王毅辉数控课设报告.doc

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1、机床数控技术课程设计姓名：王毅辉 学号：050610525指导老师：王宏涛2010.01.15目录：一：课程设计要求3二：DDA直线插补41. DDA法直线插补流程图42. DDA法直线插补源程序53. 实现的情况 9三：逐点比较法圆弧插补 101. 逐点比较法圆弧插补流程图102. 逐点比较法圆弧插补源程序113. 实现情况20四：总计与体会 20一：课程设计的要求：1.DDA法直线插补第1象限2.采用逐点比较法插补第2-1象限顺圆弧。3.具有数据输入界面，如起点、终点、圆心、半径及插补步长等4.具有插补过程的动态显示功能，如单步插补、连续插补5.插补步长可调，直线的起点、圆弧圆心在坐标系中

2、位置可调。6建议用C语言编程 ，也可以使用VB、VC等语言编程。二、DDA法直线插补1、流程图求num的值起始并初始化yi0ry有溢出-y走一步rx有溢出NY+x走一步Ynum=num-1num=0？Yry有溢出+y走一步rx有溢出NY-x走一步Ynum=num-1num=0？结束YNyi xs) And (ye ys) ThenDo While (num 0)rx = rx + Abs(vx)ry = ry + Abs(vy) If (rx 256) Then xi = xi + step rx = rx Mod 256 num = num - 1 waittime (0.01) End I

3、f If (ry 256) Then yi = yi + step ry = ry Mod 256 num = num - 1 waittime (0.01) End If picture1.Line -(xi, yi) LoopEnd IfIf (xe ys) ThenDo While (num 0)rx = rx + Abs(vx)ry = ry + Abs(vy) If (rx 256) Then xi = xi - step rx = rx Mod 256 num = num - 1 waittime (0.01) End If If (ry 256) Then yi = yi + s

4、tep ry = ry Mod 256 num = num - 1 waittime (0.01) End If picture1.Line -(xi, yi) LoopEnd IfIf (xe = xs) And (ye 0)rx = rx + Abs(vx)ry = ry + Abs(vy) If (rx 256) Then xi = xi - step rx = rx Mod 256 num = num - 1 waittime (0.01) End If If (ry 256) Then yi = yi - step ry = ry Mod 256 num = num - 1 wait

5、time (0.01) End If picture1.Line -(xi, yi) LoopEnd IfIf (xe xs) And (ye 0)rx = rx + Abs(vx)ry = ry + Abs(vy) If (rx 256) Then xi = xi + step rx = rx Mod 256 num = num - 1 waittime (0.01) End If If (ry 256) Then yi = yi + step ry = ry Mod 256 num = num - 1 waittime (0.01) End If picture1.Line -(xi, y

6、i) LoopEnd IfExit SubBUG: MsgBox (出错了重新填数值_)End Sub3、实现情况：本程序是DDA法的直线插补程序，不仅实现了第1象限的插补，同时可以实现所有象限的直线插补过程。在写程序的时候考虑到再第1象限仅仅实现从原点到某点的直线插补比较容易实现，可以将不从原点开始的直线通过坐标转移或相对坐标地法转化为从原点开始的直线，并且几个象限情况类似。故支持四个象限的直线插补三逐点比较法圆弧插补1、流程图起始E=Xe+YeE0直线走向Y左上方+Y走一步-X走一步F0f=(lym-lyn)*(xn-lxn)-(yn-lyn)*(lxm-lxn)F0+X走一步N+Y走一步

7、f=(yn-yn)*(lxm-lxn)-(lym-lyn)*(xn-lxn)左下方-Y走一步-X走一步F0f=(yn-lyn)*(lxm-lxn)-(lym-lyn)*(xn-lxn)右下方-Y走一步+X走一步F0f=(lym-lyn)*(xn-lxn)-(yn-lyn)*(lxm-lxn)EE-1结束N右上方YNYNNYY2、源程序：Private Sub Command2_Click()On Error GoTo BUG1Dim xs As Double, ys As Double, xe As Double, ye As Double, x0 As Double, y0 As Doubl

8、eDim xi As Double, yi As Double, step As Double, r As Double, F As DoubleDim num As Integerstep = Text6.Textxs = Text7.Textys = Text8.Textxe = Text9.Textye = Text10.Textx0 = Text11.Texty0 = Text12.Textr = Sqr(xs - x0) * (xs - x0) + (ys - y0) * (ys - y0)num = 0xi = xsyi = yspicture1.CurrentX = xipict

9、ure1.CurrentY = yiIf (xs - x0) = 0 And (xe - x0) = 0 And (ye - y0) = 0) Then Do While (xi - x0) = 0) Then xi = xi + step yi = yi Else xi = xi yi = yi + step End If picture1.Line -(xi, yi) waittime (i) Loop num = Abs(xe - xi) / step + Abs(ye - yi) / step Do While (num 0) F = (xi - x0) * (xi - x0) + (

10、yi - y0) * (yi - y0) - r * r If (F = 0) Then xi = xi yi = yi - step Else xi = xi + step yi = yi End If picture1.Line -(xi, yi) num = num - 1 waittime (i) LoopElseIf (xs - x0) = 0 And (xe - x0) = 0) Then num = Abs(xe - xi) / step + Abs(ye - yi) / step Do While (num 0) F = (xi - x0) * (xi - x0) + (yi

11、- y0) * (yi - y0) - r * r If (F = 0) Then xi = xi + step yi = yi Else xi = xi yi = yi + step End If picture1.Line -(xi, yi) num = num - 1 waittime (i) LoopElseIf (xs - x0) = 0 And (ys - y0) = 0 And (xe - x0) = 0 And (ye - y0) = 0) Then num = Abs(xe - xi) / step + Abs(ye - yi) / step Do While (num 0)

12、 If (F = 0) Then xi = xi yi = yi - step Else xi = xi + step yi = yi End If F = (xi - x0) * (xi - x0) + (yi - y0) * (yi - y0) - r * r picture1.Line -(xi, yi) num = num - 1 waittime (i) LoopElse: GoTo BUG2End Ifm = 0Exit SubBUG1: MsgBox (请输入数据)BUG2: MsgBox (不符合条件)End Sub3、实现情况 本程序是逐点比较法圆弧插补第2-1象限顺圆弧代码。使用起点坐标、终点坐标和圆心坐标来确定一段圆弧。对于圆心和圆点不重合的圆弧用坐标转移将原新和圆点重合，然后依照基本方法来插补。分别给出F的判别式，x、y的判别式，判定定进给方向就可以根据步长来一步一步插补了。不过稍微有些麻烦的是关于单步插补和跨象限的问题。四、总结体会：通过这次课程设计，对于数控机床的DDA法、逐点比较法有了更深的理解，进一步提高了我们动手能力和解决的问题能力。并且对于Visual Basic语言运用得更加熟练和得心应手。这次课设给了我极大的练手的机会，非常感谢老师和同学们的帮助。