpace.m_Position * m_Scale);
        worldSpace.m_Rotation = m_Rotation * localSpace.m_Rotation;
        worldSpace.m_Scale = m_Scale * (m_Rotation * localSpace.m_Scale);
        return worldSpace;
    }

    // Local = World / Parent (*this = World)
    tTransform operator / (const tTransform& parentSpace)
    {
        tTransform localSpace;
        tQuaternion parentSpaceConjugate = parentSpace.m_Rotation.conjugate(); 
        localSpace.m_Position = (parentSpaceConjugate * 
                                (m_Position - parentSpace.m_Position)) /
                                parentSpace.m_Scale;

        localSpace.m_Rotation = parentSpaceConjugate * m_Rotation;

        localSpace.m_Scale = parentSpaceConjugate *
                             (m_Scale / parentSpace.m_Scale);
        return localSpace;
    }
};

Answer 2

我回答了托马斯（Tomas）的问题，我也将在此复制，因为它也回答了您的问题。答案几乎是伪代码，因此您应该可以将其应用于您的课程。您没有指定使用（TRS或SRT）构建矩阵的顺序，因此我假设使用TRS。（您的向量是列）

transform transform::operator * (const transform &other) const
{
    // mat1 = T1 * R1 * S1; mat2 = T2 * R2 * S2
    // mat = mat1 * mat2; mat*v = mat1 * mat2 * v
    // assuming "this" is mat1, and other is mat2
    // alternatively "this" can be considered parent, and other child in a node hierarchy
    transform r;
    // R = R1 * R2
    r.orientation = orientation * other.orientation;
    // Note: I don't know how to implement inverse of quat in your lib
    // S = R2^-1 * (S1 * (R2 * S2))
    r.scale = inverse(other.orientation) * (scale * (other.orientation * other.scale));
    // T = T1 * (R1 * (S1 * T2))
    r.position = position + (orientation * (scale * other.position));
    return r;
}

您可以在此处了解如何通过四元数旋转向量： https://gamedev.stackexchange.com/questions/28395/rotating-vector3-by-a-quaternion

Answer 3

有人告诉我这在一般情况下是不可能的。参见https://gamedev.stackexchange.com/questions/167287/combine-two-translation-rotation-scale-triplets-without-matrices

问题是结构不能代表剪切力，这在组合旋转和非均匀缩放后可能需要。

请纠正我的错误。

矩阵乘以位置，四元数和比例分量

3 个答案: