如何判断曲线与直线相交，怎么求出交点
高手进来看一下，如果CAD的dwg图形上有一个曲线，在边上任取一点，往曲线上画，可以有一个交点，如果用VBA编程的话，可以选定一个曲线，再选定一个点，那么，如何判断该点可以以一个角度作的直线与曲线相交呢，交点坐标如何求呢，高手在的话，回一下贴啊，看看有没有什么好的算法，或者一些具体的函数，求助！！！！！！！！！！！！！！！！ http://www.mccad.net/object/acad2004/idh_intersectwith.htm 你这个方法我在AUTOCAD帮助中的开发文档里找到了，里面有个例子，是画线，画圆，然后求交点， 我复制过去，运行出现了错误，代码如下：Public Sub CommandButton1_Click()
UserForm1.Hide
Dim lineObj As AcadLine
Dim startPt(0 To 2) As Double
Dim endPt(0 To 2) As Double
startPt(0) = 1: startPt(1) = 1: startPt(2) = 0
endPt(0) = 5: endPt(1) = 5: endPt(2) = 0
Set lineObj = ThisDrawing.ModelSpace.AddLine(startPt, endPt)
UserForm1.Show
End Sub
Public Sub CommandButton2_Click()
UserForm1.Hide
Dim circleObj As AcadCircle
Dim centerPt(0 To 2) As Double
Dim radius As Double
centerPt(0) = 3: centerPt(1) = 3: centerPt(2) = 0
radius = 1
Set circleObj = ThisDrawing.ModelSpace.AddCircle(centerPt, radius)
ZoomAll
UserForm1.Show
End Sub
Public Sub CommandButton3_Click()
UserForm1.Hide
Dim intPoints As Variant
Set intPoints = lineObj.IntersectWith(circleObj, acExtendNone)
' Print all the intersection points
Dim I As Integer, j As Integer, k As Integer
Dim str As String
If VarType(intPoints)vbEmpty Then
For I = LBound(intPoints) To UBound(intPoints)
str = "Intersection Point[" & k & "] is: " & intPoints(j) & "," & intPoints(j + 1) & "," & intPoints(j + 2)
MsgBox str, , "IntersectWith Example"
str = ""
I = I + 2
j = j + 3
k = k + 1
Next
End If
UserForm1.Show
End Sub
提示错误如下：运行错误‘424’
在Set intPoints = lineObj.IntersectWith(circleObj, acExtendNone)行上标示错误，
版主帮忙看一下，怎么回事
intPoints = lineObj.IntersectWith(circleObj, acExtendNone) 现在又有一个新问题，想问一下，我是做了一个循环，先画线，选取一个曲线，然后求交点，删除原来的直线，重新画线，以原起点为起点，交点为端点，结果出错，程序如下：
Private Sub CommandButton1_Click()
Dim angle As Double
Dim p1 As Variant
Dim p2(0 To 2) As Double
Dim lineobj As AcadLine
Dim lineobj2 As AcadLine
Dim curveobj As AcadObject
Dim intersectpoint As Variant '定义交点
Dim temppt(0 To 2) As Double
UserForm1.hide
p1 = ThisDrawing.Utility.GetPoint(, "选取坝体上某一点")
ThisDrawing.Utility.GetEntity curveobj, pickedpoint, "请选定曲线"
For angle = 180 To 360 Step 5
p2(0) = p1(0) + 100 * Cos(angle) '计算初始直线另一端点
p2(1) = p1(1) + 100 * Sin(angle)
p2(2) = p1(2)
Set lineobj = ThisDrawing.ModelSpace.AddLine(p1, p2)
intersectpoint = lineobj.IntersectWith(curveobj, acExtendthisEntity) '得到交点
temppt(0) = intersectpoint(0)
temppt(1) = intersectpoint(1)
temppt(2) = 0
lineobj.Delete
Set lineobj = ThisDrawing.ModelSpace.AddLine(p1, temppt) '重新画线
lineobj.color = Int(255 * Rnd + 1)
lineobj.Highlight True
lineobj.Update
Next angle
UserForm1.Show
End Sub
结果：temp(0)=intersectpoint(0)上提示下标越界
问一下，怎么回事，应该怎么解决
实体间没有交点时，IntersectWith函数返回上界为－3的数组，会出现这种情况，要先判断一下
谢谢,回答,再问一个问题,如果在直线上选定一个点,能不能在选定点的同时,返回直线上的标注,同时如果有一个曲线与直线上标注的数字一样的话,选择该曲线,
版主提示一下,用什么方法,,
谢谢啦,听他们说好像不能,不知道你有没有什么高招,
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