反正切函数确定空间直线的方向角

兰州人 发表于 2006-12-25 13:48:00

问题提出：在空间坐标系中，插入一图块需要的基本条件是，插入的空间坐标点，X，Y，Z轴的方向角。       我的解决方法是用一个线段来确定插入空间坐标系中的方向角，特点是程序调试时对方向角的确认比较直观，提高工作效率。      在空间任意两点可以确定一条直线，两点坐标确定，一种方法是输入坐标，另一种是通过交互式方法确定两点坐标。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。请问各位大侠是否还有更好的几何解决方法。

highflybir 发表于 2006-12-25 14:52:00

想法很好，但是算法可能可能复杂了一点：我的建议：已知两点pt1 (x1,y1,z1),pt2 (x2,y2,z2) 得到两点的矢量 (mapcar '- pt1 pt2) 即 dx ,dy, dz和两点的距离 L 则其方向角为 acos (dx/L) ,acos (dy/L) ，acos (dz/L)acos为反余弦函数.

兰州人 发表于 2006-12-26 09:36:00

本帖最后由作者于 2008-12-21 17:45:55 编辑

谢谢楼上给的思路,其解法如下Sub ll()
Dim objLine As AcadLine
Dim objCount As Integer
Dim rotateAngular As Double, rotateDegree As Double
Dim rotateAngularX As Double, rotateAngularY As Double, rotateAngularZ As Double
With ThisDrawing.ModelSpace
objCount = .Count
For ii = 0 To objCount - 1
   Set objLine = .Item(ii)
   With objLine
   rotateAngularX = ACos(.Delta(0) / .Length)
   Xx = RadToDeg(rotateAngularX)
   rotateAngularY = ACos(.Delta(1) / .Length)
   Yy = RadToDeg(rotateAngularY)
   rotateAngularZ = ACos(.Delta(2) / .Length)
   Zz = RadToDeg(rotateAngularZ)
   End With
Next ii

End With
End Sub
Function ACos(ByVal Number As Double) As Double
If Number = 0 Then
ACos = 2 * Atn(1)
Else
ACos = Atn(-Number / Sqr(-Number * Number + 1)) + 2 * Atn(1)
End If
End Function
Function RadToDeg(Alfa As Double) As Double
RadToDeg = Alfa * 180 * Alfa / (Atn(1) * 4)
End Function

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L=SQR(dx^2+dy^2+dz^2),我用过这种方程。
方程解如下：
alfa = (x - x1) / Sqr((x - x1) ^ 2 + (y - y1) ^ 2 + (z - z1) ^ 2)
beta = (y - y1) / Sqr((x - x1) ^ 2 + (y - y1) ^ 2 + (z - z1) ^ 2)
theta = (z - z1) / Sqr((x - x1) ^ 2 + (y - y1) ^ 2 + (z - z1) ^ 2)

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反正切函数确定空间直线的方向角