求过直线端点作垂线

ningyong58

已知p1-Pa线段长度为5，夹角为10度

用Kp1-pa = deltaY/deltaX

oo-pa线段斜率为Koo-pa = -1/Kp1-pa

alfa = 10 * Atn(1)*4/180

PaX= 5*cos(Alfa) =  4.924039

PaY = 5 * Sin(Alfa) = 0.868241

Kp1-pa= 5.6713

Koo-pa = -1/ Kp1-pa = -0.763

OOy = -PaX * Koo-pa = 28.79385

 

ningyong58

2. Sub ls()
3.   Dim l As AcadLine
4.   Dim Pp(2) As Double
5.   Dim Ang, Ang1
6.   With ThisDrawing
7.     Set l = .HandleToObject("151")
8.     With l
9.       'Debug.Print l.EndPoint(0)
10.       Kab = .Delta(0) / .Delta(1)
11.       K0 = -1 / Kab
12.       y = -.EndPoint(0) / K0 + .Delta(1)
13.       Ang = -Atn(1) * 2 + 2 * .Angle
14.       'Debug.Print 90 + Ang * 180 / (Atn(1) * 4)
15.       Ang1 = -Atn(1) * 2 - 2 * .Angle
16.       Pp(0) = 0: Pp(1) = y / 2
17.     End With
18.     ''
19.     Dim objArc As AcadArc
20.     Set objArc = .ModelSpace.AddArc(Pp, y / 2, Ang1, Ang)
21.    
22.   End With
23. End Sub

Sub ls()   Dim l As AcadLine   Dim Pp(2) As Double   Dim Ang, Ang1   With ThisDrawing     Set l = .HandleToObject("151")     With l       'Debug.Print l.EndPoint(0)       Kab = .Delta(0) / .Delta(1)       K0 = -1 / Kab       y = -.EndPoint(0) / K0 + .Delta(1)       Ang = -Atn(1) * 2 + 2 * .Angle       'Debug.Print 90 + Ang * 180 / (Atn(1) * 4)       Ang1 = -Atn(1) * 2 - 2 * .Angle       Pp(0) = 0: Pp(1) = y / 2     End With     ''     Dim objArc As AcadArc     Set objArc = .ModelSpace.AddArc(Pp, y / 2, Ang1, Ang)       End With End Sub

hqso

好！ 需要这个

chenjun_nj

To ningyong58:

OOy = 5/sin(10*atn(1)*4/180)
就可以了，不要这么复杂。

水仙的错

哪位能提供个源码