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楼主 |
发表于 2008-6-15 07:21
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有没有懂VB的高手
请大家帮忙看看,为什么会在T1的那行出现“实时错误‘6’。溢出”
Private Sub Command1_Click()
Dim p As Single, Kc As Single, i As Integer, fi As Single, cgh0 As Single, cgf0 As Single, yr As Single
Dim du As Single, hu As Single, n As Single, a0 As Single, Mn As Single, Zc As Integer, b As Single
Dim cgh As Single, cgf1 As Single, cgf2 As Single, N1 As Double, N2 As Double, T1 As Double
p = Val(Text1.Text)
Kc = Val(Text2.Text)
i = Val(Text3.Text)
fi = Val(Text4.Text)
cgh0 = Val(Text5.Text)
cgf0 = Val(Text6.Text)
yr = Val(Text7.Text)
du = Val(Text8.Text)
hu = Val(Text9.Text)
n = Val(Text10.Text)
N1 = 60 * n * 1 * yr * du * hu
If N1 <= 5 * 10 ^ 5 Then KHN1 = 1.6
If N1 <= 10 ^ 6 And N1 > 5 * 10 ^ 5 Then KHN1 = 0.9 + 0.8 / (2 * 10 ^ 9 - 5 * 10 ^ 5) * n
If N1 <= 2 * 10 ^ 7 And N1 > 10 ^ 6 Then KHN1 = 0.95
If N1 <= 10 ^ 10 And N1 > 2 * 10 ^ 7 Then KHN1 = 0.88
If N1 <= 10 ^ 4 Then KFN1 = 2.5
If N1 <= 2 * 10 ^ 6 And N1 > 10 ^ 4 Then KFN1 = 1 + 1.5 / (10 ^ 6 - 10 ^ 4) * n
If N1 <= 10 ^ 7 And N1 > 2 * 10 ^ 6 Then KFN1 = 0.95
If N1 <= 10 ^ 8 And N1 > 10 ^ 7 Then KFN1 = 0.9
If N1 <= 10 ^ 9 And N1 > 10 ^ 8 Then KFN1 = 0.85
If N1 <= 10 ^ 10 And N1 > 10 ^ 9 Then KFN1 = 0.82
N2 = 60 * n * i * 1 * yr * du * hu
If N2 <= 5 * 10 ^ 5 Then KHN2 = 1.6
If N2 <= 10 ^ 9 And N2 > 5 * 10 ^ 5 Then KHN2 = 0.9 + 0.8 / (2 * 10 ^ 9 - 5 * 10 ^ 5) * n
If N2 <= 2 * 10 ^ 7 And N2 >= 10 ^ 6 Then KHN2 = 0.95
If N2 <= 10 ^ 10 And N2 > 2 * 10 ^ 7 Then KHN2 = 0.88
If N2 <= 10 ^ 4 Then KFN2 = 2.5
If N2 <= 10 ^ 6 And N2 > 10 ^ 4 Then KFN2 = 1 + 1.5 / (10 ^ 6 - 10 ^ 4) * n
If N2 <= 10 ^ 7 And N2 > 10 ^ 6 Then KFN2 = 0.95
If N2 <= 10 ^ 8 And N2 > 10 ^ 7 Then KFN2 = 0.9
If N2 <= 10 ^ 9 And N2 > 10 ^ 8 Then KFN2 = 0.85
If N2 <= 10 ^ 10 And N2 >= 10 ^ 9 Then KFN2 = 0.82
Slim = 1
cgh = (cgh1 + cgh2) / 2
cgh1 = KHN1 * cgh0 / Slim
cgh2 = KHN2 * cgh0 / Slim
cgf1 = KFN1 * cgf0 / Slim
cgf2 = KFN2 * cgf0 / Slim
ZE = 189.8
ZH = 0.00693 * β - 0.2179 * β + 0.4091
T1 = 9.55 * 1000000 * p / n
Фd = 0.5 * (1 + i) * fi
d1 = (2 * Kc * T1 * (i + 1) * (ZH * ZE / cgh) ^ 2 / (Фd * i)) ^ 1 / 3
Mn = d1 * Cos(β) / z1
β = 30
a0 = (z1 + Z2) / (2 * Cos(β)) * Mn
Zc = z1 + Z2
b = Фd * d1
Text11.Text = a0
Text12.Text = Mn
Text13.Text = Zc
Text14.Text = b
Text15.Text = cgh
Text16.Text = cgf1
Text17.Text = cgf2
End Sub |
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