适当的三角法旋转原点的位置

Question

以下两种方法中的哪一种使用正确的数学来旋转点？如果是这样，哪一个是正确的？

POINT rotate_point(float cx,float cy,float angle,POINT p) 
{ 
    float s = sin(angle); 
    float c = cos(angle); 

    // translate point back to origin: 
    p.x -= cx; 
    p.y -= cy; 

    // Which One Is Correct: 
    // This? 
    float xnew = p.x * c - p.y * s; 
    float ynew = p.x * s + p.y * c; 
    // Or This? 
    float xnew = p.x * c + p.y * s; 
    float ynew = -p.x * s + p.y * c; 

    // translate point back: 
    p.x = xnew + cx; 
    p.y = ynew + cy; 
} 

Answer 1

这取决于你如何定义angle。如果是逆时针测量（这是数学约定），那么正确的旋转是你的第一个：

// This? 
float xnew = p.x * c - p.y * s; 
float ynew = p.x * s + p.y * c; 

但如果是顺时针测量，然后第二个是正确的：

// Or This? 
float xnew = p.x * c + p.y * s; 
float ynew = -p.x * s + p.y * c; 

Answer 2

From Wikipedia

要使用的矩阵的点（x，y）与被旋转被写为一个向量，然后通过从角度计算，θ，像这样的矩阵相乘进行旋转：

其中（x'，y'）的是旋转后的点的坐标，并且x的公式'和y'可以被看作是

Answer 3

这是从我自己的矢量库中提取..

//---------------------------------------------------------------------------------- 
// Returns clockwise-rotated vector, using given angle and centered at vector 
//---------------------------------------------------------------------------------- 
CVector2D CVector2D::RotateVector(float fThetaRadian, const CVector2D& vector) const 
{ 
    // Basically still similar operation with rotation on origin 
    // except we treat given rotation center (vector) as our origin now 
    float fNewX = this->X - vector.X; 
    float fNewY = this->Y - vector.Y; 

    CVector2D vectorRes( cosf(fThetaRadian)* fNewX - sinf(fThetaRadian)* fNewY, 
          sinf(fThetaRadian)* fNewX + cosf(fThetaRadian)* fNewY); 
    vectorRes += vector; 
    return vectorRes; 
}