Proper Trigonometry For Rotating A Point Around The Origin

后端 未结 3 411
悲哀的现实
悲哀的现实 2020-12-10 12:05

Do either of the below approaches use the correct mathematics for rotating a point? If so, which one is correct?

POINT rotate_point(float cx,float cy,float          


        
相关标签:
3条回答
  • 2020-12-10 12:43

    From Wikipedia

    To carry out a rotation using matrices the point (x, y) to be rotated is written as a vector, then multiplied by a matrix calculated from the angle, θ, like so:

    https://upload.wikimedia.org/math/0/e/d/0ed0d28652a45d730d096a56e2d0d0a3.png

    where (x′, y′) are the co-ordinates of the point after rotation, and the formulae for x′ and y′ can be seen to be

    alt text

    0 讨论(0)
  • 2020-12-10 12:44

    This is extracted from my own vector library..

    //----------------------------------------------------------------------------------
    // 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;
    }
    
    0 讨论(0)
  • 2020-12-10 12:54

    It depends on how you define angle. If it is measured counterclockwise (which is the mathematical convention) then the correct rotation is your first one:

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

    But if it is measured clockwise, then the second is correct:

    // Or This?
    float xnew = p.x * c + p.y * s;
    float ynew = -p.x * s + p.y * c;
    
    0 讨论(0)
提交回复
热议问题