[求助]VBA求两条相交直线内切圆的切点?

vocabulary

各位大侠:

      已知两条相交直线的三个点P1,P2,P3,已知内切圆的半径R1,请问如何求出t1和t2两个切点的位置?

jkbanana

如下图,用解析法可求的要求的点t1, t2的坐标

vocabulary

多谢楼上的大侠,您的图我看得不是很明白,我贴出自己勉强写出来的程序,运行结果基本达到要求,但t2处有一点问题,圆弧与t2p2直线的切点处总是无法正确连接好,有0.0000017间隙,但p3t1与t1处的切点却没有问题,百思不得其解,请各位帮忙看一看程序到底哪有问题,先谢谢了!

Private Sub Fil()
 Dim Tmp As Variant
 Dim p1(0 To 2) As Double
 Dim p2(0 To 2) As Double
 Dim p3(0 To 2) As Double
 Dim pc(0 To 2) As Double
 Dim t1(0 To 2) As Double
 Dim t2(0 To 2) As Double
 Dim Angp1p2Radian As Double
 Dim Angp1p3Radian As Double
 Dim Angpct1Radian As Double
 Dim Angpct2Radian As Double
 Dim AngJJ1 As Double
 Dim AngJJ2 As Double
 Dim Dstp1pc As Double
 Dim Dstp1t1 As Double
 Dim Dstp1t2 As Double
 
 p1(0) = 5#: p1(1) = 45#: p1(2) = 0#
 p2(0) = 50#: p2(1) = 50#: p2(2) = 0#
 p3(0) = 0#: p3(1) = 0#: p3(2) = 0#
 'p1到p2的角度
 Angp1p2Radian = ThisDrawing.Utility.AngleFromXAxis(p1, p2)
 'p1到p3的角度
 Angp1p3Radian = ThisDrawing.Utility.AngleFromXAxis(p1, p3)
 '算出P2P1P3的角度的一半
 AngJJ1 = ((3.1415926 * 2 - Angp1p3Radian + Angp1p2Radian) / 2)
 Dstp1pc = 1 / Sin(AngJJ1) * dRadius
 Dstp1t1 = Cos(AngJJ1) * Dstp1pc
 Dstp1t2 = Dstp1t1
 AngJJ2 = Angp1p3Radian + AngJJ1
 Tmp = ThisDrawing.Utility.PolarPoint(p1, AngJJ2, Dstp1pc)
 pc(0) = Tmp(0)
 pc(1) = Tmp(1)
 pc(2) = Tmp(2)
 Tmp = ThisDrawing.Utility.PolarPoint(p1, Angp1p3Radian, Dstp1t1)
 t1(0) = Tmp(0)
 t1(1) = Tmp(1)
 t1(2) = Tmp(2)
 Tmp = ThisDrawing.Utility.PolarPoint(p1, Angp1p2Radian, Dstp1t2)
 t2(0) = Tmp(0)
 t2(1) = Tmp(1)
 t2(2) = Tmp(2)
 Angpct1Radian = ThisDrawing.Utility.AngleFromXAxis(pc, t1)
 Angpct2Radian = ThisDrawing.Utility.AngleFromXAxis(pc, t2)
 Call ThisDrawing.ModelSpace.AddLine(p2, t2)
 Call ThisDrawing.ModelSpace.AddArc(pc, dRadius, Angpct2Radian, Angpct1Radian)
 Call ThisDrawing.ModelSpace.AddLine(p3, t1)
End Sub

vocabulary

没有哪位兄长帮一帮小弟吗？

tongji007

我试过了 是PI的问题  取PI=3.1415927 就没问题了
AngJJ1 = ((3.1415927 * 2 - Angp1p3Radian + Angp1p2Radian) / 2.0)