三角形旋转问题

把你的问题具体为：平面直角坐标系上的一点P（a, b），把直线OP以O为轴旋转n弧度，则点P的坐标(x, y)为：
x = a*cos(n) - b*sin(n)
y = b*cos(n) + a*sin(n)

按楼上所说，旋转90度，坐标不对啊Private Sub Command1_Click()
  Form1.Scale (-110, 110)-(110, -110)
  Line (-105, 0)-(105, 0)
  Line (0, 105)-(0, -105)
  CurrentX = 105: CurrentY = 20: Print "x"
  CurrentX = 10: CurrentY = 105: Print "y"
  For i = -100 To 100 Step 20
          CurrentX = i: CurrentY = 7: Line -(i, 0)
          CurrentX = i - 5: CurrentY = -5: Print i / 10
  Next i
  CurrentX = 45: CurrentY = 35: Print "A"
  PSet (40, 30), vbRed

  Dim xx As Single, yy As Single

    For i = 1 To 360 Step 1        xx = 40 * Cos(i) - 30 * Sin(i)
        yy = 30 * Cos(i) + 40 * Sin(i)

        CurrentX = xx + 5: CurrentY = yy + 5: If i = 90 Then Print "(" & Int(xx) & "," & Int(yy) & ")"
        PSet (xx, yy), vbRed

    Next

End Sub

按楼上所说，旋转90度，坐标不对啊 Private Sub Command1_Click()
  Form1.Scale (-110, 110)-(110, -110)
  Line (-105, 0)-(105, 0)
  Line (0, 105)-(0, -105)
  CurrentX = 105: CurrentY = 20: Print "x"
  CurrentX = 10: CurrentY = 105: Print "y"
  For i = -100 To 100 Step 20
          CurrentX = i: CurrentY = 7: Line -(i, 0)
          CurrentX = i - 5: CurrentY = -5: Print i / 10
  Next i
  CurrentX = 45: CurrentY = 35: Print "A"
  PSet (40, 30), vbRed

  Dim xx As Single, yy As Single

    For i = 1 To 360 Step 1        xx = 40 * Cos(i) - 30 * Sin(i)
        yy = 30 * Cos(i) + 40 * Sin(i)

        CurrentX = xx + 5: CurrentY = yy + 5: If i = 90 Then Print "(" & Int(xx) & "," & Int(yy) & ")"
        PSet (xx, yy), vbRed

    Next

End Sub

角度转换成弧度也不对Private Const PI = 3.14159265358979Private Sub Command1_Click()
Dim DAngle As Single DAngle = PI / 180  Form1.Scale (-110, 110)-(110, -110)
  Line (-105, 0)-(105, 0)
  Line (0, 105)-(0, -105)
  CurrentX = 105: CurrentY = 20: Print "x"
  CurrentX = 10: CurrentY = 105: Print "y"
  For i = -100 To 100 Step 20
          CurrentX = i: CurrentY = 7: Line -(i, 0)
          CurrentX = i - 5: CurrentY = -5: Print i / 10
  Next iCurrentX = 45: CurrentY = 35: Print "A"
  PSet (40, 30), vbRed

  Dim xx As Single, yy As Single

    For i = 1 To 360 Step 1        xx = 40 * Cos(i * DAngle) - 30 * Sin(i * DAngle)
        yy = 30 * Cos(i * DAngle) + 40 * Sin(i * DAngle)
        CurrentX = xx + 5: CurrentY = yy + 5: If i = 90 Then Print "(" & Int(xx) & "," & Int(yy) & ")"
        PSet (xx, yy), vbRed

    Next
End Sub

x1=(X^2+y^2)^0.5*cos(atn(y/x)+Angle)
y1=(X^2+y^2)^0.5*sin(atn(y/x)+Angle)

