[求助]圆和空间任意直线垂直公式

兰州人 发表于 2007-2-17 13:04:00

要求见图，求一个通用公式，圆要与空间任意直线垂直的公式．原因是在Autocad可中画圆，在Ｘ－Ｙ平面完成，用ＶＢ处理ＵＣＳ编程方法麻烦．有个数学公式，简单明了．我现在做法比较麻烦，下面的例子用反正切方法，解绕Z轴得出公式Function RotateZ_Axis(ByVal sPoint As Variant, ByVal ePoint As Variant) As Double     Dim EntAngle As Double     Dim deltaX As Double, deltaY As Double, deltaZ As Double     deltaX = sPoint(0) - ePoint(0): deltaY = sPoint(1) - ePoint(1): deltaZ = sPoint(2) - ePoint(2):           If deltaY >= 0 And deltaX > 0 Then       EntAngle = Atn(deltaY / deltaX)     ElseIf deltaY >= 0 And deltaX < 0 Then       EntAngle = Pi + Atn(deltaY / deltaX)     ElseIf deltaY < 0 And deltaX < 0 Then       EntAngle = Pi + Atn(deltaY / deltaX)     ElseIf deltaY < 0 And deltaX > 0 Then       EntAngle = 2 * Pi + Atn(deltaY / deltaX)     End If         If deltaX = 0 Then       If deltaY > 0 Then         EntAngle = Pi / 2       ElseIf deltaY > 0 Then         EntAngle = Pi * 1.5      End If     End If             RotateZ_Axis = EntAngle End FunctionFunction RotateX_Axis(txtEnt As String) As Double     Dim Ent As AcadLine         Dim EntAngle As Double     Set Ent = ThisDrawing.HandleToObject(txtEnt)           If deltaY >= 0 And deltaX > 0 Then       EntAngle = Atn(deltaY / deltaX)     ElseIf deltaY >= 0 And deltaX < 0 Then       EntAngle = Pi + Atn(deltaY / deltaX)     ElseIf deltaY < 0 And deltaX < 0 Then       EntAngle = Pi + Atn(deltaY / deltaX)     ElseIf deltaY < 0 And deltaX > 0 Then       EntAngle = 2 * Pi + Atn(deltaY / deltaX)     End If         If deltaX = 0 Then       If deltaY > 0 Then         EntAngle = Pi / 2       ElseIf deltaY > 0 Then         EntAngle = Pi * 1.5      End If     End If             RotateZ_Axis = EntAngle End Function RotateZ_Axis,RotateX_Axis,RotateY_Axis返回的的是直线在X，Y，Z坐标轴的方向角。用L=SQR(dx^2+dy^2+dz^2)方程。    alfa = (x - x1) / Sqr((x - x1) ^ 2 + (y - y1) ^ 2 + (z - z1) ^ 2)  　绕Ｘ轴     beta = (y - y1) / Sqr((x - x1) ^ 2 + (y - y1) ^ 2 + (z - z1) ^ 2)　　绕Ｙ轴     theta = (z - z1) / Sqr((x - x1) ^ 2 + (y - y1) ^ 2 + (z - z1) ^ 2)　　绕Ｚ轴 也还需要很多判断语句请教各位大侠是否还有其它什么公式可以一次性解决．

tqr 发表于 2007-2-17 14:41:00

这样也可以：在XY平面画圆，再从圆心画平行与Z轴的直线作辅助线，然后用对齐命令使圆到位。

兰州人 发表于 2007-2-17 20:11:00

这要用到3X3的矩阵,才能编程,在线段a-b外找一点c，过c点垂直线段a-b且交于线段a点？？，才有解．通过a-c点，采用rotate3d point1,point2,Pi/2 才能实现．我现在已经找到３Ｘ３矩阵公式，形体的旋转变换有绕主轴旋转，或绕空间任一直线旋转等多种形式。若令Rθ表示绕z轴转θ角，Rβ表示绕y轴转β角，Rγ表示绕x轴转γ角，则点P绕x、y、z轴转γ、β、θ角的变换公式是 Ｒ＝RθRβRγ      |  cosθ   sinθ   0 | Rθ＝ | -sinθ   cosθ    0  |       |  0       0     1  |      |  cosβ  -sinβ   0  | Rβ＝ |   0     1     0   |      |  sinβ  cosβ    1   |     |  1     0       0   | Rγ＝ |  0    cosγ   sinγ  |      |  0     -sinγ  cosγ | 如何展开３Ｘ３矩阵公式，求Ｘ　＝　Ｙ　＝Ｚ　＝　这是我的终极目标．楼上所述，只能手动调试，不适合于编程．

