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1、一、 课设任务用DDA法插补第二象限直线。用逐点比较法插补第三到第二象限顺圆弧。二、课设要求 1、具有数据输入界面，有起点、终点、半径及插补步长等； 2、具有单步插补过程的动态显示功能； 3、插补的步长可调；三、编程语言Visual Basic四、功能说明本程序用逐点比较法插补第二象限的直线，及第三二象限的顺圆弧，可进行连续插补或单步插补。1、直线插补：用逐点比较法实现第二象限任意直线的插补，需要输入起点、终点坐标及步长。2、圆弧插补：用逐点比较法实现第三二象限的顺圆弧段的插补，需要输入起点、终点坐标、半径和步长。3、插补步长可调。4、可以单步执行所有插补动作，单击一次按钮执行一次插补。五、程
2、序内容 DDA法插补第二象限直线1、源程序： Dim A, B, C, D, E, F, G, I, J, E, F As Single*连续直线插补*Private Sub Command1_Click()If (-C + A) (D - B) Then F = -C + A Else: F = D - BG = 1Do While (2 G) = E Then I = I - E * (2 G): x2 = x2 - EJ = J + (D - B)y1 = y2If (J / (2 G) = E Then J = J - E * (2 G): y2 = y2 - EPicture1.Li
3、ne (x2 * Text6.Text, y2 * Text6.Text)-(x1 * Text6.Text, y1 * Text6.Text), vbGreenNext HPicture1.Line (C * Text6.Text, -D * Text6.Text)-(A * Text6.Text, -B * Text6.Text), vbRedEnd Sub*单步直线插补*Private Sub Command2_Click()If (-C + A) (D - B) Then F = -C + A Else: F = D - BG = 1Do While (2 G) = E And (J
4、/ (2 G) = E ThenPicture1.Line (x2 - E) * Text6.Text, (y2 - E) * Text6.Text)-(x2 * Text6.Text, y2 * Text6.Text), vbGreenI = I - E * (2 G)J = J - E * (2 G)x2 = x2 - Ey2 = y2 - EElseIf (I / (2 G) = E And (J / (2 G) E ThenPicture1.Line (x2 - E) * Text6.Text, y2 * Text6.Text)-(x2 * Text6.Text, y2 * Text6
5、.Text), vbGreenI = I - E * (2 G)x2 = x2 - EElseIf (I / (2 G) = E ThenPicture1.Line (x2 * Text6.Text, (y2 - E) * Text6.Text)-(x2 * Text6.Text, y2 * Text6.Text), vbGreenJ = J - E * (2 G)y2 = y2 - EEnd IfEnd Sub3. DDA法插补第二象限直线的流程图：YNYNYNMm-1XeXeXeYeYeYe初始化xeXe，yeYe，累加次数mM，Xe、Ye清零M=0吗？Ye有溢出吗？Xe有溢出吗？+Y向走
6、一步-X向走一步结束逐点比较法插补第三到第二象限顺圆弧1. 源程序：Dim X0, Y0, CXY, CD, CX1, CY1, CX2, CY2, CX11, CY11, CX22, CY22, ci, R, a1, b1, c1, d1, e1, x1, y1, x2, y2 As SingleCONST Pi=3.1415926*连续圆弧插补*Private Sub Command6_Click()d1 = 1 / 2 * (CX22 2 + CY22 2 - CX11 2 - CY11 2) / (CY22 - CY11)e1 = (CX11 - CX22) / (CY22 - CY1
7、1)a1 = 1 + e1 2b1 = 2 * d1 * e1 - 2 * CX11 - 2 * CY11 * e1c1 = CX11 2 + CY11 2 + d1 2 - 2 * d1 * CY11 - R 2X0 = (-b1 + Sqr(b1 2 - 4 * a1 * c1) / (2 * a1)Y0 = d1 + e1 * X0 *求得圆心If CX1 X0 And CX2 X0 Theno = Atn(CY1 - Y0) / (CX1 - X0) + Pi *起始角o1 = Atn(CY2 - Y0) / (CX2 - X0) + Pi *终止角Picture2.Circle (X
8、0, Y0), R, vbRed, o1, oElsePicture2.Circle (X0, Y0), R, vbRed, 2 / Pi, 3 * Pi / 2End If *画圆 again: *开始插补过程If CY1 Y0 ThenIf (CX1 - X0) 2 + (CY1 - Y0) 2 = R 2 ThenPicture2.Line (CX1, CY1)-(CX1 - ci, CY1)CX1 = CX1 - ciElsePicture2.Line (CX1, CY1)-(CX1, CY1 + ci)CY1 = CY1 + ciEnd IfElseIf (CX1 - X0) 2 +
9、 (CY1 - Y0) 2 (CX1 - CX2 + ci) 2 + (CY1 - CY2) 2 Or (CX1 - CX2) 2 + (CY1 - CY2) 2 (CX1 - CX2 - ci) 2 + (CY1 - CY2) 2 Or (CX1 - CX2) 2 + (CY1 - CY2) 2 (CX1 - CX2) 2 + (CY1 - CY2 + ci) 2 Or (CX1 - CX2) 2 + (CY1 - CY2) 2 (CX1 - CX2) 2 + (CY1 - CY2 - ci) 2 ThenGoTo againElseEnd IfEnd Sub*单步圆弧插补*Private
10、Sub Command7_Click()d1 = 1 / 2 * (CX22 2 + CY22 2 - CX11 2 - CY11 2) / (CY22 - CY11)e1 = (CX11 - CX22) / (CY22 - CY11)a1 = 1 + e1 2b1 = 2 * d1 * e1 - 2 * CX11 - 2 * CY11 * e1c1 = CX11 2 + CY11 2 + d1 2 - 2 * d1 * CY11 - R 2X0 = (-b1 + Sqr(b1 2 - 4 * a1 * c1) / (2 * a1)Y0 = d1 + e1 * X0 *求得圆心If CX1 X
11、0 And CX2 X0 Theno = Atn(CY1 - Y0) / (CX1 - X0) + Pi *终止角o1 = Atn(CY2 - Y0) / (CX2 - X0) + Pi *终止角Picture2.Circle (X0, Y0), R, vbRed, o1, oElsePicture2.Circle (X0, Y0), R, vbRed, 2 / Pi, 3 * Pi / 2End If *画圆If (CX1 - CX2) 2 + (CY1 - CY2) 2 (CX1 - CX2 + ci) 2 + (CY1 - CY2) 2 Or (CX1 - CX2) 2 + (CY1 - CY2) 2 (CX1 - CX2 - ci) 2 + (CY1 - CY2) 2 Or (CX1 - CX2) 2 + (CY1 - CY2) 2 (CX1 - CX2) 2 + (CY1 - CY2 + ci) 2 Or (CX1 - CX2) 2 + (CY1 - CY2) 2 (CX1 - CX2) 2 + (CY1 - CY2 - ci)
