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誰能提供一段vb在CAD中繪制多段線的實(shí)例代碼作參考！
多線段沒畫過，
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線倒是畫過，二話不說直接上代碼你參考一下
繪制邊框
Private Sub Command4_Click()
Set acadDoc = acadApp.ActiveDocument
link2
acadDoc.ActiveLayer = acadlay(0)
With adoRes
While Not .EOF
pt1(0) = !X: pt1(1) = !Y: pt1(2) = !z: pt2(0) = !X1: pt2(1) = !Y1: pt2(2) = !z1
draw pt1, pt2
.MoveNext
Wend
End With
pt1(0) = 0: pt1(1) = 0: pt1(2) = 0
pt2(0) = 20: pt2(1) = 10: pt2(2) = 0
draw pt1, pt2
'在坡
pt1(0) = 0: pt1(1) = 0: pt1(2) = 0
pt2(0) = 20: pt2(1) = 10: pt2(2) = 0
draw pt1, pt2
For c = 0 To 50 Step 10
pt1(0) = 0: pt1(1) = c: pt1(2) = 0
pt2(0) = 270: pt2(1) = c: pt2(2) = 0
draw pt1, pt2
Next c
pt1(0) = 20: pt1(1) = 0: pt1(2) = 0
pt2(0) = 20: pt2(1) = 190: pt2(2) = 0
draw pt1, pt2
pt1(0) = 0: pt1(1) = 0: pt1(2) = 0
pt2(0) = 0: pt2(1) = -10: pt2(2) = 0
draw pt1, pt2
pt1(0) = 0: pt1(1) = -10: pt1(2) = 0
pt2(0) = 270: pt2(1) = -10: pt2(2) = 0
draw pt1, pt2
pt1(0) = 270: pt1(1) = -10: pt1(2) = 0
pt2(0) = 270: pt2(1) = 0: pt2(2) = 0
draw pt1, pt2
pt3(0) = 5: pt3(1) = -7
acadDoc.ModelSpace.AddText t, pt3, 4
pt3(0) = 55: pt3(1) = -7
acadDoc.ModelSpace.AddText t1, pt3, 4
a = 0
For c = 60 To 100 Step 10
pt1(0) = 20: pt1(1) = c: pt1(2) = 0
pt2(0) = 23: pt2(1) = c: pt2(2) = 0
draw pt1, pt2
pt1(0) = 10: pt1(1) = c - 1.6: pt1(2) = 0
acadDoc.ModelSpace.AddText a, pt1, 3.2
a = a + 2
Next c
'在圖中加入地面高程
acadDoc.ActiveTextStyle = acadDoc.TextStyles.Add("楷體")
pt1(0) = 1: pt1(1) = 13: pt1(2) = 0
Set acadtext = acadDoc.ModelSpace.AddText("地面高程", pt1, 3.2)
Call rote1(acadtext, pt1)
'在圖中加入設(shè)計(jì)高程
Call rote1(acadtext, pt1)
pt1(0) = 1: pt1(1) = 23: pt1(2) = 0
Set acadtext = acadDoc.ModelSpace.AddText("設(shè)計(jì)高程", pt1, 3.2)
Call rote1(acadtext, pt1)
'在圖中加入填挖高
pt1(0) = 1: pt1(1) = 33: pt1(2) = 0
Set acadtext = acadDoc.ModelSpace.AddText("填 挖 高", pt1, 3.2)
Call rote1(acadtext, pt1)
'在圖中加入坡度/坡長(zhǎng)
pt1(0) = 0: pt1(1) = 5.5: pt1(2) = 0
Set acadtext = acadDoc.ModelSpace.AddText("坡 度", pt1, 3.2)
Call rote1(acadtext, pt1)
pt1(0) = 9: pt1(1) = 1: pt1(2) = 0
Set acadtext = acadDoc.ModelSpace.AddText("坡 長(zhǎng)", pt1, 3.2)
Call rote1(acadtext, pt1)
'在圖中加入樁號(hào)
pt1(0) = 1: pt1(1) = 43: pt1(2) = 0
Set acadtext = acadDoc.ModelSpace.AddText("樁 號(hào)", pt1, 3.2)
Call rote1(acadtext, pt1)
Call Command5_Click
End Sub
Private Sub draw(pt1, pt2)
Set AcadLine = acadDoc.ModelSpace.AddLine(pt1, pt2)
acadDoc.Regen acActiveViewport
End Sub
線倒是畫過，二話不說直接上代碼你參考一下
vb曲線擬合
老弟，公式打不出來的
一般都是用matlab搞定的，它里面有現(xiàn)成的函數(shù)供使用的
典型程序解析：
x=[0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1];%input xi data
y=[1.978 3.28 6.16 7.08 7.34 7.66 9.56 9.48 9.30 11.2]; % input yi data
n=2; % polynomial order
p=polyfit(x, y, n)% polyfit 的輸出是一個(gè)多項(xiàng)式系數(shù)的行向量（擬合二項(xiàng)式的系數(shù)）
ezplot('-9.8108*x*x+20.1293*x-0.0317')%對(duì)擬合的函數(shù)作圖
xi=linspace(0,1,100); % x-axis data for plotting
z=polyval(p, xi);% 為了計(jì)算在xi數(shù)據(jù)點(diǎn)的多項(xiàng)式值，調(diào)用MATLAB的函數(shù)polyval
plot(x,y, 'o' ,x,y,xi,z,':')%在同一個(gè)圖形里看他們的擬合程度
典型例題：對(duì)以下數(shù)據(jù)分別作二次,三次多項(xiàng)式擬合,并畫出圖形.