呵呵，弄成椭圆的了，抱歉抱歉！
同意8楼，实质可由圆的参数方程：
x = r * Cos(n)
y = r * Sin(n)
推得

据说可以用向量的旋转来解决此问题，我根据BCB的代码转换过来，好像还不行
'
'  #define   PI   3.14159265
'
'  void   DrawLineArrow(TCanvas*   Canvas,   TPoint   p0,   TPoint   p1)
'  {
'  //对于点(x0,y0)指向(x1,y1)的箭头线，以(x1,y1)为原点，先计算出(x1,y1)
'  //指向(x0,y0)的单位向量，箭头线与直线的角度为d，那么就是将这个向量
'  //旋转角度d和－d，向量旋转的公式为：
'  //(a+b*i)*(cos(d)+i*sin(d))   =   a*cos(d)-b*sin(d)   +   i*(a*sin(d)+b*cos(d))
'  //因为上面的向量以(x1,y1)为原点的单位向量，所以还需要乘以一个长度（
'  //箭头线的长度），然后加上偏移就可以了。也就是画的箭头线的顶点的坐标为
'  //x=x1+L*(a*cos(d)-b*sin(d)),   y=y1+L*(a*sin(d)+b*cos(d))。
'  //另一个顶点的坐标，将上面的sin(d)换成   -sin(d)   就可以了。
'  //上面公式中的a为(x0-x1)/L0,   b为(y0-y1)/L0，L0是(x0,y0)与(x1,y1)的长度，
'  //L是箭头线的长度(自己设定)。
'          float   a,   b,   L0,   L,   d,   x,   y;
'
'          Canvas->MoveTo(p0.x,   p0.y);
'          Canvas->LineTo(p1.x,   p1.y);   //画p0   到   p1   的直线
'
'          L0   =   sqrt(pow(p0.x-p1.x,2)+pow(p0.y-p1.y,2));   //直线的长度
'          a     =   (p0.x-p1.x)/L0;     b   =   (p0.y-p1.y)/L0;   //单位向量的两个分量
'          L   =   10.0;   //箭头线长度，这儿我写成固定的，你可以自己修改或者计算
'          d   =   30.0/180.0*PI;     //箭头线与直线的角度30度，你可以自己调整
'          for(int   n=0;   n<2;   n++)   {
'                  x   =   p1.x+L*(a*cos(d)-b*sin(n==0?d:-d));
'                  y   =   p1.y+L*(a*sin(n==0?d:-d)+b*cos(d));
'                  Canvas->MoveTo(p1.x,   p1.y);
'                  Canvas->LineTo(x,   y);   //画箭头线
'          }
'  }Private Declare Function LineTo Lib "gdi32" (ByVal hdc As Long, ByVal x As Long, ByVal y As Long) As Long
Private Declare Function MoveTo Lib "gdi32" Alias "MoveToEx" ( _
     ByVal hdc As Long, ByVal x As Long, ByVal y As Long, _
     Optional lpPoint As Any = 0) As Long

Private Type POINTAPI
        x As Long
        y As Long
End TypePrivate Const PI = 3.14159265358979Private Sub DrawLineArrow(TCanvas As PictureBox, p0 As POINTAPI, p1 As POINTAPI)Dim a%, b%, L0%, L%, d%, x%, y%
Dim n As Integer    MoveTo TCanvas.hdc, p0.x, p0.y
    '画p0   到   p1   的直线
    LineTo TCanvas.hdc, p1.x, p1.y

    '直线的长度
    L0 = Sqr((p1.x - p0.x) ^ 2 + (p1.y - p0.y) ^ 2)
    '单位向量的两个分量
    a = (p0.x - p1.x) / L0
    b = (p0.y - p1.y) / L0
    '箭头线长度10
    L = 10
    '箭头线与直线的角度30度
    d = 30 / 180 * PI

    n = 0
    x = p1.x * L * (a * Cos(d) - b * Sin(IIf(n = 0, d, -d)))
    y = p1.y * L * (a * Sin(IIf(n = 0, d, -d) + b * Cos(d)))
    MoveTo TCanvas.hdc, p1.x, p1.y
    '画箭头线
    LineTo TCanvas.hdc, x, y

    n = 1
    x = p1.x * L * (a * Cos(d) - b * Sin(IIf(n = 0, d, -d)))
    y = p1.y * L * (a * Sin(IIf(n = 0, d, -d) + b * Cos(d)))
    MoveTo TCanvas.hdc, p1.x, p1.y
    '画箭头线
    LineTo TCanvas.hdc, x, y


    n = 2
    x = p1.x * L * (a * Cos(d) - b * Sin(IIf(n = 0, d, -d)))
    y = p1.y * L * (a * Sin(IIf(n = 0, d, -d) + b * Cos(d)))
    MoveTo TCanvas.hdc, p1.x, p1.y
    '画箭头线
    LineTo TCanvas.hdc, x, y