兰州人 发表于 2007-2-18 14:00:00

计算点到圆的最近距离及交点坐标： 如果该点在圆心，因为圆心到圆周任一点的距离相等，返回UNDEFINED。 连接点P和圆心O，如果PO平行于X轴，则根据P在O的左边还是右边计算出最近点的横坐标为centerPoint.x - radius 或 centerPoint.x + radius。如果PO平行于Y轴，则根据P在O的上边还是下边计算出最近点的纵坐标为 centerPoint.y -+radius或 centerPoint.y - radius。如果PO不平行于X轴和Y轴，则PO的斜率存在且不为0，这时直线PO斜率为k = （ P.y - O.y ）/ ( P.x - O.x )。直线PO的方程为：y = k * ( x - P.x) + P.y。设圆方程为<shapetype id="_x0000_t75" stroked="f" filled="f" path="m@4@5l@4@11@9@11@9@5xe" opreferrelative="t" ospt="75" coordsize="21600,21600">
<stroke joinstyle="miter"></stroke><formulas><f eqn="if lineDrawn pixelLineWidth 0"></f><f eqn="sum @0 1 0"></f><f eqn="sum 0 0 @1"></f><f eqn="prod @2 1 2"></f><f eqn="prod @3 21600 pixelWidth"></f><f eqn="prod @3 21600 pixelHeight"></f><f eqn="sum @0 0 1"></f><f eqn="prod @6 1 2"></f><f eqn="prod @7 21600 pixelWidth"></f><f eqn="sum @8 21600 0"></f><f eqn="prod @7 21600 pixelHeight"></f><f eqn="sum @10 21600 0"></f></formulas><path oconnecttype="rect" gradientshapeok="t" oextrusionok="f"></path><lock aspectratio="t" vext="edit"></lock></shapetype><shape id="_x0000_i1025" alt="" type="#_x0000_t75" style="WIDTH: 12pt; HEIGHT: 12pt;"><imagedata ohref="http://www.vfor.net/forum/images/smilies/frown.gif" src="file:///C:\DOCUME~1\NINGYO~1\LOCALS~1\Temp\msohtml1\01\clip_image001.gif"></imagedata></shape>x - O.x ) ^2 + ( y - O.y ) ^2 = r ^2，联立两方程组可以解出直线PO和圆的交点，取其中离P点较近的交点即可。这是一种X-Y平面求交点的解法.

兰州人 发表于 2007-2-21 12:22:00

本帖最后由作者于 2007-2-21 12:51:46 编辑 采用offset平移,作平行线方法,搞定效果见图。 offset 线段a-b与线段a'-b'平行,b-b'与a-b是垂直的。现在关键问题是要找参数方程公式求解.省去建立临时线段和删除线段的无用功。Function Rotate3dEntity(ByVal EnterEintity As Object, ByVal ll As Object) As Object   Dim theta As Double   Dim x As Double, x1 As Double, y As Double, y1 As Double, z As Double, z1 As Double   Dim Point1 As Variant, Point2 As Variant   Dim l1 As Object, l2 As Object   'Dim Point1(0 To 2) As Double, Point2(0 To 2) As Double   Point1 = ll.StartPoint   Point2 = ll.EndPoint  x = Point1(0): x1 = Point2(0)   y = Point1(1): y1 = Point2(1)   z = Point1(2): z1 = Point2(2) '   theta = (z - z1) / Sqr((x - x1) ^ 2 + (y - y1) ^ 2 + (z - z1) ^ 2)   theta = Format(ArcCos(theta) - Pi / 2, "0.00")   If Format(x - x1, "0.00") = 0 And Format(y - y1, "0.00") = 0 Then    If z1 - z < 0 And Format(x - x1, "0.00") = 0# And Format(y - y1, "0.00") = 0# Then      Point2(0) = Point2(0) + 2      Point2(1) = Point2(1)      Point2(2) = Point2(2)      EnterEintity.Rotate3D ll.EndPoint, Point2, Pi    End If     ElseIf x - x1 <> 0 Or y - y1 <> 0 Then    ll.Offset 10    With obj_ModelSpace      Set l1 = .Item(.Count - 1)      Set l2 = .AddLine(ll.EndPoint, l1.EndPoint)      l1.Delete    End With    If Format(z - z1, "0.00") = 0 Then      EnterEintity.Rotate3D ll.EndPoint, l2.EndPoint, Pi / 2    ElseIf z - z1 <> 0 Then      EnterEintity.Rotate3D ll.EndPoint, l2.EndPoint, theta      If z1 - z < 0 Then        EnterEintity.Rotate3D ll.EndPoint, l2.EndPoint, Pi      End If    End If   l2.Delete  End If End Function Function lll() ' Dim Point1 As Variant, Point2 As Variant If boo = False Then    AutoCADConnect  End If  Dim ll As Object, ss As Object, ReturnEntity As Object  For Each ll In obj_ModelSpace    If ll.objectname = "AcDbLine" Then         Set ss = obj_ModelSpace.AddCone(ll.EndPoint, 3, 10)      Set ReturnEntity = Rotate3dEntity(ss, ll)    End If  Next ll  obj_Doc.Regen (0) End Function

pjf2000qq 发表于 2007-3-1 16:55:00

不好意思没有看明白

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[求助]圆和空间任意直线垂直公式