如何判断曲线与直线相交，怎么求出交点

matrix 发表于 2006-4-18 16:19:00

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！！！！！！！！！！！！！！！！

雪山飞狐_lzh 发表于 2006-4-18 21:50:00

http://www.mccad.net/object/acad2004/idh_intersectwith.htm

matrix 发表于 2006-4-19 11:07:00

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

雪山飞狐_lzh 发表于 2006-4-19 13:34:00

intPoints = lineObj.IntersectWith(circleObj, acExtendNone)

matrix 发表于 2006-4-20 10:48:00

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

雪山飞狐_lzh 发表于 2006-4-20 14:53:00

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

matrix 发表于 2006-4-20 20:16:00

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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如何判断曲线与直线相交，怎么求出交点