適當的三角法旋轉原點的位置

Question

以下兩種方法中的哪一種使用正確的數學來旋轉點？如果是這樣，哪一個是正確的？

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

這取決於你如何定義angle。如果是逆時針測量（這是數學約定），那麼正確的旋轉是你的第一個：

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

但如果是順時針測量，然後第二個是正確的：

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

Answer 2

From Wikipedia

要使用的矩陣的點（x，y）與被旋轉被寫爲一個向量，然後通過從角度計算，θ，像這樣的矩陣相乘進行旋轉：

其中（x'，y'）的是旋轉後的點的座標，並且x的公式'和y'可以被看作是

Answer 3

這是從我自己的矢量庫中提取..

//---------------------------------------------------------------------------------- 
// 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; 
}