Proper Trigonometry For Rotating A Point Around The Origin

22,210

Solution 1

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;

Solution 2

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

Solution 3

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;
}
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Joshua
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Joshua

Updated on July 26, 2022

Comments

  • Joshua
    Joshua almost 2 years

    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 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;
    }
    
  • Justin Ardini
    Justin Ardini almost 14 years
    You could save the cosf and sinf results to variables to use half as many trig function calls. :)
  • Daniel Bingham
    Daniel Bingham over 6 years
    Don't forget, that if you're working in a typical screen coordinate space that your y axis will be inverted from the mathematical standard (down is +y, up is -y) and you'll need to account for that.