x=1:16;
y=[4, 6.4, 8, 8.4, 9.28, 9.5, 9.7, 9.86, 10, 10.2, 10.32, 10.42, 10.5, 10.55, 10.58, 10.6];
源程序：二次多項(xiàng)式擬合
x=1:1:16;
y=[4, 6.4, 8, 8.4, 9.28, 9.5, 9.7, 9.86, 10, 10.2, 10.32, 10.42, 10.5, 10.55, 10.58, 10.6];
a=polyfit(x,y,2)
a =
-0.0445 1.0711 4.3252
ezplot('-0.0445*x^2+1.0711*x+4.3252')
三次多項(xiàng)式擬合
x=1:1:16;
y=[4, 6.4, 8, 8.4, 9.28, 9.5, 9.7, 9.86, 10, 10.2, 10.32, 10.42, 10.5, 10.55, 10.58, 10.6];
a=polyfit(x,y,3)
a =
0.0060 -0.1963 2.1346 2.5952
ezplot('0.0060*x^3-0.1963*x^2+2.1346*x+2.5952')
VB.NET 簡(jiǎn)單多線程
多線程一般是不推薦用的，因?yàn)榫€程之間如果有共享資源的話會(huì)引起競(jìng)爭(zhēng)，需要加鎖處理；而且線程間沒有時(shí)序關(guān)系，所以你在調(diào)試中可能會(huì)出現(xiàn)異步處理結(jié)束順序與開始處理順序不一致的情況（我在調(diào)試中已經(jīng)發(fā)現(xiàn)該問題）。
針對(duì)你提出的這個(gè)問題，采用了多線程處理，利用的是BackgroundWorker也就是異步處理控件進(jìn)行了處理。
代碼已經(jīng)經(jīng)過調(diào)試通過。歡迎交流，如有問題，留下QQ或其他聯(lián)系方式。
代碼如下，并附程序截圖。
‘---------------------------------------------------
Imports?System.ComponentModel?'導(dǎo)入異步控件命名空間
Public?Class?Form1
Private?howmany?As?Integer?=?10
Private?AnalysisNumber(0?To?howmany?-?1)?As?BackgroundWorker
Private?Sub?Button1_Click(ByVal?sender?As?System.Object,?ByVal?e?As?System.EventArgs)?Handles?Button1.Click
ListBox1.Items.Clear()
creatNewBackgroundWorker()
addHandle()
startWork()
End?Sub
Private?Sub?creatNewBackgroundWorker()
For?i?As?Integer?=?0?To?AnalysisNumber.Length?-?1
AnalysisNumber(i)?=?New?BackgroundWorker
Next
End?Sub
Private?Sub?addHandle()
For?i?As?Integer?=?0?To?AnalysisNumber.Length?-?1
AddHandler?AnalysisNumber(i).DoWork,?AddressOf?AnalysisNumber_DoWork
AddHandler?AnalysisNumber(i).RunWorkerCompleted,?AddressOf?AnalysisNumber_RunWorkerCompleted
Next
End?Sub
Private?Sub?startWork()
For?i?As?Integer?=?0?To?9
Dim?temp(0?To?9)?As?Integer
For?j?As?Integer?=?1?To?10
temp(j?-?1)?=?10?*?i?+?j
Next
AnalysisNumber(i).RunWorkerAsync(temp)
Next
End?Sub
Private?Sub?AnalysisNumber_DoWork(ByVal?sender?As?Object,?ByVal?e?As?System.ComponentModel.DoWorkEventArgs)
Dim?data?As?Integer()
data?=?CType(e.Argument,?Integer())
Dim?temp?As?Integer
For?i?As?Integer?=?0?To?data.Length?-?1
temp?=?data(i)
data(i)?=?temp?*?temp
Next
e.Result?=?data
End?Sub
Private?Sub?AnalysisNumber_RunWorkerCompleted(ByVal?sender?As?Object,?ByVal?e?As?System.ComponentModel.RunWorkerCompletedEventArgs)
Dim?data?As?Integer()
data?=?CType(e.Result,?Integer())
For?i?As?Integer?=?0?To?data.Length?-?1
ListBox1.Items.Add(data(i))
Next
End?Sub
End?Class
文章題目：vb.net擬合多段線 多段線如何擬合
轉(zhuǎn)載來于：http://ef60e0e.cn/article/doedsge.html
