C++ - 图形旋转

如果果一个点（x1,y1）旋转到（x2,y2），对应的角度旋转从θ1到θ1+θ2

sinθ1=y1/sqrt(x1*x1+ y1*y1)
cosθ1=x1/sqrt(x1*x1+ y1*y1)

sin(θ1+θ2)=sin(θ1)*cos(θ2)+cos(θ1)*sin(θ2)=y2/sqrt(x2*x2+y2*y2)

cos(θ1+θ2)=cos(θ1)*cos(θ2)-sin(θ1)*sin(θ2)=x2/sqrt(x2*x2+y2*y2)

由于sqrt(x1*x1+ y1*y1)  =  sqrt(x2*x2+y2*y2)

所以由上述推到出

y2=x1*sin(θ2)+y1*cos(θ2);

x2=x1*cos(θ2)-y1*sin(θ2);

 

 

/* ptBase:旋转的基点 ; pts[] 旋转的数组; nCount点数目

*/

void RotatePoints(const POINT& ptBase,POINT pts[],long nCount,double angle,BOOL isUseDegree)

{

     if ( fabs(angle) < 1e-6) return;

 

     if ( isUseDegree)//把度数换算成弧度

     {

        angle = angle /180 * PI;//PI = 3.1415

     }

 

     double cose = cos(angle);

     double sine = sin(angle);

 

     double dx(0.0),dy(0.0);

 

     for(int i=0;i<nCount; ++i)

     {

        dx = pts[i].x - ptBase.x;

        dy = pts[i].y - ptBase.y;

 

        double x = dx * cose - dy * sine + ptBase.x;

        double y = dx * sine + dy * cose + ptBase.y;

        pts[i].x = x;

        pts[i].y = y;

     }

}