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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、程序
2、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 Comma nd1_Click()If (-C + A) (D - B) The n F = -C + A Else: F = D - BG = 1Do While (2 A G) = E The n I = I - E * (2 a G): x2 = x2 - EJ = J + (D - B)y1 = y2If (J / (2 a G) = E The n J 二 J - E * (2 a G): y2 = y2 - ETex
3、t6.Text, y1-BPicturel line (x2 * Text6.Text, y2 * Text6.Text)-(x1 * Text6.Text), vbGree nNext HPicturel.L ine (C * Text6.Text, -D * Text6.Text)-(A * Text6.Text,* Text6.Text), vbRedEnd Sub*单步直线插补*Private Sub Comma nd2_Click()If (-C + A) (D - B) The n F = -C + A Else: F = D - BG = 1Do While (2 a G) =
4、E And (J / (2 a G) = E The nPicture1.Line (x2 - E) * Text6.Text, (y2 - E) * Text6.Text)-(x2 *Text6.Text, y2 * Text6.Text), vbGree nI = I - E * (2 a G)J = J - E * (2 a G)x2 = x2 - Ey2 = y2 - EElseIf (I / (2 a G) = E And (J / (2 a G) E The nText6.Text,Picture1 丄ine(x2 - E) * Text6.Text, y2 * Text6.Tex
5、t)-(x2y2 * Text6.Text), vbGree nI = I - E * (2 a G)x2 = x2 - EElseIf (I / (2 a G) = E The nText6.Text,Picture1 丄 i ne(x2 * Text6.Text, (y2 - E) * Text6.Text)-(x2y2 * Text6.Text), vbGree nJ = J - E * (2 a G)y2 = y2 - EEnd IfEnd Sub3. DDA法插补第二象限直线的流程图:丄Y结束逐点比较法插补第三到第二象限顺圆弧1. 源程序:Dim XO, YO, CXY, CD, C
6、X1, CY1, CX2, CY2, CX11, CY11, CX22, CY22,ci, R, a1, b1, c1, d1, e1, x1, y1, x2, y2 As Si ngleCONST Pi=3.1415926*连续圆弧插补*Private Sub Comma nd6_Click()d1 = 1 / 2 * (CX22 A 2 + CY22 A 2 - CX11 A 2 - CY11 A 2) / (CY22 - CY11)e1 = (CX11 - CX22) / (CY22 - CY11)a1 = 1 + e1 a 2b1 = 2 * d1 * e1 - 2 * CX11 -
7、2 * CY11 * e1c1 = CX11 a 2 + CY11 a 2 + d1 a 2 - 2 * d1 * CY11 - R a 2XO = (-b1 + Sqr(b1 a 2 - 4 * a1 * c1) / (2 * a1)YO = d1 + e1 * XO*求得圆心If CX1 XO And CX2 XO The n o = Atn (CY1 - YO) / (CX1 - XO) + Pi o1 = Atn (CY2 - YO) / (CX2 - XO) + Pi Picture2.Circle (XO, YO), R, vbRed, o1, o Else*起始角终止角Pictu
8、re2.Circle (XO, YO), R, vbRed, 2 / Pi, 3 * Pi / 2 End If*画圆aga in:If CY1 YO The n*开始插补过程If (CX1 - XO) A 2 + (CY1 - YO) A 2 = R a 2 The nPicture2.Line (CX1, CY1)-(CX1 - ci, CY1)CX1 = CX1 - ciElsePicture2.Line (CX1, CY1)-(CX1, CY1 + ci)CY1 = CY1 + ciEnd IfElseIf (CX1 - XO) a 2 + (CY1 - YO) a 2 (CX1 -
9、CX2 + ci) a 2 + (CY1-CY2) a 2 Or (CX1 - CX2) a 2 + (CY1 - CY2) a 2 (CX1 - CX2 - ci) a 2 + (CY1 - CY2) a 2 Or (CX1 - CX2) a 2 + (CY1 - CY2) a 2 (CX1 -CX2) A 2 + (CY1 - CY2 + ci) A 2 Or (CX1 - CX2) A 2 + (CY1 - CY2) A 2 (CX1 - CX2) a 2 + (CY1 - CY2 - ci) a 2 The nGoTo aga inElseEnd IfEnd Sub*单步圆弧插补*Pr
10、ivate Sub Comma nd7_Click()di = 1 / 2 * (CX22 a 2 + CY22 a 2 - CX11 a 2 - CY11 a 2) / (CY22 -CY11)el = (CX11 - CX22) / (CY22 - CY11)al = 1 + el a 2bi = 2 * di * el - 2 * CX11 - 2 * CY11 * elci = CX11 a 2 + CY11 a 2 + di a 2 - 2 * di * CY11 - R a 2X0 = (-bi + Sqr(bi a 2 - 4 * ai * ci) / (2 * ai)Y0 =
11、di + ei * X0*求得圆心If CXi X0 And CX2 X0 The no = Atn (CYi - Y0) / (CXi - X0) + Pi*终止角oi = Atn (CY2 - Y0) / (CX2 - X0) + Pi*终止角Picture2.Circle (X0, Y0), R, vbRed, oi, oElse*画圆Picture2.Circle (X0, Y0), R, vbRed, 2 / Pi, 3 * Pi / 2 End IfIf (CXi - CX2) a 2 + (CYi - CY2) a 2 (CXi - CX2 + ci) a 2 + (CYi -
12、CY2) a 2 Or (CXi - CX2) a 2 + (CYi - CY2) a 2 (CXi - CX2 - ci) a 2 + (CYi - CY2) a 2 Or (CXi - CX2) a 2 + (CYi - CY2) a 2 (CXi - CX2) a 2 + (CYi - CY2 + ci) a 2 Or (CXi - CX2) a 2 + (CYi - CY2) a 2 (CXi -CX2) a 2 + (CYi - CY2 - ci) a 2 The nElseGoTo cover*开始插补过程End IfIf CYi Y0 The nIf (CXi - X0) a 2 + (CYi - Y0) a 2 = R a 2 The nPicture2.Line (CXi, CYi)-(CXi - ci, CYi)CXi = CXi - ciElsePicture2.Line (CXi, CYi)-(CXi, CYi + ci)CYi = CYi + ciEnd IfElseIf (CXi - X0) a 2 + (CYi - Y
