Proper Trigonometry For Rotating A Point Around The Origin?

Question

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; }

Answer 1

To rotate a point around another point (cx, cy) by an angle, the correct mathematics involve a rotation matrix. The formulas for rotating a point (px, py) around the origin and then translating it back are as follows:

xnew = cos(angle) * (px - cx) - sin(angle) * (py - cy) + cx

ynew = sin(angle) * (px - cx) + cos(angle) * (py - cy) + cy

Comparing this to your provided code:

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;

// Rotating the point

float xnew = p.x * c - p.y * s;

float ynew = p.x * s + p.y * c;

// ... (presumably translate back)

}

The part of your code given here:

float xnew = p.x * c - p.y * s;

float ynew = p.x * s + p.y * c;

is the correct way to apply the rotation part of the transformation. This part effectively rotates the point around the origin (0,0). Since you first translate your point p to the origin with p.x -= cx; p.y -= cy;, this rotation is effectively around the point (cx, cy). After applying this rotation, you would need to translate the point back to its original location. You can do this by adding cx to xnew and cy to ynew

xnew += cx;

ynew += cy;

So, your first approach is the correct one for rotating a point around another point. Remember to translate the point back after rotating it.