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本帖最后由 leeyeafu 于 2002-12-13 14:10:10 编辑
按照版主要求,重新写了一些提示.帮我分析用VBA编的凸轮参数式绘制程序。。我不能自己修改成功。请教高手-->leeyeafu转移[br]’该程序是绘制一偏置移动从动件盘型凸轮机构的用VBA编制的程序,有关参数说明在程序中。
Option Explicit
Private Sub CommandButton1_Click()
UserForm1.Hide
Dim s, u, u1, u2, u3, u4, h, e, ro, rt, incrim, s0, beta0, beta, rou, ceta As Double 's为从动件位移 ,u为凸轮转角,u1为升程运动角,u2为远休止角,u3为回程运动角,u4为近休止角,h为升程,e为偏距,ro为基圆半径,rt为滚子半径,incrim为精度
Dim i As Integer
Dim ptpnts() As Double
Dim pt As Variant
Dim bp As Variant
Dim circleobj As AcadCircle
Dim objpoly As AcadLWPolyline
Dim objline As AcadLine
Dim pt0 As Double
Dim firstpt As Double
Const pi = 3.1415926
u1 = Val(TextBox1.Text) ’读取参数
u2 = Val(TextBox2.Text)
u3 = Val(TextBox3.Text)
u4 = Val(TextBox4.Text)
h = Val(TextBox5.Text)
e = Val(TextBox6.Text)
ro = Val(TextBox7.Text)
rt = Val(TextBox8.Text)
incrim = Val(TextBox9.Text)
i = UserForm1.ComboBox1.ListIndex
bp = ThisDrawing.Utility.GetPoint(, "请输入凸轮基圆中心:")
Set circleobj = ThisDrawing.ModelSpace.AddCircle(bp, ro) ‘绘制基圆
u = 0#
Do Until u <= 360# ‘选择运动规律
If u <= u1 Then
Select Case i
Case 0
Call isovelocity_up
Case 1
Call isoacceleration_up
Case 2
Call libration_up
End Select
s0 = Sqr((ro ^ 2) - (e ^ 2)) ’这里开始都是有关凸轮极坐标计算公式
beta0 = Atn(e / s0)
beta = Atn(e / (s0 + s))
rou = Sqr((s + s0) ^ 2 + e ^ 2)
ceta = u * pi / 180# + beta - beta0
pt = ThisDrawing.Utility.PolarPoint(bp, ceta, rou)
ElseIf u > u1 And u <= (u1 + u2) Then
s = h
s0 = Sqr(ro ^ 2 - e ^ 2)
beta0 = Atn(e / s0)
beta = Atn(e / (s0 + s))
rou = Sqr((s + s0) ^ 2 + e ^ 2)
ceta = u * pi / 180# + beta - beta0
pt = ThisDrawing.Utility.PolarPoint(bp, ceta, rou)
ElseIf u > (u1 + u2) And u <= (u1 + u2 + u3) Then
Select Case i
Case 0
Call isovelocity_down
Case 1
Call isoacceleration_down
Case 2
Call libration_down
End Select
s0 = Sqr(ro ^ 2 - e ^ 2)
beta0 = Atn(e / s0)
beta = Atn(e / (s0 + s))
rou = Sqr((s + s0) ^ 2 + e ^ 2)
ceta = u * pi / 180# + beta - beta0
pt = ThisDrawing.Utility.PolarPoint(bp, ceta, rou)
ElseIf u > (u1 + u2 + u3) Then
s = 0
beta0 = Atn(e / s0)
beta = Atn(e / (s0 + s))
rou = Sqr((s + s0) ^ 2 + e ^ 2)
ceta = u * pi / 180# + beta - beta0
pt = ThisDrawing.Utility.PolarPoint(bp, ceta, rou)
End If
If (firstpt <> "") Then
Set objline = ThisDrawing.ModelSpace.AddLine(pt0, pt)
ElseIf (firstpt = "") Then
firstpt = pt
End If
pt0 = pt
。。。。
我对这一段不知该怎样处理?
。。
If u <= u1 Then
Select Case i
Case 0
Call isovelocity_up
Case 1
Call isoacceleration_up
Case 2
Call libration_up
End Select
s0 = Sqr((ro ^ 2) - (e ^ 2))
。。。这一段也不会
改怎样处理?
u = u + incrim
Loop
End Sub
Private Sub UserForm_Initialize()
UserForm1.ComboBox1.AddItem "等速运动", 0
UserForm1.ComboBox1.AddItem "等加速和等减速运动", 1
UserForm1.ComboBox1.AddItem "简谐运动", 2
UserForm1.ComboBox1.ListIndex = 0
UserForm1.TextBox1.SetFocus
End Sub
Public Sub isovelocity_up() ‘匀速运动升程
s = h * u / u1
End Sub
Public Sub isoacceleration_up() ’等加速等减速升程
If u < ui / 2 Then
s = 2 * h * u ^ 2 / u1 ^ 2
Else
s = h - 2 * h * ((u - u1) ^ 2) / (u1 ^ 2)
End If
End Sub
Public Sub isovelocity_down() ‘匀速回程
s = h * (1 - u / u3)
End Sub
Public Sub isoacceleration_down() ’等加速等减速回程
If u < u3 / 2 Then
s = h - 2 * h * u62 / u3 ^ 2
Else
s = 2 * h * ((u - u3) ^ 2) / (u3 ^ 2)
End If
End Sub
Public Sub libration_up() ‘简谐运动升程
s = h * (1 - Cos(pi * u / u1))
End Sub
Public Sub libration_down() ’简谐回程
s = h * (1 + Cos(pi * u / u3)) / 2
End Sub
[此贴子已经被作者于2002-12-6 22:49:59编辑过] |
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