End Sub
Private Sub Command1_Click()    Dim p0 As POINTAPI, p1 As POINTAPI
    p0.x = 10: p0.y = 10
    p1.x = 120: p1.y = 130

    DrawLineArrow Picture1, p0, p1End Sub

Option ExplicitPrivate Type POINTAPI    X As Long
    Y As LongEnd TypePrivate Const ALTERNATE = 1
Private Const BLACK_BRUSH = 4Private Declare Function Polygon _
                Lib "gdi32" (ByVal hdc As Long, _
                             lpPoint As POINTAPI, _
                             ByVal nCount As Long) As Long
Private Declare Function GetStockObject Lib "gdi32" (ByVal nIndex As Long) As Long
Private Declare Function CreatePolygonRgn _
                Lib "gdi32" (lpPoint As Any, _
                             ByVal nCount As Long, _
                             ByVal nPolyFillMode As Long) As Long
Private Declare Function FillRgn _
                Lib "gdi32" (ByVal hdc As Long, _
                             ByVal hRgn As Long, _
                             ByVal hBrush As Long) As Long
Private Declare Function DeleteObject Lib "gdi32" (ByVal hObject As Long) As Long' 画一个从(x1, y1)到到(x2, y2)的箭头
Private Sub DrawArrow(ByVal pic As Object, _
                        ByVal x1 As Single, _
                        ByVal y1 As Single, _
                        ByVal x2 As Single, _
                        ByVal y2 As Single, _
                        ByVal length As Single)
Dim hBrush      As Long
Dim hRgn        As Long
Dim lpTPoint(2) As POINTAPIDim vx As Single
Dim vy As Single
Dim dist As Single
Dim i As Integer    '画直线
    pic.Line (x1, y1)-(x2, y2)    '获取直线单位向量
    vx = x2 - x1
    vy = y2 - y1
    dist = Sqr(vx ^ 2 + vy ^ 2)
    vx = vx / dist
    vy = vy / dist    '顶点
    lpTPoint(0).X = x2
    lpTPoint(0).Y = y2    ' 右顶点位置
    lpTPoint(1).X = x2 + (-vy - vx) * length
    lpTPoint(1).Y = y2 + (vx - vy) * length
    '左顶点位置
    lpTPoint(2).X = x2 + (vy - vx) * length
    lpTPoint(2).Y = y2 + (-vx - vy) * length    Polygon pic.hdc, lpTPoint(0), 3
    ' 获得黑色画刷
    hBrush = GetStockObject(BLACK_BRUSH)
    ' 为了填充颜色创建区域
    hRgn = CreatePolygonRgn(lpTPoint(0), 3, ALTERNATE)
    ' 通过创建成功就使用颜色填充
    If hRgn Then FillRgn pic.hdc, hRgn, hBrush

    DeleteObject hRgnEnd Sub
Private Sub Form_Load()
Const BARB_LEN = 10

    Me.Show
    Picture1.AutoRedraw = True
    Picture1.ScaleMode = 3      'Twip

    '画箭头
    DrawArrow Picture1, 20, 40, 100, 50, BARB_LEN
    DrawArrow Picture1, 180, 30, 20, 190, BARB_LEN
    DrawArrow Picture1, 30, 240, 320, 30, BARB_LEN
    DrawArrow Picture1, 100, 240, 270, 240, BARB_LEN
    DrawArrow Picture1, 340, 70, 340, 260, BARB_LEN
End Sub

因在本论坛不知怎样上传文件，所以将示例放在下面地址，如合你的要求，我再加以补充完善
请到网页下“旋转示例”试试
http://khjkhj.gbaopan.com/

调试易

三角形旋转问题

解决方案 »