pro中斜齿轮渐开线的方程

armstrongs

如题

chenfg0515

把这部分程序拷贝到（代码）窗口中; 注意在CAD中可以这样用，PROE中未知，希望能帮到你！

Sub jkx()
    Rem 绘制渐开线,By Sdh,2006-6-8
    Dim d As Double   '节圆直径
    Dim r As Double   '节圆半径
    Dim A As Double   '总展开角度
    Dim Ai As Double  '展开角度
    Dim Li As Double  '展开弧长
    d = 100
    A = 360
    r = d / 2
    Dim Pnt1(2) As Double
    Dim Pnt2(2) As Double
    Dim PntLst() As Double, N As Integer
    ThisDrawing.ModelSpace.AddCircle Pnt1, r
    For Ai = 0 To A * Atn(1) / 45# Step Atn(1) / 45#
        Li = r * Ai
        Pnt1(0) = r * Sin(Ai)
        Pnt1(1) = r * Cos(Ai)
        Pnt2(0) = Pnt1(0) - Li * Cos(-Ai)
        Pnt2(1) = Pnt1(1) - Li * Sin(-Ai)
        ThisDrawing.ModelSpace.AddLine Pnt1, Pnt2
        N = N + 1
        ReDim Preserve PntLst(N * 2 - 1)
        PntLst(N * 2 - 2) = Pnt2(0)
        PntLst(N * 2 - 1) = Pnt2(1)
    Next
    If N > 1 Then
        ThisDrawing.ModelSpace.AddLightWeightPolyline PntLst
    End If
End Sub

[ 本帖最后由 chenfg0515 于 2011-3-7 07:41 编辑 ]