中心旋转

Question

CPoint pointCenter,pointA,pointB;
pointA 以 pointCenter为中心旋转角度 X 后的点 pointB是多少？谢谢各位

Answer 1

我实验写一个：            //平移pointCenter到（0，0）点
            pointA.X -= pointCenter.X;
            pointA.Y -= pointCenter.Y;            //绕原点旋转pointA获得pointB的坐标
            double radin = angle * 3.1415926 / 180.0; //角度转弧度
            double sin = Math.Sin(radin);
            double cos = Math.Cos(radin);            pointB.X = pointA.X * cos - pointA.Y * sin;
            pointB.Y = pointA.X * sin + pointA.Y * cos;            //平移回去
            pointB.X += pointCenter.X;
            pointB.Y += pointCenter.Y;           

Answer 2

既然自己都知道是三角函数的问题，查一下不就好了。Google也能查出来。

Answer 3

#define M_PI       3.14159265358979323846
#define DegToRad(degree) ((degree) * (M_PI / 180.0f))    
void MyDraw(HWND hwnd)
{
RECT device;
GetClientRect(hwnd,&device);
int cx,cy;
cx=device.right;
cy=device.bottom;

if (cx<cy)
{
cy=cx;
}
else
{
cx=cy;
}
if (cx%2)
{
cy=--cx;
} HDC hdc;
hdc=GetDC(hwnd);  SetMapMode(hdc,MM_ANISOTROPIC);  int r=100;

// scale & xlate
SetWindowExtEx(hdc,2*r,2*r,NULL);
SetViewportExtEx(hdc,cx,-cy,NULL); SetWindowOrgEx  (hdc, 0, 0,NULL);
SetViewportOrgEx(hdc,cx/2,cy/2,NULL); // x,y axes
MoveToEx(hdc,-r,0,NULL);
LineTo  (hdc, r,0);
MoveToEx(hdc,0,r,NULL);
LineTo  (hdc,0,-r); // 100-unit circle
MoveToEx(hdc,r,0,NULL);
AngleArc(hdc,0,0,r,0,360); // eg. for unit circle, in page space, (1,0) counter-clockwise rotate 30 degrees, becomes (cos30,sin30) 
// we can calculate corresponding coordinate in device page by lptodp POINT p;
p.x=(int)(cos(DegToRad(30)) * r);
p.y=(int)(sin(DegToRad(30)) * r); MoveToEx(hdc,0,0,NULL);
LineTo(hdc,p.x,p.y); // (86,49) ->(   532,146) LPtoDP(hdc,&p,1);}