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 Function

Function 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

-----------------------------------------------
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

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